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Actuarial Mathematics for Life Contingent Risks
How can actuaries best equip themselves for the products and risk structures
of the future? In this new textbook, three leaders in actuarial science give a
modern perspective on life contingencies
...
The
authors then develop a more contemporary outlook, introducing multiple state
models, emerging cash flows and embedded options
...
Over 150 exercises
and solutions teach skills in simulation and projection through computational
practice
...


International Series on Actuarial Science
Christopher Daykin, Independent Consultant and Actuary
Angus Macdonald, Heriot-Watt University
The International Series on Actuarial Science, published by Cambridge
University Press in conjunction with the Institute of Actuaries and the Faculty of Actuaries, contains textbooks for students taking courses in or related to
actuarial science, as well as more advanced works designed for continuing professional development or for describing and synthesizing research
...


ACTUARIAL MATHEMATICS FOR
LIFE CONTINGENT RISKS
D AV I D C
...
D I C K S O N
University of Melbourne

M A RY R
...
WAT E R S
Heriot-Watt University, Edinburgh

CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore,
São Paulo, Delhi, Dubai, Tokyo
Cambridge University Press
The Edinburgh Building, Cambridge CB2 8RU, UK
Published in the United States of America by Cambridge University Press, New York
www
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org
Information on this title: www
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org/9780521118255
© D
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M
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R
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R
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Subject to statutory exception and to the
provision of relevant collective licensing agreements, no reproduction of any part
may take place without the written permission of Cambridge University Press
...


To
Carolann,
Vivien
and Phelim

Contents

Preface
page xiv
1
Introduction to life insurance
1
1
...
2 Background
1
1
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3
...
3
...
3
...
3
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3
...
3
...
3
...
4 Other insurance contracts
12
1
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5
...
5
...
6 Mutual and proprietary insurers
14
1
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8 Notes and further reading
15
1
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1 Summary
17
2
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3 The force of mortality
21
2
...
5 Mean and standard deviation of Tx
29
2
...
6
...
6
...
7 Notes and further reading
2
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1 Summary
3
...
3 Fractional age assumptions
3
...
1
Uniform distribution of deaths
3
...
2
Constant force of mortality
3
...
5 Survival models for life insurance policyholders
3
...
7 Select and ultimate survival models
3
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9 Select life tables
3
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11 Exercises
Insurance benefits
4
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2 Introduction
4
...
4 Valuation of insurance benefits
¯
4
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1 Whole life insurance: the continuous case, Ax
4
...
2 Whole life insurance: the annual case, Ax
(m)
4
...
3 Whole life insurance: the 1/mthly case, Ax
4
...
4
Recursions
4
...
5 Term insurance
4
...
6
Pure endowment
4
...
7
Endowment insurance
4
...
8
Deferred insurance benefits
(m)
¯
4
...
5
...
5
...
6 Variable insurance benefits
4
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8 Notes and further reading
4
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1 Summary
5
...
3
5
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4
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4
...
4
...
4
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5 Annuities payable continuously
5
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1 Whole life continuous annuity
5
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2 Term continuous annuity
5
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6
...
6
...
6
...
7 Comparison of annuities by payment frequency
5
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9 Guaranteed annuities
5
...
10
...
10
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11 Evaluating annuity functions
5
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1 Recursions
5
...
2 Applying the UDD assumption
5
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3 Woolhouse’s formula
5
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13 Functions for select lives
5
...
15 Exercises
Premium calculation
6
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2 Preliminaries
6
...
4 The present value of future loss random variable
6
...
5
...
6 Gross premium calculation
6
...
8 The portfolio percentile premium principle
6
...
9
...
9
...
9
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10 Notes and further reading
6
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1 Summary
7
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3 Policies with annual cash flows
7
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1 The future loss random variable
7
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2
Policy values for policies with annual cash flows
7
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3
Recursive formulae for policy values
7
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4 Annual profit
7
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5 Asset shares
7
...
4
...
4
...
5 Policy values with continuous cash flows
7
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1 Thiele’s differential equation
7
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2
Numerical solution of Thiele’s differential
equation
7
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7 Retrospective policy value
7
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9 Notes and further reading
7
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1 Summary
8
...
2
...
2
...
2
...
2
...
2
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3 Assumptions and notation
8
...
4
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5 Numerical evaluation of probabilities
8
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7 Policy values and Thiele’s differential equation
8
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1 The disability income model
8
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2 Thiele’s differential equation – the general case

169
170
176
176
176
176
176
182
191
196
200
203
204
205
207
207
211
213
219
220
220
220
230
230
230
230
232
232
233
234
235
239
242
243
247
250
251
255

Contents
8
...
9

9

10

11

Multiple decrement models
Joint life and last survivor benefits
8
...
1 The model and assumptions
8
...
2 Joint life and last survivor probabilities
8
...
3 Joint life and last survivor annuity and
insurance functions
8
...
4 An important special case: independent
survival models
8
...
11 Notes and further reading
8
...
1 Summary
9
...
3 The salary scale function
9
...
5 The service table
9
...
6
...
6
...
7 Funding plans
9
...
9 Exercises
Interest rate risk
10
...
2 The yield curve
10
...
3
...
4 Diversifiable and non-diversifiable risk
10
...
1 Diversifiable mortality risk
10
...
2 Non-diversifiable risk
10
...
6 Notes and further reading
10
...
1 Summary
11
...
2
...
2
...
3 Profit measures
11
...
5 Notes and further reading
11
...
1 Summary
12
...
3 Deterministic profit testing for equity-linked insurance
12
...
5 Stochastic pricing
12
...
6
...
6
...
6
...
6
...
7 Notes and further reading
12
...
1 Summary
13
...
3 The ‘no arbitrage’ assumption
13
...
5 The binomial option pricing model
13
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1 Assumptions
13
...
2 Pricing over a single time period
13
...
3 Pricing over two time periods
13
...
4 Summary of the binomial model option pricing
technique
13
...
6
...
6
...
7 Notes and further reading
13
...
1 Summary
14
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3 Guaranteed minimum maturity benefit
14
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1 Pricing
14
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2 Reserving
14
...
4
...
4
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5 Pricing methods for embedded options
14
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7 Emerging costs
14
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9 Exercises
A
Probability theory
A
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1
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1
...
1
...
1
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2 The central limit theorem
A
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3
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3
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3
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4 Conditional expectation and conditional variance
A
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1 Numerical integration
B
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1 The trapezium rule
B
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2 Repeated Simpson’s rule
B
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3 Integrals over an infinite interval
B
...
3 Notes and further reading
C
Simulation
C
...
2 Simulation from a normal distribution
C
...
1 The Box–Muller method
C
...
2 The polar method
C
...

New and innovative products have been developed at the same time as we have
seen vast increases in computational power
...

Given the changes occurring in the interconnected worlds of finance and life
insurance, we believe that this is a good time to recast the mathematics of life
contingent risk to be better adapted to the products, science and technology that
are relevant to current and future actuaries
...
The
material is presented with a certain level of mathematical rigour; we intend for
readers to understand the principles involved, rather than to memorize methods
or formulae
...

We start from a traditional approach, and then develop a more contemporary perspective
...

Through the focus on realistic contracts and assumptions, we aim to foster a
general business awareness in the life insurance context, at the same time as
we develop the mathematical tools for risk management in that context
...
Using multiple state models
allows a single framework for a wide range of insurance, including benefits
which depend on health status, on cause of death benefits, or on two or more
lives
...
Pension mathematics has some specialized
concepts, particularly in funding principles, but in general this chapter is an
application of the theory in the preceding chapters
...
In this chapter we explore the crucially important
difference between diversifiable and non-diversifiable risk
...

In Chapter 11 we continue the cash flow approach, developing the emerging
cash flows for traditional insurance products
...
This is
much richer information that the actuary can use to assess profitability and to
better manage portfolio assets and liabilities
...
The real risks for such products can only be assessed
taking the random variation in potential outcomes into consideration, and we
demonstrate this with Monte Carlo simulation of the emerging cash flows
...
Option theory is the mathematics
of valuation and risk management of financial guarantees
...

In Chapter 14 we apply option theory to the embedded options of financial
guarantees in insurance products
...

The material in this book is designed for undergraduate and graduate programmes in actuarial science, and for those self-studying for professional
actuarial exams
...
We also assume familiarity
with the binomial, uniform, exponential, normal and lognormal distributions
...
We also assume

xvi

Preface

that readers have completed an introductory level course in the mathematics of
finance, and are aware of the actuarial notation for annuities-certain
...
Spreadsheets are ubiquitous tools in actuarial practice, and it is
natural to use them throughout, allowing us to use more realistic examples,
rather than having to simplify for the sake of mathematical tractability
...
To keep the computation requirements reasonable, we have ensured that every example and exercise can be
completed in Microsoft Excel, without needing any VBA code or macros
...
It will be very useful for anyone working through
the material of this book to construct their own spreadsheet tables as they
work through the first seven chapters, to generate mortality and actuarial
functions for a range of mortality models and interest rates
...

One of the advantages of spreadsheets is the ease of implementation of numerical integration algorithms
...

The material in this book is appropriate for two one-semester courses
...
Chapters 8–14 introduce more contemporary material
...
Chapter
9, on pension mathematics, is not required for subsequent chapters, and could
be omitted if a single focus on life insurance is preferred
...
Particular thanks go to Carole Bernard, Phelim
Boyle, Johnny Li, Ana Maria Mera, Kok Keng Siaw and Matthew Till
...
These visits significantly shortened
the time it has taken to write the book (to only one year beyond the original
deadline)
...
1 Summary
Actuaries apply scientific principles and techniques from a range of other disciplines to problems involving risk, uncertainty and finance
...
Because pension liabilities are similar in many ways
to life insurance liabilities, we also describe some common pension benefits
...
How to answer such questions, and solve the resulting
problems, is the subject of the following chapters
...
2 Background
The first actuaries were employed by life insurance companies in the early
eighteenth century to provide a scientific basis for managing the companies’
assets and liabilities
...
The modelling of mortality became a
topic of both commercial and general scientific interest, and it attracted many
significant scientists and mathematicians to actuarial problems, with the result
that much of the early work in the field of probability was closely connected
with the development of solutions to actuarial problems
...
If the named life insured died
during the year that the contract was in force, the insurer would pay a predetermined lump sum, the sum insured, to the policyholder or his or her estate
...
Each year the premium
would increase as the probability of death increased
...
Over a large number
of contracts, the premium income each year should approximately match the
claims outgo
...
This method is still used for group life insurance, where an employer
purchases life insurance cover for its employees on a year-to-year basis
...
The problem with assessmentism was that the annual increases
in premiums discouraged policyholders from renewing their contracts
...
This was much more popular with policyholders, as they would not
be priced out of the insurance contract just when it might be most needed
...
However, a problem for the
insurer was that the longer contracts were more complex to model, and offered
more financial risk
...
In particular, it
became necessary to incorporate financial considerations into the modelling of
income and outgo
...
Over, say, a 30-year contract, it becomes a very important part
of the modelling and management of risk
...
This was a requirement in law that the person
contracting to pay the life insurance premiums should face a financial loss on
the death of the insured life that was no less than the sum insured under the
policy
...

Subsequently, insurance policies tended to be purchased by the insured life,
and in the rest of this book we use the convention that the policyholder who
pays the premiums is also the life insured, whose survival or death triggers the
payment of the sum insured under the conditions of the contract
...
A life table summarizes a survival model by specifying the
proportion of lives that are expected to survive to each age
...
Edmund Halley, famous for his astronomical
calculations, used mortality data from the city of Breslau in the late seventeenth
century as the basis for his life table, which, like Graunt’s, was constructed by

1
...
Halley took the work two steps
further
...
That is, given the probability that a newborn life survives to each subsequent age, it is possible to infer the probability
that a life aged, say, 20, will survive to each subsequent age, using the condition
that a life aged zero survives to age 20
...
A whole life annuity is a
contract paying a level sum at regular intervals while the named life (the annuitant) is still alive
...

This book continues in the tradition of combining models of mortality with
models in finance to develop a framework for pricing and risk management of
long-term policies in life insurance
...
However, there have been many changes since
the first long-term policies of the late eighteenth century
...
3 Life insurance and annuity contracts
1
...
1 Introduction
The life insurance and annuity contracts that were the object of study of the early
actuaries were very similar to the contracts written up to the 1980s in all the
developed insurance markets
...
The reasons for the changes
include:
• Increased interest by the insurers in offering combined savings and insurance

•
•

•
•

products
...
Many modern
contracts combine the indemnity concept with an opportunity to invest
...

Policyholders have become more sophisticated investors, and require more
options in their contracts, allowing them to vary premiums or sums insured,
for example
...

The risk management techniques in financial products have also become
increasingly complex, and insurers have offered some benefits, particularly

4

Introduction to life insurance
financial guarantees, that require sophisticated techniques from financial
engineering to measure and manage the risk
...

Different countries have different names and types of contracts; we have tried
to cover the major contract types in North America, the United Kingdom and
Australia
...
We usually use the term ‘insurance’
when the benefit is paid as a single lump sum, either on the death of the policyholder or on survival to a predetermined maturity date
...
) An annuity is a benefit in the
form of a regular series of payments, usually conditional on the survival of the
policyholder
...
3
...
We describe each in a little more detail here
...
Term insurance
allows a policyholder to provide a fixed sum for his or her dependents in
the event of the policyholder’s death
...

Decreasing term insurance indicates that the sum insured and (usually) premiums decrease over the term of the contract
...

Renewable term insurance offers the policyholder the option of renewing the policy at the end of the original term, without further evidence
of the policyholder’s health status
...


1
...

Whole life insurance pays a lump sum benefit on the death of the policyholder whenever it occurs
...
This avoids
the problem that older lives may be less able to pay the premiums
...

This is a mixture of a term insurance benefit and a savings element
...
Endowment insurance is obsolete in many jurisdictions
...
The policyholder (who is the home owner) paid interest
on the mortgage loan, and the principal was paid from the proceeds on
the endowment insurance, either on the death of the policyholder or at
the final mortgage repayment date
...
It is hard for small investors to achieve good rates of return on
investments, because of heavy expense charges
...

With-profit insurance
Also part of the traditional design of insurance is the division of business
into ‘with-profit’ (also known, especially in North America, as ‘participating’,
or ‘par’ business), and ‘without profit’ (also known as ‘non-participating’ or
‘non-par’)
...
In North America, the with-profit
arrangement often takes the form of cash dividends or reduced premiums
...
Reversionary bonuses are awarded during the term of the contract;
once a reversionary bonus is awarded it is guaranteed
...
Reversionary bonuses

6

Introduction to life insurance
Table 1
...

Year
1
2
3

...


...
5%
2
...


...


5%
6%
6%

...


...
00
4620
...
20

...


...
Reversionary and terminal bonuses are
determined by the insurer based on the investment performance of the invested
premiums
...
At
the end of the first year of the contract a bonus of 2% on the sum insured and
5% on previous bonuses is declared; in the following two years, the rates are
2
...
Then the total guaranteed sum insured increases each year as
shown in Table 1
...

If the policyholder dies, the total death benefit payable would be the original
sum insured plus reversionary bonuses already declared, increased by a terminal
bonus if the investment returns earned on the premiums have been sufficient
...
However, the traditional with-profit contract
is designed primarily for the life insurance cover, with the savings aspect a
secondary feature
...
3
...
g
...
Additional flexibility also allows policyholders to
purchase less insurance when their finances are tight, and then increase the
insurance coverage when they have more money available
...

Universal life insurance combines investment and life insurance
...
Some

1
...
Premiums are flexible, as long as they are sufficient to pay for the designated sum insured under the term insurance part
of the contract
...
Universal life is a common
insurance contract in North America
...
Premiums are used to purchase units (shares) of an investment
fund, called the with-profit fund
...
The shares will not decrease
in value
...

After some poor publicity surrounding with-profit business, and, by
association, unitized with-profit business, these product designs were
withdrawn from the UK and Australian markets by the early 2000s
...

Equity-linked insurance has a benefit linked to the performance of an
investment fund
...
The first is where the
policyholder’s premiums are invested in an open-ended investment company style account; at maturity, the benefit is the accumulated value of
the premiums
...
In some cases, there is
also a guaranteed minimum maturity benefit payable
...
In Canada they are known as segregated
fund policies and always carry a maturity guarantee
...
(The use of the term ‘annuity’
for these contracts is very misleading
...
)
The second form of equity-linked insurance is the Equity-Indexed
Annuity (EIA) in the USA
...
At
maturity, the policyholder receives a proportion of the return on a specified
stock index, if that is greater than the guaranteed minimum return
...
EIAs are much less popular with
consumers than variable annuities
...
3
...
Brokers who connect individuals to an appropriate insurance product have, since the earliest times, played
an important role in the market
...

Brokers, or other financial advisors, are often remunerated through a commission system
...
Typically, there is a higher percentage paid on the first premium
than on subsequent premiums
...
Some
advisors may be remunerated on a fixed fee basis, or may be employed by one
or more insurance companies on a salary basis
...

Insurers may use television advertising or other telemarketing methods to sell
direct to the public
...
For example, often the sum insured
is smaller
...
Another mass marketed insurance contract is loan or
credit insurance, where an insurer might cover loan or credit card payments in
the event of the borrower’s death, disability or unemployment
...
3
...
Selling life insurance policies is a
competitive business and life insurance companies (also known as life offices)
are constantly considering ways in which to change their procedures so that they
can improve the service to their customers and gain a commercial advantage
over their competitors
...

For a given type of policy, say a 10-year term insurance, the life office will
have a schedule of premium rates
...
An applicant’s risk level
is assessed by asking them to complete a proposal form giving information on

1
...
The
life insurer may ask for permission to contact the applicant’s doctor to enquire
about their medical history
...

The process of collecting and evaluating this information is called underwriting
...

On the basis of the application and supporting medical information, potential life insurance policyholders will generally be categorized into one of the
following groups:
• Preferred lives have very low mortality risk based on the standard infor-

mation
...

The preferred life category is common in North America, but has not yet
caught on elsewhere
...

• Normal lives may have some higher rated risk factors than preferred lives
(where this category exists), but are still insurable at standard rates
...

• Rated lives have one or more risk factors at raised levels and so are not
acceptable at standard premium rates
...
An example might be someone having a family history of
heart disease
...
This category would also include lives
with hazardous jobs or hobbies which put them at increased risk
...

Within the first three groups, applicants would be further categorized according to the relative values of the various risk factors, with the most fundamental
being age, gender and smoking status
...
Another 2–3% may be accepted at non-standard rates

10

Introduction to life insurance

because of an impairment, or a dangerous occupation, leaving around 2–3%
who will be refused insurance
...
Term insurance is generally more strictly underwritten than
whole life insurance, as the risk taken by the insurer is greater
...
Under, say, 10-year term insurance, it is assumed that the majority
of contracts will expire with no death benefit paid
...
Since high sum insured contracts carry more risk than low sum
insured, high sums insured would generally trigger more rigorous underwriting
...
Often, direct
marketed contracts are sold with relatively low benefit levels, and with the
attraction that no medical evidence will be sought beyond a standard questionnaire
...
By keeping the underwriting relatively light, the expenses of writing new business can be kept low, which is an
attraction for high-volume, low sum insured contracts
...
Insurers have
a phrase for this – that both insurer and policyholder may assume ‘utmost
good faith’ or ‘uberrima fides’ on the part of the other side of the contract
...
If
it appears that the policyholder held back information, or submitted false or
misleading information, the insurer may not pay the full sum insured
...
3
...
In traditional contracts the regular
premium is generally a level amount throughout the term of the contract; in
more modern contracts the premium might be variable, at the policyholder’s
discretion for investment products such as equity-linked insurance, or at the
insurer’s discretion for certain types of term insurance
...
Monthly premiums are common as it is convenient for policyholders
to have their outgoings payable with approximately the same frequency as their
income
...
3 Life insurance and annuity contracts

11

An important feature of all premiums is that they are paid at the start of each
period
...
The premiums will be paid at the start of the contract, and
then at the start of each subsequent year provided the policyholder is alive
...
Similarly, if the premiums are monthly, then the first monthly
instalment will be paid at t = 0, and the final premium will be paid at the start
11
of the final month at t = 9 12 years
...
)

1
...
7 Life annuities
Annuity contracts offer a regular series of payments
...
The recipient is
called an annuitant
...
If the annuity is paid for some maximum period,
provided the annuitant survives that period, it is called a term life annuity
...

Buying a whole life annuity guarantees that the income will not run out before
the annuitant dies
...
The annuity is ‘life
contingent’, by which we mean the annuity is paid only if the policyholder
survives to the payment dates
...
If the policyholder dies
soon after the annuity commences, there may be some minimum payment
period, called the guarantee period, and the balance would be paid to the
policyholder’s estate
...
This might, for example, be used to convert a lump sum
retirement benefit into a life annuity to supplement a pension
...

Regular Premium Deferred Annuity (RPDA) The RPDA offers a deferred
life annuity with premiums paid through the deferred period
...

Joint life annuity A joint life annuity is issued on two lives, typically a
married couple
...

Last survivor annuity A last survivor annuity is similar to the joint life
annuity, except that payment continues while at least one of the lives
survives, and ceases on the second death of the couple
...
One is designated as the annuitant, and one the insured
...
On the death
of the insured life, if the annuitant is still alive, the annuitant receives an
annuity for the remainder of his or her life
...
4 Other insurance contracts
The insurance and annuity contracts described above are all contingent on death
or survival
...
These are known as morbidity risks
...
For someone in regular employment, the employer may cover salary for a period, but if the
sickness continues the salary will be decreased, and ultimately will stop
being paid at all
...
Income protection policies replace
at least some income during periods of sickness
...

Critical illness insurance Some serious illnesses can cause significant
expense at the onset of the illness
...
Critical illness
insurance pays a benefit on diagnosis of one of a number of severe conditions, such as certain cancers or heart disease
...

Long-term care insurance This is purchased to cover the costs of care in
old age, when the insured life is unable to continue living independently
...


1
...
The pension plan is usually sponsored by an employer
...
5 Pension benefits

13

sums or annuity benefits or both on retirement, or deferred lump sum or annuity
benefits (or both) on earlier withdrawal
...
The benefits therefore depend on the
survival and employment status of the member, and are quite similar in nature
to life insurance benefits – that is, they involve investment of contributions long
into the future to pay for future life contingent benefits
...
5
...
For
example, suppose an employee reaches retirement age with n years of service
(i
...
membership of the pension plan), and with pensionable salary averaging S
in, say, the final three years of employment
...
The formula may be interpreted as a pension
benefit of, say, 2% of the final average salary for each year of service
...
The contributions are invested, and the accumulated contributions must be enough, on
average, to pay the pensions when they become due
...
The
employee and employer pay a predetermined contribution (usually a fixed percentage of salary) into a fund, and the fund earns interest
...
In the UK most of the proceeds must be converted to an annuity
...


1
...
2 Defined benefit pension design
The age retirement pension described in the section above defines the pension
payable from retirement in a standard final salary plan
...

Many employees leave their jobs before they retire
...
Employees may have the option of taking a lump sum with the

14

Introduction to life insurance

same value as the deferred pension, which can be invested in the pension plan
of the new employer
...
Such benefits might include a lump
sum, often based on salary and sometimes service, as well as a pension for the
employee’s spouse
...
6 Mutual and proprietary insurers
A mutual insurance company is one that has no shareholders
...
All profits are distributed to the
with-profit policyholders through dividends or bonuses
...
The participating policyholders are not owners,
but have a specified right to some of the profits
...

Many early life insurance companies were formed as mutual companies
...

Although it would appear that a mutual insurer would have marketing advantages, as participating policyholders receive all the profits and other benefits of
ownership, the advantages cited by companies who have demutualized include
increased ability to raise capital, clearer corporate structure and improved
efficiency
...
7 Typical problems
We are concerned in this book with developing the mathematical models and
techniques used by actuaries working in life insurance and pensions
...
Premiums must be sufficient to pay benefits;
the assets held must be sufficient to pay the contingent liabilities; bonuses to
policyholders should be fair
...
The sum insured may not be paid for 30 years or more
...

To ensure this, the actuary needs to model the survival probabilities of the
policyholder, the investment returns likely to be earned and the expenses likely

1
...
The actuary may take into consideration
the probability that the policyholder decides to terminate the contract early
...
Then,
when all of these factors have been modelled, they must be combined to set a
premium
...
This is called the valuation process
...

The problems are rather more complex if the insurance also covers morbidity
risk, or involves several lives
...

The actuary may also be involved in decisions about how the premiums
are invested
...
The way the underlying investments are
selected can increase or mitigate the risk of insolvency
...

The pensions actuary working with defined benefit pensions must determine
appropriate contribution rates to meet the benefits promised, using models that
allow for the working patterns of the employees
...
When one company with a pension plan takes
over another, the actuary must assist with determining the best way to allocate
the assets from the two plans, and perhaps how to merge the benefits
...
8 Notes and further reading
A number of essays describing actuarial practice can be found in Renn
(ed
...
This book also provides both historical and more contemporary
contexts for life contingencies
...
Anyone interested in the history of probability and
actuarial science will find these interesting, and remarkably modern
...
9 Exercises
Exercise 1
...
2 Explain why an insurer might demand more rigorous evidence of
a prospective policyholder’s health status for a term insurance than for a whole
life insurance
...
3 Explain why premiums are payable in advance, so that the first
premium is due now rather than in one year’s time
...
4 Lenders offering mortgages to home owners may require the borrower to purchase life insurance to cover the outstanding loan on the death of
the borrower, even though the mortgaged property is the loan collateral
...

(b) Describe how this term insurance might differ from the standard term
insurance described in Section 1
...
2
...
5 Describe the difference between a cash bonus and a reversionary bonus for with-profit whole life insurance
...
6 It is common for insurers to design whole life contracts with
premiums payable only up to age 80
...
7 Andrew is retired
...

He is considering the following options for using the money:
(a) Purchase an annuity from an insurance company that will pay a level amount
for the rest of his life
...

(c) Purchase a 20-year annuity certain
...

(e) Invest the capital and draw $40 000 per year to live on
...
1 Summary
In this chapter we represent the future lifetime of an individual as a random
variable, and show how probabilities of death or survival can be calculated
under this framework
...
We introduce the curtate future lifetime
random variable
...
We explain why this
function is useful and derive its probability function
...
2 The future lifetime random variable
In Chapter 1 we saw that many insurance policies provide a benefit on the
death of the policyholder
...
In order to estimate the time at which
a death benefit is payable, the insurer needs a model of human mortality, from
which probabilities of death at particular ages can be calculated, and this is the
topic of this chapter
...
Let (x) denote a life aged x, where x ≥ 0
...
This means
that x + Tx represents the age-at-death random variable for (x)
...

Then Fx (t) represents the probability that (x) does not survive beyond age
x + t, and we refer to Fx as the lifetime distribution from age x
...

Thus, Sx (t) represents the probability that (x) survives for at least t years, and
Sx is known as the survival function
...

To see this, consider T0 and Tx for a particular individual who is now aged x
...
This individual could have died before reaching age x – the probability of
this was Pr[T0 < x] – but has survived
...
If the individual dies within t years from now, then
Tx ≤ t and T0 ≤ x + t
...
We
achieve this by making the following assumption for all x ≥ 0 and for all t > 0
Pr[Tx ≤ t] = Pr[T0 ≤ x + t|T0 > x]
...
1)

This is an important relationship
...
1) to give
Pr[Tx ≤ t] =

Pr[x < T0 ≤ x + t]
,
Pr[T0 > x]

that is,
Fx (t) =

F0 (x + t) − F0 (x)

...
2)

S0 (x + t)
,
S0 (x)

(2
...
2 The future lifetime random variable

19

which can be written as
S0 (x + t) = S0 (x) Sx (t)
...
4)

This is a very important result
...

Note that Sx (t) can be thought of as the probability that (0) survives to at least
age x + t given that (0) survives to age x, so this result can be derived from the
standard probability relationship
Pr[A and B] = Pr[A|B] Pr[B]
where the events here are A = [T0 > x + t] and B = [T0 > x], so that
Pr[A|B] = Pr[T0 > x + t|T0 > x],
which we know from (2
...

Similarly, any survival probability for (x), for, say, t + u years can be split
into the probability of surviving the first t years, and then, given survival to age
x + t, subsequently surviving another u years
...


(2
...
4), we also know survival probabilities for our individual from
any future age x
...
5) takes this a stage further
...

Any survival function for a lifetime distribution must satisfy the following
conditions to be valid
...
Sx (0) = 1; that is, the probability that a life currently aged x
survives 0 years is 1
...
limt→∞ Sx (t) = 0; that is, all lives eventually die
...
The survival function must be a non-increasing function of t; it
cannot be more likely that (x) survives, say 10
...
5 years, (x) must first survive 10 years
...
5), for all ages
greater than x
...
Sx (t) is differentiable for all t > 0
...

Assumption 2
...

Assumption 3
...

These last two assumptions ensure that the mean and variance of the distribution of Tx exist
...
These three extra assumptions are valid for all
distributions that are feasible for human lifetime modelling
...
1 Let
F0 (t) = 1 − (1 − t/120)1/6

for 0 ≤ t ≤ 120
...

Solution 2
...
9532
...
2), the required probability is
F30 (20) =

F0 (50) − F0 (30)
= 0
...

1 − F0 (30)

(c) From formula (2
...
9395
...
3 The force of mortality

21

We remark that in the above example, S0 (120) = 0, which means that under
this model, survival beyond age 120 is not possible
...
In general, if there is a limiting age, we use
the Greek letter ω to denote it
...

2
...
We denote the force of mortality at age x by µx and define it as
µx = lim

dx→0+

1
Pr[T0 ≤ x + dx | T0 > x]
...
6)

From equation (2
...

dx

(2
...

The force of mortality is best understood by noting that for very small dx,
formula (2
...


(2
...
For example, suppose we
have a male aged exactly 50 and that the force of mortality at age 50 is 0
...
A small value of dx might be a single day, or 0
...
Then the
approximate probability that (50) dies on his birthday is 0
...
00274 =
1
...

We can relate the force of mortality to the survival function from birth, S0
...
7) gives
µx =
=

1
S0 (x) − S0 (x + dx)
lim
S0 (x) dx→0+
dx
1
d
− S0 (x)
...

S0 (x) dx

(2
...

dt
dt

So, it follows from equation (2
...

S0 (x)

We can also relate the force of mortality function at any age x + t, t > 0,
to the lifetime distribution of Tx
...
Then
d (x + t) = dt and so
µx+t = −

1
d
S0 (x + t)
S0 (x + t) d (x + t)

=−

1
d
S0 (x + t)
S0 (x + t) dt

=−

1
d
(S0 (x)Sx (t))
S0 (x + t) dt

=−

S0 (x) d
Sx (t)
S0 (x + t) dt

=

−1 d
Sx (t)
...

Sx (t)

(2
...
3 The force of mortality

23

This relationship gives a way of finding µx+t given Sx (t)
...
9) to develop a formula for Sx (t) in terms of the force of mortality
function
...
9) we have
µx = −

d
log S0 (x),
dx

and integrating this identity over (0, y) yields
y

µx dx = − (log S0 (y) − log S0 (0))
...


(2
...
In other words, the force of mortality
function fully describes the lifetime distribution, just as the function S0 does
...

Example 2
...
1, let
F0 (x) = 1 − (1 − x/120)1/6
for 0 ≤ x ≤ 120
...

Solution 2
...

720 − 6x

24

Survival models

As an alternative, we could use the relationship
µx = −
=

d
d
log S0 (x) = −
dx
dx

1
1
log(1 − x/120) =
6
720(1 − x/120)

1

...
3 Let µx = Bcx , x > 0, where B and c are constants such that
0 < B < 1 and c > 1
...

Derive an expression for Sx (t)
...
3 From equation (2
...


x

Writing cr as exp{r log c},
x+t

x+t

Bcr dr = B

x

exp{r log c}dr
x

=
=

B
exp{r log c}
log c

x+t
x

B
cx+t − cx ,
log c

giving
Sx (t) = exp

−B x t
c (c − 1)
...

At first sight this seems reasonable, but as we will see in the next chapter, the
force of mortality for most populations is not an increasing function of age over
the entire age range
...

Example 2
...
0003 and c = 1
...
Plot the results and comment on the features of
the graphs
...
3 The force of mortality

25

Solution 2
...

log c

S20 (t) = exp

The probability density function is found from (2
...

S20 (t)
log c

Figure 2
...
2
shows the corresponding probability density functions
...

First, we see an effective limiting age, even though, in principle there is no
age to which the survival probability is exactly zero
...
1, we
see that although Sx (t) > 0 for all combinations of x and t, survival beyond age
120 is very unlikely
...
For survival functions that give a more realistic representation
of human mortality, this ordering can be violated, but it usually holds at ages
of interest to insurers
...

Looking at Figure 2
...
For ages
20 and 50, the densities have their respective maximums at (approximately)

1
0
...
8
0
...
6
0
...
4
0
...
2
0
...
1 Sx (t) for x = 20 (bold), 50 (solid) and 80 (dotted)
...
07
0
...
05
0
...
03
0
...
01
0
0

10

20

30

40

50
Time, t

60

70

80

90

100

Figure 2
...


t = 60 and t = 30, indicating that death is most likely to occur around age
80
...
A
further point to note about these density functions is that although each density
function is defined on (0, ∞), the spread of values of fx (t) is much greater for
x = 20 than for x = 50, which, as we will see in Table 2
...

2

2
...
Actuarial science has developed its own notation, International
Actuarial Notation, that encapsulates the probabilities and functions of greatest
interest and usefulness to actuaries
...
We summarize the relevant actuarial
notation in this section, and rewrite the important results developed so far in
this chapter in terms of actuarial functions
...
12)

= Pr[Tx ≤ t] = 1 − Sx (t) = Fx (t),

(2
...


(2
...
4 Actuarial notation

27

So
• t px is the probability that (x) survives to at least age x + t,
• t qx is the probability that (x) dies before age x + t,
• u |t qx is the probability that (x) survives u years, and then dies in the sub-

sequent t years, that is, between ages x + u and x + u + t
...


We may drop the subscript t if its value is 1, so that px represents the probability
that (x) survives to at least age x + 1
...
In actuarial terminology qx is called the mortality rate
at age x
...
5),
from (2
...


1 d
x p0
p0 dx
x

(2
...
16)

Similarly,
µx+t = −
µx+t =
t px

1 d
d
t px ⇒
t px = −t px µx+t ,
dt
t px dt

fx (t)
⇒ fx (t) = t px µx+t
Sx (t)
t

= exp −

µx+s ds

from (2
...
11)
...
17)
(2
...
19)

0

As Fx is a distribution function and fx is its density function, it follows that
t

Fx (t) =

fx (s)ds,

0

which can be written in actuarial notation as
t qx

t

=

s px

µx+s ds
...
20)

0

This is an important formula, which can be interpreted as follows
...
The probability that (x) is alive at time s is s px ,

28

Survival models

Time

0

Age

x

Event



Probability

s

x+s
(x) survives s years

s+ds

x+s+ds
(x)
dies


©

t

x+t

©

µx+s ds

s px

Figure 2
...
Thus
s px µx+s ds can be interpreted as the probability that (x) dies between ages x + s
and x + s + ds
...

We can illustrate this event sequence using the time-line diagram shown in
Figure 2
...

This type of interpretation is important as it can be applied to more
complicated situations, and we will employ the time-line again in later chapters
...
20) becomes
1

qx =

s px

µx+s ds
...
Thus
1

qx ≈

µx+s ds ≈ µx+1/2 ,

0

where the second relationship follows by the mid-point rule for numerical
integration
...
5 As in Example 2
...
Calculate both qx and µx+1/2 for x = 20 and for x = 110,
and comment on these values
...
5 Mean and standard deviation of Tx

29

Solution 2
...
00167 and q110 = 0
...
2, µ20 1 = 0
...
01754
...

2

2
...
We also call this the complete expectation of life
...
17) and (2
...

dt

(2
...


0

We can now use (2
...


0

In Section 2
...

0

(2
...


(2
...
For some lifetime distributions we are able to integrate directly
...
22) and (2
...
The variance
of Tx can then be calculated as
◦

2
V [Tx ] = E Tx − ex

2


...
6 As in Example 2
...
Calculate ex and V[Tx ] for (a) x = 30 and (b) x = 80
...
6 As S0 (x) = (1 − x/120)1/6 , we have
t px

=

S0 (x + t)
t
= 1−
S0 (x)
120 − x

1/6


...
Technically, we have

t px

=

1−

t
120−x

1/6

for x + t ≤ 120,
for x + t > 120
...
22) is 120 − x, and
120−x

◦

ex =
0

1−

t
120 − x

1/6

dt
...
5 Mean and standard deviation of Tx

31

We make the substitution y = 1 − t/(120 − x), so that t = (120 − x)(1 − y),
giving
1

◦

ex = (120 − x)

y1/6 dy

0

= 6 (120 − x)
...
143 and e80 = 34
...

Under this model the expectation of life at any age x is 6/7 of the time to
age 120
...
Using equation (2
...


Again, we substitute y = 1 − t/(120 − x) giving
2
E Tx = 2(120 − x)2

= 2(120 − x)2

1

(y1/6 − y7/6 ) dy

0

6
6
−

...
056515) = ((120 − x) (0
...

So V[T30 ] = 21
...
5092
...

2
A feature of the model used in Example 2
...
For
example, when we model mortality using Gompertz’ law, there is no explicit
◦
formula for ex and we must use numerical integration to calculate moments of
Tx
...


32

Survival models
◦

Table 2
...
Values of ex , SD[Tx ] and expected
age at death for the Gompertz model with
B = 0
...
07
...
938
62
...
703
43
...
252
26
...
550
13
...
848
5
...
152

18
...
579
16
...
841
14
...
746
10
...
449
6
...
246
2
...
938
72
...
703
73
...
752
76
...
550
83
...
848
95
...
152

◦

Table 2
...
0003 and c = 1
...
For this survival model, 130 p0 = 1
...

◦
We see that ex is a decreasing function of x, as it was in Example 2
...
In
◦
that example ex was a linear function of x, but we see that this is not true in
Table 2
...

2
...
6
...
The curtate future lifetime random variable is defined as the integer
part of future lifetime, and is denoted by Kx for a life aged x
...

We can think of the curtate future lifetime as the number of whole years lived
in the future by an individual
...
provided that (x) is alive at these times
...
6 Curtate future lifetime

33

the number of payments made equals the number of complete years lived after
time 0 by (x)
...

We can find the probability function of Kx by noting that for k = 0, 1, 2,
...
Thus
for k = 0, 1, 2,
...

The expected value of Kx is denoted by ex , so that ex = E[Kx ], and is referred to
as the curtate expectation of life (even though it represents the expected curtate
lifetime)
...


(2
...

Similarly,
∞
2
E[Kx ] =

k 2 ( k px −

k+1 px )

k=0

= (1 px − 2 px ) + 4(2 px − 3 px ) + 9(3 px − 4 px ) + 16(4 px − 5 px ) + · · ·
∞

=2

∞

k k px −
k=1

k px
k=1

∞

=2

k k px − e x
...
For more realistic
models, such as Gompertz’, we can calculate the values easily using Excel or
other suitable software
...

◦

2
...
2 The complete and curtate expected future lifetimes, ex and ex
As the curtate future lifetime is the integer part of future lifetime, it is natural
◦
to ask if there is a simple relationship between ex and ex
...


j

j=0

If we approximate each integral using the trapezium rule for numerical
integration (see Appendix B), we obtain
j+1
t px

dt ≈

j

1
2 j px

+ j+1 px ,

and hence
◦

∞

ex ≈

∞
1
2 j px

+ j+1 px =

j=0

1
2

+

j px
...

2

(2
...
Table 2
...
0003 and c = 1
...
Values of ex were calculated
by applying formula (2
...
1
...
25) is
a very good approximation in this particular case for younger ages, but is less
accurate at very old ages
...


2
...
2
...
0003
and c = 1
...

◦

x

ex

ex

0
10
20
30
40
50
60
70
80
90
100

71
...
723
52
...
992
34
...
192
19
...
058
8
...
944
2
...
938
62
...
703
43
...
752
26
...
550
13
...
848
5
...
152

2
...

A simple extension of Gompertz’ law is Makeham’s law (Makeham, 1860),
which models the force of mortality as
µx = A + Bcx
...
26)

This is very similar to Gompertz’law, but adds a fixed term that is not age related,
that allows better for accidental deaths
...

In recent times, the Gompertz–Makeham approach has been generalized
further to give the GM(r, s) (Gompertz–Makeham) formula,
µx = h1 (x) + exp{h2 (x)},
r
s
where h1 and h2 are polynomials in x of degree r and s respectively
...
(1988)
...

In Section 2
...
A further
significant point is that when mortality data are analysed, the force of mortality

36

Survival models

is a natural quantity to estimate, whereas the lifetime distribution is not
...

An excellent summary article on this topic is Macdonald (1996)
...


2
...
1 Let F0 (t) = 1 − (1 − t/105)1/5 for 0 ≤ t ≤ 105
...


Exercise 2
...

(a)
(b)
(c)
(d)
(e)
(f)

What is the implied limiting age ω?
Verify that the function G satisfies the criteria for a survival function
...

Determine the survival function for a life aged 20
...

Calculate the force of mortality at age 50
...
3 Calculate the probability that a life aged 0 will die between ages
19 and 36, given the survival function
S0 (x) =

1√
100 − x,
10

0 ≤ x ≤ 100 (= ω)
...
4 Let
1
C
C
Dx −
S0 (x) = exp − Ax + Bx2 +
2
log D
log D
where A, B, C and D are all positive
...
8 Exercises
(a)
(b)
(c)
(d)

37

Show that the function S0 is a survival function
...

Derive a formula for µx
...
00005,
(i)
(ii)
(iii)
(iv)
(v)

B = 0
...
0003,

D = 1
...


Calculate t p30 for t = 1, 5, 10, 20, 50, 90
...

Calculate t |10 q30 for t = 1, 10, 20
...

◦
Calculate ex for x = 70, 71, 72, 73, 74, 75, using numerical integration
...
5 Let F0 (t) = 1 − e−λt , where λ > 0
...

Show that µx = λ
...

What conclusions do you draw about using this lifetime distribution to
model human mortality?

Exercise 2
...
99, px+1 = 0
...
95 and qx+3 =
0
...

2 px

Exercise 2
...


Exercise 2
...
001 x

2

for x ≥ 0,

38

Survival models

find expressions for, simplifying as far as possible,
(a) f0 (x), and
(b) µx
...
9 Show that
d
t px = t px (µx − µx+t )
...
10 Suppose that Gompertz’ law applies with µ30 = 0
...
000344
...

Exercise 2
...


(a) Show that under Makeham’s law
t px

= st g c

x (ct −1)

where s = e−A and g = exp{−B/ log c}
...
27)

and 10 p70
...
1


...
12 (a) Construct a table of px for Makeham’s law with parameters
A = 0
...
00035 and c = 1
...
You should set
the parameters so that they can be easily changed, and you should keep the
table, as many exercises and examples in future chapters will use it
...

(c) Use the table to calculate e70
...

Exercise 2
...

(a) Show that, if * represents smokers’ mortality, and the ‘unstarred’ function
represents non-smokers’ mortality, then
∗
t px

= (t px )2
...
8 Exercises

39

(b) Calculate the difference between the life expectancy of smokers and nonsmokers aged 50, assuming that non-smokers mortality follows Gompertz’
law, with B = 0
...
07
...

Hint: You will need to use numerical integration for parts (b) and (c)
...
14 (a) Show that
◦

◦

ex ≤ ex+1 + 1
...

(c) Explain (in words) why
1
◦
ex ≈ ex +
...
15 (a) Show that
o

ex =

1
S0 (x)

∞

S0 (t)dt,

x

where S0 (t) = 1 − F0 (t), and hence, or otherwise, prove that
d o
o
ex = µx ex − 1
...
What about
dx a
dx
(b) Deduce that

a

Hint:

g(t)dt ?
x

o

x + ex
is an increasing function of x, and explain this result intuitively
...
1 (a) 0
...
8586
(c) 0
...
2 (a)
(c)
(d)
(e)
(f)
2
...
1
2
...
6

2
...
10
2
...
13

0
...
28
45
...
18
90
0
...
1169
0
...
9976, 0
...
0047, 0
...
0349, 0
...
046, 12
...
544, 13
...
98
(b) 0
...
96939
(d) 0
...
03031
(d) 0
...
08218
0
...
339
(d) 9
...
432
(c) 125
...
9672,
0
...
1082
12
...
498,

0
...
3812,

3
...
499,
11
...
009,
11
...
533
11
...
11 (smokers)

3
Life tables and selection

3
...
For a life table tabulated at integer ages
only, we show, using fractional age assumptions, how to calculate survival
probabilities for all ages and durations
...

We then consider life tables appropriate to individuals who have purchased
particular types of life insurance policy and discuss why the survival probabilities differ from those in the corresponding national life table
...
We define a select survival model and we derive some formulae
for such a model
...
2 Life tables
Given a survival model, with survival probabilities t px , we can construct the life
table for the model from some initial age x0 to a maximum age ω
...
Let lx0 be an arbitrary positive number
(called the radix of the table) and, for 0 ≤ t ≤ ω − x0 , define
lx0 +t = lx0 t px0
...


(3
...
This interpretation is more
natural if lx is an integer, and follows because the number of survivors to age
x + t is a random variable with a binomial distribution with parameters lx and
t px
...
Then the number of survivors to age
x + t is a binomial random variable, Lt , say, with parameters lx and t px
...

We always use the table in the form ly /lx which is why the radix of the table is
arbitrary – it would make no difference to the survival model if all the lx values
were multiplied by 100, for example
...
1) we can use the lx function to calculate survival probabilities
...
For example,
q30 = 1 −

l31
l30 − l31
=
l30
l30

(3
...

l40

(3
...
In practice, it is very common for a life table to be presented, and in
some cases even defined, at integer ages only
...

It is usual for a life table, tabulated at integer ages, to show the values of dx ,
where
dx = lx − lx+1 ,
in addition to lx , as these are used to compute qx
...
4) we have
dx = lx 1 −

lx+1
lx

= lx (1 − px ) = lx qx
...
4)

3
...
1
...

x

lx

dx

30
31
32
33
34
35
36
37
38
39

10 000
...
22
9 927
...
35
9 839
...
29
9 734
...
56
9 607
...
08

34
...
10
41
...
81
50
...
17
60
...
49
72
...
11

We can also arrive at this relationship if we interpret dx as the expected number
of deaths in the year of age x to x + 1 out of lx lives aged exactly x, so that,
using the binomial distribution again
dx = lx qx
...
5)

Example 3
...
1 gives an extract from a life table
...


Solution 3
...
4),
l40 = l39 − d39 = 9 453
...

(b) From equation (3
...
97
= 0
...

=
l30
10 000

(c) From equation (3
...
17
=
= 0
...

l35
9 789
...
2),
5 q30

=

l30 − l35
= 0
...

l30

(e) This probability is 5 | q30
...
3),
5 | q30

=

l35 − l36
d35
=
= 0
...

l30
l30
2

3
...
However, a life table tabulated at integer ages only
does not contain all the information in the corresponding survival model, since
values of lx at integer ages x are not sufficient to be able to calculate probabilities involving non-integer ages, such as 0
...
5
...
Specifically, we need to
make some assumption about the probability distribution for the future lifetime
random variable between integer ages
...

It may be specified in terms of the force of mortality function or the survival or
mortality probabilities
...


3
...
1 Uniform distribution of deaths
The uniform distribution of deaths (UDD) assumption is the most common
fractional age assumption
...

UDD1
For integer x, and for 0 ≤ s < 1, assume that
s qx

= sqx
...
6)

UDD2
Recall from Chapter 2 that Kx is the integer part of Tx , and define a new
random variable Rx such that
Tx = Kx + Rx
...
3 Fractional age assumptions

45

The UDD2 assumption is that, for integer x, Rx ∼ U(0, 1), and Rx is
independent of Kx
...
First,
assume that UDD1 is true
...

This proves that Rx ∼ U(0, 1)
...
This proves that UDD1 implies UDD2
...
Then for
integer x, and for 0 ≤ s < 1,
s qx

= Pr[Tx ≤ s]
= Pr[Kx = 0 and Rx ≤ s]
= Pr[Rx ≤ s] Pr[Kx = 0]

46

Life tables and selection

as Kx and Rx are assumed independent
...


(3
...

An immediate consequence is that
lx+s = lx − s dx

(3
...
This follows because
s qx

=1−

lx+s
lx

and substituting s qx for s qx gives
s

dx
lx − lx+s
=

...
Thus, we assume that lx+s is a linearly decreasing function of s
...
6) with respect to s, we obtain
d
s qx = qx ,
ds

0≤s≤1

and we know that the left-hand side is the probability density function for Tx
at s, because we are differentiating the distribution function
...
9)

for 0 ≤ s < 1
...

Since qx is constant with respect to x, and s px is a decreasing function of
s, we can see that µx+s is an increasing function of s, which is appropriate
for ages of interest to insurers
...
3 Fractional age assumptions

47

with discontinuities occurring at integer ages, as illustrated in Example 3
...

Although this is undesirable, it is not a serious drawback
...
2 Given that p40 = 0
...


0
...
2

under the

Solution 3
...
7), that for
fractional of a year s, s qx = s qx , requires x to be an integer
...
4 q40
...
4 q40
...
4 p40
...
6 p40
0
...
6
l40
...
6q40
1 − 0
...
108 × 10−4
...
3 Use the life table in Example 3
...
7 q33 and (b) 1
...
5
...
3 (a) We note first that
1
...
7 p33 = 1 − (p33 ) (0
...


We can calculate p33 directly from the life table as l34 /l33 = 0
...
7 p34 = 1 − 0
...
996424 under UDD, so that 1
...
008192
...
7 q33
...
7 q33
...
7 p33
...
2
l33
...
2d35
l33 − 0
...
008537
...
4 Under the assumption of a uniform distribution of deaths, calculate lim µ40+t using p40 = 0
...
999429
...
4 From formula (3
...
Setting x = 40
yields
lim µ40+t = q40 /p40 = 5
...
71 × 10−4
...
5 Given that q70 = 0
...
011670, calculate 0
...
6
assuming a uniform distribution of deaths
...
5 As deaths are assumed to be uniformly distributed between ages
70 to 71 and ages 71 to 72, we write
0
...
6

=

0
...
6

+ (1 −

0
...
6 ) 0
...


Following the same arguments as in Solution 3
...
4 q70
...
191 × 10−3 ,
1 − 0
...
3 q71 = 0
...
501 × 10−3 , we obtain 0
...
6 = 7
...
2

3
...
2 Constant force of mortality
A second fractional age assumption is that the force of mortality is constant
between integer ages
...
We can obtain the value of µ∗ by
x
x
using the fact that
1

px = exp −

µx+s ds
...
Further, under the assumption of a constant force of mortality,
x
for 0 ≤ s < 1 we obtain
s px

s

= exp −
0

∗

µ∗ du = e−µx s = (px )s
...

x

3
...

The assumption of a constant force of mortality leads to a step function for
the force of mortality over successive years of age
...
However, if
the true force of mortality increases slowly over the year of age, the constant
force of mortality assumption is reasonable
...
6 Given that p40 = 0
...

Solution 3
...
4 q40
...
4 p40
...
4 q40
...
4 = 2
...

2

Example 3
...
010413 and q71 = 0
...
7 q70
...

Solution 3
...
5 we write
0
...
6

=

0
...
6

+ (1 −

0
...
6 ) 0
...
4 q70
...
4 = 4
...
515 × 10−3 , giving 0
...
6 = 7
...


0
...
3 =
2

Note that in Examples 3
...
5 and in Examples 3
...
7 we have used two
different methods to solve the same problems, and the solutions agree to five
decimal places
...
The reason for this can be seen from the following approximations
...


In other words, the approximation to t qx is t times the approximation to qx ,
which is what we obtain under the uniform distribution of deaths assumption
...
4 National life tables
Life tables based on the mortality experience of the whole population of a
country are regularly produced for many countries in the world
...
2
...

Australian Life Tables
2000–02

English Life Table 15
1990–92

US Life Tables
2002

x

Males

Females

Males

Females

Males

Females

0
1
2
10
20
30
40
50
60
70
80
90
100

567
44
31
13
96
119
159
315
848
2 337
6 399
15 934
24 479

466
43
19
8
36
45
88
202
510
1 308
4 036
12 579
23 863

814
62
38
18
84
91
172
464
1 392
3 930
9 616
20 465
38 705

632
55
30
13
31
43
107
294
830
2 190
5 961
15 550
32 489

764
53
37
18
139
141
266
570
1 210
2 922
7 028
16 805
−

627
42
28
13
45
63
149
319
758
1 899
4 930
13 328
−

tables are usually produced for males and for females and possibly for some
other groups of individuals, for example on the basis of smoking habits
...
2 shows values of qx × 105 , where qx is the probability of dying
within one year, for selected ages x, separately for males and females, for the
populations of Australia, England & Wales and the United States
...

The relevant years are indicated in the column headings for each of the three
life tables in Table 3
...
Data at the oldest ages are notoriously unreliable
...

For all three national life tables and for both males and females, the values
of qx follow exactly the same pattern as a function of age, x
...
1 shows
the US 2002 mortality rates for males and females; the graphs for England
& Wales and for Australia are similar
...
Also, although the
information plotted consists of values of qx for x = 0, 1,
...
) We note the following
points from Table 3
...
1
...
Mortality rates immediately following

birth, perinatal mortality, are high due to complications arising from the

3
...
1

0
...
001

0
...
1 US 2002 mortality rates, male (dotted) and female (solid)
...
The value of qx
does not reach this level again until about age 55
...
1
...

In Figure 3
...

Not only is this feature of mortality for young adult males common for different populations around the world, it is also a feature of historical populations
in countries such as the UK where mortality data has been collected for some
time
...

Mortality rates increase from age 10, with the accident hump creating a
relatively large increase between ages 10 and 20 for males, a more modest
increase from ages 20 to 40, and then steady increases from age 40
...
The one exception is the Australian Life Table, where q100
is slightly higher for a female than for a male
...


52

Life tables and selection
0
...
30

Mortality rates

0
...
20
0
...
10
0
...
00
50

60

70

Age

80

90

100

Figure 3
...

• The Gompertz model introduced in Chapter 2 is relatively simple, in that

it requires only two parameters and has a force of mortality with a simple
functional form, µx = Bcx
...
We can see from Figure 3
...
However,
the mortality rates at later ages are rather better behaved, and the Gompertz
model often proves useful over older age ranges
...
2 shows the older
ages US 2002 Males mortality rate curve, along with a Gompertz curve fitted
to the US 2002 Table mortality rates
...

A final point about Table 3
...
This is because values of µx are not published for any
ages for the US Life Tables
...


3
...
5 Survival models for life insurance policyholders

53

Table 3
...
Values of the force of mortality ×105
from English Life Table 15 and CMI (Table A14)
for UK males who purchase a term insurance
policy at age 50
...
We could use a national life table, say English Life Table 15, so that,
for example, we could assume that the probability this man dies before age
51 is 0
...
2
...
The mortality of different types of life insurance policyholders is investigated separately, and life tables appropriate for these groups
are published
...
3 shows values of the force of mortality (×105 ) at two-year intervals
from age 50 to age 60 taken from English Life Table 15, Males (ELTM 15), and
from a life table prepared from data relating to term insurance policyholders
in the UK in 1999–2002 and which assumes the policyholders purchased their
policies at age 50
...

Hereafter we refer to this working paper as CMI, and further details are given
at the end of this chapter
...
2
...
3 is the difference between the two sets of
values
...

There are at least three reasons for this difference
...

(a) The data on which the two life tables are based relate to different calendar
years; 1990–92 in the case of ELTM 15 and 1999–2002 in the case of CMI
...
However, this explains only a small part of the differences in Table
3
...
An interim life table for England & Wales based on population data from
2002–2004, gives the following values for males: µ50 = 391 × 10−5 and
µ60 = 1008 × 10−5
...
3
...
3 are
needed
...
3 is that
ELTM 15 is a life table based on the whole male population of England
& Wales, whereas CMI (Table A14) is based on the experience of males
who are term insurance policyholders
...
This is true
in the case of the population of England and Wales, where social class,
defined in terms of occupation, has a significant effect on mortality
...
Given that people who purchase term insurance
policies are likely to be among the better paid people in the population, we
have an explanation for a large part of the difference between the values in
Table 3
...

CMI (Table A2) shows values of the force of mortality based on data from males
in the UK who purchased whole life or endowment insurance policies
...
3 for term insurance policyholders and hence
much lower than the values for the whole population
...


3
...
3 taken from CMI are values
based on data for males who purchased term insurance at age 50
...
Values for ages at purchase 50, 52, 54 and 56 are shown in Table 3
...

There are two significant features of the values in Table 3
...

(a) Consider the row of values for age 56
...

The only difference is that they purchased their policies at different ages
...
For example, for a male who purchased his policy at age 56, the value

3
...
4
...

Age at purchase of policy
x

50

52

54

56

50
52
54
56
58
60
62
64
66

78
152
240
360
454
573
725
917
1159

—
94
186
295
454
573
725
917
1159

—
—
113
227
364
573
725
917
1159

—
—
—
136
278
448
725
917
1159

is 0
...
00360
...
These values, all equal to
0
...

These features are due to life insurance underwriting, which we described in
Chapter 1
...

The important point for this discussion is that the mortality rates in CMI
are based on individuals accepted for insurance at normal premium rates, i
...

individuals who have passed the required health checks
...
This
explains a major part of the difference between the mortality rates in Table 3
...

When this man reaches age 56, we can no longer be certain he is in good health
– all we know is that he was in good health six years ago
...
This explains the differences between the values of the force of mortality
at age 56 in Table 3
...


56

Life tables and selection

The effect of passing the health checks at an earlier age eventually wears
off, so that at age 62, the force of mortality does not depend on whether the
policy was purchased at age 52, 54 or 56
...
However,
note that these rates, 0
...
01664) from
ELTM 15
...
In fact the population is made
up of many heterogeneous lives, and the effect of initial selection is only one
area where actuaries have tried to manage the heterogeneity
...
These
preferred lives tend to be from higher socio-economic groups
...


3
...
For example, for a
given survival model and a given term t, t px , the probability that an individual
currently aged x will survive to age x + t, depends only on the current age x
...

The difference between an aggregate survival model and the survival model
for term insurance policyholders discussed in Section 3
...
e
...

This leads us to the following definition
...

(a) Future survival probabilities for an individual in the group depend on the
individual’s current age and on the age at which the individual joined the
group
...
The initial selection effect
is assumed to have worn off after d years
...
An individual
who enters the group at, say, age x, is said to be selected, or just select, at age
x
...
7 Select and ultimate survival models

57

probabilities is called the select period for the model
...

Going back to the term insurance policyholders in Section 3
...
A select survival
model is appropriate in this case because passing the health checks at age x
indicates that the individual is in good health and so has lower mortality rates
than someone of the same age who passed these checks some years ago
...
4 that the select period, d , for this group is less than or
equal to six years
...
6
...
Select periods typically range from one year to
15 years for life insurance mortality models
...
6, being selected at age x
meant that the mortality rate for the individual was lower than that of a term
insurance policyholder of the same age who had been selected some years
earlier
...
8 shows
...
8 Consider men who need to undergo surgery because they are
suffering from a particular disease
...
If
they do survive for a year, then they are fully cured and their future mortality
follows the Australian Life Tables 2000–02, Males, from which you are given
the following values:
l60 = 89 777,

l61 = 89 015,

l70 = 77 946
...

Solution 3
...

Being selected at age x means having surgery at age x
...


58

Life tables and selection

(a) The probability of surviving to age 61 is 0
...
Given that he survives to age
61, the probability of surviving to age 70 is
l70 /l61 = 77 946/89 015 = 0
...

Hence, the probability that this individual survives from age 60 to age 70 is
0
...
8757 = 0
...

(b) Since this individual has already survived for one year following surgery,
his mortality follows the Australian Life Tables 2000–02, Males
...
8682
...

Hence, his probability of surviving to age 70 is 0
...

2
Selection is not a feature of national life tables since, ignoring immigration, an
individual can enter the population only at age zero
...
We can see from Tables 3
...
4 that its effect on
the force of mortality can be considerable
...

The select period may be different for different survival models
...

In the next section we introduce notation and develop some formulae for
select survival models
...
8 Notation and formulae for select survival models
A select survival model represents an extension of the ultimate survival model
studied in Chapter 2
...
For a select survival model, probabilities of
survival depend on current age and (within the select period) age at selection, i
...
age at joining the group
...
This means that, provided we
fix and specify the age at selection, we can adapt the notation and formulae

3
...
This leads to the following
definitions:
S[x]+s (t) = Pr[a life currently aged x + s who was select at age x survives to
age x + s + t],
t q[x]+s = Pr[a life currently aged x + s who was select at age x dies before age
x + s + t],
t p[x]+s

= 1 − t q[x]+s ≡ S[x]+s (t),

µ[x]+s is the force of mortality at age x + s for an individual who was select
at age x,
µ[x]+s = limh→0+

1−S[x]+s (h)
h


...


This formula is derived precisely as in Chapter 2
...

For a select survival model with a select period d and for t ≥ d , that is, for
durations at or beyond the select period, the values of µ[x−t]+t , s p[x−t]+t and
u|s q[x−t]+t do not depend on t, they depend only on the current age x
...
For values of t < d , we refer to, for example,
µ[x−t]+t as being in the select part of the survival model and for t ≥ d we refer
to µ[x−t]+t (≡ µx ) as being in the ultimate part of the survival model
...
9 Select life tables
For an ultimate survival model, as discussed in Chapter 2, the life table {lx }
is useful since it can be used to calculate probabilities such as t |u qx for nonnegative values of t, u and x
...
Suppose we wish to construct this table for a select survival model for
ages at selection from, say, x0 (≥ 0)
...

The construction in this section is for a select life table specified at all ages
and not just at integer ages
...

First we consider the survival probabilities of those individuals who were
selected at least d years ago and hence are now subject to the ultimate part
of the model
...
For these people,

60

Life tables and selection

future survival probabilities depend only on their current age and so, as in
Chapter 2, we can construct an ultimate life table, {ly }, for them from which we
can calculate probabilities of surviving to any future age
...
For y ≥ x0 + d we define
ly =

(y−x0 −d ) px0 +d lx0 +d
...
10)

Note that (y−x0 −d ) px0 +d = (y−x0 −d ) p[x0 ]+d since, given that the life was select
at least d years ago, the probability of future survival depends only on the
current age, x0 + d
...


(3
...


(x−x0 −d ) px0 +d

lx0 +d

lx0 +d

This shows that within the ultimate part of the model we can interpret ly as the
expected number of survivors to age y out of lx lives currently aged x (< y),
who were select at least d years ago
...
10) defines the life table within the ultimate part of the model
...
We do this for a
life select at age x by ‘working backwards’ from the value of lx+d
...
12)

which means that if we had l[x]+t lives aged x + t, selected t years ago, then the
expected number of survivors to age x + d is lx+d
...

Example 3
...
13)

l[x]+s

...
14)

=

and
s−t p[x]+t

=

3
...
9 First,
=

y−x−d p[x]+d d −t p[x]+t

=

y−x−d px+d d −t p[x]+t

=

ly lx+d
lx+d l[x]+t

=

ly
,
l[x]+t

=

d −t p[x]+t / d −s p[x]+s

=

lx+d l[x]+s
l[x]+t lx+d

=

y−x−t p[x]+t

l[x]+s
,
l[x]+t

which proves (3
...
Second,
s−t p[x]+t

which proves (3
...


2

This example, together with formula (3
...
For example, suppose we have
l[x]+t individuals currently aged x + t who were select at age x
...
13) that ly is the expected number of survivors to age y
...
14) shows that l[x]+s can be interpreted as the
expected number of survivors to age x + s out of l[x]+t lives currently aged x + t
who were select at age x
...
10 Write an expression for 2 |6 q[30]+2 in terms of l[x]+t and ly for
appropriate x, t, and y, assuming a select period of five years
...
10 Note that 2 |6 q[30]+2 is the probability that a life currently aged
32, who was select at age 30, will die between ages 34 and 40
...


62

Life tables and selection
Table 3
...
An extract
from the US Life Tables,
2002, Females
...

l[30]+2

1−

Note that l[30]+10 ≡ l40 since 10 years is longer than the select period for this
survival model
...
11 A select survival model has a select period of three years
...
Some lx
values for this table are shown in Table 3
...

You are given that for all ages x ≥ 65,
p[x] = 0
...
998,

p[x−2]+2 = 0
...


Calculate the probability that a woman currently aged 70 will survive to age 75
given that
(a)
(b)
(c)
(d)

she was select at age 67,
she was select at age 68,
she was select at age 69, and
she is select at age 70
...
11 (a) Since the woman was select three years ago and the select
period for this model is three years, she is now subject to the ultimate part of
the survival model
...
9 Select life tables

63

where the ls are taken from US Life Tables, 2002, Females
...
8913
...


We can calculate l[68]+2 by noting that
l[68]+2 p[68]+2 = l[68]+3 = l71 = 79 026
...
997
...
9058
...


We can calculate l[69]+1 by noting that
l[69]+1 p[69]+1 p[69]+2 = l[69]+3 = l72 = 77 410
...
998 and p[69]+2 = 0
...
Hence,
l[69]+1 = 77 799 and so
5 p[69]+1

= 71 800/77 799 = 0
...


(d) In this case, the required probability is
5 p[70]

= l[70]+5 /l[70] = l75 /l[70] = 71 800/l[70]
...
997 × 0
...
999) = 76 122
...
9432
...
6
...


Age, x

Duration 0
q[x]

Duration 1
q[x−1]+1

Duration 2+
qx

60
61
62
63
···
70
71
72
73
74
75

0
...
003856
0
...
004779
···
0
...
011858
0
...
015184
0
...
019664

0
...
005059
0
...
006304
···
0
...
015868
0
...
020302
0
...
026196

0
...
005351
0
...
006781
···
0
...
017832
0
...
022759
0
...
029048

Example 3
...

It has a select period of two years
...
6
...

Solution 3
...
Since the select period is two years q[x−t]+t ≡ qx for
t ≥ 2
...
Select life tables, tabulated at integer ages,
can be set out in different ways – for example, each row could relate to a fixed
age at selection – so care needs to be taken when using such tables
...
989481 × 0
...
979855 × 0
...
932447
...
9 Select life tables

65

(b) We calculate 3 q[60]+1 as
3 q[60]+1

= q[60]+1 + p[60]+1 q62 + p[60]+1 p62 q63
= q[60]+1 + (1 − q[60]+1 ) q62 + (1 − q[60]+1 ) (1 − q62 ) q63
= 0
...
994941 × 0
...
994941 × 0
...
006781
= 0
...


(c) We calculate 2 |q73 as
2 |q73

= 2 p73 q75
= (1 − q73 ) (1 − q74 ) q75
= 0
...
974288 × 0
...
027657
...
13 A select survival model has a two-year select period and is
specified as follows
...
00022, B = 2
...
124
...
92−s µx+s
...
, 82,
(b) l[x]+1 for x = 20, 21,
...
, 80
...
13 First, note that
t px

= exp −At −

B x t
c (c − 1)
log c

and for 0 ≤ t ≤ 2,
t p[x]

t

= exp −
0

µ[x]+s ds

= exp 0
...
9t
ct − 0
...
9)
log(0
...


(3
...
7
...


x

20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49

l[x]

99 995
...
04
99 944
...
81
99 892
...
69
99 838
...
20
99 781
...
69
99 721
...
36
99 656
...
23
99 586
...
01
99 509
...
12
99 424
...
52
99 327
...
69
99 217
...
42
99 090
...
67
98 940
...
38
98 764
...
15

l[x]+1

lx+2

x+2

x

l[x]

l[x]+1

lx+2

x+2

99 973
...
40
99 922
...
43
99 869
...
38
99 814
...
70
99 756
...
70
99 694
...
48
99 627
...
96
99 554
...
73
99 474
...
02
99 384
...
62
99 283
...
72
99 166
...
80
99 030
...
40
98 869
...
94
98 679
...
40

100 000
...
04
99 949
...
98
99 897
...
08
99 843
...
86
99 787
...
71
99 727
...
83
99 663
...
26
99 593
...
75
99 517
...
75
99 433
...
29
99 338
...
88
99 229
...
41
99 104
...
94
98 957
...
50
98 783
...
88
98 576
...
24

20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51

50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80

98 552
...
98
98 297
...
81
97 987
...
07
97 607
...
05
97 142
...
22
96 568
...
34
95 858
...
51
94 981
...
11
93 895
...
63
92 551
...
58
90 891
...
62
88 846
...
25
86 339
...
49
83 282
...
30
79 584
...
70
75 153
...
67
98 318
...
79
98 013
...
44
97 640
...
18
97 182
...
80
96 617
...
48
95 920
...
80
95 057
...
72
93 989
...
38
92 667
...
03
91 034
...
15
89 023
...
03
86 555
...
37
83 545
...
54
79 901
...
44
75 531
...
22

98 326
...
77
98 022
...
20
97 651
...
17
97 195
...
59
96 634
...
75
95 940
...
43
95 082
...
73
94 020
...
05
92 706
...
88
91 082
...
96
89 082
...
84
86 627
...
46
83 632
...
34
80 006
...
35
75 657
...
31
70 507
...
, 82
...
, 80
...
7
...
, 80
...
11 Exercises

67

3
...
4 are drawn from the following sources:
• Australian Life Tables 2000–02 were produced by the Australian Government

Actuary (2004)
...

• US Life Tables 2002 were prepared in the Division of Vital Statistics of the

National Center for Health Statistics in the US – see Arias (2004)
...
Findings on mortality and morbidity experience of UK policyholders are
published via a series of formal reports and working papers
...

In Section 3
...
Coleman and
Salt (1992) give a very good account of this variability in the UK population
...


3
...
1 Sketch the following as functions of age x for a typical (human)
population, and comment on the major features
...

Exercise 3
...


Age, x

lx

52
53
54
55
56
57
58
59
60

89 948
89 089
88 176
87 208
86 181
85 093
83 940
82 719
81 429

68

Life tables and selection

Calculate
(a)
(b)
(c)
(d)
(e)
(f)

0
...
4

assuming UDD (fractional age assumption),
assuming constant force of mortality (fractional age assumption),
p52
...
7
5
...
4 assuming constant force of mortality,
3
...
5 q52
...
2 |2
...
4 assuming constant force of mortality
...
2 q52
...
3 Table 3
...
Note carefully the layout – each row relates to a
fixed age at selection
...

Table 3
...
Extract from a (hypothetical)
select life table
...


...


15 930
15 508
15 050

...


...


...


15 286
14 816
14 310

...


...


...


12 576
11 928
11 250
10 542
9 812
9 064

82
83
84
85
86
87

80
81
82
83
84
85

Exercise 3
...
It has a select period
of five years
...
9
...
11 Exercises

69

Table 3
...
Mortality rates for female non-smokers who have term
insurance policies
...
003974
0
...
004704
0
...
005870
0
...
007381
0
...
009281

0
...
005411
0
...
006651
0
...
008361
0
...
010514
0
...
005984
0
...
007229
0
...
009043
0
...
011370
0
...
014299

0
...
007663
0
...
009481
0
...
011919
0
...
014988
0
...
007994
0
...
009754
0
...
012216
0
...
015360
0
...
019316

0
...
010599
0
...
013318
0
...
016742
0
...
021053
0
...

Exercise 3
...
It has a select period of
five years
...
10
...
8 q[70]+0
...

Table 3
...
Mortality rates for female smokers who have term
insurance policies
...
010373
0
...
012458
0
...
015308
0
...
018714
0
...
013099
0
...
015825
0
...
019446
0
...
023772
0
...
015826
0
...
019192
0
...
023584
0
...
028830
0
...
018552
0
...
022559
0
...
027721
0
...
033888
0
...
021279
0
...
025926
0
...
031859
0
...
038946
0
...
026019
0
...
032133
0
...
039486
0
...
048270
0
...
6 A select survival model has a select period of three years
...
01601,
2 |q[50]

= 0
...
96411,

2 |3 q[50]+1

= 0
...


Exercise 3
...
For those who
have lived in country A for at least five years the force of mortality at each age
is 50% greater than that of US Life Tables, 2002, Females, at the same age
...
11
...
11
...

Age, x
30
31
32
33
34
35
···
40

lx
98 424
98 362
98 296
98 225
98 148
98 064
···
97 500

Calculate the probability that an employee posted to country A at age 30 will
survive to age 40 if she remains in that country
...
8 A special survival model has a select period of three years
...
Functions without an asterisk
are taken from the Canada Life Tables 2000–02, Males
...


3
...
12
...

Age, x

lx

15
16
17
18
19
20
21
22
23
24
25
26

...


...


...


62
63
64
65

87 503
86 455
85 313
84 074

A life table, tabulated at integer ages, is constructed on the basis of the special
∗
survival model and the value of l25 is taken as 98 363 (i
...
l26 for Canada Life
Tables 2000–02, Males)
...
12
...

∗
∗
∗
∗
∗
(b) Calculate 2 |38 q[21]+1 , 40 p[22] , 40 p[21]+1 , 40 p[20]+2 , and 40 p22
...
9 (a) Show that a constant force of mortality between integer ages
implies that the distribution of Rx , the fractional part of the future life time,
conditional on Kx = k, has the following truncated exponential distribution
for integer x, for 0 ≤ s < 1 and for k = 0, 1,
...
16)

where µ∗ = − log px+k
...
16) holds for k = 0, 1, 2,
...

Exercise 3
...
15)
...
2 (a) 0
...
001917
(c) 0
...
935423
(e) 0
...
030950
3
...
66177
(b) 0
...
08993
3
...
987347
(b) 0
...
010514
(d) 0
...
5 (a) 0
...
055008
(c) 0
...
6 0
...
7 0
...
8 (a) The values are as follows:
x
20
21
22
(b) 0
...
872587,

0
...
875937,

0
...


4
Insurance benefits

4
...
In particular, we consider whole life, term and endowment insurance
...
The functions we develop for traditional benefits will also
be useful when we move to modern variable contracts
...
We introduce a new random variable, Kx , which we use
to value benefits which depend on the number of complete periods of length
1/m years lived by a life (x)
...

We also introduce the actuarial notation for the expected values of the present
value of insurance benefits
...
2 Introduction
In the previous two chapters, we have looked at models for future lifetime
...
Because of the dependence on death or survival, the
timing and possibly the amount of the benefit are uncertain, so the present value
of the benefit can be modelled as a random variable
...

We generally assume in this chapter (and in the following three chapters) that
the interest rate is constant and fixed
...
In Chapter 10 we introduce
more realistic term structures, and consider some models of interest that allow
for uncertainty
...
In the case of a death benefit, working
in continuous time means that we assume that the death payment is paid at the
exact time of death
...

Clearly both assumptions are impractical; it will take time to process a payment after death, and annuities will be paid at most weekly, not every moment
(though the valuation of weekly payments is usually treated as if the payments
were continuous, as the difference is very small)
...
In addition, when the survival model being
used is in the form of a life table with annual increments (that is, lx for integer x),
it is simplest to use annuity and insurance present value functions that assume
payments are made at integer durations only
...
It is then straightforward to adapt the results from
continuous time analysis to discrete time problems
...
3 Assumptions
To perform calculations in this chapter, we require assumptions about mortality
and interest
...

Throughout this chapter we illustrate the results with examples using the
same survival model, which we call the Standard Ultimate Survival Model:
Makeham’s law with A = 0
...
7 × 10−6
c = 1
...
This model is the ultimate part of the model used
in Example 3
...
We will also assume a constant rate of interest
...
However, it is
a convenient assumption from a pedagogical point of view, is often accurate

4
...

It is also convenient to work with other interest theory functions that are in
common actuarial and financial use
...

We use
v=

1
1+i

as a shorthand for discounting
...
The force of interest per year is denoted δ where
δ = log(1 + i),

1 + i = eδ ,

and

v = e−δ ;

δ is also known as the continuously compounded rate of interest
...

The nominal rate of interest compounded p times per year is denoted i(p)
where
i(p) = p (1 + i)1/p − 1 ⇔ 1 + i = 1 + i(p) /p

p


...


4
...
4
...

Since the present value of a future payment depends on the payment date, the
present value of the benefit payment is a function of the time of death, and is
therefore modelled as a random variable
...

We start by considering the value of a benefit of amount $1 payable following
the death of a life currently aged x
...

We first assume that the benefit is payable immediately on the death of (x)
...
Although in practice there would normally be
a short delay between the date of a person’s death and the time at which an
insurance company would actually pay a death benefit (due to notification of
death to the insurance company and legal formalities) the effect is slight and
we will ignore that delay here
...

We are generally most interested in the expected value of the present value
random variable for some future payment
...
It is also commonly referred to as the Actuarial
Value
...
In actuarial notation, we denote this expected value by Ax , where
the bar above A denotes that the benefit is payable immediately on death
...


(4
...
We use the time-line
format that was introduced in Section 2
...
1
...
The probability that (x) is alive at
time s is s px , and the probability that (x) dies between ages x + s and x + s + ds,
having survived to age x + s, is, loosely, µx+s ds, provided that ds is very small
...

Now we can integrate (that is, sum the infinitesimal components of) the
product of present value and probability over all the possible death intervals s

4
...
1 Time-line diagram for continuous whole life insurance
...

Similarly, the second moment (about zero) of the present value of the death
benefit is
E[Z 2 ] = E[(e−δ Tx )2 ] = E[e−2δ Tx ]
∞

=

e−2δt t px µx+t dt

0
2¯

= Ax

(4
...

The variance of the present value of a unit benefit payable immediately on
death is
¯
¯
V[Z] = V[e−δ Tx ] = E[Z 2 ] − E[Z]2 = 2 Ax − Ax

2


...
3)

Now, if we introduce a more general sum insured, S, say, then the EPV of the
death benefit is
¯
E[SZ] = E[Se−δ Tx ] = S Ax
and the variance is
V[SZ] = V[Se−δ Tx ] = S 2

¯x
Ax − A 2
...
Suppose we are interested in the

78

Insurance benefits

probability Pr[Z ≤ 0
...
We can rearrange this into a probability
for Tx :
Pr[Z ≤ 0
...
5]
= Pr[−δ Tx ≤ log(0
...
5)]
= Pr[δ Tx > log(2)]
= Pr[Tx > log(2)/δ]
= u px
where u = log(2)/δ
...
This makes sense because the benefit is more expensive to the
insurer if it is paid early, as there has been little opportunity to earn interest
...


4
...
2 Whole life insurance: the annual case, Ax
Suppose now that the benefit of $1 is payable at the end of the year of death of
(x), rather than immediately on death
...
Recall that Kx measures
the number of complete years of future life of (x)
...
For example, if (x) lived for 25
...

We again use Z to denote the present value of the whole life insurance benefit
of $1, so that Z is the random variable
Z = v Kx +1
...

In Chapter 2 we derived the probability function for Kx , Pr[Kx = k] = k |qx ,
so the EPV of the benefit is
Ax = E[v Kx +1 ] =

∞

v k+1 k |qx = vqx + v 2 1 |qx + v 3 2 |qx + · · ·
...
4)

k=0

Each term on the right-hand side of this equation represents the EPV of a
death benefit of $1, payable at time k conditional on the death of (x) in (k −1, k]
...
4 Valuation of insurance benefits
Time

0

1

2

79
3
……

Amount

$1

$1

$1

Discount

v

v2

v3

Probability

qx

1 |qx

2 |qx

Figure 4
...


In fact, we can always express the EPV of a life-contingent benefit by considering each time point at which the benefit can be paid, and summing over all
possible payment times the product of
(1) the amount of the benefit,
(2) the appropriate discount factor, and
(3) the probability that the benefit will be paid at that time
...
6
...
2
...


k=0

Just as in the continuous case, we can calculate the second moment about zero
of the present value by an adjustment in the rate of interest from i to (1+i)2 −1
...
5)

k=0

and so the variance of the present value of a benefit of S payable at the end of
the year of death is
S2

2

Ax − (Ax )2
...
6)

4
...
3 Whole life insurance: the 1/mthly case, A(m)
x
In Chapter 2 we introduced the random variable Kx , representing the curtate
future lifetime of (x), and we saw in Section 4
...
2 that the present value of an

80

Insurance benefits

insurance benefit payable at the end of the year of death can be expressed in
terms of Kx
...
The
most common values of m are 2, 4 and 12, corresponding to half years, quarter
(4)
years and months
...

Symbolically, if we let
denote the integer part (or floor) function, then
(m)
Kx =

1
mTx
...
7)

For example, suppose (x) lives exactly 23
...
Then
(2)
(4)
(12)
8
Kx = 23, Kx = 23
...
5, and Kx = 23 12 = 23
...

(m)
(m)
Note that Kx is a discrete random variable
...

(m)
The probability function for Kx can be derived from the associated
1 2
probabilities for Tx
...
,
(m)
Pr[Kx = k] = Pr k ≤ Tx < k +

1
m

= k | 1 qx = k px −
m

1
k+ m px


...
3 we show the time-line for the mthly benefit
...

Suppose, for example, that m = 12
...


2/m

3/m
……

Amount

$1

$1

$1

Discount

v 1/m

v 2/m

v 3/m

Probability

1 qx
m

1 | 1 qx
m m

2 | 1 qx
m m

Figure 4
...


4
...

Similarly, for any m,
A(m) = v 1/m 1 qx + v 2/m 1 | 1 qx + v 3/m 2 | 1 qx + v 4/m 3 | 1 qx + · · ·
x
m

∞

v

=
k=0

m m

m m

m m

k+1
m k 1
m m

| qx
...
8)
(4
...
We have
(12)

E[Z 2 ] = E[v 2(Kx

+1/12)

(12)

] = E[(v 2 )Kx

+1/12

] = 2 A(12) ,
x

so the variance is
2 (12)
Ax

− (A(12) )2
...
4
...
Nevertheless, the annual insurance
function Ax is still useful
...

Using the annual life table in a spreadsheet, we can calculate the values of
Ax using backwards recursion
...
We assume all lives expire by age ω, so that qω−1 = 1
...
This means that any life attaining age ω − 1 may be
treated as certain to die before age ω, in which case we know that Kω−1 = 0
and so
Aω−1 = E[v Kω−1 +1 ] = v
...


(4
...

The intuition for equation (4
...

We can use the same approach for mthly benefits; now the recursion will give
(m)
(m)
Ax in terms of A 1
...
We have
A(m) = v 1/m 1 qx + v 2/m 1 px 1 qx+ 1 + v 3/m 2 px 1 qx+ 2 + · · ·
x
m

=v

1/m

1
m

m

qx + v

1/m

1
m

m

px v

m

1/m

m

1
m

qx+ 1 + v

m

2/m

m

m

1
m

px+ 1

1
m m

qx+ 2 + · · · ,
m

giving the recursion formula
A(m) = v 1/m 1 qx + v 1/m 1 px A
x
m

m

(m)
1
x+ m


...
1 Using the Standard Ultimate Survival Model from Section 4
...
, 100
...

Solution 4
...
13 and so values of t px can
be calculated as explained in the solution to that example
...
4 Valuation of insurance benefits

83

Table 4
...
Spreadsheet results for Example 4
...

x

Ax

x

Ax

x

Ax

20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46

0
...
05144
0
...
05622
0
...
06147
0
...
06725
0
...
07359
0
...
08054
0
...
08817
0
...
09653
0
...
10569
0
...
11571
0
...
12665
0
...
13859
0
...
15161
0
...
16577
0
...
18114
0
...
19780
0
...
21582
0
...
23524
0
...
25613
0
...
27852
0
...
30243
0
...
32785
0
...
35477
0
...
38313
0
...
41285
0
...
44379
0
...
47580

74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100

0
...
50868
0
...
54217
0
...
57599
0
...
60984
0
...
64336
0
...
67622
0
...
70806
0
...
73853
0
...
76735
0
...
79423
0
...
81897
0
...
84143
0
...
86153
0
...
99996, which is indeed close to 1
...
, A20 , starting from A129 = v
...
, 100, are given in Table 4
...

2
Example 4
...
3,
and an interest rate of 5% per year effective, develop a spreadsheet of values
of A(12) for x starting at age 20, in steps of 1/12
...
2 For this example, we follow exactly the same process as for the
previous example, except that we let the ages increase by 1/12 year in each

84

Insurance benefits
Table 4
...
Spreadsheet results for Example 4
...

x

(12)

1 px
12

1 qx
12

Ax

20

0
...
000021

0
...
999979

0
...
05051

0
...
000021

0
...


...


50

0
...


...

0
...
000021

...


...
000096

0
...


...

0
...


...

129 10
12

0
...


...

0
...
000097

...


...
586045

0
...


...

0
...
We construct a column of values of

1
12

1
12

px using

px = exp −A/12 − Bcx (c1/12 − 1)/ log(c)
...
Then set A
(12)

for all the other values of Ax

(12)
11
129 12

= v 1/12 , and

use the recursion

A(12) = v 1/12
x

1
12

qx + v 1/12

1
12

px A

(12)
1
...
2
...
1 and 4
...
In
Example 4
...
99996, which is sufficiently close to 1 to
justify us starting our recursive calculation by setting A129 = v
...
2,
our recursive calculation started from A129 11 = v 1/12
...
58960, which is certainly not close to 1
...
1, we are
replacing the exact calculation
A129 = v (q129 + p129 A130 )

4
...
Similarly, for Example 4
...
As the value of A(12) is very close to 1, it follows that
130
129 12

v 1/12

1
12

q129 11 +
12

1
12

(12)

p129 A130

can by approximated by v 1/12
...
3 Using the Standard Ultimate Survival Model specified in Section
4
...

¯
Solution 4
...
25%, and for parts (b) and (c) we replace each Ax by Ax and Ax ,
respectively
...
3
...
1
and 4
...

We can make the following observations about these values
...
This is because the death benefit is payable soonest under
(a) and latest under (c)
...
This occurs because the smaller the value of x, the longer the expected
time until payment of the death benefit is
...
And further, as we
get to very old ages, the standard deviation decreases in absolute terms, as the
possible range of payout dates is reduced significantly
...
That is to be expected, as the difference arises
from the change in the value of money in the period between the moment of
death and the end of the month of death, a relatively short period
...
3
...
3
...
Dev
...
Dev
...
Dev
...
4
...

In the continuous case, the benefit is payable immediately on death
...


¯1
The EPV of this benefit is denoted Ax:n in actuarial notation
...

Then
¯1
Ax:n =

n

e−δt t px µx+t dt

(4
...

1
The annual case, Ax:n
Next, we consider the situation when a death benefit of 1 is payable at the end
of the year of death, provided this occurs within n years
...
4 Valuation of insurance benefits

87

random variable for the benefit is now
Z=

v Kx +1
0

if Kx ≤ n − 1,
if Kx ≥ n
...


(4
...
The present value
random variable for the benefit is
(m)

v Kx
Z=
0

(m)

1
+m

if Kx ≤ n −
(m)
if Kx ≥ n
...


(4
...
4 Using the Standard Ultimate Survival Model as specified in
(4)
¯
Section 4
...


x:10

x:10

¯
Solution 4
...
11) with n = 10 to calculate A 1 (using
x:10
numerical integration), and formulae (4
...
12), with m = 4 and n = 10
(4)
to calculate A 1 and A1
...
4, and we observe that values in each case
increase as x increases, reflecting the fact that the probability of death in a 10year period increases with age for the survival model we are using
...
3, for the same reason – the
ordering reflects the fact that any payment under the continuous benefit will be
paid earlier than a payment under the quarterly benefit
...

2

88

Insurance benefits
Table 4
...
EPVs of term insurance benefits
...
00214
0
...
04356
0
...
00213
0
...
04329
0
...
00209
0
...
04252
0
...
4
...
For
example, a 10-year pure endowment with sum insured $10 000, issued to (x),
will pay $10 000 in 10 years if (x) is still alive at that time, and will pay nothing if
(x) dies before age x + 10
...
However,
pure endowment valuation functions turn out to be very useful
...


There are two ways to denote the EPV of the pure endowment benefit using
1
actuarial notation
...
The ‘1’ over the term subscript
indicates that the term must expire before the life does for the benefit to be
paid
...
A more convenient standard actuarial notation for the
EPV of the pure endowment is n Ex
...


(4
...


(4
...


4
...

4
...
7 Endowment insurance
An endowment insurance provides a combination of a term insurance and a
pure endowment
...

Traditional endowment insurance policies were popular in Australia, North
America and the UK up to the 1990s, but are rarely sold these days in these
markets
...
Also, companies operating in these
territories will be managing the ongoing liabilities under the policies already
written for some time to come
...

We first consider the case when the death benefit (of amount 1) is payable
immediately on death
...

Thus, the EPV of the benefit is
n

E[Z] =
0

e−δn t px µx+t dt

n
n

=

∞

e−δt t px µx+t dt +

e−δt t px µx+t dt + e−δn n px

0
1
¯1
= Ax:n + Ax:n

and in actuarial notation we write
1
¯
¯
Ax:n = A1 + Ax:n
...
16)

Similarly, the expected value of the squared present value of the benefit is
n
0

¯
which we denote 2 Ax:n
...

The EPV of the benefit is then
n−1
1
v k+1 k |qx + v n P[Kx ≥ n] = Ax:n + v n n px ,

(4
...


(4
...

k=0

Finally, when the death benefit is payable at the end of the 1/mth year of death,
the present value of the benefit is
(m)

v Kx
Z= n
v

(m)

1
+m

(m)

= v min(Kx

if Kx ≤ n −
(m)
if Kx ≥ n

1
+ m ,n)

1
m

,


...


(4
...
5 Using the Standard Ultimate Survival Model as specified in
(4)
¯
Section 4
...


4
...
5
...

x
20
40
60
80

(4)
x:10

0
...
61508
0
...
68502

Ax:10

0
...
61504
0
...
68292

¯
Ax:10

0
...
61494
0
...
67674

A

¯
Solution 4
...
4
...
5
...

The reason for this is that the probability of surviving 10 years is large (10 p20 =
0
...
9425) and so for each value of x, the benefit is payable after
10 years with a high probability
...
6139, and as time 10 years
(4)
¯
is the latest possible payment date for the benefit, the values of Ax:10 , A
x:10
and Ax:10 must be greater than this for any age x
...
4
...
Suppose a benefit of $1 is
payable immediately on the death of (x) provided that (x) dies between ages
x + u and x + u + n
...


This random variable describes the present value of a deferred term insurance
...

The actuarial notation for the EPV of the deferred term insurance benefit is
¯1
u |Ax:n
...


(4
...


(4
...
22)

which follows from formula (4
...


0

Thus, the EPV of a deferred term insurance benefit can be found by differencing
the EPVs of term insurance benefits for terms u + n and u
...
21)
...
If the life survives u years, to the end
¯ 1
of the deferred period, then the EPV at that time of the term insurance is Ax+u:n
...

Multiplying by v u x
Our main interest in this EPV is as a building block
...


(4
...


¯
4
...

In particular, it is useful to note that
1
Ax = Ax:n + n |Ax

(4
...

1
This relationship can be used to calculate Ax:n for integer x and n given a table
of values of Ax and lx
...
5 Relating Ax , Ax and A(m)
x
We mentioned in the introduction to this chapter that, even though insurance
contracts with death benefits payable at the end of the year of death are very
unusual, functions like Ax are still useful
...

(4)
In Table 4
...
3, with interest at 5% per
year effective
...
6 that, over a very wide range of ages, the ratios of
(4)
Ax to Ax and Ax to Ax are remarkably stable, giving the appearance of being
independent of x
...


4
...
1 Using the uniform distribution of deaths assumption
¯
The difference between Ax and Ax depends on the lifetime distribution between
ages y and y + 1 for all y ≥ x
...
3
...


94

Insurance benefits
Table 4
...
Ratios of A(4) to Ax
x
and Ax to Ax , Standard Ultimate
Survival Model
...
0184
1
...
0184
1
...
0198
1
...
0246
1
...
0246
1
...
0261
1
...
9), that under UDD, we have for 0 < s < 1, and
for integer y, s py µy+s = qy
...

δ

Because eδ = 1 + i, under the assumption of UDD we have
i
¯
A x = Ax
...
25)

This exact result under the UDD assumption gives rise to the approximation
i
¯
Ax ≈ A x
...
26)

The same approximation applies to term insurance and deferred insurance,
which we can show by changing the limits of integration in the proof above
...


¯
4
...
27)
(m)

and the right-hand side is used as an approximation to Ax
...
27) is left as an exercise for the reader
...
The endowment insurance combines the death and survival benefits, so we need to split
off the death benefit before applying one of the approximations
...


(4
...
5
...
The only difference between these benefits is
(4)
the timing of the payment
...
Since the insured
life (x) dies in the year of age x + Kx to x + Kx + 1, under the end year of death
benefit (valued by Ax ), the sum insured is paid at time Kx + 1
...
If the deaths occur evenly over the year (the same assumption as we use in the UDD approach), then, on average, the benefit is paid at
time Kx + 5/8, which is 3/8 years earlier than the end of year of death benefit
...
Assuming deaths occur uniformly over the year, then on average the benefit is paid at
Kx + 13/24, which is 11/24 years earlier than the end year of death benefit
...

So we have the resulting approximation
A(m) ≈ qx v
x

m+1
2m

+ 1 |qx v 1+

∞

=

k |qx v

m+1
2m

+ 2 |qx v 2+

k+ m+1
2m

k=0

= (1 + i)

m−1
2m

∞
k |qx v
k=0

k+1


...


(4
...
29), to give
the approximation
¯
Ax ≈ (1 + i)1/2 Ax
...
30)

This is explained by the fact that, if the benefit is paid immediately on death,
and lives die uniformly through the year, then, on average, the benefit is paid
half-way through the year of death, which is half a year earlier than the benefit
valued by Ax
...
Hence, for an endowment insurance using the claims acceleration approach
we have
1
¯
Ax:n ≈ (1 + i)1/2 Ax:n + v n n px
...
31)

Note that both the UDD and the claims acceleration approaches give values
(m)
(m)
for Ax or Ax such that the ratios Ax /Ax = i/i(m) and Ax /Ax = i/δ are
independent of x
...
0186 and i/δ = 1
...
0185 and (1 + i)1/2 = 1
...
The values in Table 4
...


4
...


This approach works for the EPV of any traditional benefit – that is, where
the future lifetime is the sole source of uncertainty
...

The approach can be justified technically using indicator random variables
...
6 Variable insurance benefits

97

x dies in the interval (k, k + 1]
...


In this example, Pr[E is True ] = k |qx , so the expected value of the indicator
random variable is
E[I (E)] = 1(k |qx ) + 0(1 − k |qx ) = k |qx ,
and, in general, the expected value of an indicator random variable is the
probability of the indicator event
...

We can write the present value random variable as
1 000 I (Tx ≤ 10)v 10 + 2 000 I (10 < Tx ≤ 20)v 20
and the EPV is then
1 000 10 qx v 10 + 2 000 10 |10 qx v 20
...
Here we
consider indicators of the form
I (t < Tx ≤ t + dt)
for infinitesimal dt, with associated probability
E[I (t < Tx ≤ t + dt)] = Pr[t < Tx ≤ t + dt]
= Pr[Tx > t] Pr[Tx < t + dt|Tx > t]
≈ t px µx+t dt
...
That is, the benefit is exactly equal to the
number of years lived by an insured life from age x to his or her death
...

To find the EPV of this benefit, we note that the payment may be made at
any time, so we consider all the infinitesimal intervals (t, t + dt), and we sum
over all these intervals by integrating from t = 0 to t = ∞
...
The amount is t, the discount factor is e−δt
...

So, we can write the EPV of this benefit as
∞

t e−δt t px µx+t dt
...
32)

0

¯¯
In actuarial notation we write this as (I A)x
...

An alternative approach to deriving equation (4
...

Then any moment of Z can be found from
∞

E[Z k ] =

t e−δt

k

t px

µx+t dt
...

If the policy term ceases after a fixed term of n years, the EPV of the death
benefit is
¯¯ 1
(I A)x:n =

n

t e−δt t px µx+t dt
...
We present several in the following examples
...
6 Consider an n-year term insurance policy issued to (x) under
which the death benefit is k +1 if death occurs between ages x +k and x +k +1,
for k = 0, 1, 2,
...

(a) Derive a formula for the EPV of the benefit using the first approach
described, that is multiplying together the amount, the discount factor and
the probability of payment, and summing for each possible payment date
...

Solution 4
...
, n − 1
...
6 Variable insurance benefits

99

is $(k + 1)
...

k=0
1
In actuarial notation the above EPV is denoted (IA)x:n
...

k=0

(b) We must go back to first principles
...


So
n−1

E[Z 2 ] =

v 2(k+1) (k + 1)2 k |qx ,
k=0

and the variance is
n−1
1
v 2(k+1) (k + 1)2 k |qx − (IA)x:n

V[Z] =

2


...
7 A whole life insurance policy offers an increasing death benefit
payable at the end of the quarter year of death
...
Derive an
expression for the EPV of the death benefit
...
7 First, we note that the possible payment dates are 1/4, 2/4, 3/4, …
...
Third, corresponding to
the possible payment dates, the discount factors are v 1/4 , v 2/4 ,
...

4 4

100

Insurance benefits

Hence, the EPV, which is denoted (IA(4) )x , can be calculated as
1

1
4

2

3

qx v 4 + 1 | 1 qx v 4 + 2 | 1 qx v 4 + 3 | 1 qx v 1
4 4

+2

1 | 1 qx
4

+3
=

2 | 1 qx
4

4 4

v

11
4

v

21
4

(4)
A 1
x:1

4 4

+ 1 1 | 1 qx v

12
4

+ 2 1 | 1 qx v

22
4

4 4

4 4

(4)
+ 2 1 |A 1
x:1

3

+ 1 2 | 1 qx v 1 4 + 1 3 | 1 qx v 2
4 4

4 4

+ 2 2 | 1 qx v

23
4

4 4

(4)
+ 3 2 |A 1
x:1

+ 2 3 | 1 qx v 3 + · · ·
4 4

+ ···
...
This is important in practice because compound
reversionary bonuses will increase the sum insured as a geometric progression
...
8 Consider an n-year term insurance issued to (x) under which the
death benefit is paid at the end of the year of death
...
Thus,
if death occurs between ages x + k and x + k + 1, the death benefit is (1 + j)k
for k = 0, 1, 2,
...
Derive a formula for the EPV of this death benefit
...
8 The amount of benefit is 1 if the benefit is paid at time 1, (1 + j)
if the benefit is paid at time 2, (1 + j)2 if the benefit is paid at time 3, and so
on, up to time n
...
33)

where
i∗ =

1+i
i−j
−1=

...
In most practical situations, i > j so that i∗ > 0
...
8 Notes and further reading

101

Example 4
...

(a) Derive an expression for the EPV of the death benefit if the policy is an
n-year term insurance
...

Solution 4
...

1+j

(b) Similarly, if the policy is a whole life insurance rather than a term insurance,
then the EPV of the death benefit would be
∞

¯
(1 + j)t v t t px µx+t dt = (Ax ) i∗

0

where
i∗ =

1+i
− 1
...
7 Functions for select lives
Throughout this chapter we have developed results in terms of lives subject to
ultimate mortality
...

All of the above development equally applies to lives subject to select mortality
...
Similarly, A[x]:n denotes the EPV of a benefit of
1 payable at the end of the year of death of a select life [x] should death occur
within n years, or after n years if [x] survives that period
...
8 Notes and further reading
The Standard Ultimate Survival Model incorporates Makeham’s law as its survival model
...
11
...


0

This integral cannot be evaluated analytically but can be evaluated numerically
...
7
...
Functions
¯
such as Ax can still be evaluated numerically but, since the integrand has to
be evaluated numerically, the procedure may be a little more complicated
...
18 for an example
...
18 has been
derived from data for UK whole life and endowment insurance policyholders
(non-smokers), 1999–2002
...


4
...
1 You are given the following table of values for lx and Ax , assuming
an effective interest rate of 6% per year
...
00
99 737
...
91
99 154
...
91
98 485
...
151375
0
...
165386
0
...
180505
0
...


Exercise 4
...

x

4
...
3 A with-profit whole life insurance policy issued to a life aged
exactly 30 has a basic sum insured of $100 000
...

Using the Standard Ultimate Survival Model, with interest at 5% per year,
calculate the EPV of this benefit
...
4 (a) Show that
n−2

Ax:n =

v k+1 k |qx + v n n−1 px
...
17) and comment on the differences
...
5 Show that
(m)

(m)

(IA(m) )x = A(m) + vpx Ax+1 + 2 px v 2 Ax+2 + · · ·
x
and explain this result intuitively
...
6 (a) Derive the following recursion formula for an n-year increasing term insurance:
(IA)1 = vqx + vpx (IA) 1
x:n

x+1:n−1

+A1

x+1:n−1


...

(c) You are given that (IA)50 = 4
...
00558, A51 = 0
...
06
...

Exercise 4
...
25, Ax+20 = 0
...
55 and
¯
i = 0
...
Calculate 10 000Ax:20 using
(a) claims acceleration, and
(b) UDD
...
8 Show that
n
1
1
(IA)x:n = (n + 1)Ax:n −

A1

x:k

k=1

and explain this result intuitively
...
9 Show that Ax is a decreasing function of i, and explain this result
by general reasoning
...
10 Calculate A70 given that
A50:20 = 0
...
14996,

A50 = 0
...


Exercise 4
...

Under a term insurance issued to a life aged x, let Y denote the present value
of a unit sum insured, payable at the moment of death within the n-year term
...
3,

V[X ] = 0
...
8,

E[Y ] = 0
...

Exercise 4
...
, then
under the assumption of a constant force of mortality between integer ages,
¯
Ax =

∞

v t t px
t=0

νx+t (1 − vpx+t )

...
13 Let Z1 denote the present value of an n-year term insurance
benefit, issued to (x)
...

Express the covariance of Z1 and Z2 in actuarial functions, simplified as far
as possible
...
14 You are given the following excerpt from a select life table
...
9 Exercises

105

(c) the probability that the present value of the benefit described in (b) is less
than or equal to $85 000
...
15 (a) Describe in words the insurance benefits with the present
values given below
...

if Tx ≤ 5,
if 5 < Tx ≤ 15,
if Tx > 15
...

(c) Derive expressions in terms of standard actuarial functions for the expected
values of Z1 and Z2
...

Exercise 4
...
0001, B =
0
...
075
...

(a) Use Excel and backward recursion in parts (i) and (ii)
...

(4)
(ii) Construct a table of values of Ax for x = 50, 50
...
5,
...
)
(4)
(4)
(iii) Hence, write down the values of A50 , A100 , A50 and A100
...

(c) Compare your estimated values for the A(4) functions (from (b)) with your
accurate values (from (a))
...

Exercise 4
...
Use the Standard Ultimate Survival Model,
with interest at 5% per year, in the following
...

(b) Calculate the standard deviation of the present value of the benefit
...
Assuming that the insurer
invests all funds at exactly 5% per year effective, what is the probability

106

Insurance benefits
that the policy benefit has greater value than the accumulation of the single
premium?

Exercise 4
...
5 × 10−4 , B = 5
...
00085, D = 1
...

(a) Calculate t p60 for t = 0, 1/40, 2/40,
...

Hint: Use the repeated Simpson’s rule
...

60:2
Hint: Use the repeated Simpson’s rule
...
1 (a) 0
...
137503
(c) 0
...
215182
4
...
47
4
...
07307
4
...
44
(b) 5 507
...
10 0
...
11 0
...
14 (a) 0
...
71
(c) 0
...
16 (a) (iii) 0
...
87508,
0
...
89647
(b) 0
...
89453
4
...
83
(b) $239
...
04054
4
...
999031, p60 = 0
...
991903
(b) 0
...
1 Summary
In this chapter we derive expressions for the valuation and analysis of life
contingent annuities
...

We consider how to calculate annuity valuation functions
...
Where we are calculating benefits
payable more frequently than annual (monthly or weekly, say) using only an
integer age life table, a very common situation in practice, then some approximation is required
...


5
...
The payments
are normally made at regular intervals and the most common situation is that
the payments are of the same amount
...
The present value
of a life annuity is a random variable, as it depends on the future lifetime;
however, we will use some results and notation from the valuation of annuitiescertain, where there is no uncertainty in the term, so we start with a review of
these
...
3 Review of annuities-certain

Recall that, for integer n,
an = 1 + v + v 2 + · · · + v n−1 =
¨

1 − vn
d

(5
...
Also
¨
an = v + v 2 + v 3 + · · · + v n = an − 1 + v n
denotes the present value of an annuity-certain of 1 payable annually in arrear
for n years
...
2)

denotes the present value of an annuity-certain payable continuously at rate 1
per year for n years
...


(5
...


5
...
If the annuity is to be paid throughout
the annuitant’s life, it is called a whole life annuity
...

Annual annuities are quite rare
...
However, the annual annuity is still important
in the situation where we do not have full information about mortality between
integer ages, for example because we are working with an integer age life table
...


5
...


5
...
1 Whole life annuity-due
Consider first an annuity of 1 per year payable annually in advance throughout the lifetime of an individual now aged x
...
The first payment occurs
immediately, the second in one year from now, provided that (x) is alive then,
and payments follow at annual intervals with each payment conditional on the
survival of (x) to the payment date
...
1 we show the payments and
associated probabilities and discount functions in a time-line diagram
...
, k,
for a total of k + 1 payments
...
This means that, for k = 0, 1, 2,
...
Thus, using equation (5
...

d

There are three useful ways to derive formulae for calculating the expected
value of this present value random variable
...
4
...
For the expected value of Y , which

Time

0

1

2

3

...
1 Time-line diagram for whole life annuity-due
...

d

That is,
ax =
¨

1 − Ax

...
4)

This is a useful approach, as it also immediately gives us the variance of Y as
1 − v Kx +1
d

V[Y ] = V

1
V[v Kx +1 ]
d2
2 A − A2
x
x

...
5)

Secondly, we may use the indicator random variable approach from Section
4
...
The condition for the payment at k, say, is that (x) is alive at age x + k, that
is, that Tx > k
...
6)

and the EPV of the annuity is the sum of the expected values of the individual
terms
...


(5
...
However, this approach does not lead to
¨
useful expressions for the variance and higher moments of Y
...
6) are dependent random variables
...

¨
k=0

(5
...
4 Annual life annuities

111

This is less used in practice than equations (5
...
7)
...
7) and (5
...
7) the summation is taken over the possible payment dates, and in equation
(5
...

Example 5
...
7) and (5
...


k=0

k=0

Solution 5
...

k=t

Then
∞

∞

k

ak+1 k |qx =
¨

v t k |qx
k=0 t=0

k=0

= qx + (1 + v) 1 |1 qx + (1 + v + v 2 ) 2 |1 qx
+ (1 + v + v 2 + v 3 ) 3 |1 qx + · · ·
...


2

112

Annuities
5
...
2 Term annuity-due

Now suppose we wish to value a term annuity-due of 1 per year
...

Thus, payments are made at times k = 0, 1, 2,
...
The present value of this annuity is Y , say, where
Y =

aKx +1
¨
an
¨

if Kx = 0, 1,
...


that is
Y = amin(K
¨

=

x +1, n)

1 − v min(Kx +1, n)

...

¨
We have seen the random variable v min(Kx +1, n) before, in Section 4
...
7, where
the EPV Ax:n is derived
...

d

(5
...
2
...

Using Figure 5
...


Amount

1

1

1

1

1

Discount

1

v

v2

v3

v n−1

Probability

1

px

2 px

3 px

n−1 px

Figure 5
...


n

5
...


(5
...
8) above
...


5
...
3 Whole life immediate annuity
Now consider a whole life annuity of 1 per year payable in arrear, conditional
on the survival of (x) to the payment dates
...
The actuarial notation for the EPV of this
annuity is ax , and the time-line for the annuity cash flow is shown in Figure 5
...

Let Y ∗ denote the present value random variable for the whole life immediate
annuity
...

We can see from this expression and from the time-line, that the difference
in present value between the annuity-due and the immediate annuity payable
in arrear is simply the first payment under the annuity-due, which, under the
annuity-due, is assumed to be paid at time t = 0 with certainty
...


Amount

1

1

1

Discount

v

v2

v3

Probability

px

2 px

3 px

Figure 5
...


114

Annuities

So, if Y is the random variable for the present value of the whole life annuity
payable in advance, and Y ∗ is the random variable for the present value of the
whole life annuity payable in arrear, we have Y ∗ = Y − 1, so that E[Y ∗ ] =
E[Y ] − 1, and hence
ax = ax − 1
...
11)

Also, from equation (5
...

d2

x

5
...
4 Term immediate annuity
The EPV of a term immediate annuity of 1 per year is denoted ax:n
...
, n, conditional on the
survival of the annuitant
...
4
...


(5
...
10) and (5
...

Amount

1

1

1

1

1

Discount

v

v2

v3

v n−1

vn

Probability

px

2 px

3 px

n−1 px

n px

Figure 5
...


5
...

¨

(5
...


5
...
5
...
Consider now the case when the annuity is payable
continuously at a rate of 1 per year as long as (x) survives
...
If payments were daily, for an annuity of 1 per
year, the daily payment would be 1/365
...

The EPV is denoted ax
...

¯
Analogous to the annual annuity-due, we can derive formulae for the EPV
of the annuity in three different ways
...

δ

That is,
ax =
¯

¯
1 − Ax

...
14)

116

Annuities

Using this formulation for the random variable Y , we can also directly derive
the variance for the continuous annuity present value from the variance for the
continuous insurance benefit
1 − v Tx
δ

V[Y ] = V

=

2A
¯

x

¯x
− A2

...
For each such interval,
the amount is dt, the discount factor is e−δt and the probability of payment is
t px , giving
∞

ax =
¯

e−δt t px dt
...
15)

0

We remark that this EPV can also be derived using indicator random variables
by expressing the present value as
∞

Y =

e−δt I(Tx > t) dt
...
15) is illustrated in Figure 5
...
The interval is so small that payments
can be treated as being made exactly at t
...

¯

t

t+dt

...
5 Time-line diagram for continuous whole life annuity
...
5 Annuities payable continuously

117

We can evaluate this using integration by parts, noting that if we differentiate
equation (5
...

¯
dt t
Then
∞

ax =
¯
0

at
¯

d
(−t px )dt
dt

= − at t px |∞ −
¯
0
∞

=

∞
t px

e−δt dt

0

e−δt t px dt
...

¯

5
...
2 Term continuous annuity
The term continuous life annuity present value random variable
amin(T
¯

x ,n)

1 − v min(Tx ,n)
δ

=

has EPV denoted by ax:n
...

Using results for endowment insurance functions from Section 4
...
7, we have
ax:n =
¯

¯
1 − Ax:n

...
16)

Using the indicator random variable approach we have
n

ax:n =
¯

e−δt t px dt ,

(5
...

¯
¯

One way to understand the difference between the second and third approaches
is to see that in the second approach we integrate over the possible payment
dates, and in the third approach we integrate over the possible dates of death
...


118

Annuities
5
...
6
...
Premiums are more commonly payable monthly, quarterly, or
sometimes weekly
...

We can define the present value of an annuity payable m times per year in
(m)
terms of the random variable Kx , which was introduced in Section 4
...
3
...

We will also use the formula for the present value of an 1/mthly annuity
(m)
certain
...
It is important to remember that
(m)
an is an annual factor, that is, it values a payment of 1 per year, and therefore
¨
(m)

for valuing annuities for other amounts, we need to multiply the an factor by
¨
the annual rate of annuity payment
...
Each monthly payment is $1000
...
If K60 = 0, then (60) died
in the first month, there was a single payment made at t = 0 of $1000, and the
present value is
(12)

...

2/12

= 12 000 a
¨

Continuing, we see that the present value random variable for this annuity can
be written as
12 000

1
12

+

1
1 12
12 v

+

2
1 12
12 v

+ ··· +

(12)

1 Kx
12 v

= 12 000 a
¨

(12)
(12)

K60 +1/12


...
6 Annuities payable m times per year
Time

0

1/m

2/m

3/m

119
4/m

...
6 Time-line diagram for whole life 1/mthly annuity-due
...
6
...

Figure 5
...

The present value random variable for this annuity is
(m)

a
¨

(m)
(m)

1
Kx + m

1 − v Kx
=
d (m)

1
+m


...


(5
...

v
m
m

(5
...
Similar to the annual annuity case, the only difference in
the whole life case is the first payment, of $1/m, so that the EPV of the 1/mthly
immediate annuity is
(m)
ax = ax −
¨ (m)

1
m


...
20)

120

Annuities

Time

0

1/m

2/m

3/m

4/m

n-1

n

1/m

0


...
7 Time-line diagram for term life 1/mthly annuity-due
...
6
...
Consider
now an annuity of total amount 1 per year, payable in advance m times per year
throughout the lifetime of (x) for a maximum of n years, with each payment
being 1/m
...
7
...


d (m)
(m)

The EPV of this annuity is denoted by ax:n and is given by
¨
(m)

(m)

ax:n =
¨

1 − E[v min(Kx
d (m)

1
+ m ,n)

]

so that
(m)
ax:n
¨

(m)

=

1 − Ax:n
d (m)


...
21)

Using the indicator random variable approach we find that
(m)

mn−1

ax:n =
¨
r=0

1 r/m
r px
...
22)

For the 1/mthly term immediate annuity, by comparison with the 1/mthly
annuity-due, the difference is the first payment under the annuity due, with EPV

5
...

m

121
1 n
m v n px ,

(5
...
3) for the annuity-certain
...
19) and (5
...
7) and (5
...
Also, by letting m → ∞ in equations (5
...
22) we obtain equations (5
...
17) for continuous annuities, ax
¯
and ax:n
...


5
...
1 we show values for ax , ax , ax , ax and ax for x = 20, 40, 60
and 80, using the Standard Ultimate Survival Model from Section 4
...
Using equations (5
...
20), (5
...
19) and (5
...
1
...

We also have, for each age, the ordering
(4)
¯
¨ (4) ¨
ax < ax < ax < ax < ax
...

• While the life is alive, the payments in each year sum to 1 under each annuity,

but on average, the payments under the annuity-due are paid earlier
...

• In the year that (x) dies, the different annuities pay different amounts
...
Under the annual immediate
annuity, in the year of death no payment is made as the life does not survive
to the payment date at the year end
...

For example, suppose the life dies after seven months
...
1
...

¯ ¨
¨

x

ax

20
40
60
80

18
...
458
13
...
548

(4)

ax
¯

19
...
829
14
...
917

19
...
954
14
...
042

ax

(4)

ax
¨

19
...
079
14
...
167

ax
¨
19
...
458
14
...
548

the year
...
The first year’s final
payment, due at time 3/4, is not made, as the life does not survive to that
date
...
Under the quarterly immediate annuity, the life collects payments at
times 1/4, 1/2, and misses the two payments due at times 3/4 and 1
...

This second point explains why we cannot make a simple interest adjustment to
relate the annuity-due and the continuous annuity
...
There is no difference in the amount of the payment, only
(4)
in the timing
...

We also note from Table 5
...
We will see in
¨
¯
2
Section 5
...
3 that there is indeed a way of calculating an approximation to ax
¯
from ax , but it involves an extra adjustment term to ax
...
2 Using the Standard Ultimate Survival Model, with 5% per year
(4)
(4)
, ax:10 , a
¨
¨
and ax:10 for x = 20, 40,
¯
interest, calculate values of ax:10 , a
x:10
x:10
60 and 80, and comment
...
2 Using equations (5
...
12), (5
...
23) and (5
...
2
...
That means,
for example, that the second term in equation (5
...
8 Deferred annuities

123

Table 5
...
Values of ax:10 , a(4) , ax:10 , a(4) and ax:10
...
711
7
...
534
6
...
855
7
...
691
6
...
904
7
...
743
6
...
952
7
...
796
6
...
099
8
...
956
6
...
The present value of an annuity certain provides an upper bound
for each set of values
...
722 and
(4)
x:10

(4)

< a = 7
...

¨
10
Due to the differences in timing of payments, and in amounts for lives who
die during the 10-year annuity term, we have the same ordering of annuity
values by payment frequency for any age x:
a
¨

(4)
x:10

ax:10 < a

(4)
x:10

< ax:10 < a
¯
¨

< ax:10
...
8 Deferred annuities
A deferred annuity is an annuity under which the first payment occurs at some
specified future time
...
This is an annuity-due deferred
u years
...
Note that we have used the format u |
...
Figure 5
...

Combining Figure 5
...
2, we can see that the combination of the payments under a u-year temporary annuity-due and a u-year deferred annuity-due gives the same sequence
of payments as under a lifetime annuity in advance, so we obtain
a
¨
ax:u + u |¨ x = ax ,
¨

(5
...
25)

or, equivalently,
= ax − ax:u
...



...
8 Time-line diagram for deferred annual annuity-due
...

¯
¯

Summing the EPVs of the individual payments for the deferred whole life
annuity-due gives
a
u |¨ x

= v u u px + v u+1 u+1 px + v u+2 u+2 px + · · ·
= v u u px 1 + v px+u + v 2 2 px+u + · · ·

so that
a
u |¨ x

= v u u px ax+u = u Ex ax+u
...
26)

We see again that the pure endowment function acts like a discount function
...
For
example, for a deferred term immediate annuity,
u |ax:n

= u Ex ax+u:n ,

and for an annuity-due payable 1/mthly,
a(m)
u |¨ x

(m)

= u Ex ax+u
...
27)

This result can be helpful when working with tables
...
Then, using equations (5
...
26), we have
ax:n = ax − v n n px ax+n
...
28)

5
...

¨ (m) ¨ (m)
¨

(5
...
3 Let Y1 , Y2 and Y3 denote present value random variables for a
u-year deferred whole life annuity-due, a u-year term annuity-due and a whole
life annuity-due, respectively
...
Assume annual
payments
...
3 The present value random variable for a u-year deferred whole
life annuity-due, with annual payments is
Y1 =
=

if Kx ≤ u − 1,
if Kx ≥ u,

0
v u aKx +1−u
¨

if Kx ≤ u − 1,
if Kx ≥ u
...
30)

From Section 5
...
2 we have
Y2 =

aKx +1
¨
au
¨

if Kx ≤ u − 1,
if Kx ≥ u
...


We use deferred annuities as building blocks in later sections, noting that an
n-year term annuity, with any payment frequency, can be decomposed as the
sum of n deferred annuities, each with term 1 year
...


(5
...
9 Guaranteed annuities
A common feature of pension benefits is that the pension annuity is guaranteed
to be paid for some period even if the life dies before the end of the period
...

Suppose an annuity-due of 1 per year is paid annually to (x), and is guaranteed
for a period of n years
...
, n − 1, but is paid only if (x) is alive at age x + k
for k = n, n + 1,
...
30), and
a
¨
E[Y1 ] = n |¨ x = n Ex ax+n
...

¨
¨
¨

(5
...
9 shows the time-line for an n-year guaranteed unit whole life annuitydue
...

We can derive similar results for guaranteed benefits payable 1/mthly; for
example, a monthly whole life annuity-due guaranteed for n years has EPV
(12)
x:n

a
¨

Time

0

1

(12)

= an
¨

(12)

+ n Ex ax+n
...

Amount

1

1

1

Discount

1

v1

v2

Probability

1

1

1


...
9 Time-line diagram for guaranteed annual annuity-due
...
10 Increasing annuities

127

Example 5
...
She can opt to take a lower
benefit, with a 10-year guarantee
...
Calculate the revised benefit using the
Standard Ultimate Survival Model, with interest at 5% per year
...
4 Let B denote the revised monthly benefit
...

The resulting equation of EPVs is usually called an equation of value
...
0870, and
¨
a
¨

(12)

65:10

(12)
10

=a
¨

(12)

+ 10 p65 v 10 a75 = 13
...

¨

Thus, the revised monthly benefit is B = $978
...
So the pension plan member
can gain the security of the 10-year guarantee at a cost of a reduction of $21
...

2

5
...

Some of the annuities which arise in actuarial work are not level
...
For these annuities, we are generally interested in determining the EPV, and are rarely concerned with higher
moments
...

The best approach for calculating the EPV of non-level annuities is to use
the indicator random variable, or time-line, approach – that is, sum over all
the payment dates the product of the amount of the payment, the probability of
payment (that is, the probability that the life survives to the payment date) and
the appropriate discount factor
...
10
...
Consider an increasing annuity-due where
the amount of the annuity is t + 1 at times t = 0, 1, 2,
...
The time-line is shown in Figure 5
...


128

Annuities
Time

0

1

2

3

4

...
10 Time-line diagram for arithmetically increasing annual annuity-due
...

¨
From the diagram we see that
∞

(I a)x =
¨

v t (t + 1) t px
...
33)

t=0

Similarly, if the annuity is payable for a maximum of n payments rather than for
the whole life of (x), the EPV, denoted by (I a)x:n in standard actuarial notation,
¨
is given by
n−1

(I a)x:n =
¨

v t (t + 1) t px
...
34)

t=0

If the annuity is payable continuously, with the payments increasing by 1 at
each year end, so that the rate of payment in the tth year is constant and equal
to t, for t = 1, 2,
...
By analogy with formula (5
...

a
m=0

We also have standard actuarial notation for the continuous annuity under which
the rate of payment at time t > 0 is t; that is, the rate of payment is changing
¯¯
continuously
...
For every infinitesimal interval,
life annuity, and (I ¯
(t, t + dt), the amount of annuity paid, if the life (x) is still alive, is t dt, the
probability of payment is t px and the discount function is e−δ t = v t
...
11
...
10 Increasing annuities
Time

0

t

129

t+dt

...
11 Time-line diagram for increasing continuous whole life annuity
...


(5
...
10
...

The approach is similar to the geometrically increasing insurance benefit which
was considered in Examples 4
...
9
...
5 Consider an annuity-due with annual payments where the amount
of the annuity is (1 + j)t at times t = 0, 1, 2,
...
Derive an expression for the EPV of this benefit, and simplify as far
as possible
...
5 First, consider the time-line diagram in Figure 5
...

By summing the product of
• the amount of the payment at time t,
• the discount factor for time t, and
• the probability that the payment is made at time t,

over all possible values of t, we obtain the EPV as
n−1

(1 + j)t v t t px = ax:n i∗
¨
t=0

130
Time

Annuities
0

1

2

3

4
……

(1 + j)2

(1 + j)

(1 + j)3

(1 + j)4

Amount

1

Discount

1

v1

v2

v3

v4

Probability

1

1 px

2 px

3 px

4 px

Figure 5
...


where ax:n i∗ is the EPV of a term annuity-due evaluated at interest rate i∗ where
¨
i∗ =

i−j
1+i
−1=

...
11 Evaluating annuity functions
If we have full information about the survival function for a life, then we can use
summation or numerical integration to compute the EPV of any annuity
...
In this section we consider how to evaluate the EPV of 1/mthly and
continuous annuities given only the EPVs of annuities at integer ages
...
We present two methods that
¨
are commonly used, and we explore the accuracy of these methods for a fairly
typical (Makeham) mortality model
...

5
...
1 Recursions
¨
In a spreadsheet, with values for t px available, we may calculate ax using
a backward recursion
...
First, we set aω−1 = 1
...
is
ax = 1 + v px ax+1
¨
¨
since
ax = 1 + v px + v 2 2 px + v 3 3 px + · · ·
¨
= 1 + v px 1 + v px+1 + v 2 2 px+1 + · · ·
¨
= 1 + v px ax+1
...
36)

5
...
is

ax =
¨ (m)

1
m

1

+ vm

1
m

(m)
1
...
37)

We can calculate EPVs for term annuities and deferred annuities from the whole
life annuity EPVs, using, for example, equations (5
...
26)
...
Note, however, that Woolhouse’s formula, which is described in
Section 5
...
3, gives an excellent approximation to 1/mthly and continuous
annuity EPVs
...
11
...
The indication from Table 4
...

From Section 4
...
1 recall the results from equations (4
...
26) that,
under the UDD assumption,
A(m) =
x

i
Ax
i(m)

i
¯
A x = Ax
...
4), (5
...
14) in this chapter that for
any survival model
ax =
¨

1 − Ax
,
d

ax =
¨ (m)

(m)

1 − Ax
d (m)

and ax =
¯

Now, putting these equations together we have
ax =
¨ (m)
=
=

(m)

1 − Ax
d (m)
1−

i
A
i(m) x
(m)
d

i(m) − iAx
i(m) d (m)

using UDD

¯
1 − Ax

...
4)
i(m) d (m)
id
i − i(m)
= (m) (m) ax − (m) (m)
¨
i d
i d
=

= α(m) ax − β(m)
¨
where
α(m) =

id
i(m) d (m)

i − i(m)

...
38)

For continuous annuities we can let m → ∞, so that
ax =
¯

id
i−δ
a − 2
...
29) we have,
(m)
ax:n = ax − v n n px ax+n
¨ (m) ¨ (m)
¨

a
= α(m)¨ x − β(m) − v n n px (α(m)¨ x+n − β(m))
a
= α(m) ax − v n px ax+n − β(m) 1 − v
¨
¨
n

n

using UDD

n px

= α(m) ax:n − β(m) 1 − v n n px
...
It can be shown (see Exercise
5
...

2m

(5
...
11
...

(m)
It is based on the Euler–Maclaurin formula and expresses ax in terms of ax
...
It gives a
series expansion for the integral of a function, assuming that the function is
differentiable a certain number of times
...
40)
2
12
720

5
...

To obtain our approximations we apply this formula twice to the function
g(t) = v t t px , in each case ignoring second and higher order derivatives of g,
which is reasonable as the function is usually quite smooth
...

First, let h = 1
...
40) becomes
∞

g(k) −
k=0

1
1
+ g (0) =
2 12

∞

v k k px −
k=0

= ax −
¨

1
1
−
(δ + µx )
2 12

1
1
−
(δ + µx )
...
41)

Second, let h = 1/m
...
40) becomes
1
m

∞

g(k/m)−
k=0

1
1
1
g (0) =
+
2m 12m2
m

∞

k

v m k px −
k=0

= ax −
¨ (m)

m

1
1
−
(δ + µx )
2m 12m2

1
1
−
(δ + µx )
...
42)
2m 12m2

Since each of (5
...
42) approximates the same quantity, ax , we can
¯
(m)
obtain an approximation to ax by equating them, so that
¨
ax −
¨ (m)

1
1
1
1
−
(δ + µx ) ≈ ax − − (δ + µx )
...

2m
12m2

(5
...
43) gives the first three terms of Woolhouse’s
formula, and this is the basis of our actuarial approximations
...
29),
(m)

(m)

ax:n = ax − v n n px ax+n
¨
¨ (m)
¨
and applying formula (5
...

12m2

= ax:n −
¨

(5
...
43) and (5
...
41)), so that
ax ≈ ax −
¯
¨

1
1
− (δ + µx )
2 12

(5
...

2
12
(m)

An important difference between the approximation to ax:n based on Wool¨
house’s formula and the UDD approximation is that we need extra information
for the Woolhouse approach, specifically values for the force of mortality
...
44) is often omitted (leading to the same
approximation as equation (5
...

If the integer age information available does not include values of µx , then
we may still use Woolhouse’s formula
...
46)

and the results for the illustrations given in the next section are almost identical
to where the exact value of the force of mortality is used
...
12 Numerical illustrations

135

In fact, Woolhouse’s formula (with three terms) is so accurate that even if the
full force of mortality curve is known, it is often a more efficient way to calculate
annuity values than the more direct formulae with comparable accuracy
...

¨ (m)
x
In Section 2
...
2 we saw an approximate relationship between the complete
expectation of life and the curtate expectation of life, namely
◦

ex ≈ ex +

1
2


...
45) gives a refinement of this
approximation, namely
◦

ex ≈ ex +

1
2

−

1
12 µx


...
12 Numerical illustrations
In this section we give some numerical illustrations of the different methods
of computing ax:n
...
3 shows values of a(12) for x = 20, 30,
...
05
...
1, while Table 5
...
3
...
37);
denotes the approximation to the EPV based on the uniform distribution
of deaths assumption;
denotes the approximation to the EPV based on Woolhouse’s formula,
using the first two terms only;
denotes the approximation to the EPV based on Woolhouse’s formula,
using all three terms, including the exact force of mortality;
denotes the approximation to the EPV based on Woolhouse’s formula,
using all three terms, but using the approximate force of mortality
estimated from integer age values of px
...
Also, note that the
inclusion of the third term is important for accuracy; the two-term Woolhouse

136

Annuities
Table 5
...
Values of a(12) for i = 0
...

¨
x:10

x

Exact

UDD

W2

W3

W3*

20
30
40
50
60
70
80
90
100

6
...
4630
6
...
4295
6
...
0991
5
...
8975
2
...
4655
6
...
4550
6
...
3482
6
...
3989
3
...
0699

6
...
4679
6
...
4344
6
...
1044
5
...
9117
2
...
4655
6
...
4550
6
...
3485
6
...
4003
3
...
0497

6
...
4630
6
...
4295
6
...
0990
5
...
8975
2
...
4
...
05
...
5770
14
...
4663
14
...
4275
11
...
2889
4
...
4425

14
...
5505
14
...
2024
13
...
5104
8
...
9281
2
...
5792
14
...
4684
14
...
4295
11
...
2938
4
...
4656

14
...
5506
14
...
2028
13
...
5117
8
...
9242
2
...
5770
14
...
4663
14
...
4275
11
...
2889
4
...
4424

formula is the worst approximation
...
In this case approximations based on Woolhouse’s formula are
superior, provided the three-term version is used
...
Using Woolhouse’s formula requires only
the integer age table, of 100 rows, and the accuracy all the way up to age 100
is excellent, using the exact or approximate values for µx
...

5
...
Just

5
...
Thus, for example, the EPV of an n-year term annuity
payable continuously at rate 1 per year to a life who is aged x + k and who was
select at age x is a[x]+k:n , with
¯
¯
¯
A[x]+k:n = 1 − δ a[x]+k:n
...


t=0

5
...
In this paper he also showed that his
theory applied to joint-life annuities, a topic we discuss in Chapter 8
...
The Euler–Maclaurin formula was derived independently (about
130 years before Woolhouse’s paper) by the famous Swiss mathematician Leonhard Euler and by the Scottish mathematician Colin Maclaurin
...
(1994)
...
15 Exercises
When a calculation is required in the following exercises, unless otherwise
stated you should assume that mortality follows the Standard Ultimate Survival
Model as specified in Section 4
...

Exercise 5
...

(a)

Y1 =

aTx
¯
a15
¯

if Tx ≤ 15,
if Tx > 15
...


Exercise 5
...


This is called a Family Income Benefit
...

¯
¯
(c) Explain the answer in (b) by general reasoning
...
3 Given that a50:10 = 8
...
8277, and
¨
0
...
4 Given that a60 = 10
...
756, a62 = 10
...
06, calculate 2 p60
...
5 You are given the following extract from a select life table
...
15 Exercises

139

(f) the probability that the present value of an annuity-due of 1 per year issued
to a select life aged 40 is less than 3
...

Exercise 5
...

Thus
µx = (µA + µB )/2
x
x
for all x
...


Will this approximation overstate or understate the true value of ax ?
Exercise 5
...
Obtain the formula
¨
¨
(IA)x = ax − d (I a)x
by writing down the present value random variables for
(a) an increasing annuity-due to (x) with payments of t + 1 at times t =
0, 1, 2,
...
, if the
death of (x) occurs between ages x + t − 1 and x + t
...

d

Exercise 5
...

(a) Show that

V[aH ] =

2A

x:n+1

− Ax:n+1
d2

2


...



...
9 Consider the random variables Y = aTx and Z = v Tx
...

(b) Derive an expression for the covariance, in terms of standard actuarial
functions
...

Exercise 5
...

dx ¨

Exercise 5
...


Age

Number of
annuitants

60
70
80

40
30
10

Each annuity has an annual payment of $10 000 as long as the annuitant
survives
...
Calculate
(a) the expected present value of the total outgo on annuities,
(b) the standard deviation of the present value of the total outgo on annuities, and
(c) the 95th percentile of the distribution of the present value of the total outgo
on annuities using a Normal approximation
...
12 Consider the quantities α(m) and β(m) in formula (5
...
By
expressing i, i(m) , d and d (m) in terms of δ, show that
α(m) ≈ 1

and β(m) ≈

m−1

...
13 Using a spreadsheet, calculate the mean and variance of the
present value of
(a) an arithmetically increasing term annuity-due payable to a life aged 50
for at most 10 years under which the payment at time t is t + 1 for t =
0, 1,
...
03t for t = 0, 1,
...


5
...
14 Using a spreadsheet, calculate the mean and variance of the
present value of
(a) a whole life annuity-due to a life aged 65, with annual payments of 1, and
(b) a whole life annuity-due to a life aged 65, with annual payments of 1 and a
guarantee period of 10 years
...

Exercise 5
...

Use Jensen’s inequality to show that
ax ≤ aE[Tx ]
...
3 4
...
4 0
...
5 (a) 3
...
45057
(c) 8
...
16305
(e) 119
...
00421
5
...
13 (a) 40
...
057
(b) 9
...
32965
5
...
550,
12
...
814,
8
...
1 Summary
In this chapter we discuss principles of premium calculation for insurance
policies and annuities
...
We next introduce the present value
of future loss random variable
...
We look at how we can use the future loss random
variable to determine when a contract moves from loss to profit or vice versa
...
The chapter concludes with a discussion of how a premium can be
calculated when the insured life is subject to some extra level of risk
...
2 Preliminaries
An insurance policy is a financial agreement between the insurance company
and the policyholder
...
The premiums will also need to reimburse
the insurance company for the expenses associated with the policy
...
In this case we refer to a net premium (also, sometimes, a
risk premium or mathematical premium)
...

The premium may be a single payment by the policyholder – a single premium – or it may be a regular series of payments, possibly annually, quarterly,
monthly or weekly
...
3 Assumptions

143

people receive their salaries monthly and it is convenient to have payments made
with the same frequency as income is received
...

A key feature of any life insurance policy is that premiums are payable in
advance, with the first premium payable when the policy is purchased
...
In this case, a person could
purchase the policy and then withdraw from the contract at the end of the first
year before paying the premium then due
...

Regular premiums for a policy on a single life cease to be payable on the
death of the policyholder
...
The premium paying term may
be the same as the term of the policy, but it could be shorter
...

As we discussed in Chapter 1, premiums are payable to secure annuity benefits as well as life insurance benefits
...
Immediate annuities are always
purchased by a single premium
...
Or, a person aged 65 might secure a monthly
annuity from an insurance company by payment of a single premium
...
However, there are other methods of calculating
premiums and we discuss one of these, the portfolio percentile principle
...
This approach is discussed in Chapter 11
...
3 Assumptions
As in Chapter 4, unless otherwise stated, we use a standard set of assumptions
for mortality and interest in the numerical examples in this chapter
...
13

144

Premium calculation
Table 6
...
Annuity values using the Standard Select Survival Model
...
96732
19
...
87165
19
...
76647
19
...
65087
19
...
52389
19
...
38449
19
...
23156
19
...
06390
18
...
88024
18
...
67927
18
...
45956
18
...
21969
18
...
95814
17
...
67340
17
...
36397
17
...
02835

19
...
87095
19
...
76574
19
...
65012
19
...
52310
19
...
38365
19
...
23066
19
...
06292
18
...
87917
18
...
67807
18
...
45822
18
...
21815
18
...
95637
17
...
67135
17
...
36156
17
...
02551
16
...
87070
19
...
76549
19
...
64985
19
...
52282
19
...
38336
19
...
23034
19
...
06258
18
...
87880
18
...
67766
18
...
45776
18
...
21763
18
...
95577
17
...
67065
17
...
36074
17
...
02453
16
...
66060

22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52

51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80

16
...
66514
16
...
27303
16
...
85091
15
...
39789
15
...
91340
14
...
39724
14
...
84972
13
...
27169
12
...
66467
12
...
03093
11
...
37350
11
...
69629
10
...
00402
9
...
30225
8
...
59722

16
...
46908
16
...
06137
15
...
62302
15
...
15325
14
...
65165
14
...
11822
13
...
55351
13
...
95864
12
...
33547
12
...
68661
11
...
01550
10
...
32644
9
...
62458
9
...
91584
8
...
20681

16
...
26762
16
...
84443
15
...
39012
15
...
90407
14
...
38606
14
...
83632
13
...
25568
12
...
64561
12
...
00830
11
...
34678
11
...
66491
10
...
96740
9
...
25981
8
...
54841
8
...
84008

53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82

with an interest rate of 5% per year effective
...
00022, B = 2
...
124
...
92−s µx+s
for 0 ≤ s ≤ 2
...
7 and values of Ax at an effective rate of interest of 5%

6
...
1
...

Example 6
...
, 80
...

¨
Solution 6
...
13
...
Annuity values can then be calculated recursively
¨
using
¨
ax = 1 + v px ax+1 ,
¨
a[x]+1 = 1 + v p[x]+1 ax+2 ,
¨
¨
a[x] = 1 + v p[x] a[x]+1
...
1
...
4 The present value of future loss random variable
The cash flows for a traditional life insurance contract consist of the insurance or
annuity benefit outgo (and associated expenses) and the premium income
...
So we can model the future outgo less future income with the random
variable that represents the present value of the future loss
...
When
0
expenses are included, then the premiums are the gross premiums, and the
g
random variable is referred to as the gross future loss, denoted L0
...

In cases where the meaning is obvious from the context, we will drop the n or
g superscript
...
2 An insurer issues a whole life insurance to [60], with sum insured
S payable immediately on death
...
The net annual premium is P
...

Solution 6
...


Since both terms of the random variable depend on the future lifetime of the
same life, [60], they are clearly dependent
...
No premium
is payable on reaching age 80
...
This distribution can be
used to find a suitable premium for a given benefit, or an appropriate benefit
for a specified premium
...
This is a method of selecting an appropriate premium using a given loss
distribution
...


6
...
5
...

Thus, expenses are not a part of net premium calculation
...

We start by stating the equivalence principle
...
That means that
E[Ln ] = 0
0
which implies that
E[PV of benefit outgo − PV of net premium income] = 0
...


(6
...
5 The equivalence principle

147

The equivalence principle is the most common premium principle in traditional
life insurance, and will be our default principle – that is, if no other principle is
specified, it is assumed that the equivalence principle is to be used
...
3 Consider an endowment insurance with term n years and sum
insured S payable at the earlier of the end of the year of death or at maturity,
issued to a select life aged x
...

Derive expressions in terms of S, P and standard actuarial functions for
(a)
(b)
(c)
(d)

the net future loss, Ln ,
0
the mean of Ln ,
0
the variance of Ln , and,
0
the annual net premium for the contract
...
3 (a) The future loss random variable is
¨
Ln = Sv min(K[x] +1,n) − P amin(K
0

[x] +1,n)


...

(c) Expanding the expression above for Ln gives
0
1 − v min(K[x] +1,n)
d
P
v min(K[x] +1,n) − ,
d

Ln = Sv min(K[x] +1,n) − P
0
= S+

P
d

which isolates the random variable v min(K[x] +1,n)
...


(d) Setting the EPVs of the premiums and benefits to be equal gives the net
premium as
P=S

A[x]:n

...
2)
2

148

Premium calculation

Furthermore, using formula (6
...

¨
Example 6
...
Premiums are payable monthly throughout the deferred
period
...

(a) Write down the net future loss random variable in terms of lifetime random
variables for [x]
...

(c) Assume now that, in addition, the contract offers a death benefit of S payable
immediately on death during the deferred period
...

Solution 6
...
Then
⎧
(12)
if T[x] ≤ n,
¨
⎪0 − 12P a (12) 1
⎪
⎪
K[x] + 12
⎨
Ln =
0
⎪ n (12)
(12)
⎪X v a
⎪
¨ (12)
− 12P an
¨
if T[x] > n
...

¨ (12)
By equating these EPVs we obtain the premium equation which gives
(12)

P=

X v n n p[x] a[x]+n
¨
(12)

12¨ [x]:n
a

=

(12)
¨
n E[x] X a[x]+n

...
5 The equivalence principle

149

(c) We now have

Ln =
0

⎧ T
(12)
¨
⎪Sv [x] − 12P a (12) 1
⎪
⎪
K[x] + 12
⎨
⎪ n (12)
⎪X v a
⎪
¨ (12)
⎩

if T[x] ≤ n,
(12)

1
K[x] + 12 −n

− 12P an
¨

if T[x] > n
...


12¨ [x]:n
a

2

Example 6
...
Usually, the premium calculation does not require the
identification of the future loss random variable
...

Example 6
...
Using the Standard Select Survival
Model, with interest at 5% per year, calculate the total amount of net premium
payable in a year if premiums are payable (a) annually, (b) quarterly, and (c)
monthly, and comment on these values
...
5 Let P denote the total amount of premium payable in a year
...


Using Tables 3
...
1, we have
a[45]:20 = a[45] −
¨
¨

l65 20
v a[65] = 12
...

¨
l[45]

From this we get
¨
A[45]:20 = 1 − d a[45]:20 = 0
...

Hence, for m = 1 the net premium is P = $2 965
...


150

Premium calculation
Table 6
...
Annuity values and premiums
...
69859
12
...
69859

3 022
...
16
3 022
...
64512
12
...
64512

3 034
...
94
3 034
...
11
...
Table 6
...
The ordering of these premiums
for m = 1, 4, 12 reflects the ordering of EPVs of 1/mthly annuities which we
observed in Chapter 5
...


6
...
There are three main types
of expense associated with policies – initial expenses, renewal expenses and
termination expenses
...
There are two major types of initial expenses – commission to agents for
selling a policy and underwriting expenses
...

Underwriting expenses may vary according to the amount of the death benefit
...

Renewal expenses are normally incurred by the insurer each time a premium
is payable, and in the case of an annuity, they are normally incurred when an

6
...
These costs arise in a variety of ways
...

Renewal expenses also cover the ongoing fixed costs of the insurer such as
staff salaries and rent for the insurer’s premises, as well as specific costs such
as annual statements to policyholders about their policies
...
Often, per policy renewal
costs are assumed to be increasing at a compound rate over the term of the
policy, to approximate the effect of inflation
...
Generally these expenses are small, and are largely
associated with the paperwork required to finalize and pay a claim
...

Where allowance is made, it is usually proportional to the benefit amount
...

The equivalence principle applied to the gross premiums and benefits states
that the EPV of the gross future loss random variable should be equal to zero
...

In other words, under the equivalence premium principle,
EPV of benefits + EPV of expenses = EPV of gross premium income
...
3)
We conclude this section with three examples in each of which we apply the
equivalence principle to calculate gross premiums
...
6 An insurer issues a 25-year annual premium endowment insurance with sum insured $100 000 to a select life aged 30
...
5% of each subsequent premium
...


152

Premium calculation

(a) Write down the gross future loss random variable
...

Solution 6
...
Then
g

L0 = S v min(T[x] ,n) + 2000 + 0
...
025P amin(K
¨

[x] +1,n)

− P amin(K
¨

[x] +1,n)

=Sv

min(T[x] ,n)

+ 2000 + 0
...
975P amin(K
¨

[x] +1,n)


...
5% of the second and subsequent premiums are more conveniently written
as 2
...
5% of the first premium
...

(b) We may look separately at the three parts of the gross premium equation
of value
...
73113 P
...
7 and 6
...

¨
The EPV of all expenses is
2000+0
...
025P a[30]:25 = 2000 + 0
...
025 × 14
...
843278P
...
298732 = 29 873
...

Thus, the equivalence principle gives
P=

29 873
...
04
...
73113 − 0
...
7 Calculate the monthly gross premium for a 10-year term insurance
with sum insured $50 000 payable immediately on death, issued to a select life

6
...
7 Let P denote the monthly premium
...
To find the EPV of premium related expenses, we can
[55]:10
apply the same idea as in the previous example, noting that initial expenses
apply to each premium in the first year
...
09 × 12P a
¨

(12)
[55]:10

+ 0
...
The EPV of the insurance benefit
¯
is 50 000A 1
and so the equivalence principle gives
[55]:10

(12)
[55]:10

12P 0
...
09¨
a

= 7
...
9773 and A

1
[55]:10


...
024954,

giving P = $18
...

Calculating all the EPVs exactly gives the same answer for the premium to
four significant figures
...
8 Calculate the gross single premium for a deferred annuity of
$80 000 per year payable monthly in advance, issued to a select life now aged
50 with the first annuity payment on the life’s 65th birthday
...
Assume that the renewal expense will
be $20 on the first anniversary of the issue date, and that expenses will increase
with inflation from that date at the compound rate of 1% per year
...

Solution 6
...
The renewal expense on the tth policy anniversary

154

Premium calculation

is 20 1
...
so that the EPV of renewal expenses is
∞

1
...
01
20
1
...
01t v t t p[50]
t=1
∞

vjt t p[50]
t=1

20
(¨ [50] j − 1)
a
1
...
01v = 1/(1 + j), that is j = 0
...
The EPV of the deferred annuity is
(12)
a
80 000 15 |¨ [50] , so the single premium is
1 000 +

20
(12)
(¨ [50] j − 1) + 80 000 15 |¨ [50]
...
01
(12)

As a[50] j = 19
...
04129, the single premium is $484 669
...
6 and 6
...
In Example 6
...
04 and the expenses
at time 0 are $2 000 plus 50% of the first premium, a total of $3146
...
Similarly, in Example 6
...
88 and the total expenses in the first year are $500 plus 10%
of premiums in the first year
...
This situation is common in
practice, especially when initial commission to agents is high, and is referred
to as new business strain
...
From time to time
insurers get into financial difficulties through pursuing an aggressive growth
strategy without sufficient capital to support the new business strain
...
These early expenses are gradually paid off by the
expense loadings in future premiums
...


6
...
Since
writing business generally involves a loan from shareholder or participating

6
...
In traditional insurance, we often load for profit implicitly, by margins
in the valuation assumptions
...
The extra income from the invested premiums will contribute to profit
...
Some will be paid as dividends to
shareholders, if the company is proprietary
...
For a term insurance,
we might use a slightly higher overall mortality rate than we expect
...

More modern premium setting approaches, which use projected cash flows,
are presented in Chapter 11, where more explicit allowance for profit is
incorporated in the methodology
...
Although we calculate a premium assuming a given survival model, for each individual policy the
experienced mortality rate in any year can take only the values 0 or 1
...
For the actual profit from a group of policies to be reliably close to the
expected profit, we need to sell a large number of individual contracts, whose
future lifetimes can be regarded as statistically independent, so that the losses
and profits from individual policies are combined
...
Let
us suppose that the life is subject to a mortality of rate of 0
...

Then, using the equivalence principle, the premium is
P = 1 000 × 0
...
05 = 9
...

The future loss random variable is
Ln =
0

1 000v − P = 942
...
52
if Tx > 1,

with probability 0
...
99
...
01 × 942
...
99 × (−9
...
99, and the probability of
loss is 0
...
The balance arises because the profit, if the policyholder survives

156

Premium calculation

the year, is small, and the loss, if the policyholder dies, is large
...
01
...

The insurer would expect to make a (small) profit on 99 of them
...
If one life dies, there is no profit or loss
...
Let D denote the number of deaths in the
portfolio, so that D ∼ B(100, 0
...
The probability that the profit on the whole
portfolio is greater than or equal to zero is
Pr[D ≤ 1] = 0
...
In fact, as the number of policies
issued increases, the probability of profit will tend, monotonically, to 0
...
On
the other hand, while the probability of loss is increasing with the portfolio size,
the probability of very large aggregate losses (relative, say, to total premiums)
is much smaller for a large portfolio, since there is a balancing effect from
diversification of the risk amongst the large group of policies
...
For this policy
¨
¨
L0 = Sv K[x] +1 + I + e aK[x] +1 − P aK[x] +1 ,
g

where K[x] denotes the curtate future lifetime of [x]
...
We can
use the future loss random variable to find the minimum future lifetime for
the policyholder in order that the insurer makes a profit on this policy
...

g

6
...


(6
...
4)
by τ , so that the contract generates a profit for the insurer if K[x] + 1 > τ
...
Thus, if τ denotes the integer part of τ , then
the insurer makes a profit if the life survives at least τ years, the probability
of which is τ p[x]
...
Then we find that P = $498
...
4) we find that there is a profit if K[30] + 1 > 52
...
Thus, there is a profit if
the life survives for 52 years, the probability of which is 52 p[30] = 0
...

Figure 6
...
We see that large losses occur in the
early years of the policy, and even larger profits occur if the policyholder dies
at an advanced age
...
For example, if the policyholder dies in the first policy year, the loss
to the insurer is $100 579, and the probability of this loss is q[30] = 0
...

Similarly, a profit of $308 070 arises if the death benefit is payable at time 80,

600000
500000
400000

Profit

300000
200000
100000
0
0

10

20

30

40

50

60

70

80

90

-100000
-200000

Year

Figure 6
...
7
...
00023
...

Example 6
...
Premiums
are payable annually throughout the term of the policy
...
The insurer allows for a compound reversionary
bonus of 2
...
The death benefit is payable at the end of the year of death
...

g

(a) Derive an expression for the future loss random variable, L0 , for this policy
...

(c) Let L0 (k) denote the present value of the loss on the policy given that
K[30] = k for k ≤ 24 and let L0 (25) denote the present value of the loss on
the policy given that the policyholder survives to age 55
...
, 25
...

g
(e) Calculate V[L0 ]
...
9 (a) First, we note that if the policyholder’s curtate future lifetime,
K[30] , is k years where k = 0, 1, 2,
...
025)K[30] v K[30] +1
...
Thus, if P
denotes the annual premium,
L0 = 250 000(1
...
39P − 0
...


(b) The EPV of the premiums, less premium expenses, is
0
...
5838P
...
025t ) if the policyholder dies in the tth
policy year, the EPV of the death benefit is
24

v t+1 t |q[30] (1
...
025 [30]:25 j

where 1 + j = (1 + i)/(1
...
02439
...
37

6
...
3
...
9
...


...

23
24
≥ 25

233 437
218 561

...


...
02525 = 134 295
...
39P
...
44
...
, 24, the present value of the loss
is the present value of the death benefit payable at time k +1 less the present
value of k + 1 premiums plus the present value of expenses
...
025k ) v k+1 + 1 200 + 0
...
99P ak+1
...
02525 ) v 25 + 1 200 + 0
...
99P a25
...
3
...
Hence the probability of a
profit is 24 p[30] = 0
...

Note that this probability is based on the assumption that future expenses
and future interest rates are known and will be as in the premium basis
...
552
k=0
g

which is equal to the variance as E[L0 ] = 0
...
1
...

Example 6
...
The first annuity payment will take
place 10 years from issue, and payments will be annual
...
Ignoring expenses, and using the Standard Select Survival
Model with interest at 5% per year, calculate
(a) the single premium,
(b) the probability the insurance company makes a profit from this policy, and
(c) the probability that the present value of the loss exceeds $100 000
...
10 (a) Let P denote the single premium
...
03t−10 ) t p[55] = $546 812
...
Then
L0 (k) =

−P
¨
−P + 50 000v 10 ak−9 j

for k = 0, 1,
...
,

(6
...
05/1
...
019417
...
5) shows that
¨
L0 (k) is an increasing function of k for k ≥ 10
...
Using formula (6
...
0510 P/50 000
...
7 Profit

161

Writing ak−9 j = (1 − v k−9 )/dj where dj = j/(1 + j), this condition
¨
becomes
vjk−9 > 1 − dj 1
...
0510 P/50 000 /δj
...
55, and so there will be a profit if
the policyholder dies before age 86
...
41051
...
68
...
38462
...
2 shows L0 (k) for k = 1, 2,
...
We can see that the loss is constant
for the first 10 years at −P and then increases due to annuity payments
...
1, longevity results in large losses to the insurer
...

2

400000
300000

Present value of loss

200000
100000
0
0
–100000

5

10

15

20

25

30

35
40
45
Year of death, k

–200000
–300000
–400000
–500000
–600000

Figure 6
...
10
...
8 The portfolio percentile premium principle

The portfolio percentile premium principle is an alternative to the equivalence
premium principle
...
By ‘identical’ we mean that the policies have the same premium, benefits, term, and so on, and that the policyholders are all subject to the same survival
model
...

Suppose we know the sum insured for these policies, and wish to find an
appropriate premium
...
Let N denote the number of policies in the portfolio
and let L0,i represent the future loss random variable for the ith policy in the
portfolio, i = 1, 2, 3,
...
The total future loss in the portfolio is L, say, where
N

L=

N

L0,i ;

E[L] =

i=1

E[L0,i ] = N E[L0,1 ];
i=1
N

V[L] =

V[L0,i ] = N V[L0,1 ]
...
)
The portfolio percentile premium principle sets a premium so that there is a
specified probability, say α, that the total future loss is negative
...

Now, if N is sufficiently large (say, greater than around 30), the central limit
theorem tells us that L is approximately normally distributed, with mean E[L] =
N E[L0,1 ] and variance V[L] = N V[L0,1 ]
...


6
...
However, as illustrated in the next example, both the mean
and variance of L are functions of P
...
11 An insurer issues whole life insurance policies to select lives
aged 30
...

Initial expenses, incurred at the issue of the policy, are 15% of the total of the
first year’s premiums
...

Assume the Standard Select Survival Model with interest at 5% per year
...

(b) Calculate the monthly premium using the portfolio percentile principle,
such that the probability that the future loss on the portfolio is negative is
95%
...

Solution 6
...
Then the EPV of premiums is
(12)

12P a[30] = 227
...

¨
The EPV of benefits is
(12)

100 000A[30] = 7 866
...
15 × 12P + 0
...
8826P
...
39 per month
...
15 × 12P − 0
...


and its expected value can be calculated using the solution to part (a) as
E[L0,i ] = 7 866
...
18P
...
96 × 12P
d (12)

(12)

1

v K[30] + 12 + 0
...
96 × 12P
d (12)

164

Premium calculation
so that
V[L0,i ] = 100 000 +

0
...
59P)2 (0
...
59P) (0
...

The future loss random variable for the portfolio of policies is L =
10 000
i=1 L0,i , so
E[L] = 10 000(7866
...
18P)
and
V[L] = 10 000 (100 000 + 236
...
0053515)
...
18P − 7 866
...
59P) (0
...
95
...
645) = 0
...
18P − 7 866
...
645
(100 000 + 236
...
073154)
which gives P = $36
...

2
Note that the solution to part (b) above depends on the number of policies in
the portfolio (10 000) and the level of probability we set for the future loss
being negative (0
...
If the portfolio had n policies instead of 10 000, then the
equation we would have to solve for the premium, P, is
√
n(216
...
18)
= 1
...

(6
...
59P) (0
...
4 shows some values of P for different values of n
...
In fact, as n → ∞, P → $36
...
The reason for this is that as n → ∞ the
insurer diversifies the mortality risk
...


6
...
4
...

n

P

1 000
2 000
5 000
10 000
20 000

38
...
74
37
...
99
36
...
9 Extra risks
As we discussed in Section 1
...
5, when an individual wishes to effect a life
insurance policy, underwriting takes place
...
There are different ways in
which we can model the extra mortality risk in a premium calculation
...
9
...
In such circumstances
we refer to the individual as an impaired life, and the insurer may compensate
for this extra risk by treating the individual as being older
...
This approach to modelling extra risk involves no new
ideas in premium calculation – for example, we could apply the equivalence
principle in our calculation, and we would simply change the policyholder’s
age
...

6
...
2 Constant addition to µx
Individuals can also be deemed to be ineligible for standard rates if they regularly participate in hazardous pursuits, for example parachuting
...
The application of this approach leads to
some computational shortcuts for the following reason
...
Then
t p[x]

t

= exp −

µ[x]+s ds = exp −

0

t
0

µ[x]+s + φ ds = e−φt t p[x]
...
7)

t=0

where j denotes calculation at interest rate j = eφ+δ − 1
...

Now suppose that the impaired life has curtate future lifetime K[x]
...

d
d

So
¨
¨
A[x]:n = 1 − d a[x]:n = 1 − d a[x]:n j
...
8)

It is important to note here that for the insurance benefit we cannot just change
the interest rate
...
8), the annuity is evaluated at rate j, but the
function d uses the original rate of interest, that is d = i/(1 + i)
...
8) for any insurance factors
...


Example 6
...
01
...
9 Extra risks

167

subsequent premiums
...

Solution 6
...
Then by applying formula
(6
...
05e0
...
06055
...
38P + 0
...

¨
The EPV of the benefit is 200 000A
, where the dash denotes extra
[30]:20
mortality and the interest rate is i = 0
...
Using formula (6
...

¨

= 12
...
05/1
...
84
...
425158

6
...
3 Constant multiple of mortality rates
A third method of allowing for extra mortality is to assume that lives are subject
to mortality rates that are higher than the standard lives’ mortality rates
...
1q[x]+t where the superscript again denotes
extra mortality risk
...
A computational disadvantage
of this approach is that we have to apply approximations in calculating EPVs
if payments are at other than annual intervals
...

Example 6
...
Assume that each year throughout the 10-year term the life is subject to
mortality rates that are 10% higher than for a standard life of the same age
...
Use
the UDD assumption where appropriate, and use the Standard Select Survival
Model with interest at 5% per year
...
13 Let P denote the total premium per year
...
0002

and
β(12) =

i − i(12)
= 0
...

i(12) d (12)

As the initial expenses are 1000 plus 50% of the first premium, which is
we can write the EPV of expenses as
1000 +

1
12 P,

0
...

+ 0
...

50:10
δ

is
9

a
¨

50:10

=

v t t p50
t=0

where
t−1
t p50

=

(1 − 1
...


(6
...
9) as standard lives are subject to select
(12)
¯
mortality
...
0516, a
¨
= 7
...
01621,
50:10
50:10
50:10
which give P = $345
...
76
...
10 Notes and further reading

169

Table 6
...
Spreadsheet calculations for Example 6
...

(2)

t |q50

(4)
vt

(5)
v t+1

(6)
(2) × (4)

(7)
(3) × (5)

1
...
9989
0
...
9959
0
...
9921
0
...
9875
0
...
9819

0
...
0014
0
...
0018
0
...
0022
0
...
0027
0
...
0033

1
...
9524
0
...
8638
0
...
7835
0
...
7107
0
...
6446

0
...
9070
0
...
8227
0
...
7462
0
...
6768
0
...
6139

1
...
9513
0
...
8603
0
...
7774
0
...
7018
0
...
6329

0
...
0013
0
...
0015
0
...
0016
0
...
0018
0
...
0020

Total

0
1
2
3
4
5
6
7
8
9

(3)

t p50

(1)
t

8
...
0158

Table 6
...

Column (2) was created from the original mortality rates using formula (6
...
1q50+t )
...
Note that this must then by multiplied by i/δ to get
2


...
10 Notes and further reading
The equivalence principle is the traditional approach to premium calculation,
and we apply it again in Chapter 7 when we consider the possibility that a
policy may terminate for reasons other than death
...
We have seen one in Section 6
...

A modification of the equivalence principle which builds an element of profit
into a premium calculation is to select a profit target amount for each policy,
, say, and set the premium to be the smallest possible such that E[L0 ] ≤
...


170

Premium calculation

Besides the premium principles discussed in this chapter, there is one further
important method of calculating premiums
...

The international actuarial notation for premiums may be found in Bowers
et al
...
We have omitted it in this work because we find it has no particular
benefit in practice
...
11 Exercises
When a calculation is required in the following exercises, unless otherwise
stated you should assume that mortality follows the Standard Select Survival
Model as specified in Section 6
...

Exercise 6
...
A select individual aged 41 purchased a three-year
term insurance with a net premium of $350 payable annually
...


[x]

l[x]

[40] 100 000
[41] 99 802
[42] 99 597

l[x]+1

l[x]+2

l[x]+3

lx+4

x+4

99 899
99 689
99 471

99 724
99 502
99 628

99 520
99 283
99 030

99 288
99 033
98 752

44
45
46

Use an effective rate of interest of 6% per year to calculate
(a) the sum insured, assuming the equivalence principle,
(b) the standard deviation of L0 , and
(c) Pr[L0 > 0]
...
2 Consider a 10-year annual premium term insurance issued to a
select life aged 50, with sum insured $100 000 payable at the end of the year of
death
...

(b) Calculate the net annual premium
...
3 Consider a 20-year annual premium endowment insurance with
sum insured $100 000 issued to a select life aged 35
...
11 Exercises

171

expenses of 3% of the second and subsequent premiums
...

(a)
(b)
(c)
(d)

Write down an expression for the gross future loss random variable
...

Calculate the standard deviation of the gross future loss random variable
...


Exercise 6
...
The basic sum insured is $200 000 payable
at the end of the month of death, and the premium term is 25 years
...
5% per year, vesting on each policy anniversary, initial expenses of 60% of the annual premium, renewal expenses of 2
...

Calculate the annual premium
...
5 A select life aged exactly 40 has purchased a deferred annuity
policy
...
The
initial annuity payment will be $50 000, and each subsequent payment will be
2% greater than the previous one
...
Calculate the gross monthly premium allowing for initial
expenses of 2
...

Exercise 6
...
7 A life insurer is about to issue a 30-year deferred annuity-due with
annual payments of $20 000 to a select life aged 35
...

(a) Calculate the single premium for this annuity
...
Calculate the revised
premium
...

Exercise 6
...
Let L0 be the
net future loss random variable with the premium determined by the equivalence principle
...
5
...
75, calculate V[L∗ ]
...
9 Calculate both the net and gross premiums for a whole life insurance issued to a select life aged 40
...

Premiums are payable annually in advance for a maximum of 20 years
...
0001, B = 0
...
075
2 year select period, q[x] = 0
...
9qx+1
6% per year effective
30% of the first year’s premium
plus 3% of all premiums after the first year
On each premium date an additional expense
starting at $10 and increasing at a compound rate
of 3% per year

Exercise 6
...
Level premiums are paid monthly
in advance throughout the term
...

Assume the sum insured and claim expenses are paid one month after the date
of death, and use claims acceleration
...
11 Exercises

173

Exercise 6
...

Calculate the initial annual gross premium
...
12 For a whole life insurance with sum insured $150 000 paid at
the end of the year of death, issued to (x), you are given:
(i) 2 Ax = 0
...
0653, and
(iii) the annual premium is determined using the equivalence principle
...

0
Exercise 6
...
Show that at rate of interest i,
j
¯
¯j
Ax = Ax + φ ax ,
¯

where j is a rate of interest that you should specify
...
14 A life insurer is about to issue a 25-year annual premium endowment insurance with a basic sum insured of $250 000 to a life aged exactly 30
...
The office allows for a compound reversionary bonus of 2
...
The death benefit is payable at the end
of the year of death
...
Show
that
L0 = 250 000Z1 +

0
...
39P − 0
...
025)K[30]
v 25 (1
...


(b) Using the equivalence principle, calculate P
...
Hence calculate
V[L0 ] using the value of P from part (b)
...

Hint: recall the standard results from probability theory, that for random
variables X and Y and constants a, b and c, V[X + c] = V[X ], and
V[aX + bY ] = a2 V[X ] + b2 V[Y ] + 2ab Cov[X , Y ],
with Cov[X , Y ] = E[XY ] − E[X ]E[Y ]
...
15 An insurer issues a 20-year endowment insurance policy to (40)
with a sum insured of $250 000, payable at the end of the year of death
...

(a) Calculate the premium using the equivalence principle
...

(c) Assuming 10 000 identical, independent contracts, estimate the 99th
percentile of the net future loss random variable using the premium in (a)
...
1 (a) $216 326
...
03
(c) 0
...
2 (b) $178
...
3 (b) $3 287
...
10
(d) 0
...
4 $3 262
...
5 $2 377
...
6 $212
...
7 (a) $60 694
...
30
6
...
6875
6
...
40 (net),
6
...
30

$1 431
...
11 Exercises
6
...
13
6
...
72
6
...
44
(c) $0
...
30251,
146 786 651
...
98297
6
...
84
(b) 0,
$14 485
(c) $33 696

$0
...
09020,

0
...
1 Summary
In this chapter we introduce the concept of a policy value for a life insurance
policy
...

We start by considering the case where all cash flows take place at the start or
end of a year
...
We also show how to calculate the profit from a policy
in any year and we introduce the asset share for a policy
...
We also consider policy alterations
...
2 Assumptions
In almost all the examples in this chapter we assume the Standard Select Survival
Model specified in Example 3
...
We
assume, generally, that lives are select at the time they purchase their policies
...
This means that the life table in Table 3
...
1 on page 83 and the whole life
annuity values in Table 6
...

7
...
3
...
In this chapter
we are concerned with the estimation of future losses at intermediate times
176

7
...
We therefore extend the future
loss random variable definition, in net and gross versions
...
The present value of future net
loss random variable is denoted Ln and the present value of gross future loss
t
g
random variable is denoted Lt , where
Ln = Present value, at time t, of future benefits
t
− Present value, at time t, of future net premiums
and
g

Lt = Present value, at time t, of future benefits
+ Present value, at time t, of future expenses
− Present value, at time t, of future gross premiums
...

Note that the future loss random variable Lt is defined only if the contract is
still in force t years after issue
...
The important features of
this example for our present purposes are that premiums are payable annually
and the sum insured is payable at the end of the year of death, so that all cash
flows are at the start or end of each year
...
1 Consider a 20-year endowment policy purchased by a life aged
50
...

The basis used by the insurance company for all calculations is the Standard
Select Survival Model, 5% per year interest and no allowance for expenses
...
33
...

Solution 7
...
8456
¨

and A[50]:20 = 0
...


The equation of value for P is
P a[50]:20 − 500 000 A[50]:20 = 0,
¨

(7
...
33
...
The
policyholder will then be aged 60 and the select period for the survival
model, two years, will have expired eight years ago
...
Hence, the formulae for
¨
60
Ln and Ln are
10
11
Ln = 500 000 v min(K60 +1,10) − P amin(K
¨
10

60 +1,10)

and
Ln = 500 000 v min(K61 +1,9) − P amin(K
¨
11


...
9555
¨

and a61:9 = 7
...

¨
11
2
We are now going to look at Example 7
...
At the time when
the policy is issued, at t = 0, the future loss random variable, Ln , is given by
0
Ln = 500 000 v min(K[50] +1,20) − P amin(K
¨
0

[50] +1,20)


...
1)
...


7
...
The policyholder may have died before time 10
...
In this case the
insurer no longer has any liability with respect to this policy
...
In this case the calculation in part
(b) shows that the future loss random variable, Ln , has a positive expected
10
value ($190 339) so that future premiums (from time 10) are not expected to
be sufficient to provide the future benefits
...

Speaking generally, when a policy is issued the future premiums should be
expected to be sufficient to pay for the future benefits and expenses
...

The amount needed to cover this shortfall is called the policy value for the
policy at time t
...
For
example, in the first year the premium of $15 114
...
04
...
1, for each year
except the last the premium exceeds the EPV of the benefits, that is
P > 500 000 v q[50]+t

for t = 0, 1,
...


The final year is different because
P = 15 114
...

Note that if the policyholder is alive at the start of the final year, the sum
insured will be paid at the end of the year whether or not the policyholder
survives the year
...
1 shows the excess of the premium over the EPV of the benefit
payable at the end of the year for each year of this policy
...
2 shows the corresponding values for a 20-year term insurance
issued to (50)
...
1
...
1 EPV of premiums minus claims for each year of a 20-year endowment
insurance, sum insured $500 000, issued to (50)
...
2 EPV of premiums minus claims for each year of a 20-year term insurance,
sum insured $500 000, issued to (50)
...
These assets are needed in the later
years when the premium is not sufficient to pay for the expected benefits
...
This explains the concept of a policy value – we need to hold
capital during the term of a policy to meet the liabilities in the period when

7
...
We give a formal definition
of a policy value later in this section
...
1
...
1, to independent
lives all aged 50
...
In
other words, interest is earned on investments at 5% every year, the mortality
of the group of policyholders follows precisely the Standard Select Survival
Model and there are no expenses
...
Some policyholders will have died, so that their sum
insured of $500 000 will have been paid at the end of the year in which they
died, and some policyholders will still be alive
...
The accumulation
to time 10 at 5% interest of all premiums received (not including the premiums
due at time 10) minus all sums insured which have been paid is
NP 1
...
059 + · · · + 9 p[50] 1
...
059 + 1 |q[50] 1
...
0510 NP 1 + p[50] 1
...
05−9
− 1
...
05−1 + 1 |q[50] 1
...
05−10
= 1
...

(Note that, using the values in part (a) of Example 7
...
0566
¨
¨
¨
1
[50]:10

A

= 1 − d a[50]:10 − v 10 10 p[50] = 0
...
)
¨

So, if the experience over the first 10 years follows precisely the assumptions
set out in Example 7
...
The number of policyholders still alive at that time will be
10 p[50] N and so the share of this fund for each surviving policyholder is

182

Policy values

186 634N /(10 p[50] N ) = $190 339
...
This is not a coincidence! This happens in this example because the
premium was calculated using the equivalence principle, so that the EPV of
the profit was zero when the policies were issued, and we have assumed the
experience up to time 10 was exactly as in the calculation of the premium
...
We can prove that this is true in this case by
manipulating the equation of value, equation (7
...
0510
P a[50]:10 − 500 000A 1
¨
[50]:10
10 p[50]

= 500 000A60:10 − P a60:10
...
2)

The left-hand side of equation (7
...

For this example, the proof that the total amount needed by the insurer at
time 10 for all policies still in force is precisely equal to the amount of the fund
built up, works because
(a) the premium was calculated using the equivalence principle,
(b) the expected value of the future loss random variable was calculated using
the premium basis, and
(c) we assumed the experience followed precisely the assumptions in the
premium basis
...


7
...
2 Policy values for policies with annual cash flows
In general terms, the policy value for a policy in force at duration t (≥ 0) years
after it was purchased is the expected value at that time of the future loss random
variable
...


7
...

t
There is a standard actuarial notation associated with policy values for certain
traditional contracts
...
(Interested readers can consult the references in Section 7
...
)
Intuitively, the policy value at time t represents the amount the insurer should
have in its investments at that time in respect of a policy which is still in force,
so that, together with future premiums, the insurer can, in expectation, exactly
pay future benefits and expenses
...


An important element in the financial control of an insurance company is the
calculation at regular intervals, usually at least annually, of the sum of the policy
values for all policies in force at that time and also the value of all the company’s
investments
...
This process is called a valuation
of the company
...

In the literature, the terms reserve, prospective reserve and prospective
policy value are sometimes used in place of policy value
...

The precise definitions of policy value are as follows
...
1 The gross premium policy value for a policy in force at duration
t (≥ 0) years after it was purchased is the expected value at that time of the
gross future loss random variable on a specified basis
...

Definition 7
...
The premiums used in the calculation are the net premiums
calculated on the policy value basis using the equivalence principle, not the
actual premiums payable
...

1
...
3 we restrict ourselves to policies where the cash
flows occur only at the start or end of a year since these policies have some
simplifying features in relation to policy values
...
1
and 7
...


184

Policy values

2
...
These assumptions, called the policy value basis, may differ
from the assumptions used to calculate the premium, that is, the premium
basis
...
A net premium policy value can be regarded as a special case of a gross
premium policy value
...

4
...
See Example
7
...
This is a vestige of a time before modern computers, when easy
calculation was a key issue – using a net premium policy value allowed the
use of computational shortcuts
...
We make more use of gross, rather than net, premium
policy values in this book
...

5
...
It is the usual practice
to regard a premium and any premium-related expenses due at that time as
future payments and any insurance benefits (i
...
death or maturity claims)
and related expenses as past payments
...

6
...
Note also that if the premium is
calculated using the equivalence principle and the policy value basis is the
same as the premium basis, then 0 V = E[L0 ] = 0
...
For an endowment insurance which is still in force at the maturity date, the
policy value at that time must be sufficient to pay the sum insured, S, say,
so in this case n− V = S and n V = 0, where n− denotes the moment before
time n
...
In the discussion following Example 7
...
3
...
Broadly speaking,

7
...
3
...

Example 7
...

The sum insured of $100 000 is payable at the end of the year of death
...

(a) Calculate the gross premium policy value five years after the inception of
the contract, assuming that the policy is still in force, using the following
basis:
Survival model: Standard Select Survival Model
Interest: 5% per year effective
Expenses: 12
...
2 We assume that the life is select at age 50, when the policy is
purchased
...
Note
that a premium due at age 55 is regarded as a future premium in the calculation
of a policy value
...
875 × 1 300 aK55 +1 ,
g

so
5V

g

g

= E[L5 ] = 100 000 A55 − 0
...
35
...
At 4% per year,
P = 100 000

A[50]
= $1321
...

a[50]
¨

So, at 4% per year,
a
Ln = 100 000v K55 +1 − 1321
...
31¨ 55 = $6704
...

a

Notice in this example that the net premium calculation ignores expenses,
but uses a lower interest rate, which provides a margin, implicitly allowing
for expenses and other contingencies
...
3 A woman aged 60 purchases a 20-year endowment insurance with
a sum insured of $100 000 payable at the end of the year of death or on survival
to age 80, whichever occurs first
...
The insurer uses the following basis for the calculation of policy
values:
Survival model: Standard Select Survival Model
Interest: 5% per year effective
Expenses: 10% of the first premium, 5% of subsequent premiums, and
$200 on payment of the sum insured
Calculate 0 V , 5 V , 6 V and 10 V , that is, the gross premium policy values for
this policy at times t = 0, 5, 6 and 10
...
3 You should check the following values, which will be needed for
the calculation of the policy values:
a[60]:10 = 7
...
41004,

a65:5 = 4
...
51140,

a66:4 = 3
...
53422,

A70:10 = 0
...

At time 0, when the policy is issued, the future loss random variable, allowing
for expenses as specified in the policy value basis, is
L0 = 100 200v min(K[60] +1,20) + 0
...
95 P amin(K
¨

[60] +1,10)

where P = $5 200
...
95 a[60]:10 − 0
...

¨

Similarly,
L5 = 100 200v min(K65 +1,15) − 0
...
3 Policies with annual cash flows

187

so that
5V

= E[L5 ] = 100 200A65:15 − 0
...
95 P amin(K

66 +1,4)

so that
6V

= E[L6 ] = 100 200A66:14 − 0
...

¨

Finally, as no premiums are payable after time 9,
L10 = 100 200v min(K70 +1,10)
so that
10 V

= E[L10 ] = 100 200A70:10 = $63 703
...
3, the initial policy value, 0 V , is greater than zero
...
This
sounds uncomfortable but is not uncommon in practice
...

For example, the insurer may assume an interest rate of 6% in the premium
calculation, but, for policy value calculations, assumes investments will earn
only 5%
...

Example 7
...
The annuity will be paid annually for life, with the first payment on his 60th birthday
...
Level premiums of $11 900 are payable
annually for at most 10 years
...
The insurer uses
the following basis for the calculation of policy values:
Survival model: Standard Select Survival Model
Interest: 5% per year
Expenses: 10% of the first premium, 5% of subsequent premiums,
$25 each time an annuity payment is paid, and $100 when a death
claim is paid
Calculate the gross premium policy values for this policy at the start of the
policy, at the end of the fifth year, and at the end of the 15th year, just before
and just after the annuity payment and expense due at that time
...
4 We are going to need the following values, all of which you should
check:
a[50]:10 = 8
...
5268,
¨

v 5 5 p55 = 0
...
01439,

1
[50]:10

55:5

a65 = 13
...
60196,

(IA)

(IA) 1

a60 = 14
...
08639,

A1

55:5

= 0
...
03302
...

At the inception of the contract, the EPV of the death benefit is
P (IA)

1
[50]:10

,

the EPV of the death claim expenses is
100A

1
[50]:10

,

the EPV of the annuity benefit and associated expenses is
¨
10 025 v 10 10 p[50] a60 ,
and the EPV of future premiums less associated expenses is
0
...
05P ,
¨
so that
0V

= P(IA)

1
[50]:10

+ 100A

1
[50]:10

+ 10 025v 10 10 p[50] a60
¨

− (0
...
05)P
¨
= $485
...
, 10P depending on whether the life
dies in the 6th, 7th,
...
We can write this benefit as a
level benefit of 5P plus an increasing benefit of P, 2P,
...


7
...
95P a55:10 , so
¨
that
5V

= P(IA) 1

55:5

+ 5 PA 1

55:5

+ 100A 1

55:5

+ 10 025 v 5 5 p55 a60 − 0
...

Once the premium payment period of 10 years is completed, there are no future
premiums to value, so the policy value is the EPV of the future annuity payments
and associated expenses
...

2

We can make two comments about Example 7
...

1
...
3, 0 V > 0, which implies that the valuation basis is more
conservative than the premium basis
...
In Example 7
...
This makes sense if we
regard the policy value at any time as the amount of assets being held at that
time in respect of a policy still in force
...
Immediately after making
these payments, the insurer’s assets will have reduced by $10 025, and the
new policy value is 15+ V
...
1 and for the term insurance with the same sum
insured and term
...
1 and 7
...
Figures 7
...
4, respectively, show the policy values
...
3 we see that the policy
values build up over time to provide the sum insured on maturity
...
4 the policy values increase then decrease
...
3 Policy values for each year of a 20-year endowment insurance, sum insured
$500 000, issued to (50)
...
4 Policy values for each year of a 20-year term insurance, sum insured
$500 000, issued to (50)
...
In Figure 7
...
00, which is a small amount compared with the sum insured of $500 000
...


7
...
3
...
These formulae can be useful in the calculation
of policy values in some cases – we give an example at the end of this section to
illustrate this point – and they also provide an understanding of how the policy
value builds up and how profit emerges while the policy is in force
...
1 and 7
...

Example 7
...
1 and for t = 0, 1,
...
3)

where P =$15 114
...
1
...
5 From the solution to Example 7
...
, 19,
= 500 000 A[50]+t:20−t − P a[50]+t:20−t
...


Rearranging, multiplying both sides by (1 + i) and recognizing that
t+1 V

= 500 000 A[50]+t+1:19−t − P a[50]+t+1:19−t
¨
2

gives equation (7
...

We comment on Example 7
...

Example 7
...
4 and for t = 1, 2,
...
95P)(1 + i) = ((t + 1)P + 100) q[50]+t + p[50]+t

t+1 V

(7
...
4
...
6 For Example 7
...
, 9, t V has the same form as
5 V , that is
tV

= P(IA)

1
[50]+t:10−t

+ (tP + 100)A

1
[50]+t:10−t

+ 10 025v 10−t 10−t p[50]+t a60 − 0
...

¨
¨

192

Policy values

Recall that recurrence relations for insurance and annuity functions can be
derived by separating out the EPV of the first year’s payments, so that
¨
a[x]+t:n−t = 1 + vp[x]+t a[x]+t+1:n−t−1 ,
¨
1
A[x]+t:n−t = vq[x]+t + vp[x]+t A 1

[x]+t+1:n−t−1

and
1
(IA)[x]+t:n−t = vq[x]+t + vp[x]+t (IA) 1

[x]+t+1:n−t−1

+A1

[x]+t+1:n−t−1


...
, 9,
tV

= P vq[50]+t + vp[50]+t (IA)

1
[50]+t+1:10−t−1

+ (tP + 100) vq[50]+t + vp[50]+t A
+ 10 025 vp[50]+t v 10−t−1

+A

1
[50]+t+1:10−t−1

1
[50]+t+1:10−t−1

¨
10−t−1 p[50]+t+1 a60

− 0
...
95P
+ v p[50]+t P(IA)

1
[50]+t+1:10−t−1

+ ((t + 1)P + 100) A

1
[50]+t+1:10−t+1

+ 10 025 10−t−1 p[50]+t+1 v 10−t−1 a60 − 0
...

¨
¨

Notice that the expression in curly braces, { }, is
rearranging,

t+1 V ,

so, substituting and

( t V + 0
...


(7
...
3) and (7
...
Such formulae always exist but the precise form
they take depends on the details of the policy being considered
...
3) and (7
...
3 Policies with annual cash flows

193

the coming year, t to t + 1, and EPVs of payments from t + 1 onwards
...

Consider a policy issued to a life (x) where cash flows – premiums, expenses
and claims – can occur only at the start or end of a year
...
Consider the
(t + 1)st year, and let
Pt
et
St+1
Et+1
tV
t+1 V

denote the premium payable at time t,
denote the premium-related expense payable at time t,
denote the sum insured payable at time t + 1 if the policyholder dies in
the year,
denote the expense of paying the sum insured at time t + 1,
denote the gross premium policy value for a policy in force at time t,
and
denote the gross premium policy value for a policy in force at time t +1
...
The
quantities et , Et , q[x]+t and it are all as assumed in the policy value basis
...

Note that Lt involves present values at time t whereas Lt+1 involves present
values at time t + 1
...


Lt =

Taking expected values, we have
tV

= E[Lt ] = q[x]+t (1 + it )−1 (St+1 + Et+1 ) − (q[x]+t + p[x]+t )(Pt − et )
+ p[x]+t (1 + it )−1 E[Lt+1 ] ,

which, after a little rearranging and recognizing that
important equation

t+1 V

= E[Lt+1 ], gives the

( t V + Pt − et )(1 + it ) = q[x]+t (St+1 + Et+1 ) + p[x]+t t+1V
...
6)

Equation (7
...
3) and (7
...


194

Policy values

For policies with cash flows only at the start/end of each year, the recursive formulae always have the same general form
...
6)
...
Recall that t V is the expected value on the policy value basis of the
future loss random variable, assuming the policyholder is alive at time t
...

Now add to t V the net cash flow received by the insurer at time t as assumed
in the policy value basis
...
6) this is Pt − et ; in Example 7
...
33; in Example 7
...
05P
...
There are no further cash flows until the end of the year
...
This gives
the amount of the insurer’s assets at the end of the year before any further
cash flows (assuming everything is as specified in the policy value basis)
...
6), (7
...
4)
...
With
probability p[x]+t the policyholder will be alive, and with probability q[x]+t
the policyholder will die in the year (where these probabilities are calculated
on the policy value basis)
...
The expected
amount the insurer needs for the policy being considered above is given by
the right-hand side of equation (7
...
3) and (7
...
5 and 7
...
For the general policy and both examples, this is precisely the
amount the insurer will have (given our assumptions)
...
We assumed that at time t the insurer had sufficient
assets to expect (on the policy value basis) to break even over the future
course of the policy
...


7
...
6), (7
...
4)
...
95P)(1 + i) =

t+1 V

+ q[x]+t ((t + 1)P q50+t −

t+1 V ) ,

t+1 V ) ,

(7
...


The left-hand sides of these formulae are unchanged – they still represent the amount
of assets the insurer is assumed to have at time t +1 in respect of a policy which was
in force at time t
...

• For each policy in force at time t the insurer needs to provide the policy value,
t+1 V ,

at time t + 1, whether the life died during the year or not
...
5 and
(t + 1)P − t+1 V in Example 7
...

This extra amount required to increase the policy value to the death benefit
is called the Death Strain At Risk (DSAR), or the Sum at Risk or the Net
Amount at Risk, at time t + 1
...
This is an important measure of the insurer’s risk
if mortality exceeds the basis assumption, and is useful in determining risk
management strategy, including reinsurance – which is the insurance that an
insurer buys to protect itself against adverse experience
...
In more complicated examples, in particular where the
benefits are defined in terms of the policy value, this may not be possible
...
6), can be very useful as
the following example shows
...
7 Consider a 20-year endowment policy purchased by a life aged 50
...
A sum insured of $700 000 is payable at the end of the term if the life
survives to age 70
...


196

Policy values

The policy value basis used by the insurance company is as follows:
Survival model: Standard Select Survival Model
Interest: 3
...

Solution 7
...
6) becomes
( t V + P) × 1
...
, 19,

where P = $23 500
...

Putting t = 19 in the above equation gives:
( 19 V + P) × 1
...

Tidying this up and noting that St+1 = t V , we can work backwards as follows:
19 V

= (p69 × 700 000 − 1
...
035 − q69 ) = 652 401,

18 V

= (p68 ×

19 V

− 1
...
035 − q68 ) = 606 471,

17 V

= (p67 ×

18 V

− 1
...
035 − q67 ) = 562 145,

16 V

= (p66 ×

17 V

− 1
...
035 − q66 ) = 519 362,

15 V

= (p65 ×

16 V

− 1
...
035 − q65 ) = 478 063
...


2

7
...
4 Annual profit
Consider a group of identical policies issued at the same time
...
These cash flows depend
on mortality, interest, expenses and, for participating policies, bonus rates
...
If the
assumptions are not met, then the value of the insurer’s assets at time t + 1 may
be more than sufficient to pay any benefits due at that time and to provide a policy
value of t+1 V for those policies still in force
...
If the insurer’s assets at time t + 1 are not sufficient to
pay any benefits due at that time and to provide a policy value of t+1 V for those
policies still in force, the insurer will have made a loss in the year
...
3 Policies with annual cash flows

197

In general terms:
• Actual expenses less than the expenses assumed in the policy value basis will

be a source of profit
...

• Actual mortality less than the mortality assumed in the policy value basis
can be a source of either profit or loss
...

• Actual bonus or dividend rates less than the rates assumed in the policy value
basis will be a source of profit
...

Example 7
...
3 to women aged 60
...
In the following year,
•
•
•
•

expenses of 6% of each premium paid were incurred,
interest was earned at 6
...


(a) Calculate the profit or loss on this group of policies for this year
...

Solution 7
...
From Example 7
...
If the insurer’s assets were worth less (resp
...
profits) have been made in previous years
...

Now consider the cash flows in the 6th year
...
06P at time 5
...
94 P = $3395 551
...
5%, giving the
value of the insurer’s assets at time 6, before paying any death claims and
expenses and setting up policy values, as
(100 5 V + 100 × 0
...
065 = $3616 262
...

A policy value equal to 6 V (calculated in Example 7
...
Hence, the total
amount the insurer requires at the end of the year is
100 250 + 99 6 V = $3597 342
...
94 P) × 1
...


(b) In this example the sources of profit and loss in the sixth year are as
follows
...
5%, is higher than the rate assumed in the policy value basis
...

(iii) Mortality: The probability of dying in the year for any of these policyholders is q65 (= 0
...
Hence, out of 100 policyholders alive
at the start of the year, the insurer expects 100 q65 (= 0
...
In
fact, one died
...

Since the overall profit is positive, (i) has had a greater effect than (ii) and (iii)
combined in this year
...

Interest: If expenses at the start the start of the year had been as assumed in
the policy value basis, 0
...
05 × (100 5 V + 100 × 0
...


7
...
065 × (100 5 V + 100 × 0
...

Hence, there was a profit of $51 011 attributable to interest
...
That
is, we look at the loss arising from the expense experience given that the
interest rate earned is 6
...

The expected expenses on this basis, valued at the year end, are
100 × 0
...
065 + 100 q65 × 200 = $27 808
...
06P × 1
...

The loss from expenses, allowing for the actual interest rate earned in the
year but allowing for the expected, rather than actual, mortality, was
33 376 − 27 808 = $5568
...
5%) and actual expenses, and look
at the difference between the expected cost from mortality and the actual
cost
...

This gives a total profit of
51 011 − 5568 − 26 524 = $18 919
which is the amount calculated earlier
...
At each
step we assume that factors not yet considered are as specified in the policy value
basis, whereas factors already considered are as actually occurred
...

However, we could follow the same principle, building from expected to actual,
one basis element at a time, but change the order of the calculation as follows
...
06 − 0
...
05 + 100 q65 × (250 − 200) = $5490
...
065 − 0
...
94 P) = $50 933
...

This gives a total profit of
−5490 + 50 933 − 26 524 = $18 919
which is the same total as before, but with (slightly) different amounts of profit
attributable to interest and to expenses
...
The analysis
of surplus will indicate if any parts of the valuation basis are too conservative or too
weak; it will assist in assessing the performance of the various managers involved
in the business, and in determining the allocation of resources, and, for participating
business it will help to determine how much surplus should be distributed
...
3
...
3
...
1, that if the three conditions, (a),
(b) and (c), at the end of the section were fulfilled, then the accumulation of the
premiums received minus the claims paid for a group of identical policies issued
simultaneously would be precisely sufficient to provide the policy value required
for the surviving policyholders at each future duration
...
In practice, the invested premiums may have
earned a greater or smaller rate of return than that used in the premium basis, the
expenses and mortality experience will differ from the premium basis
...

It is of practical importance to calculate the share of the insurer’s assets
attributable to each policy in force at any given time
...
3 Policies with annual cash flows

201

the asset share of the policy at that time and it is calculated by assuming the policy
being considered is one of a large group of identical policies issued simultaneously
...
At any given
time, the accumulated fund divided by the (notional) number of survivors gives the
asset share at that time for each surviving policyholder
...

The policy value at duration t represents the amount the insurer needs to have at
that time in respect of each surviving policyholder; the asset share represents (an
estimate of) the amount the insurer actually does have
...
9 Consider a policy identical to the policy studied in Example 7
...
Suppose that over the
past five years the insurer’s experience in respect of similar policies has been as
follows
...

Year
Interest %

1
4
...
6

3
5
...
9

5
4
...

• Expenses at the start of each year after the year in which a policy was issued

were 6% of the premium
...

• The mortality rate, q[50]+t , for t = 0, 1,
...
0015
...

Solution 7
...
As we will see, the value of N does
not affect our final answers
...
, 5
...
We adopt the convention that ASt does not include the premium
and related expense due at time t
...
Note that for our policy, using the policy value basis
specified in Example 7
...

The premiums minus expenses received at time 0 are
0
...

This amount accumulates to the end of the year with interest at 4
...

A notional 0
...
0015 × (11 900 + 120) N = 18 N
which leaves
10 601 N − 18 N = 10 582 N
at the end of the year
...
9985 N policyholders are still surviving at the start of
the second year, AS1 , the asset share for a policy surviving at the start of the second
year, is given by
AS1 = 10 582N /(0
...

These calculations, and the calculations for the next four years, are summarized
in Table 7
...
You should check all the entries in this table
...
99854 × 0
...
99854 N policyholders are alive at the start of the fifth year, a fraction 0
...

Note that the figures in Table 7
...

2
We make the following comments about Example 7
...


1
...
The only purpose of this notional group of N identical policies issued
simultaneously is to simplify the presentation
...
4 Policy values with 1/mthly cash flows

203

Table 7
...
Asset share calculation for Example 7
...

Fund at Cash flow Fund at end
Death
Fund at
start of
at start of year before claims and end of
Year, t
year
of year death claims expenses
year
Survivors
1
2
3
4
5

0
10 582 N
22 934 N
35 805 N
49 170 N

10 115 N
11 169 N
11 152 N
11 136 N
11 119 N

10 601 N
22 970 N
35 859 N
49 241 N
63 123 N

18 N
36 N
54 N
71 N
89 N

10 582 N
22 934 N
35 805 N
49 170 N
63 034 N

0
...
99852 N
0
...
99854 N
0
...
The experience of the insurer over the five years has been close to the assumptions in the policy value basis specified in Example 7
...
The actual interest
rate has been between 4
...
6%; the rate assumed in the policy value
basis is 5%
...
The
actual mortality rate is comparable to the rate in the policy value basis, e
...

0
...
99252 is close to 5 p[50] = 0
...

As a result of this, the asset share, AS5 (= $63 509), is reasonably close
to the policy value, 5 V (= $65 470) in this example
...
4 Policy values for policies with cash flows at discrete
intervals other than annually
Throughout Section 7
...
This simplified the presentation and the calculations in
the examples
...
The definition of a policy value
from Definitions 7
...
2 can be directly applied to policies with more frequent
cash flows
...

The following example illustrates these points
...
10 A life aged 50 purchases a 10-year term insurance with sum insured
$500 000 payable at the end of the month of death
...


204

Policy values

Calculate the (gross premium) policy values at durations 2
...
5 years
using the following basis
...
10 To calculate 2
...

Note that the premium and related expense due at time t = 2
...
Note also that from duration 2
...

Hence
(12) 1

= 500 000A

2
...
75: 7
...
9 × 4 × P a
¨

(4)
52
...
25

= $3 091
...
75: 7
...
01327

and

(4)
52
...
25

a
¨

= 2
...


Similarly
3V

(12) 1

= 500 000A

53: 7

− 0
...
94,
where
(12) 1

A

53: 7

= 0
...
91446,

and
6
...
5: 3
...
008532 = $4 265
...


2

7
...
1 Recursions
We can derive recursive formulae for policy values for policies with cash flows at
discrete times other than annually
...
75 V and 3 V in Example 7
...
We

7
...
We can use a recurrence relationship to generate the policy value at each
month end, allowing for premiums only every third month
...
75 V + 460 − 0
...
050
...
0833 q52
...
0833 p52
...
8333 V
and similarly
2
...
050
...
0833 q52
...
0833 p52
...
9167 V

2
...
050
...
0833 q52
...
0833 p52
...


7
...
2 Valuation between premium dates
All of the calculations in the sections above considered policy values at a premium
date, or after premiums have ceased
...
The principle when
valuing between premium dates is the same as when valuing on premium dates, that
is, the policy value is the EPV of future benefits plus expenses minus premiums
...
We demonstrate this in the following
example, which uses the same contract as Example 7
...

Example 7
...
10, calculate the policy
value after (a) 2 years and 10 months and (b) 2 years and 9
...

Solution 7
...
8333:7
...
0132012 = 6600
...


(m)1
(m)
and a x:n are defined only if n is an integer
¨
x:n
(12)
(4)
A 1
is well defined, but a
¨
52
...
1333
52
...
1333

Note that the functions A

1
multiple of m , so that
is not
...
9)(4P)v 0
...
1667 p52
...
9)(4P)(1
...
86

So the policy value is 2
...
72
...
We know
that the EPV of benefits minus premiums at 2 years and 10 months is

206

Policy values
2
...


One-half of a month earlier, we know that the life must either
survive the time to the month end, in which case the EPV of future benefits
less premiums is 2
...
0417 , or the life will die, in which case the
EPV of benefits less premiums is S v 0
...
Allowing for the appropriate
probabilities of survival or death, the value at t = 2
...
7917 V

= 0
...
7917 S v 0
...
0417 p52
...
0417 2
...
99
...

It is interesting to note here that it would not be appropriate to apply simple
interpolation to the two policy values corresponding to the premium dates before
and after the valuation date, as we have, for example,
2
...
02,

2
...
99

and

3V

= $3357
...


The reason is that the function t V is not smooth if premiums are paid at discrete
intervals, since the policy value will jump immediately after each premium payment by the amount of that payment
...
Immediately after the premium
payment, it is no longer included, so the policy value increases by the amount of
the premium
...
5 we show the policy values at all durations for the policy in Examples
7
...
11
...
In the second half of the contract, after the premium
payment term, the policy value is run down
...
1 and 7
...

A reasonable approximation to the policy value between premium dates can
usually be achieved by interpolating between the policy value just after the previous premium and the policy value just before the next premium
...

k
k

In the example above, this would give approximate values for 2
...
8333 V
of $3480
...
99, respectively, compared with the accurate values of
$3480
...
72, respectively
...
5 Continuous cash flows

207

6000

Policy Value ($)

5000
4000
3000
2000
1000
0
0

1

2

3

4
5
6
Duration (Years)

7

8

9

10

Figure 7
...
11
...
5 Policy values with continuous cash flows
7
...
1 Thiele’s differential equation
We have seen in Section 7
...
These ideas extend to policies where regular
payments – premiums and/or annuities – are payable continuously and sums insured
are payable immediately on death
...
This is a continuous time version of the
recursion equation (7
...
3
...
Recall that for the
discrete case
( t V + Pt − et )(1 + it ) =

t+1 V

+ q[x]+t (St+1 + Et+1 −

t+1 V )
...
7)

Our derivation of Thiele’s differential equation is somewhat different to the derivation of equation (7
...
However, once we have completed the derivation, we explain
the link with this equation
...
Suppose this policy
has been in force for t years, where t ≥ 0
...


Et
µ[x]+t
δt
tV

We assume that Pt , et , St , µ[x]+t and δt are all continuous functions of t and that
et , Et , µ[x]+t and δt are all as assumed in the policy value basis
...
3
...
Thus, if v (t) denotes the present value of a payment of 1 at time t, we have
t

v (t) = exp −

δs ds
...
8)

0

As t V represents the difference between the EPV of benefits plus benefit-related
expenses and the EPV of premiums less premium-related expenses, we have
tV

∞

=
0

v (t + s)
(St+s + Et+s ) s p[x]+t µ[x]+t+s ds
v (t)
∞

−
0

v (t + s)
(Pt+s − et+s ) s p[x]+t ds
...
Changing the variable of integration to r = t + s gives
tV

∞

=
t

∞ v (r)
v (r)
(Sr + Er ) r−t p[x]+t µ[x]+r dr −
(Pr − er ) r−t p[x]+t dr
...
9)
We could use formula (7
...
However, we
are instead going to turn this identity into a differential equation
...
There exist numerical techniques to solve differential equations, one of
which is discussed in the next section
...

2
...
For such policies it is usually the case that we are unable to calculate
policy values using numerical integration, and we must calculate policy

7
...
The following development of
Thiele’s differential equation sets the scene for the next chapter
...
9) into a differential equation, we note that
r−t p[x]+t

=

r p[x]
t p[x]

so that
tV

=

∞

1
v (t) t p[x]

t

v (r) (Sr + Er ) r p[x] µ[x]+r dr −

∞
t

v (r) (Pr − er ) r p[x] dr ,

which we can write as
v (t) t p[x] t V =

∞
t

v (r) (Sr + Er ) r p[x] µ[x]+r dr −

∞
t

v (r) (Pr − er ) r p[x] dr
...
10)
Differentiation of equation (7
...
First, differentiation of the right-hand side yields
− v (t) (St + Et ) t p[x] µ[x]+t + v (t) (Pt − et ) t p[x]
= v (t) t p[x] Pt − et − (St + Et ) µ[x]+t
...
11)

Differentiation of the left-hand side is most easily done in two stages, applying the
product rule for differentiation at each stage
...

tV + tV
dt
dt
dt
Next,
d
d
d
v (t) t p[x] = v (t) t p[x] + t p[x]
v (t)
...
8)
d
v (t) = −δt exp −
dt

t
0

δs ds = −δt v (t)
...
10) is
d
d
v (t) t p[x] t V = v (t) t p[x]
t V − t V v (t) t p[x] µ[x]+t + t p[x] δt v (t)
dt
dt
d

...
11) yields Thiele’s differential equation, namely
d
t V = δt t V + Pt − et − (St + Et − t V ) µ[x]+t
...
12)

Formula (7
...
The left-hand side of the formula,
d t V /dt, is the rate of increase in the policy value at time t
...
The amount

of interest earned in the time interval t to t + h is δt t V h (+o(h)), so that the
rate of increase at time t is δt t V
...
If there were annuity payments at time t, this would
decrease the policy value at the rate of the annuity payment (plus any annuityrelated expenses)
...
The expected extra amount payable in the time interval t to t + h
is µ[x]+t h (St + Et − t V ) and so the rate of decrease at time t is µ[x]+t (St +
Et − t V )
...

We can also relate formula (7
...
7) assuming that for some very
small value h,
d
1
( t+h V − t V ) ,
tV ≈
dt
h

(7
...


Remembering that h is very small, the interpretation of the left-hand side is that it
is the accumulation from time t to time t + h of the policy value at time t plus the

7
...
(Note that for very small h, sh ≈ h
...
The
probability of death in the interval (t, t + h) is approximately hµ[x]+t
...
5
...
The key to this is to apply equation (7
...
This leads to
t+h V

− t V = h(δt t V + Pt − et − µ[x]+t (St + Et − t V ))
...
14)

The smaller the value of h, the better this approximation is likely to be
...
But we always know the value of t V as t approaches the end of the
policy term since, in the limit, it is the amount that should be held in respect of
a policyholder who is still alive
...

Using the endowment policy with term n years and sum insured S as an example,
formula (7
...
Another application of formula (7
...


212

Policy values

This method for the numerical solution of a differential equation is known as
Euler’s method
...
7
...
12 Consider a 20-year endowment insurance issued to a life aged 30
...
Premiums are payable continuously at a
constant rate of $2500 per year throughout the term of the policy
...


(a) Evaluate 10 V
...
05 years to calculate

10 V
...
04 per year
Solution 7
...

a

Using numerical integration or the three-term Woolhouse formula, we get
a40:10 = 8
...

(b) For this example, δt = 0
...
95 =
0
...
Hence
100 000 − V19
...
05 × (0
...
95 + 2 500 − 0
...
95 ))
and so
V19
...


7
...
9 , V19
...
, we arrive at
10 V

= 46 635
...
Using a value of h = 0
...
2
We remarked earlier that a useful feature of setting up and numerically solving a
differential equation for policy values is that the numerical solution gives policy
values at a variety of durations
...
In part (a)
we wrote down an expression for 10 V and evaluated it using numerical integration
...
05, as a by-product of our backwards recursive
calculation of 10 V we also obtained values of 10+h V , 10+2h V
...

Other major advantages of Thiele’s equation arise from its versatility and flexibility
...
We
can also use the equation to solve numerically for the premium given the benefits,
interest model and boundary values for the policy values
...
6 Policy alterations
A life insurance policy is a contract between an individual, the policyholder, and the
insurance company
...
So far in this book we have assumed that
the terms of the contract are never broken or altered in any way
...
Typical changes might be:

(1) The policyholder wishes to cancel the policy with immediate effect
...
This will be the case if the policy has
a significant investment component – such as an endowment insurance,
or a whole life insurance
...
A policy which is cancelled at the request of the
policyholder before the end of its originally agreed term, is said to lapse or
to be surrendered, and any lump sum payable by the insurance company
for such a policy is called a surrender value or a cash value
...


214

Policy values

In the US and some other countries, insurers are required to offer cash
surrender values on certain contract types once they have been in force
for one or two years
...

The allowance for zero cash values for early surrenders reflects the need of
the insurers to recover the new business strain associated with issuing the
policy
...
Any policy for which no
further premiums are payable is said to be paid-up, and the reduced sum
insured for a policy which becomes paid-up before the end of its original
premium paying term is called a paid-up sum insured
...

(4) Many other types of alteration can be requested: reducing or increasing
premiums; changing the amount of the benefits; converting a whole life
insurance to an endowment insurance; converting a non-participating policy to a with-profit policy; and so on
...

If the change was not part of the original terms of the policy, and if it has been
requested by the policyholder, it could be argued that the insurance company is
under no obligation to agree to it
...
In the US the non-forfeiture
law states that, for investment type policies, each of (1), (2) and (3) would generally
be available on pre-specified minimum terms
...

For policies with pre-specified cash surrender values, let Ct denote the cash
surrender value at duration t
...

Starting points for the calculation of Ct could be the policy value at t, t V , if it
is to be calculated in advance, or the policy’s asset share, ASt , when the surrender
value is not pre-specified
...
Recall also that if the policy value basis is close to
the actual experience, then t V will be numerically close to ASt
...
6 Policy alterations

215

Setting Ct equal to either ASt or t V could be regarded as over-generous to the
policyholder for several reasons, including:

(1) It is the policyholder who has requested that the contract be changed
...
Another implication of the fact
that the policyholder has called for alteration is that the policyholder may
be acting on knowledge that is not available to the insurer
...
This is called anti-selection or selection against the
insurer
...

(3) The alteration may, at least in principle, cause the insurance company to
realize assets it would otherwise have held, especially if the alteration is a
surrender
...
Under US non-forfeiture law, the insurer has six months to
pay the cash surrender value, so that it is not forced to sell assets at short
notice
...

For alterations other than cash surrenders, we can apply Ct as if it were a single
premium, or an extra preliminary premium, for the future benefits
...


(7
...
15) depends on the basis used for the calculation, that is, the assumptions concerning the survival model, interest rate, expenses and future bonuses (for
a with profits policy)
...

The rationale behind equation (7
...

Example 7
...
4 and 7
...
You are given
that the insurer’s experience in the five years following the issue of this policy is
as in Example 7
...
At the start of the sixth year, before paying the premium then
due, the policyholder requests that the policy be altered in one of the following
three ways
...

(b) No more premiums are paid and a reduced annuity is payable from age 60
...

(c) Premiums continue to be paid, but the benefit is altered from an annuity
to a lump sum (pure endowment) payable on reaching age 60
...
There
is an expense of $100 associated with paying the sum insured at the new
maturity date
...

Solution 7
...
4 and 7
...


Hence, the amount C5 to be used in equation (7
...
9 × AS5 − 200 = 56 958,
(ii) 0
...

(a) The surrender values are the cash values C5 , so we have
(i) $56 958,
(ii) $58 723
...
In this case, equation (7
...

¨

Using values calculated for the solution to Example 7
...
6 Policy alterations

217

(i) X = $4859,
(ii) X = $5012
...
Equation (7
...
95 × 11 900¨ 55:5 = 11 900 (IA)

1
55:5

+ 100A

1
55:5

+ 5A

1
55:5

+ v 5 5 p55 (S + 100)

which we solve using the two different values for C5 to give
(i) S = $138 314,
(ii) S = $140 594
...
14 Ten years ago a man now aged 40 purchased a with-profit whole
life insurance
...
Premiums of $1500 were payable annually for life
...

The insurer uses the following basis for the calculation of policy values and
policy alterations
...
2% per year at
the start of each policy year, including the first
...
You are given that the actual bonus rate declared in each of the
past 10 years has been 1
...


(a) Calculate the revised sum insured, to which future bonuses will be added,
assuming the premium now due has not been paid and the bonus now due
has not been declared
...
6%
...
14 (a) Before the declaration of the bonus now due, the sum insured

for the original policy is
200 000 × 1
...


218

Policy values
Hence, the policy value for the original policy,
10 V

10 V ,

is given by

= 234 405A40 j − P a40
¨

where P = 1500 and the subscript j indicates that the rate of interest to be
used is 3
...
05/1
...
0375494
...
Then, using equation (7
...

¨

(7
...
No further health checks are carried out at the
time of a policy alteration and so the policyholder is now assumed to be
subject to the ultimate part of the survival model
...
19569,
A40:20 j = 0
...
4578,
¨
a40:20 = 12
...

¨

Hence
S = $76 039
...
The term A40 j used in the
calculation of 10 V assumed the bonus to be declared at time 10 would be
1
...
012, in
the 12th year would be 234 405 × 1
...
Given that the bonus
declared at time 10 is 1
...
016
(this value is known) and 234 405 × 1
...
012 (this is an assumed
value since it assumes the bonus declared at the start of the 12th year will
be 1
...
Hence
10+ V

= (1
...
012) × 234 405A40 j − Pa40
= (1
...
012) × 234 405A40 j − P a40 + P
...
Equation (7
...
012) A40:20 j − Pa40:19
= (S /1
...
7 Retrospective policy value

219

and hence
S = $77 331
...
14, the sum insured payable in the 11th year is S × 1
...
The difference between these values is
not due to rounding – the timing of the request for the alteration has made a (small)
difference to the sum insured offered by the insurer for the endowment insurance
...
In Example 7
...
012 if there were no charge for
making the alteration and the bonus rate declared in the 11th year were the same as
the rate assumed in the reserve basis (and the full policy value is still used in the
calculation of the revised benefit)
...
7 Retrospective policy value
Our definition of a policy value is based on the future loss random variable
...
3
...
Since prospective means looking to the future,
this name has some merit
...
This is precisely the
calculation detailed in the final part of Section 7
...
1 in respect of the policy studied
in Example 7
...
2) is a formula for the
retrospective policy value (at duration 10) for this particular policy
...
These conditions are conditions (a) and (b) at the end of Section
7
...
1 – note that our condition (c) has already been used to calculate the retrospective
policy value
...
3
...

(2) The retrospective policy value equals the asset share if the experience follows precisely the assumptions in the policy value basis
...
Since the asset share represents the amount the
insurer actually has at time t in respect of a policy still in force, it is a more
useful quantity than the retrospective policy value
...
8 Negative policy values

In all our examples in this chapter, the policy value was either zero or positive
...
In fact, negative policy values are
not unusual in the first few months of a contract, after the initial expenses have
been incurred, and before sufficient premium is collected to defray these expenses
...
If we consider the policy value equation
tV

= EPV at t of Future Benefits + Expenses − EPV at t of Future Premiums,

then we can see that, since the future benefits and premiums must both have nonnegative EPVs, the only way for a negative policy value to arise is if the future
benefits are worth less than the future premiums
...
Allowing them to be entered as assets
(negative liabilities) ignores the policyholder’s option to lapse the contract, in which
case the excess premium will not be received
...
If the policyholder lapses then the policyholder will
have benefitted from the higher benefits in the early years without waiting around
to pay for the benefit in the later years
...


7
...
Thiele
(1838–1910)
...

Euler’s method for the numerical solution of a differential equation has the advantages that it is relatively simple to implement and it relates to the recursive formulae
for policy values for policies with annual cash flows
...
See
Burden and Faires (2001)
...
(1986) refer to retrospective policy
values
...


7
...
10 Exercises

221

specified in Example 3
...
9, and that the equivalence principle is used
for the calculation of premiums
...
1 You are given the following extract from a select life table with fouryear select period
...


[x]

l[x]

l[x]+1

l[x]+2

l[x]+3

lx+4

x+4

[40]
[41]
[42]

100 000
99 802
99 597

99 899
99 689
99 471

99 724
99 502
99 628

99 520
99 283
99 030

99 288
99 033
98 752

44
45
46

The basis for all calculations is an effective rate of interest of 6% per year, and no
expenses
...
59
...

(c) Calculate the sum insured for a three-year endowment insurance for a select
life age 41, with the same premium as for the term insurance, P = $323
...

(d) Calculate the mean and standard deviation of the present value of future
loss random variable, L1 , for the endowment insurance
...

Exercise 7
...
Premiums are paid annually in advance and the death benefit is paid
at the end of the year of death
...

Calculate the net premium policy value at t = 1 using the premium basis
...

Explain why the gross premium policy value is less than the net premium
policy value
...
5% per year
...

(f) Calculate the asset share per policy at the end of the first year of the contract
if experience exactly follows the premium basis
...
0012, the interest
rate earned on assets was 10%, and expenses followed the premium basis,
except that there was an additional initial expense of $25 per policy
...

(i) Analyse the surplus in (h) into components for interest, mortality and
expenses
...
3 A whole life insurance with reduced early sum insured is issued to a
life age 50
...

The benefit payable at the end of the year of death in any subsequent year is
$20 000
...
The
insurer calculates premiums and policy values using the standard select survival
model, with interest at 6% per year and no expenses
...

(ii) Write down an expression for the policy value at time 2, 2 V , in terms
of the premium P and standard actuarial functions
...

(b) Calculate 2
...

4
Exercise 7
...

The return of all premiums paid, without interest, at the moment of death,
in the event of death within the first 30 years
...


7
...

(b) Write down an expression in terms of annuity and insurance functions for
the net annual premium rate, P, for this contract
...

(d) Write down an expression for 5 V , the net premium policy value at time 5
for the contract, in terms of annuity and insurance functions, and the net
annual premium rate, P
...
5 An insurer issues a 20-year term insurance policy to (35)
...
The basis for calculating premiums
and policy values is:
Survival model:
Interest:
Expenses:

Standard Select Survival Model
5% per year effective
Initial:
$200 plus 15% of the first premium
Renewal: 4% of each premium after the first

(a) Show that the premium is $91
...

(b) Show that the policy value immediately after the first premium payment is
0+ V

= −$122
...


(c) Explain briefly why the policy value in (b) is negative
...
At what duration does the policy value first become
strictly positive?
(e) Suppose now that the insurer issues a large number, N say, of identical contracts to independent lives, all aged 35 and all with sum insured
$100 000
...

Exercise 7
...
8 in the order: mortality, interest, expenses
...


224

Policy values

Exercise 7
...
Premiums and policy values are calculated assuming:
Survival model:
Interest:
Expenses:

Standard Select Survival Model
5% per year effective
None
...
40 per year
...
42
...
0004, with B = 2
...
124
as before
...

(d) Explain why the policy value does not change very much
...
00022 but that the interest assumption
changes from 5% per year to 4% per year
...

(f) Explain why the policy value has changed considerably
...
That is, the paid up sum insured after k
years of premiums have been paid, out of the original total of 20 years,
should be S × k/20, where S is the original sum insured
...

Using a spreadsheet, calculate the EPV of the proportionate paid-up sum
insured at each year end, and compare these graphically with the policy
values at each year end, assuming the original basis above is used for each
...

Exercise 7
...

Premiums of $P per year are payable continuously throughout the policy term, and
the sum insured of $S is paid immediately on death
...


(b) Assume the life is aged 55 at issue, and that premiums are $1200 per year
...
44
...
10 Exercises
Mortality:
Interest:
Expenses:

225

Standard Select Survival Model
5% per year effective
None

(c) Calculate the standard deviation of Ln , Ln and Ln
...

Exercise 7
...

Derive the following formula for the net premium policy value at time t years,
where t is a premium date:
⎛
tV

= S ⎝1 −

(12)

a[x]+t:n−t
¨
(12)

a[x]:n
¨

⎞
⎠
...
10 A life aged 50 buys a participating whole life insurance policy with
sum insured $10 000
...

The premium is payable annually in advance
...

The premium basis is:
Initial expenses:
Renewal expenses:
Interest:
Survival model:

22% of the annual gross premium plus $100
5% of the gross premium plus $10
4
...
63 per year
...

(c) Assume the insurer earns interest of 5
...
Calculate the dividend
payable each year assuming
(i) the policy is still in force at the end of the year,
(ii) experience other than interest exactly follows the premium basis, and
(iii) that 90% of the profit is distributed as dividends to policyholders
...

(e) What would be a reasonable surrender benefit for lives surrendering their
contracts at the end of the first year?
Exercise 7
...
The sum
insured is payable at the end of the year of death or on survival to the maturity date
...
Premiums are
paid annually in advance
...
71
...

(c) Just before the fifth premium is due the policyholder requests that all future
premiums, including the fifth, be reduced to one half their original amount
...

Calculate the revised death benefit
...
12 An insurer issues a whole life insurance policy to a life aged 40
...
In subsequent years
the death benefit is $50 000
...

Basis for premiums and policy values:
Survival model:
Interest:
Expenses:

(a)
(b)
(c)
(d)
(e)

Standard Select Survival Model
6% per year effective
None

Calculate the premium for the contract
...

Calculate the policy value at t = 3
...

The insurer issued 1000 of these contracts to identical, independent lives
aged 40
...
In the following year
there were four further deaths in the cohort, and the rate of interest earned
on assets was 5
...
Calculate the profit or loss from mortality and interest
in the year
...
13 A 20-year endowment insurance issued to a life aged 40 has level
premiums payable continuously throughout the term
...
The sum insured payable immediately on death within the term is
$20 000 if death occurs within the first 10 years and t V if death occurs after t years,
10 ≤ t < 20, where t V is the policy value calculated on the premium basis
...
10 Exercises

227

Premium basis:
Survival model:
Interest:
Expenses:

Standard Select Survival Model
δt = 0
...
001t per year
None

(a) Write down Thiele’s differential equation for t V , separately for 0 < t < 10
and 10 < t < 20, and give any relevant boundary conditions
...
05
...

Exercise 7
...
The single premium was paid on 1 June 2008
...

Let t V denote the policy value after t years
...
Write down and explain
a recurrence relation between t V and t+1 V for t = 0, 1,
...

(b) Suppose the benefit is payable at the end of every h years, where
h < 1
...
, 20 − h
...

(d) Show that
tV

¯
=A

1
[60]+t:20−t

is the solution to the differential equation in (c)
...
15 An insurer issues identical deferred annuity policies to 100 independent lives aged 60 at issue
...
Level premiums are payable annually throughout the deferred period
...

The basis for premiums and policy values is:
Survival model:
Interest:
Expenses:

Standard Select Survival Model
6% per year
None

228

Policy values

(a) Calculate the premium for each contract
...

(c) Calculate the death strain at risk in the third year of the contract, for each
contract still in force at the start of the third year
...

(e) Two years after the issue date, 97 policies remain in force
...
Calculate the total mortality profit in the third year, assuming
all other experience follows the assumptions in the premium basis
...
Calculate the total mortality profit in the 13th year
...
16 Consider Example 7
...
Calculate the policy values at intervals of
h = 0
...


Answers to selected exercises
7
...
2 (a)
(b)
(c)
(e)
(f)
(g)
(h)
(i)
7
...
5 (a)
(b)
(d)

$323
...
68,
$11 663
...
26
$342
...
73
$469
...
39
$132
...
38
$132
...
10
−$107
...
28,
−$86
...
50
(iii) $185
...
78
$588
...
37
−$122
...
53,
4+ V = $55
...
95,
V = $326
...

7
...
04,
$51 011
...
00
7
...
40
(b) $26 131
...
41

7
...
8
7
...
11
7
...
13

7
...
16

(e) $36 575
...
56,
$6 078
...
46,
$70 070
...
32,
$16 240
...
98
(b) Selected values: 5 V = $509
...
77
(c) Selected values: Bonus at t = 5: $4
...
96
(d) $263
...
68
(c) $14 565
...
07
(c) $863
...
58
(e) −$4 476
...
73
(c) Selected values: 5 V = $10 400
...
21,
15 V = $40 387
...
25
(c) $23 671
...
83
(e) −$61 294
...
21
Selected values: 0
...
56,
1 V = $15 369
...
5 V = $30 962
...
28

229

8
Multiple state models

8
...
We then introduce several other multiple state
models which are useful as models for different types of life insurance policies
...
3
...
4 we discuss the derivation of
formulae for probabilities and in Section 8
...
This is extended in Section 8
...
7 to the numerical evaluation of policy values
...


8
...
They are a natural tool for many important areas of
practical interest to actuaries
...
In this section we illustrate some of the
uses of multiple state models using a number of examples which are common
in actuarial practice
...
2
...
This is a
230

8
...
1 The alive–dead model
...

We can represent this model diagrammatically as shown in Figure 8
...
Our
individual is, at any time, in one of two states, ‘Alive’ and ‘Dead’
...
Transition from state 0
to state 1 is allowed, as indicated by the direction of the arrow, but transitions
in the opposite direction cannot occur
...

We can use this multiple state model to reformulate our survival model as
follows
...
For each t ≥ 0 we
define a random variable Y (t) which takes one of the two values 0 and 1
...
The set of random variables {Y (t)}t≥0
is an example of a continuous time stochastic process
...
For all t, Y (t) is either 0 or 1, and Tx is connected to this model
as the time at which Y (t) jumps from 0 to 1, that is
Tx = max{t : Y (t) = 0}
...
1 captures all the survival/mortality information for an individual that is necessary for calculating
insurance premiums and reserves for policies where payments – premiums,
benefits and expenses – depend only on whether the individual is alive or dead
at any given age, for example a term insurance or a whole life annuity
...
We introduce
more examples of such models in the remainder of Section 8
...
3
...
Each state represents the
status of an individual or a set of individuals
...
2 The accidental death model
...

8
...
2 Term insurance with increased benefit on accidental death
Suppose we are interested in a term insurance policy under which the death
benefit is $100 000 if death is due to an accident during the policy term and
$50 000 if it is due to any other cause
...
1 is
not sufficient for this policy since, when the individual dies – that is, transfers
from state 0 to state 1 – we do not know whether death was due to an accident,
and so we do not know the amount of the death benefit to be paid
...
2
...
Hence, for example, the event ‘Y (t) = 1’ indicates that the
individual, who is aged x at time t = 0, has died from an accident before age x+t
...
2 is an extension of the model in Figure 8
...
In both
cases an individual starts by being alive, that is, starts in state 0, and, at some
future time, dies
...
Notice that it is the benefits provided by the insurance
policy which determine the nature of the appropriate model
...
2
...
3 shows a model appropriate for a policy which provides some or all
of the following benefits:
• an annuity while permanently disabled,
• a lump sum on becoming permanently disabled, and,
• a lump sum on death,

8
...
3 The permanent disability model
...
4 The disability income insurance model
...
An important feature of this model is
that disablement is permanent – there is no arrow from state 1 back to state 0
...
2
...
Figure 8
...
It could also be used when there are lump sum payments on
becoming sick or dying
...
4 differs from that
in Figure 8
...

This model illustrates an important general feature of multiple state models
which was not present for the models in Figures 8
...
2 and 8
...
This feature
is the possibility of entering one or more states many times
...


234

Multiple state models
8
...
5 The joint life and last survivor model

A joint life annuity is an annuity payable until the first death among a group of
lives
...
In principle, and occasionally in practice, the group could consist
of three or more lives
...

A common benefit design is an annuity payable at a higher rate while both
partners are alive and at a lower rate following the first death
...
This could be viewed as a last survivor annuity for
the lower amount, plus a joint life annuity for the difference
...
A pension plan may offer a reversionary annuity benefit as part of
the pension package, payable to the pension plan member’s spouse for their
remaining lifetime after the member’s death
...
All of these benefits may be valued using the model
represented in Figure 8
...

Let x and y denote the ages of the husband and wife, respectively, when
the annuity or insurance policy is purchased
...

The multiple state models introduced above are all extremely useful in an
insurance context
...
5 The joint life and last survivor model
...
3 Assumptions and notation

235

in this chapter
...

8
...
We have a finite set
of n + 1 states labelled 0, 1,
...
These states represent different conditions for
an individual (as in Figures 8
...
3 and 8
...
5)
...
, n, and we interpret the event Y (t) = i to mean that the individual is in
state i at age x + t, or, more generally as for the model in Figure 8
...
The set of random variables
{Y (t)}t≥0 is then a continuous time stochastic process
...
Note that in these examples there is a natural
starting state for the policy, which we always label state 0
...
For example, a policy providing
an annuity during periods of sickness in return for premiums payable while
healthy, as described in Section 8
...
4 and illustrated in Figure 8
...

Assumption 1
...

Intuitively, this means that the probabilities of future events for the process are
completely determined by knowing the current state of the process
...
This property, that probabilities
of future events depend on the present but not on the past, is known as the
Markov property
...

Assumption 1 was not made explicitly for the models represented by Figures
8
...
2 since it was unnecessary given our interpretation of these models
...
Assumption 1 is more interesting
in relation to the models in Figures 8
...
4
...
4) we know that Y (t) = 1, so that

236

Multiple state models

we know that the individual is sick at time t
...
In practice, we might believe that the probability of recovery in,
say, the next week would depend on how long the current sickness has already
lasted
...
It is
important to understand the limitations of any model and also to bear in mind
that no model is a perfect representation of reality
...
We assume that for any positive interval of time h,
Pr[2 or more transitions within a time period of length h] = o(h)
...

h

Intuitively, a function is o(h) if, as h converges to 0, the function converges to
zero faster than h
...
This
assumption is unnecessary for the models in Figures 8
...
2 since in both
cases only one transition can ever take place
...
3, 8
...
5
...

In Chapter 2 we introduced the standard actuarial notation for what we are
now calling the alive–dead model, as shown in Figure 8
...
For multiple state models more complicated than that in Figure 8
...
We introduce the following notation for a general
multiple state model to be used throughout this chapter and in later chapters
...
1)

ii
t px

= Pr [ Y (x + s) = i for all s ∈ [0, t] | Y (x) = i] ,

(8
...
3 Assumptions and notation

237

ij

so that t px is the probability that a life aged x in state i is in state j at age x + t,
ii
where j may be equal to i, while t px is the probability that a life aged x in state
i stays in state i throughout the period from age x to age x + t
...


h

(8
...
For all states i and j and all ages x ≥ 0, we assume that t px
is a differentiable function of t
...
Consequences of this assumption are that the limit in the
ij
definition of µx always exists and that the probability of a transition taking
place in a time interval of length t converges to 0 as t converges to 0
...
These assumptions
are not too restrictive in practice
...

In terms of the alive–dead model represented by Figure 8
...

10
• t px = 0 since backward transitions, ‘Dead’ to ‘Alive’, are not permitted in

this model
...

• µ01 is the same as µx , the force of mortality at age x
...

ij
Terminology: We refer to µx as the force of transition or transition intensity
between states i and j at age x
...
3) is to write for h > 0
ij
h px

ij

= h µx + o(h)
...
4)

From this formulation we can say that for small positive values of h
ij
h px

ij

≈ h µx
...
5)

This is equivalent to formula (2
...


238

Multiple state models

ii
Example 8
...
Write down an inequality linking these two probabilities and
explain why
ii
t px

ii
= t px + o(t)
...
6)

ii
Solution 8
...
1) and (8
...
The important
ii
distinction is that t px includes the possibility that the process leaves state i
between ages x and x + t, provided it is back in state i at age x + t
...
This applies to all
the states in the models illustrated in Figures 8
...
2, 8
...
4 and 8
...
4
...


The difference between these two probabilities is the probability of those paths
where the process makes two or more transitions between ages x and x + t so
that it is back in state i at age x + t
...
This gives us formula (8
...

2
Example 8
...


(8
...
2 First note that 1 − h px is the probability that the process does
leave state i at some time between ages x and x + h, possibly returning to state
i before age x + h
...
Using formula (8
...
7)
...
4 Formulae for probabilities

239

8
...
This is the same approach
as we adopted in Chapter 2, where we assumed the force of mortality, µx , was
known and derived formula (2
...

The fact that all probabilities can be expressed in terms of the transition
ij
intensities is important
...
, n, i = j} are fundamental quantities which determine everything we need
to know about a multiple state model
...
19) from Chapter 2, and is valid for
ii
any multiple state model
...

For any state i in a multiple state model,
ii
t px

⎧
⎨

= exp −
⎩

n

t

0 j=0,j=i

ij
µx+s ds

⎫
⎬
⎭


...
8)

ii
We can derive this as follows
...

This is the probability that the individual/process stays in state i throughout the
time period [0, t + h], given that the process was in state i at age x
...

ii
ii
The probabilities of these two sub-events are t px and h px+t , respectively, and,
using the rules for conditional probabilities, we have
ii
t+h px

ii
ii
= t px h px+t
...
2, this can be rewritten as
⎛
⎞
n

ii
ii
t+h px = t px ⎝1 − h

µx+t + o(h)⎠
...


=−
j=0,j=i

Integration over (0, t) gives
log

ii
t px − log

t

ii
0 px = −

n
ij

0 j=0,j=i

µx+r dr
...

ij

ii
Since 0 px = 1, this proves (8
...

We comment on this result after the next example
...
3 Consider the model for permanent disability illustrated in Figure
8
...
Explain why, for x ≥ 0 and t, h > 0,
01
t+h px

01
11
00
= t px h px+t + t px h µ01 + o(h)
...
9)

Hence show that
d
dt

01
t px exp

t
0

µ12 ds
x+s

00
= t px µ01 exp
x+t

t
0

µ12 ds ,
x+s

(8
...


(8
...
11)
...
3 To derive (8
...
The lefthand side of (8
...
4 Formulae for probabilities

241

x + t + h
...
Either:
01
• the life was disabled at age x + t (probability t px ) and remained disabled
11
between ages x + t and x + t + h (probability h px+t ), or,
00
• the life was healthy at age x + t (probability t px ) and then became disabled

between ages x + t and x + t + h (probability h µ01 + o(h))
...
9)
...
)
Using Example 8
...
9) can be rewritten as
01
t+h px

01
00
= t px (1 − h µ12 ) + t px h µ01 + o(h)
...
12)

Rearranging, dividing by h and letting h → 0 gives
d 01
01 12
00 01
t p + t px µx+t = t px µx+t
...

x+s

Integrating both sides of this equation from t = 0 to t = u, and noting that
01
0 px = 0, we have
01
u px exp

u
0

µ12 ds =
x+s

u
0

00 01
t px µx+t

Finally, dividing both sides by exp
formula (8
...

x+s

and noting that, using

µ12 ds ,
x+s

we have formula (8
...

The intuitive derivation of (8
...
We can illustrate this
event sequence using the time-line in Figure 8
...


242

Multiple state models

Time

0

t

Age

x

x+t

¢

t+dt

x+t+dt

¡
¢

Event

x+u

¢
¡

¡

µ01 dt
x+t

00
t px

Probability

u

in state 0
for t years

transition
to state 1

11
u−t px+t

in state 1
for u − t years

Figure 8
...
Since the age at transfer, x + t, can be anywhere
01
between x and x + u, the total probability, u px , is the ‘sum’(i
...
integral) of
these probabilities from t = 0 to t = u
...
8) and Example 8
...

(1) As we have already noted, formula (8
...
19)
in Chapter 2 for t px
...
3 we could have replaced t px by t px for i = 0, 1,
since, for the disability insurance model, neither state 0 nor state 1 can be
re-entered once it has been left
...
1
...
8) and Example
8
...
In particular,
in both cases we wrote down an expression for the probability of being in
the required state at age x + t + h by conditioning on the state occupied at
age x + t
...
An obvious question for us is, ‘Can this
method be applied to a general multiple state model to derive formulae for
probabilities?’ The answer is, ‘Yes’
...
4
...


8
...
1 Kolmogorov’s forward equations
Let i and j be any two, not necessarily distinct, states in a multiple state model
which has a total of n + 1 states
...
13)

8
...


(8
...
14) gives a set of equations for a Markov process known as
Kolmogorov’s forward equations
...
8) and in Example 8
...
We consider the probability of being in the required
state, j, at age x + t + h, and condition on the state of the process at age x + t:
either it is already in state j, or it is in some other state, say k, and a transition
to j is required before age x + t + h
...


k=0,k =j

Using formulae (8
...
7) and (8
...


k=0,k =j

Rearranging the right-hand side of this expression gives (8
...
Further
rearranging, dividing by h and letting h → 0 gives (8
...

In the following section we give several examples of the application of the
Kolmogorov forward equations as we use them to calculate probabilities for
some of the models described in Section 8
...


8
...
In
some cases, the probabilities can be calculated directly from formulae in terms
of integrals, as the following example shows
...
4 Consider the permanent disability model illustrated in
Figure 8
...

(a) Suppose the transition intensities for this model are all constants, as follows
µ01 = 0
...


µ02 = 0
...

x
x

244

Multiple state models

(b) Now suppose the transition intensities for this model are as follows
µ01 = a1 + b1 exp{c1 x},
x
µ02 = a2 + b2 exp{c2 x},
x
µ12 = µ02 ,
x
x
where
a1 = 4 × 10−4 , b1 = 3
...
138155,
a2 = 5 × 10−4 , b2 = 7
...
087498
...


Solution 8
...
See the solution to Example 8
...

(a) Using formula (8
...
0279 + 0
...
0508t},

0

(8
...
508} = 0
...


Similarly
11
10−t p60+t

= exp{−0
...
11) as
01
10 p60

10

=
0

10

=

00 01
11
t p60 µ60+t 10−t p60+t

dt

exp{−0
...
0279 × exp{−0
...
19363
...
5 Numerical evaluation of probabilities

245

(b) In this case
00
t p60

t

= exp −
0

(µ01 + µ02 ) dr
60+r
60+r

= exp − (a1 + a2 )t +

b1 60 c1 c1 t
b2
e
(e − 1) + e60 c2 (ec2 t − 1)
c1
c2

and
11
t p60

t

= exp −
0

µ12 dr
60+r

= exp − a2 t +

b2 60 c2 c2 t
e
(e − 1)
c2


...
58395
...
11) gives an integrand that cannot be integrated
analytically
...
20577
...
In other cases numerical
integration can be used
...
4 part (b) to
01
calculate a more complicated probability, 10 p60 – deriving an integral formula
for the probability which can then be integrated numerically – is not tractable
except in the simplest cases
...
These conditions are met by the permanent disability model,
illustrated in Figure 8
...
4, but are not met, for example,
by the disability income insurance model illustrated in Figure 8
...
This means, for example, that t px is the sum
of the probabilities of exactly one transition (0 to 1), plus three transitions
(0 to 1, then 1 to 0, then 0 to 1 again), plus five transitions, and so on
...

Euler’s method, introduced in Chapter 6, can be used to evaluate probabilities
for all models in which we are interested
...
13) and we illustrate it by applying it in the following example
...
5 Consider the disability income insurance model illustrated in
Figure 8
...
Suppose the transition intensities for this model are as follows
µ01 = a1 + b1 exp{c1 x},
x
µ10 = 0
...
4, part (b) (though this is
a different model)
...
13) with a step size of h = 1/12
00
years (we use a monthly time step, because this generates values of t p60 and
01 for t = 0, 1, 2,
...
6)
...
5 For this particular model, formula (8
...


As in Chapter 6, we choose a small step size h, ignore the o(h) terms and regard
the resulting approximations as exact formulae
...


By choosing successively t = 0, h, 2h,
...

10 60
h p60 h 60 2h p60 2h p60
These calculations are very well suited to a spreadsheet
...
1
...

2

8
...
1
...


t

µ01
60+t

µ02
60+t

µ10
60+t

µ12
60+t

00
t p60

01
t p60

0

0
...
01495

0
...
01495

1
...
00000

1
12
2
12
3
12

0
...
01506

0
...
01506

0
...
00118

0
...
01517

0
...
01517

0
...
00238


...


...
01469

...


...
01527

...


...
00147

...


...
01527

...


...
99266

...


...
00358

...


...


...


0
...


...


0
...


...


0
...


...


0
...


...


0
...


...


0
...


...


11
9 12

0
...
03492

0
...
03492

0
...
20061

10

0
...
03517

0
...
03517

0
...
20263

Note that the implementation of Euler’s method in this example differs in two
respects from the implementation in Example 7
...
This is determined
by the boundary conditions for the differential equations
...
This is a typical feature of applying Euler’s method to the calculation of probabilities for multiple state models
...


8
...
The next stage in our study of multiple state models
is to calculate premiums and policy values for a policy represented by such a
model and to show how we can evaluate them
...
We implicitly use the indicator function approach,
which leads directly to intuitive formulae for the expected present values, but
does not give higher moments
...
The notation used
in this chapter generalizes the notation introduced in Chapters 4 and 5
...

We wish to value an annuity of 1 per year payable continuously while the life
is in some state j (which may be equal to i)
...

Similarly, if the annuity is payable at the start of each year, from the current
time, conditional on the life being in state j, given that the life is currently in
state i, the expected present value is
ij

∞

ij

ax =
¨

v k k px
...

For insurance benefits, the payment is usually conditional on making a transition
...

Suppose a unit benefit is payable immediately on each future transfer into
state k, given that the life is currently in state i (which may be equal to k)
...

ij

jk

(8
...
In order to
transfer into state k in (t, t + dt), the life must be in some state j that is not
k immediately before (the probability of two transitions in infinitesimal time
ij
being negligible), with probability t px , then transfer from j to k during the
jk
interval (t, t + dt), with probability (loosely) µx+t dt
...
6 Premiums

249

Summing (that is, integrating) over all the possible time intervals gives
equation (8
...

Other benefits and annuity designs are feasible; for example, a lump sum
benefit might be paid on the first transition from healthy to sick, or premiums
may be paid only during the first sojourn in state 0
...

In general, premiums are calculated using the equivalence principle and we
assume that lives are in state 0 at the policy inception date
...
6 An insurer issues a 10-year disability income insurance policy
to a healthy life aged 60
...
5
...

(a) Premiums are payable continuously while in the healthy state
...
A
death benefit of $50 000 is payable immediately on death
...
The sickness benefit of $20 000
per year is payable monthly in arrear, if the life is in the sick state at the
payment date
...

Solution 8
...

The computation of the EPV of the benefits requires numerical integration
...
1
...
1
...
Then the EPV of the premium
income is
P a00
¯

60:10

10

=P
0

00
e−δ t t p60 dt

a00
¯
60:10

and numerical integration gives
= 6
...

Next, the EPV of the sickness benefit is
20 000 a01
¯

60:10

10

= 20 000
0

01
e−δ t t p60 dt,

a01
¯
60:10

and numerical integration gives
= 0
...

Last, the EPV of the death benefit is
¯
50 000 A02

60:10

10

= 50 000
0

e−δ t

00 02
t p60 µ60+t

01
+ t p60 µ12
60+t dt
...
16231
...
65
...
1
...
5980,

and to find the EPV of the sickness benefit we require
a

(12) 01
60:10

=

1
12

1

1
12

01
p60 v 12 +

2

2
12

01
p60 v 12 +

3

3
12

01
p60 v 12 + · · · +

01 10
10 p60 v

= 0
...

Note that the premiums are payable in advance, so that the final premium
11
payment date is at time 9 12
...

The death benefit is unchanged from part (a), so the premium is $3257
...
43 per month
...
8, 8
...
10 for the models that are most common
in actuarial applications
...
7 Policy values and Thiele’s differential equation
The definition of the time t policy value for a policy modelled using a multiple
state model is exactly as in Chapter 7 – it is the expected value at that time
of the future loss random variable – with one additional requirement
...
We can express this formally as follows: a
policy value is the expected value at that time of the future loss random variable
conditional on the policy being in a given state at that time
...


8
...

This additional feature was not necessary in Chapter 7 since all policies
discussed in that, and earlier, chapters were based on the ‘alive–dead’ model
illustrated in Figure 8
...

As in Chapter 7, a policy value depends numerically on the basis used in its
calculation, that is
(a) the transition intensities between pairs of states as functions of the
individual’s age,
(b) the force of interest,
(c) the assumed expenses, and
(d) the assumed bonus rates in the case of participating policies
...
To
establish some ideas we start by considering a particular example represented
by the disability income insurance model, Figure 8
...
We then consider the
general case
...
7
...
Premiums are
payable continuously throughout the term at rate P per year while the life is
healthy, an annuity benefit is payable continuously at rate B per year while the
life is sick, and a lump sum, S, is payable immediately on death within the term
...
4, is appropriate
...
For simplicity we
ignore expenses in this section, but these could be included as extra ‘benefits’
or negative ‘premiums’ provided only that they are payable continuously at
a constant rate while the life is in a given state and/or are payable as lump
sums immediately on transition between pairs of states
...

Example 8
...


(8
...
18)

and
d
(1)
= δ t V (1) − B − µ10
tV
x+t
dt

tV

(0)

− t V (1) − µ12 S − t V (1)
...
19)

(c) Suppose that
x = 40, n = 20, δ = 0
...
1 µ01 ,
x
x
µ02 = a2 + b2 exp{c2 x},
x
µ12 = µ02 ,
x
x
where a1 , b1 , c1 , a2 , b2 and c2 are as in Example 8
...

(i) Calculate 10 V (0) , 10 V (1) and 0 V (0) for n = 20 using Euler’s method
with a step size of 1/12 years given that
(1) P = $5 500, and
(2) P = $6 000
...

Solution 8
...
17)
...
20)

where the annuity and insurance functions are defined as in Section 8
...


8
...
18) by differentiating formula (8
...
To do this it is helpful to
think of t V (0) as the amount of cash the insurer is holding at time t, given
that the policyholder is in state 0 and that, in terms of expected values, this
amount is exactly sufficient to provide for future losses
...
Consider what
happens between times t and t + h
...

s

Recall that eδh = 1 + δ h + o(h) and sh = (eδh − 1)/δ = h + o(h), so that
¯
tV

(0) δh

e

+ P¯h = t V (0) (1 + δh) + Ph + o(h)
...
This amount is a policy value of t+h V (0) and possible
extra amounts of
(i) S− t+h V (0) if the policyholder dies: the probability of which is h µ02 +
x+t
o(h), and
(ii) t+h V (1) − t+h V (0) if the policyholder falls sick: the probability of
which is h µ01 + o(h)
...


Rearranging, dividing by h and letting h → 0 gives formula (8
...

Formula (8
...

(c) (i) Euler’s method for the numerical evaluation of t V (0) and t V (1) is
based on replacing the differentials on the left-hand sides of formulae
(8
...
19) by discrete time approximations based on a step size
h, which are correct up to o(h)
...

tV
dt
Putting this into formula (8
...
This is not ideal since the starting values
for using Euler’s method are n V (0) = 0 = n V (1) and so we will be
working backwards, calculating successively policy values at durations
n − h, n − 2h,
...
For this reason, it is more convenient to have

254

Multiple state models
formulae for t−h V (0) and
achieve this by writing

t−h V

(1)

in terms of t V (0) and t V (1)
...


Putting these expressions into formulae (8
...
19), multiplying
through by h, rearranging and ignoring terms which are o(h), gives the
following two (approximate) equations
(0)

Vt−h = t V (0) (1 − δh) − Ph + hµ01 ( t V (1) − t V (0) )
x+t
+ hµ02 (S − t V (0) )
x+t

(8
...

x+t

(8
...
, V10 , V10 ,
...

(1) For n = 20, h = 1/12 and P = $5500, we get
(0)

V10 = $18 084,

(1)

V10 = $829 731,

(0)

V0

= $3815
...


(ii) Let P ∗ be the premium calculated using the equivalence principle
...
Using the results in
(0)
part (i) and assuming V0 is (approximately) a linear function of P,
we have
0 − 3815
P ∗ − 5500
≈
6000 − 5500
−2617 − 3815
so that P ∗ ≈ $5797
...
7 Policy values and Thiele’s differential equation

255

Using Solver or Goal Seek in Excel, setting 0 V (0) to be equal to
zero, by varying P, the equivalence principle premium is $5796
...

Using the techniques of Example 8
...
8535,

a01
¯

40:20

= 0
...
08521,

and hence an equivalence principle premium of $5772
...
The difference arises because we are using two different approximation methods
...
If, in this
example, the individual is in state 0 at time 10, then it is quite likely that no
benefits will ever be paid and so only a modest policy value is required
...
In this case, a substantial policy value is required
...

8
...
2 Thiele’s differential equation – the general case
Consider an insurance policy issued at age x and with term n years described
by a multiple state model with n + 1 states, labelled 0, 1, 2,
...
Let
ij

µy
δt
(i)
Bt
(ij)

St

denote the transition intensity between states i and j at age y,
denote the force of interest per year at time t,
denote the rate of payment of benefit while the policyholder is in state i,
and
denote the lump sum benefit payable instantaneously at time t on
transition from state i to state j
...
Note that premiums
are included within this model as negative benefits and expenses can be included
as additions to the benefits
...

For i = 0, 1,
...


(8
...
23) can be interpreted in exactly the same way as formula (7
...

At time t the policy value for a policy in state i, t V (i) , is changing as a result of
interest being earned at rate δt t V (i) , and
(i)
benefits being paid at rate Bt
...
The intensity of such a jump is µx+t and the effect on
the policy value will be
(ij)

a decrease of St as the insurer has to pay any lump sum benefit contingent
on jumping from state i to state j,
a decrease of t V (j) as the insurer has to set up the appropriate policy value
in the new state, and
an increase of t V (i) as this amount is no longer needed
...
23) can be derived more formally by writing down an integral
equation for t V (i) and differentiating it
...
3
...
23) to calculate policy values exactly as we did in
Example 8
...
We choose a small step size h and replace the left-hand side by
( t V (i) −

t−h V

(i)

+ o(h))/h
...
, n, in terms of the policy values at duration t
...
, h, 0
...
8 Multiple decrement models
Multiple decrement models are special types of multiple state models which
occur frequently in actuarial applications
...
Figure 8
...
The accidental
death model, illustrated in Figure 8
...

Calculating probabilities for a multiple decrement model is relatively easy
since only one transition can ever take place
...
8 Multiple decrement models
µ01

x
E

t
t 
µx
t  02
s

t
t
t µ0n
x
t
t
”
t

Alive
0

257

Exit
1

Exit
2

...


...


...
7 A general multiple decrement model
...
, n and j = 0, 1,
...


0

00 0i
s px µx+s ds,

Assuming we know the transition intensities as functions of x, we can calculate
00
0i
t px and t px using numerical or, in some cases, analytic integration
...
We discuss the general point after completing
the example
...
8 A 10-year term insurance policy is issued to a life aged 50
...
No benefit is
payable if the policyholder lapses, that is, cancels the policy during the term
...

(a) The force of interest is 2
...

The force of mortality is given by µx = 0
...
0005(x − 50)
...

No allowance is made for expenses
...
5% per year
...
002 + 0
...

The transition intensity for lapses is a constant equal to 0
...

No allowance is made for expenses
...
8 (a) Since lapses are being ignored, an appropriate model for this
policy is the ‘alive–dead’ model shown in Figure 8
...

The annual premium rate, P, calculated using the equivalence principle,
is given by
P = 200 000

¯
A01

50:10

a00
¯

50:10

where
¯
A01

50:10

a00
¯

50:10

10

=
0

10

=
0

00
e−δt t p50 µ01 dt,
50+t

00
e−δt t p50 dt

and
00
t p50

= exp{−0
...
00025t 2 }
...
03807/8
...
49
...
8
...
2 –
a single starting state and two exit states – but with different labels for the


 
 

Active
0

Dead
1

 
d

d
‚
d

Lapsed
2

Figure 8
...


8
...
Using this model, the formula for the premium, P, is still
P = 200 000

¯
A01

50:10

a00
¯

50:10

but now
00
t p50

= exp{−0
...
00025t 2 },

which gives
P = 200 000 × 0
...
9269 = $834
...


2

We make the following observations about Example 8
...

(1) The premium allowing for lapses is a little lower than the premium which
does not allow for lapses
...
The insurer will make
a profit from any lapses in this example because, without allowing for
lapses, the policy value at any duration is positive and a lapse (with no
benefit payable) releases this amount as profit to the insurer
...

(2) In practice, the insurer may prefer not to allow for lapses when pricing
policies if, as in this example, this leads to a higher premium
...

Where lapses are used to reduce the premium, the business is called lapse
supported
...

(3) Note that two different models were used in the example to calculate a
premium for the policy
...

(4) The two models used in this example are clearly different, but they are
connected
...
8 has more exit
states; the connections between the models are that the single exit state in
Figure 8
...
8 and the transition
intensity into this state, µ01 , is the same in the two models
...
For the model in
Figure 8
...
002r − 0
...
002 + 0
...
9 Independent single decrement model, exit j
...
8 it is
t

exp{−0
...
00025r 2 } (0
...
0005r)dr
...
8, we interpret
‘dies’ as dying before lapsing
...
If we increase this intensity, the probability of dying
(before lapsing) decreases, as more lives lapse before they die
...
When working with a multiple decrement model we are
often interested in a simpler model with only one of the exit states and with the
same transition intensity into this state
...
Using the notation
in Figure 8
...
9
...
7, and
0j
t px

t

=

s

exp −

0

0

00
t px

0j

0j

µx+u du µx+s ds

= exp −

for j = 0 and
t
0

0j

µx+u du

for the related single decrement model in Figure 8
...
The first two of these
probabilities are called the dependent probabilities of exiting by decrement j

8
...
The purpose of
identifying the independent probabilities is usually associated with changing
assumptions
...
9 Joint life and last survivor benefits
8
...
1 The model and assumptions
In this section we consider the valuation of benefits and premiums for an insurance policy where these payments depend on the survival or death of two lives
...
Policies relating to three or more lives also
exist, but are far less common
...
It is often the case in practice that the two lives are social
partners, but they need not be; for example, they may be business partners
...

We need to evaluate probabilities of survival/death for our two lives, and
these probabilities must come from a model
...
10 and has the same structure as the model in Figure 8
...

Our model incorporates the following assumptions and notational changes
...
If the partner is alive, the intensity depends on the exact age of
the partner, as well as the age of the life being considered, and our notation
is adjusted appropriately
...
10 The joint life and last survivor model
...
However, if one partner, say the husband, has died, the
intensity of mortality for the wife depends on her current age, and the fact
that her husband has died, but not on how long he has been dead
...

(2) Our notation for probabilities for this model differs slightly from our usual
notation for multiple state models and is consistent with the notation we
0i
adopt for transition intensities
...

For this latter probability, the exact age at which the wife died is assumed
to be irrelevant and so is not part of the notation
...
9
...
We therefore need to introduce this new notation,
which is consistent with the notation of Chapters 2, 4 and 5
...

t pxy
t qxy
1
t qxy
2
t qxy
t pxy
t qxy

00
= Pr[(x) and (y) are both alive in t years] = t pxy
...

= Pr[(x) dies before (y) and within t years]
...

00
01
02
= Pr[at least one of (x) and (y) is alive in t years] = t pxy + t pxy + t pxy
...

= Pr[(x) and (y) are both dead in t years] = t pxy

We refer to the right subscript, x y or x y as a status
...


8
...
The standard
actuarial notation µx+t:y+t denotes the total force of transition out of state 0 at
time t, that is
02
µx+t:y+t = µ01
x+t:y+t + µx+t:y+t
...
24)

1
The ‘1’ over x in, for example, t qxy indicates that we are interested in the
1
probability of (x) dying first
...

Note that in cases where it makes the notation clearer, we put a colon between
the ages in the right subscript
...

The probabilities listed above do not all correspond to t pij type probabilities
...

1
02
Example 8
...

integral equation for t qxy
2
(b) Write down an integral equation for t qxy
...
9 (a) The probability t q xy is the probability that (x) dies within t
years, and that (y) is alive at the time of (x)’s death
...
So the first probability allows for the possibility
that (y) dies after (x), within t years, and the second does not
...


2
(b) The probability t q xy is the probability that the husband dies within t years
and that the wife is already dead when the husband dies, conditional on
the husband and wife both being alive now, time 0, and aged x and y,
respectively
...
10, the process must move
into state 1 and then into state 3 within t years, given that it starts in state 0
at time 0
...

2

264

Multiple state models
8
...
3 Joint life and last survivor annuity and insurance functions

We consider the EPVs, using a constant force of interest δ per year, of the
following payments
...

axy
¯

¯
Axy
axy
¯

Joint life annuity: a continuous payment at rate 1 per year while both
husband and wife are still alive
...
If there is a maximum period, n years, for the annuity,
then we refer to a ‘temporary joint life annuity’ and the notation for the
EPV is axy:n
...

¨
Joint life insurance: a unit payment immediately on the death of the first
to die of the husband and wife
...
In multiple state model notation
we have
axy = axy + axy + axy
...
In multiple state model notation
we have
¯
¯ xy
Axy = A03
...
In multiple state
model notation we have
ax|y = axy
...
If there is a time limit on this
payment, say n years, then it is called a ‘temporary contingent insurance’
¯1
and the notation for the EPV is Axy:n
...

ay
¯
¯
Ax

Single life annuity: a continuous payment at unit rate per year while the
wife is still alive
...


8
...
For example, the EPV of a monthly joint life annuity-due
(12)
would be denoted axy
...


By manipulating the probabilities in the integrands in these formulae we can
derive the following important formulae
axy = ax + ay − axy
¯
¯
¯
¯

(8
...

¯

(8
...
25) and (8
...

• The payment stream for the last survivor annuity is equivalent to continuous

payments at unit rate per year to both husband and wife while each of them is
alive minus a continuous payment at unit rate per year while both are alive
...

• If we pay one unit per year continuously while the wife is alive but take this
amount away while the husband is also alive, we have a continuous payment
at unit rate per year while the wife is alive but the husband is dead
...


266

Multiple state models

For the EPVs of the lump sum payments we have the following formulae:
∞

¯
Ay =
0

∞

¯
Ax =
0

∞

¯
Axy =
0

∞

¯
Axy =
0

¯ xy
A1 =
¯
A1 =
xy:n

∞
0
n
0

00
02 23
e−δt ( t pxy µ01
x+t:y+t + t pxy µy+t ) dt,

00
01 13
e−δt ( t pxy µ02
x+t:y+t + t pxy µx+t ) dt,
00
02
e−δt t pxy (µ01
x+t:y+t + µx+t:y+t ) dt,
01
02
e−δt (t pxy µ13 + t pxy µ23 ) dt,
x+t
y+t

00
e−δt t pxy µ02
x+t:y+t dt,

00
e−δt t pxy µ02
x+t:y+t dt
...
27)

which can be explained in the same way as formulae (8
...
26) by
considering cash flows
...
14) extends to
¯
the joint life case, so that

axy =
¯

¯
1 − Axy

...
28)

The proof of this is left to Exercise 8
...

The formulae for EPVs have been written in terms of probabilities derived
from our model
...
9 Joint life and last survivor benefits

267

so that using formula (8
...
29)

µ13 ds ,
x+s
µ23 ds ,
y+s

and, for example,
t

=

01
t pxy

0

00 01
11
s pxy µx+s:y+s t−s px+s ds
...
30)

Assuming as usual that we know the transition intensities, probabilities for
the model can be evaluated either by starting with Kolmogorov’s forward
equations, (8
...
29) and (8
...

Example 8
...


23
02
13
(b) Now suppose that µ01
x+t:y+t = µx+t and µx+t:y+t = µy+t for all t ≥ 0
...
31)

(ii) the probability that the husband is alive and the wife is dead at
time t is
t

exp −
0

µ13 ds
x+s

t

1 − exp −
0

µ23 ds
y+s

,

(8
...
33)

(iv) the probability that both the husband and the wife are dead at time t is
t

1 − exp −
0

µ13 ds
x+s

t

1 − exp −
0

µ23 ds
y+s


...
34)

Solution 8
...

The total probability of these events, integrating over the time of death
of (x), is
t
0

00 02
s pxy µx+s:y+s ds +

t
0

01
s pxy

µ13 ds
x+s

where
00
s pxy

s

= exp −
0

01
s pxy

s

=
0

02
(µ01
x+u:y+u + µx+u:y+u )du ,

00 01
11
u pxy µx+u:y+u s−u px+u du,

and
11
s−u px+u

(b)

s−u

= exp −
0

µ13
x+u+r dr
...

00
(i) The required probability is t pxy , which can be written as
00
t pxy

t

= exp −
0

t

= exp −
0

t

= exp −
0

02
(µ01
x+s:y+s + µx+s:y+s ) ds

µ01
x+s:y+s ds exp −
µ13 ds exp −
x+s

t
0

t
0

µ02
x+s:y+s ds

µ23 ds
...
9 Joint life and last survivor benefits

269

(ii) The probability that the husband is alive and the wife is dead at time
01
t is t pxy
...

(iii) The probability that the husband is alive at time t is
00
t pxy

01
+ t pxy
...

x+s

03
(iv) The required probability, t pxy , can be written as
03
t pxy

00
01
02
= 1 − t pxy − t pxy − t pxy
...
31), (8
...
34)
...
9
...

A special case of this model, which is important because it is often used in
practice, makes the following simplifying assumptions, which were used in
part (b) of Example 8
...

These equivalencies tell us that, with these assumptions, the mortality of each
life does not depend on whether the partner is still alive
...
This independence is
illustrated in formulae (8
...
32), (8
...
34) where probabilities of
joint events are the product of the probabilities of events for each life separately
and probabilities for the separate lives are derived from the two individual
‘alive–dead’ models for the husband and wife
...
11, where we show that each transition depends only on the
single life force of mortality
...
11 The independent joint life and last survivor model
...
9 Joint life and last survivor benefits

271

In particular, in standard actuarial notation, assuming independence of the
two lives means that
t pxy

= t px t py

(8
...
31),
t pxy

= 1 − (1 − t px )(1 − t py )

(8
...
34), and
µm f = µm + µfy
...
37)

Example 8
...
Level premiums are payable monthly
for at most 10 years but only if both are alive
...
If both lives
survive 10 years, an annuity of $50 000 per year is payable monthly in advance
while both are alive, reducing to $30 000 per year, still payable monthly, while
only one is alive
...

Calculate the monthly premium on the following basis:
Survival model: Both lives are subject to the Standard Select Survival
Model and may be considered independent with respect to mortality
...

Interest: 5% per year effective
...

Solution 8
...
10(b) to write the probability that they both
survive t years as
t p[55] t p[50]

and the probability that, for example, the husband dies within t years but the
wife is still alive as
(1 − t p[55] ) t p[50]
where each single life probability is calculated using the Standard Select
Survival Model in Example 3
...


272

Multiple state models

Let P denote the annual amount of the premium
...
7786 P
...


200 000
t=0

To find the EPV of the annuities we note that if both lives are alive at time 10
years, the EPV of the payment at time t/12 years from time 10 is
t

50 000 v 12

t
12

t

p65

t
12

t

+ 30 000 v 12
=v

t
12

t
12

p60 + 30 000 v 12
p60 1 −

30 000(

t
12

t
12

p65 +

t
12

p65 1 −

t
12

p60

p65
t
12

p60 ) − 10 000

t
12

p65

t
12

p60
...

Hence the monthly premium, $P/12, is given by
P/12 = (7 660 + 411 396)/(12 × 7
...
41
...


(8
...
However, we have noted in earlier chapters that
it is sometimes the case in practice that the only information available to us to
calculate the EPV of an annuity payable more frequently than annually is a life

8
...
In Section 5
...
To illustrate, consider the annuity EPVs in equation (8
...
These
can be approximated from the corresponding annual values using UDD as
(12)

¨
a65 ≈ α(12) a65 − β(12)
¨
= 1
...
5498 − 0
...
0860,
a60
¨ (12)

≈ α(12) a60 − β(12)
¨
= 1
...
9041 − 0
...
4405,

a65:60 ≈ α(12) a65:60 − β(12)
¨ (12)
¨
= 1
...
3738 − 0
...
9097,
and
(12)
[55]:[50]:10

a
¨

≈ α(12) a[55]:[50]:10 − β(12)(1 −
¨

10 p[55] 10 p[50] v

10

)

= 1
...
9716 − 0
...
41790
= 7
...

The approximate value of the monthly premium is then
P/12 ≈

32 715 + 0
...
0860 + 14
...
9097)
12 × 7
...
33
...
It is,
¨
however, true that
a
axy ≈ α(m)¨ xy − β(m)
...
39)

Our calculations above illustrate the general point that this approximation is
usually very accurate
...
17
...
18 we illustrate how Woolhouse’s formula can be applied to
find the EPV of a joint life annuity payable m times per year
...
10 Transitions at specified ages

A feature of all the multiple state models considered so far in this chapter
is that transitions take place in continuous time so that the probability of a
transition taking place in a time interval of length h converges to 0 as h converges
to 0
...
3
...

The following example illustrates such a situation and the solution shows how
this feature can be incorporated in our calculation of probabilities and EPVs
...
12 The employees of a large corporation can leave the corporation
in three ways: they can withdraw from the corporation, they can retire or they
can die while they are still employees
...
12 illustrates this set-up
...

• The force of mortality depends on the individual’s age but not on whether the

individual is an employee, has withdrawn or is retired, so that for all ages x
µ03 ≡ µ13 ≡ µ23 = µx , say
...
Hence
µ02 =
x

µ02
0

for x < 60,
for x ≥ 60
...
It is assumed that 40% of employees reaching exact age 60,
61, 62, 63 and 64 retire at that age and 100% of employees who reach age
65 retire immediately
...
12 A withdrawal/retirement model
...
10 Transitions at specified ages

275

The corporation offers the following benefits to the employees:
• For those employees who die while still employed, a lump sum of $200 000

is payable immediately on death
...

Show that the EPVs, calculated using a constant force of interest δ per year,
of these benefits to an employee currently aged 40 can be written as follows,
¯
where A65 and 25 E40 are standard single life functions based on the force of
mortality µx
...
6k k−1 |A 1

60:1


...
6k k |A60 + 0
...


0
...
12 First, note that Ax , the EPV of a unit payment immediately on
the death of a life now aged x, does not depend on whether this individual is
still an employee, has withdrawn or has retired
...

The novel feature in this example is the non-zero probability of transitions
at specified ages, in this case retirement on birthdays from ages 60 to 65
...
3) is infinite at these specified ages
...
For example, the probability of surviving in
00
employment to just before age 60 from age 40 is, say, 20−p40 , where
00
20−p40

20

= exp −
0

20

= exp −

(µ02 + µ03 ) dt
40+t
µ40+t dt

0

=

20 p40 e

−20µ02


...
6 20−p40
...
So, the
probability of surviving in employment from age 40 to just before age 61 is
00
21−p40

=

00
20+p40

p60
...
6 21−p40 = 0
...

Consider the benefit on death after retirement
...
, 65
...

The probability that an employee currently aged 40 will retire at 60 is
00
20− p40

× 0
...
4 20 p40 e−20µ02
...
4 20 p40 e−20µ02 = 150 000 e−20µ 0
...

02

The probability of retiring at age 61 is the product of the probabilities of the
following events:
surviving in employment to age 60− ,
not retiring at age 60,
surviving from age 60+ to age 61− , and
retiring at age 61
...
6 × 1 p60 × 0
...
6 × 0
...


8
...
6 × 1 p60 × 0
...
6 × 1 p62 × 0
...
6 × 1 p64

25 p40 e

−20µ02

0
...


Hence the EPV of the benefit payable on death after retirement is
4

150 000 exp{−20µ02 }

20+k

0
...
4 exp −
0

k=0
25

+ 0
...
4

¯
¯
0
...
65 5 |A60
...
As with the death after retirement benefit, we need to split
the probabilities after age 60 into up to, at and after the year end exits
...
6
40+t

00
+ e−21δ 20− p40 × 0
...
6

1
0

1
0

00
e−δt t p60+ µ03 dt
60+t

00
e−δt t p61+ µ03 dt
61+t

+ ···
00
+ e−24δ 20− p40 × 0
...
6 × 1 p61 × 0
...
6
1

× 1 p63 × 0
...
6k k−1 |A 1

60:1


...
For other examples these assumptions may not hold
...


8
...
The processes of interest to actuaries are
time-inhomogeneous since the transition intensities are functions of time/age
...

Rolski et al
...

Andrei Andreyevich Markov (1865–1922) was a Russian mathematician best
known for his work in probability theory
...
He made many fundamental contributions to probability theory and is generally credited with putting
probability theory on a sound mathematical basis
...
Hoem (1988) provides a very comprehensive
treatment of the mathematics of such models
...
See, for
example, Macdonald et al
...

Norberg (1995) shows how to calculate the kth moment, k = 1, 2, 3,
...
He also reports that the transition intensities used in part (b) of Example
8
...

In Section 8
...
They are also in many insurance-related contexts the natural quantities
to estimate from data
...

We can extend multiple state models in various ways
...
This breaks the
Markov assumption and the new process is known as a semi-Markov process
...
4) where the intensities of recovery and death from the sick state could be
assumed to depend on how long the individual had been sick, as well as on
current age
...
See
CMI (1991)
...
12 Exercises

279

As noted at the end of Chapter 7, there are more sophisticated ways of solving
systems of differential equations than Euler’s method
...

For a discussion on how to use mathematical software to tackle the problems
discussed in this chapter see Dickson (2006)
...
12 Exercises
Exercise 8
...
2
...
6 × 10−5 and c = 1
...

(a) Calculate
00
(i) 10 p30 ,
(ii)

01
10 p30 ,

and

02
(iii) 10 p30
...
The basic sum insured is $100 000, but
the death benefit doubles to $200 000 if death occurs as a result of an
accident
...
Premiums
are payable continuously throughout the term
...


Exercise 8
...


Healthy
0

'

E

 
 
d
  ‚
© d
  d

Sick
1

d
c
Dead
2

'

c
Critically ill
3

280

Multiple state models

The transition intensities are as follows:
µ01 = a1 + b1 exp{c1 x},
x
µ12 = µ02 ,
x
x
µ10 = 0
...
2µ02 ,
x
x
µ03 = 0
...
5 × 10−6 , c1 = 0
...
6 × 10−5 , c2 = 0
...

1
00
(a) Using Euler’s method with a step size of 12 , calculate values of t p30 for
1 2
t = 0, 12 , 12 ,
...

(b) An insurance company issues a policy with term 35 years to a life aged 30
which provides a death benefit, a disability income benefit, and a critical
illness benefit as follows:
• a lump sum payment of $100 000 is payable immediately on the life
becoming critically ill,
• a lump sum payment of $100 000 is payable immediately on death,
provided that the life has not already been paid a critical illness benefit,
• a disability income annuity of $75 000 per year payable whilst the life is
disabled
...

(i) Calculate the monthly premium for this policy on the following basis:
Transition intensities: as in (a)
Interest: 5% per year effective
Expenses: Nil
Use numerical integration with the repeated Simpson’s rule with
1
h = 12
...

Use Thiele’s differential equation to solve for the total premium per
1
year, using Euler’s method with a step size of h = 12
...


Exercise 8
...
7
...


8
...


n

∞

=
j=0, j=i 0
∞

+
0

281

Explain why

v(t + s) (ij)
St+s + t+s V (j)
v(t)

ii
s px+t

ij

µx+t+s ds

v(t + s) (i) ii
Bt+s s px+t ds
...
5
...

Exercise 8
...
10
...

δ

Exercise 8
...
13 illustrates the common shock model
...

An insurance company issues a joint life insurance policy to a married couple
...
The policy provides a benefit of
$500 000 immediately on the death of the husband provided that he dies first
...
Premiums are payable annually in advance while
both lives are alive for at most 30 years
...
0001, B = 0
...
075 and D = 0
...


Husband Alive
Wife Alive
0

c
Husband Dead
Wife Alive
2

E

Husband Alive
Wife Dead
1







s

E

c
Husband Dead
Wife Dead
3

Figure 8
...


282

Multiple state models

Exercise 8
...
e
...
7
with n = 2), let µ01 = µ and µ02 = θ for 0 ≤ x ≤ 1
...

(b) Let θ = nµ
...

Exercise 8
...
8
...
e
...

x
(a) Show that if withdrawal does not affect the transition intensity to state 1
(i
...
that µ21 = µ01 ), then the probability that an individual aged x is dead
x
x
by age x + t is the same as that under the ‘alive–dead’ model with the
transition intensity µ01
...
8 An insurer prices critical illness insurance policies on the basis of
a double decrement model, in which there are two modes of decrement – death
(state 1) and becoming critically ill (state 2)
...
0001, B = 0
...
075, and µ02 = 0
...
On
x
x
the basis of interest at 4% per year effective, calculate the monthly premium,
payable for at most 20 years, for a life aged exactly 30 at the issue date of a
policy which provides $50 000 immediately on death, provided that the critical
illness benefit has not already been paid, and $75 000 immediately on becoming
critically ill, should either event occur within 20 years of the policy’s issue date
...

Exercise 8
...
The transition intensities are
µ01 = 0
...
01,
x
µ03 = A + Bcx ,
x
where A = 0
...
0004 and c = 1
...
New recruits join only at exact
age 25
...
12 Exercises

283

(a) Calculate the probability that a new recruit
(i) is transferred before age 27,
(ii) dies aged 27 last bithday, and
(iii) is in active service at age 28
...
This sum is provided by a levy on all recruits in active service on the first and second
anniversary of joining
...

(c) Those who are transferred enter an elite force
...
5µ03 , but are subject
x
to no other decrements
...

Exercise 8
...
0001,
B = 0
...
075
...


(a)
(b)

and

Exercise 8
...

(a) Show that t pxy = t pw for w = log(cx + cy )/ log c
...

x:y
c
Exercise 8
...
Smith is subject to
Gompertz’ law of mortality with B = 0
...
07, and Jones is subject
to a force of mortality at all ages x of Bcx + 0
...
Calculate the probability
that Jones dies before reaching age 50 and before Smith dies
...

Exercise 8
...
0001,
B = 0
...
075
...

¯
(f) A25:30

(a)
(b)
(c)
(d)
(e)

Exercise 8
...
They effect
an insurance policy which provides $100 000, payable at the end of the year of
Bob’s death, provided Bob dies after Mike
...
Calculate
(a) the net annual premium, and
(b) the net premium policy value after 10 years (before the premium then due
is payable) if
(i) only Bob is then alive, and
(ii) both lives are then alive
...
0003 and c = 1
...
15 Ryan is entitled to an annuity of $100 000 per year at retirement,
paid monthly in advance, and the normal retirement age is 65
...

(a) Calculate the EPV of the annuity at Ryan’s retirement date
...

(c) Calculate the revised annual amount of the annuity (payable in the the first
year) if Ryan chooses the benefit in part (b), with a ‘pop-up’ – that is, the
annuity reverts to the full $100 000 on the death of Lindsay if Ryan is still
alive
...
)
Basis:
Male mortality before and after widowerhood:
Makeham’s law, A = 0
...
0004 and c = 1
...
12 Exercises

285

Female survival before widowhood:
Makeham’s law, A = 0
...
00025 and c = 1
...
0001, B = 0
...
072
Interest:
5% per year effective
Exercise 8
...
They are
about to effect an insurance policy that pays $100 000 immediately on the first
death
...
0001, B = 0
...
075
Female survival: Makeham’s law, with A = 0
...
0003 and c =
1
...

Exercise 8
...

(a) As an EPV, what does
m

v t/m

(t−1)/m pxy

−

t/m pxy

t=1

represent?
(m)
(b) Write down an expression for Axy in summation form by considering the
insurance benefit as comprising a series of deferred one year term insurances
1
with the benefit payable at the end of the m th of a year in which the first
death of (x) and (y) occurs
...

Show that under the UDD assumption,
(t−1)/m pxy

−

t/m pxy

=

1
m − 2t + 1
(1 − pxy ) +
q x qy
m
m

286

Multiple state models
and that
m

v t/m

(t−1)/m pxy

−

t/m pxy

t=1

= (1 − pxy )

iv
i(m)

m

+ qx qy

v t/m
t=1

m − 2t + 1

...


Exercise 8
...

dt
(b) Use Woolhouse’s formula to show that
axy ≈ axy −
¨ (m) ¨

m − 1 m2 − 1
δ + µx:y
...
19 Consider a husband (x) and wife (y)
...

Let Txy denote the time to failure of the joint life status, xy, and let Txy denote
the time to failure of the last survivor status, xy
...
10
...

(c) The force of mortality associated with the joint life status is µxy , defined in
formula (8
...
Show that
¯
Axy = E v Txy
...


Exercise 8
...
12 Exercises

287

sum of $10 000, payable immediately on death, plus an annuity of $5000 per
year payable continuously throughout the lifetime of the surviving spouse
...
Premiums are payable
continuously until the first death
...
353789, A65 = 0
...
512589 at 4% per year effective rate of interest
...

(a) Calculate the EPV of the lump sum death benefits, at 4% per year interest
...

(c) Calculate the annual rate of premium, at 4% per year interest
...

(i) Write down an expression for the policy value at that time assuming
that both lives are still surviving
...

(iii) Write down Thiele’s differential equation for the policy value assuming (1) both lives are still alive, and (2) only the wife is alive
...
21 Consider Example 8
...
02,

where A = 0
...
0004 and c = 1
...

A corporation contributes $10 000 to a pension fund when an employee joins
the corporation and on each anniversary of that person joining the corporation,
provided the person is still an employee
...

Exercise 8
...
Semesters are half
a year in length and the probability that a student progresses from one semester
of study to the next is 0
...
9 in the second year, 0
...
98 in the fourth
...
Students who fail in any semester may not continue in the degree
...
For the first semester
the tuition fee is $10 000
...
Assume that the student is subject
to a constant force of mortality between integer ages x and x + 1 of 5x × 10−5

288

Multiple state models

for x = 19, 20, 21 and 22, and that there are no means of leaving the course
other than by death or failure
...
23 An insurance company sells 10-year term insurance policies with
sum insured $100 000 payable immediately on death to lives aged 50
...

Survival: Makeham’s law, with A = 0
...
0004 and c = 1
...
5% of each premium (including the first)
Value the death benefit using the UDD assumption
...
1 (a)

8
...
5
8
...
9

8
...
12
8
...
979122
(ii) 0
...
000099
(b) (i) $206
...
15
00
(a) 35 p30 = 0
...
56
(ii) $2498
...
88
$4 948
...
01
(a) (i) 0
...
003234
(iii) 0
...
24
(c) 0
...
886962
(b) 0
...
001505
(d) 0
...
567376
(a) 15
...
9670
(c) 1
...
12 Exercises
(d)
(e)
(f)
8
...
15

8
...
20

8
...
22
8
...
2493
0
...
0440
$243
...
42
(ii) $2817
...
78
(a) $5440
...
16
(c) $2470
...
33
$53 285
...
95

289

9
Pension mathematics

9
...
We discuss the difference between defined benefit (DB)
and defined contribution (DC) pension plans
...

We then define the service table, which is a summary of the multiple state
model appropriate for a pension plan
...


9
...
Employers
sponsor plans for a number of reasons, including
• competition for new employees;
• to facilitate turnover of older employees by ensuring that they can afford to
•
•
•
•

retire;
to provide incentive for employees to stay with the employer;
pressure from trade unions;
to provide a tax efficient method of remunerating employees;
responsibility to employees who have contributed to the success of the
company
...
If competition for new employees is the most important factor, for

290

9
...
Ensuring turnover of older employees, or
rewarding longer service might lead to a different benefit design
...

The defined contribution pension plan specifies how much the employer will
contribute, as a percentage of salary, into a plan
...
The contributions are accumulated in a
notional account, which is available to the employee when he or she leaves the
company
...

The defined benefit plan specifies a level of benefit, usually in relation to
salary near retirement (final salary plans), or to salary throughout employment (career average salary plans)
...
If the investment or
demographic experience is adverse, the contributions can be increased; if experience is favourable, the contributions may be reduced
...

The benefit under a DB plan, and the target under a DC plan, are set by consideration of an appropriate replacement ratio
...
The
target for the plan replacement ratio depends on other post retirement income,
such as government benefits
...
Employer sponsored plans often target
50%–70% as the replacement ratio for an employee with a full career in the
company
...
3 The salary scale function
The contributions and the benefits for most employer sponsored pension plans
are related to salaries, so we need to model the progression of salaries through

292

Pension mathematics

an individual’s employment
...
The value of sx0 can be set
arbitrarily, and then for any x, y (≥ x0 ) we define
sy
salary received in year of age y to y + 1
=
sx
salary received in year of age x to x + 1
where we assume the individual remains in employment throughout the period
from age x to y + 1
...
Future changes
in salary cannot usually be predicted with the certainty a deterministic salary
scale implies
...

Salaries usually increase as a result of promotional increases and inflation
adjustments
...

Example 9
...
Members’ salaries are increased each year, six months before the
valuation date
...
Calculate his predicted final average salary
assuming retirement at age 65
...
Calculate her predicted final average salary
assuming retirement at age 65
...
04y , and
(ii) the integer age salary scale in Table 9
...

Solution 9
...
The predicted final
average salary in the three years to age 65 is then
75 000

s62 + s63 + s64
3 s34

which gives $234 019 under assumption (i) and $201 067 under assumption (ii)
...
3 The salary scale function

293

Table 9
...
Salary scale for Example 9
...

x

sx

x

sx

x

sx

x

sx

30
31
32
33
34
35
36
37
38
39

1
...
082
1
...
260
1
...
461
1
...
674
1
...
894

40
41
42
43
44
45
46
47
48
49

2
...
115
2
...
333
2
...
539
2
...
730
2
...
897

50
51
52
53
54
55
56
57
58
59

2
...
035
3
...
139
3
...
234
3
...
332
3
...
432

60
61
62
63
64

3
...
536
3
...
643
3
...
5 to 55
...

3 s54
...
Under assumption (ii) we need to
estimate s54
...
5 = (s54 + s55 )/2 = 3
...


2

Example 9
...
Salaries are revised continuously
...
03y , estimate
(a) the employee’s salary between ages 50 and 51, and
(b) the employee’s annual rate of salary at age 51
...

Solution 9
...
The information we are given in this example is that the employee’s current
annual rate of salary is $50 000 and that salaries are increased continuously
...
We make the reasonable assumption that
the current annual rate of salary is approximately the earnings between ages
39
...
5 (assuming the employee remains in employment until at least
age 40
...


294

Pension mathematics

(a) Given our assumption, the estimated earnings between ages 50 and 51 are
given by
50 000 ×

s50
= 50 000 × 1
...
5 = $68 196
...
5

(b) We assume that the annual rate of salary at age 51 is approximately the
earnings between ages 50
...
5
...
Hence, the estimated salary rate at age 51 is given by
50 000 ×

s50
...
0311 = $69 212
...
5

2

9
...


With this information we can set a contribution rate that will be adequate if
experience follows all the assumptions
...
The following example illustrates these points
...
3 An employer establishes a DC pension plan
...

The contribution rate is set using the following assumptions
...

At age 65, the employee is married, and the age of his wife is 61
...

The salary scale is given by sy = 1
...

Contributions are payable monthly in arrear at a fixed percentage of the salary
rate at that time
...

Annuities purchased at retirement are priced assuming an interest rate of 5
...


9
...
0004, B = 4 × 10−6 , c = 1
...

• Female survival: Makeham’s law, with A = 0
...
135
...


Consider a male new entrant aged 25
...

(b) Assume now that the contribution rate will be 5
...
5% per year
...

Solution 9
...

Mortality is not relevant here, as in the event of the member’s death, the
fund is paid out anyway; the DC fund is more like a bank account than an
insurance policy
...

Suppose the initial salary rate is $S
...
Then the annual
salary rate at age x > 25 is S(1
...
, 40, is
c
S(1
...
Hence, the accumulated amount
of contributions at retirement is
⎞

⎛
cS
12

480
k=1

⎜
k
k
⎜
1
...
140− 12 = cS ⎜
⎝

1
...
0440
12

1
...
04

1
12

−1

⎟
⎟
⎟ = 719
...

⎠

The salary received in the year prior to retirement, under the assumptions, is
s64
S = 1
...
5 S = 4
...

s24
...
65 × 4
...
0601S
...
0601S per year to the member, plus
a reversionary benefit of 0
...
0601S per year to his wife, is
(12)

+ 0
...
0601S

65

(12)
m f

65|61

where the m and f scripts indicate male and female mortality, respectively
...
5% per year,
we have
(12)

am
¨

65
(12)

a
¨

m f

= 10
...
9194,

61

a
¨

(12)
m f

65:61

∞

=
k=0

1 k
v 12
12

k
12

p

m

65+k

k
12

p

f

61+k

= 10
...
1)

giving
a
¨

(12)
m f

= 3
...


65|61

Note that we can write the joint life survival probability in formula (9
...
9
...

Hence, the value of the benefit at retirement is
3
...
5222 + 0
...
9128) = 39
...

Equating the accumulation of contributions to age 65 with the EPV of the
benefits at age 65 gives
c = 5
...

(b) We now repeat the calculation, using the actual experience rather than
estimates
...
5%, and solve for the
amount of benefit funded by the accumulated contributions, as a proportion
of the final year’s salary
...
5 The service table

297

The accumulated contributions at age 65 are now 28
...
5% per year interest are
(12)

am
¨

65

= 11
...
4730,

61

(12)

am
¨

f

= 10
...


65:61

Thus, the EPV of a benefit of X per year to the member and of 0
...
1867X
...
0185S
...
8703S
...
0185S
= 29
...

6
...
This is true for both
DC and DB benefits
...
In the DB case, the risk is
usually taken by the employer, whose contributions are usually adjusted when
the difference becomes apparent
...


9
...
There are several reasons why a member
might exit the plan
...
At later ages, employees may be offered a range
of ages at which they may retire with the pension that they have accumulated
...

In a DC plan, the benefit on exit is the same, regardless of the reason for the
exit, so there is no need to model the member employment patterns
...

In the UK it is common on the death in service of a member for the pension
plan to offer both a lump sum and a pension benefit for the member’s surviving

298

Pension mathematics

spouse
...
There may be a contingent spouse’s benefit
...

For example, if an employer offers a generous benefit on disability (or ill
health) retirement, that is worth substantially more than the benefit that the
employee would have been entitled to if they had remained in good health, then
it is necessary to model that exit and to value that benefit explicitly
...
On the other hand, if there is no benefit on
death in service (for example, because of a separate group life arrangement),
then to ignore mortality before retirement would overstate the liabilities within
the pension plan
...

It is not a realistic assumption, but it simplifies the calculation and is appropriate
if it does not significantly over or under estimate the liabilities
...
It is reasonable to ignore withdrawals if the effect on the valuation of benefits is small,
compared with allowing explicitly for withdrawals
...
For example,
in a final salary plan, if withdrawal benefits are increased in line with inflation,
the value of withdrawal and age benefits will be similar
...
An additional consideration is that withdrawals
are notoriously unpredictable, as they are strongly affected by economic and
social factors, so that historical trends may not provide a good indicator of
future exit patterns
...
1
...
10
...
4 A pension plan member is entitled to a lump sum benefit on death
in service of four times the salary paid in the year up to death
...
5 The service table

299

Member
0

ˆ
ˆˆ
    ˆˆˆˆ
 
ˆˆ


ˆˆ
~


A

©
 
z
Withdrawn
1

Disability
Retirement
2

Age
Retirement
3

Died in Service
4

Figure 9
...


Assume the appropriate multiple decrement model is as in Figure 9
...
1
⎪
⎨
0
...
02 for 45 ≤ x < 60,
⎪
⎩
0
for x ≥ 60,
µ02 ≡ µi = 0
...
1

for x < 60,
for 60 < x < 65
...
For transitions to state 4,
µ04 ≡ µd = A + Bcx ; with A = 0
...
7 × 10−6 , c = 1
...

x
x
(This is the Standard Ultimate Survival Model from Section 4
...
)
(a) For a member aged 35, calculate the probability of retiring at age 65
...

Solution 9
...
To calculate this, we need to consider separately the
periods before and after the jump in the withdrawal transition intensity, and
before and after the exact age retirements at age 60
...
05 + 0
...
597342
...
02 + 0
...
597342 × 0
...
425370
...
7 25− p35 = 0
...


For 25 < t < 30,
00
00
t p35 = 25+ p35 exp

−

t−25
0

i
d
µr
60+s + µ60+s + µ60+s ds

= 0
...
1 + 0
...
297759 × 0
...
175879
...
1759
...

All withdrawals occur by age 60
...

The probability of withdrawal by age 45 is
01
10 p35

10

=
0

00 w
t p35 µ35+t

10

dt = 0
...
05 × 7
...
3908
...
5 The service table

301

The probability of withdrawal between ages 45 and 60 is
15

= 0
...
597342 × 0
...
597342 × 0
...
7560 = 0
...


So, the total probability of withdrawal is 0
...

We calculate the probability of disability retirement similarly
...
001
0

00
t p35 dt

= 0
...
8168 = 0
...
597342
0

00 i
t p45 µ45+t
15

= 0
...
001
0

dt

00
t p45 dt

= 0
...
001 × 12
...
0076
...
297759
0

00 i
t p60 µ60+t

dt

= 0
...
001 × 3
...
0012
...
0078 + 0
...
0012 = 0
...

The probability of age retirement is the sum of the probabilities of exact
age retirements and retirements between ages 60 and 65
...
3

25− p35

= 0
...
1759
...
297759
0

00 r
t p60 µ60+t

dt

= 0
...
1 × 3
...
1159
...
4194
...

We use numerical integration for all these calculations
...
0040,

and the probability of death in the next 15 years is
00
04
10 p35 15 p45

15

= 0
...
0120
...
297759
0

00 r
t p60 µ60+t

dt

= 0
...
016323
= 0
...

So the total death in service probability is 0
...

We can check our calculations by summing the probabilities of exiting by
each mode
...
5432 + 0
...
4194 + 0
...

2
Often the multiple decrement model is summarized in tabular form at integer
ages, in the same way that a life table summarizes a survival model
...
We start at some minimum integer
entry age, x0 , by defining an arbitrary radix, for example, lx0 = 1 000 000
...
5 The service table

303

the model of Figure 9
...
)
00 01
wx0 +k = lx0 k px0 px0 +k ,
00 02
ix0 +k = lx0 k px0 px0 +k ,
00 03
rx0 +k = lx0 k px0 px0 +k ,
00 04
dx0 +k = lx0 k px0 px0 +k ,
00
lx0 +k = lx0 k px0
...
We can interpret
lx0 +k as the expected number of lives who are still plan members at age x0 + k
out of lx0 members aged exactly x0
...
These interpretations are precisely in line with those for a life table – see
Section 3
...

Note that, using the law of total probability, we have the following identity
for any integer age x
lx = lx−1 − wx−1 − ix−1 − rx−1 − dx−1
...
2)

A service table summarizing the model in Example 9
...
2
from age 20 with the radix l20 = 1000 000
...
The value of lx shown in the table is then calculated recursively from age 20
...
2) holds for each row of the table
...
2 for the simple reason that all values have been rounded to
the nearer integer
...
In all subsequent calculations based on Table 9
...

Having constructed a service table, the calculation of the probability of any
event between integer ages can be performed relatively simply
...
4
...
2, is
l65 /l35 = 38 488/218 834 = 0
...


304

Pension mathematics
Table 9
...
Pension plan service table
...
5432,
retires in ill health is (i35 + i36 + · · · + i64 )/l35 =
(213 + 203 + · · · + 45 + 41)/218 834 = 0
...
4194,
218 834
dies in service is (d35 + d36 + · · · + d64 )/l35 =
(83 + 84 + · · · + 214 + 215)/218 834 = 0
...


9
...
5 Employees in a pension plan pay contributions of 6% of their
previous month’s salary at each month end
...
4, and
(b) the values in Table 9
...

Other assumptions:
Salary Scale: Salaries increase at 4% per year continuously
Interest: 6% per year effective
Solution 9
...
06 × 100 000
12

299
k=1

k

k
12

k

00
p35 (1
...
04) v

360

+
k=301

0
...
04) 12 v 12

k

00 12
k p35 vj
12

360

+

25−

00
p35 vj25

+
k=301

k

k
12

00
p35 vj12

= 6 000 × 13
...
02/1
...
01923 and where we have separated out the
term relating to age 60 to emphasize the point that contributions would be
paid by all employees reaching ages 60 and 65, even those who retire at
those ages
...

12

This approximation will work for the monthly multiple decrement annuity,
00(12)
which we will denote a x :n , provided that the decrements, in total, are
approximately UDD
...
We can take account of this by splitting the annuity into two parts,
up to age 60− and from age 60+ , and applying a UDD-style adjustment to
each part as follows:
a

00(12)
35:30

=a

00(12)
35:25

+

l60+ 25 00(12)
v a
l35 j 60+ :5

≈ α(12) a00
¨

35:25

+
As a00
¨

35:25

l60+ 25
v
l35 j

− β(12) +

1
12

α(12) a00 +
¨

60 :5

= 13
...


= 3
...


2

Using the service table and the UDD-based approximation has resulted in a
relative error of the order of 0
...
This demonstrates again
that the service table summarizes the underlying multiple decrement model
sufficiently accurately for practical purposes
...

However, just as in Section 8
...
4, if we were to assume a uniform distribution
of decrements in each of the related single decrement models, we would find
that there is not a uniform distribution of the overall decrements
...

It is very common in pension plan valuation to use approximations, primarily
because of the long-term nature of the liabilities and the huge uncertainty in the
parameters of the models used
...
While this argument
is valid, one needs to ensure that the approximation methods do not introduce
potentially significant biases in the final results, for example, by systematically
underestimating the value of liabilities
...
6 Valuation of benefits
9
...
1 Final salary plans
In a DB final salary pension plan, the basic annual pension benefit is equal to
n SFin α

9
...
01 and 0
...
For an employee who
has been a member of the plan all her working life, say n = 40 years, this
typically gives a replacement ratio in the range 40%–80%
...

Consider a member who is currently aged y, who joined the pension plan at
age x (≤ y) and for whom the normal retirement age is 60
...
This estimate is calculated using her
current salary and an appropriate salary scale
...

The first part is related to her past service, and is called the accrued benefit
...
Note that both parts use an estimate
ˆ
of the final average salary at retirement, SFin
...
If this were to happen, the final benefit would be
based on the member’s past service at the wind-up of the pension plan; in this
sense, the accrued benefits (also known as the past service benefits) are already
secured
...

In valuing the plan liabilities then, modern valuation approaches often consider only the accrued benefits, even when the plan is valued as a going
concern
...
6 The pension plan in Example 9
...
5% of final average salary for each year of service, where final average
salary is defined as the earnings in the three years before retirement
...
2 to estimate the EPV of the accrued age retirement pension for
a member aged 55 with 20 years of service, whose salary in the year prior to
the valuation date was $50 000
...

Other assumptions:
Salary scale: From Table 9
...
6 Age retirement can take place at exact age 60, at exact age 65, or at
any age in between
...
, 64) take place at age x + 0
...
This
is a common assumption in pensions calculation and is a similar approach
to the claims acceleration approach for continuous benefits in Section 4
...
The
assumption considerably simplifies calculations for complex benefits, as it converts a continuous model for exits into a discrete model, more suitable for
efficient spreadsheet calculation, and the inaccuracy introduced is generally
small
...
Then the projected final average
salary is
50 000

zy
s54

where
zy = (sy−3 + sy−2 + sy−1 )/3
and where we use the values in Table 9
...
5
...

If the member retires at exact age 60, the accrued benefit, based on 20 years’
past service and an accrual rate of 1
...
015 = $15 922
...

s54

To value this we need to use life annuity values from the age at exit
...
1 with a UDD adjustment gives similar results
...
6 Valuation of benefits

309

The EPV of the accrued age retirement pension is then
50 000 × 0
...
5 5
...
5 6
...
5 +
¨
v a61
...
5 9
...
5 +
¨
v a65
¨
l55 s54
l55 s54

= $137 508
...

2
Withdrawal pension
When an employee leaves employment before being eligible to take an immediate pension, the usual benefit (subject to some minimum period of employment)
in a DB plan is a deferred pension
...
Note that Final Average Salary here is based on earnings in the
years immediately preceding withdrawal
...
If the deferred benefit is not increased during
the deferred period, then inflation, even at relatively low levels, will have a
significant effect on the purchasing power of the pension
...
Such adjustments are called cost of living
adjustments, or COLAs
...

Some plans outside the UK do not guarantee any COLA but apply increases on
a discretionary basis
...
7 A final salary pension plan offers an accrual rate of 2%, and
the normal retirement age is 65
...
Pensions are paid monthly in
advance for life from age 65, with no spouse’s benefit, and are guaranteed for
five years
...

(b) On death during deferment, a lump sum benefit of five times the accrued
annual pension, including a COLA of 3% per year, is paid immediately
...


310

Pension mathematics

Basis:
Service table: From Table 9
...
1
Post–withdrawal survival: Standard Ultimate Survival Model
Interest: 5% per year effective
Solution 9
...
There are no ‘exact age’ withdrawals, unlike
age retirements, so if the member withdraws between ages x and x + 1
(x = 35, 36,
...
5
...
6
...
4459 + 0
...
5451
= 13
...

(i) If the member withdraws between integer ages 35 + t and 35 + t + 1,
the accrued withdrawal pension, with no COLA, payable from age 65,
is estimated to be
100 000 ×

z35+t+0
...
02
...
5
(12)
× 10 × 0
...
5 is
100 000 ×

z35+t+0
...
02 a
¨
( 29
...
5 ) v 29
...
6 Valuation of benefits

311

where the t px factor is for survival only, not for the multiple decrement,
as we are applying it to a life who has just withdrawn
...
Applying this probability and
the discount factor v t+0
...
02
l35 s34

24

w35+t z35+t+0
...
5−t p35+t+0
...

(ii) To allow for a COLA at 3% per year during deferement, the above
formula for the EPV of the accrued withdrawal benefit must be adjusted
by including a term 1
...
5−t , so that it becomes
100 000 × 10 × 0
...
5 1
...
5−t a(12) ( 29
...
5 ) v 30
¨
65:5

t=0

which is $88 853
...
The estimated initial annual
accrued pension is
0
...
5
s34

and the sum insured on death before age 65 is five times this annual
amount increased by the COLA
...
02 × 10 × 100 000 v t+0
...
5 w35+t ¯ 1
A
35+t+0
...
5) j
s34
l35

= 5 × 0
...
01813
= $1813
where the subscript j indicates that the rate of interest used to calculate the
term insurances is j = 0
...
03
...
However, as for the future service benefit, future
salary increases are not guaranteed and there is a case for omitting them from
the accrued liabilities
...


312

Pension mathematics

The approach which uses salaries projected to the exit date is called the
projected unit method
...
Each
has its adherents
...
That is, if the pension calculation uses
the average of three year earnings to retirement, the current unit valuation could
use the average of the three year earnings to the valuation date
...
For simplicity,
the valuation in a current unit approach may use the current salary at valuation
...
6
...
Suppose a plan member retires at
age xr with n years of service and total pensionable earnings during their service
of (TPE)xr
...
So a CAE plan
with an accrual rate of α would provide a pension benefit on retirement at age
xr, for a member with n years of service, of
αn

(TPE)xr
= α (TPE)xr
...
The methods available for valuing such benefits are the
same as for a final salary benefit
...
The accrual principle is the same
...

Example 9
...
The pension benefit is paid monthly in advance
for life, guaranteed for five years, with no spousal benefit
...
The multiple decrement model in
Example 9
...

Consider a member now aged 35 who has 10 years of service, with total past
earnings of $525 000
...
6 Valuation of benefits

313

(a) Write down an integral formula for an accurate calculation of the EPV of
his accrued age and withdrawal benefits
...
2 to estimate the EPV of his accrued age and withdrawal
benefits
...
8 (a) The EPV of the accrued age retirement benefit is
25−

0
...
3 25 p35 v 25 a
¨

(12)

60:5

(12)

35+t:5

dt +

00 30 (12)
¨
30 p35 v a
65:5

where the second and fourth terms allow for the exact age retirements
...
04 (TPE)35 v 30 a

30

(12)

65:5

0

00 01
t p35 µ35+t 30−t p35+t

dt

where the survival probability 30−t p35+t is calculated using a mortality
assumption appropriate for members who have withdrawn
...
04 (TPE)35
(12)
(12)
(12)
r60− v 25 a
¨
+ r60+ v 25
...
5 a
¨
+ ···
l35
60:5
60
...
5:5
+ r64 v 29
...
5:5

+ r65 v 30 a
¨

(12)

65:5

= $31 666
...

The EPV of the accrued withdrawal benefit is
0
...
5 p35
...
5 p36
...
5 p59
...

2

314

Pension mathematics
9
...

The employer’s contribution is set at the regular actuarial valuations, and is
expressed as a percentage of salary
...
The nature of the pension
plan is that there is no need for the funding to be constant, as contributions can
be adjusted from time to time
...
Nevertheless, because the employer will have an interest
in smoothing its contributions, there is some incentive for the funding to be
reasonably smooth and predictable
...
The reserve refers to
the assets set aside to meet the accrued liabilities as they fall due in the future
...
We denote
this reserve t V
...

We then set the funding level for the year to be the amount required to be
paid such that, together with the fund value at the start of the year, the assets
are exactly sufficient to pay the expected cost of any benefits due during the
year, and to pay the expected cost of establishing the new actuarial liability at
the year end
...
These are simplifying
assumptions that make the development of the principles and formulae clearer,
but they can be relaxed quite easily
...
3)

that is
00
Ct = v 1 px t+1 V + EPV of benefits for mid-year exits − t V
...


9
...
3) is interpreted as follows: the start of year actuarial
liability plus normal contributions must be sufficient, on average, to pay for the
benefits if the member exits during the following year, or to fund the value of
the actuarial liability at the year end if the member remains in employment
...
3
...

Example 9
...
His salary in the year
to valuation was $50 000
...

•
•
•
•
•
•

Accrual rate: 1
...

Interest rate: 5% per year effective
...

Mortality before and after retirement follows the Standard Ultimate Survival
Model from page 74
...
9 (a) Using the projected unit credit approach, the funding and
valuation are based on projected final average earnings, so
SFin = 50 000s64 /s49 = 50 000(1
...

The actuarial liability is the value at the start of the year of the accrued
benefits, which is
0V

(12)

= 0
...

¨
(12)

(Note that a65 = 13
...
)
¨
The value at the start of the following year of the accrued benefits, assuming
the member is still alive, is
1V

(12)

= 0
...
015 × 21 × SFin × 15 p50 × v 15 × a65 =
¨

21
0V
...
3% of salary in the previous year
...

(b) Using the traditional unit credit approach, the valuation at time t is based
on the final average earnings at time t
...
04 =
$52 000
...
Then
(12)

0V

= 0
...
015 × 21 × S50 × 14 p51 × v 14 × a65
...
015 × 21 × S50 × 15 p50 × v 15 × a65 = 0 V
¨

Hence
C = 0V

21 S50
− 1 = 8 335
20 S49

or 16
...

We can decompose the normal contribution here as
C = 0V

S50
S50 1

...

20 S49

9
...
The first term
is required here because the TUC valuation does not allow for future salary
increases, so they must be funded year by year, through the contributions,
as the salaries increase each year
...
In fact, under both funding approaches the contribution rate tends
to increase as the member acquires more service, and gets closer to retirement
...
Ultimately, if all the assumptions in the basis are realized, both methods
generate exactly the same amount at the normal retirement age for surviving
(12)
members, specifically 0
...

In the example above, we showed how the PUC and TUC funding plans
allow for the normal contribution to fund the extra year of accrual (and the
salary increase, in the TUC case)
...
However, if the employee leaves before the year end, then the normal
contribution only has to fund the additional accrual up to exit
...

We explore this in the following example
...
10 A pension plan offers a pension benefit of $1 000 for each year
of service, with fractional years counting pro-rata
...
Value the accrued age retirement benefit and determine the
normal contribution rate payable in respect of age retirement benefits using the
following plan information and valuation assumptions
...

The pension is paid monthly in advance for life
...

The unit credit funding method is used
...


Assumptions:
Exits follow the service table given in Table 9
...

Interest rate: 6% per year effective
...

Survival after retirement follows the Standard Ultimate Survival Model from
page 74
...
10 Apart from the different pension benefit, this example differs
from the previous one because we need to allow for mid-year exits
...
We have
0V

= 1000 × 35 ×
+

r61 0
...
5 (12)
r63 2
...
5 +
¨
v a62
...
5
¨
l61
l61
l61

r64 3
...
5 +
¨
v a65
¨
l61
l61

= 345 307
and
00
v p61 1 V = 1000 × 36 ×

+

r62 1
...
5 (12)
v a62
...
5
¨
l61
l61

r64 3
...
5 +
¨
v a65
¨
l61
l61

= 312 863
...

The EPV of the benefits for lives exiting by age retirement in the middle of
the valuation year is
1000 × 35
...
5 (12)
v a61
...

¨
l61

Hence, the normal contribution required at the start of the year is C where
0V

00
+ C = EPV benefits to mid-year exits + p61 v 1 V

giving
C = 41 723 + 312 863 − 345 307 = 9278
...
9 Notes and further reading

319

9
...
The presentation has been relatively simplified to
bring out some of the major concepts, in particular, accruals funding principles
...
The difference between the normal contribution and the actual
contribution paid represents a paying down of surplus or deficit
...

For more information on pension plan design and related issues, texts such as
McGill et al
...


9
...
5, the calculations are based on the exact values underlying Table 9
...
Using the
integer-rounded values presented in Table 9
...

The Standard Ultimate Survival Model is the model specified in Section 4
...

Exercise 9
...
Salary increases take place on 1 July
every year
...
Using the salary scale in Table 9
...

Exercise 9
...
1 and a valuation date
of 1 January
...
Given that final average salary is defined as the average
salary in the four years before retirement, calculate the member’s expected
final average salary assuming retirement at age 60
...
Salaries are increased on average half-way
through each year
...


320

Pension mathematics

Exercise 9
...
One of the plan benefits is a
death in service benefit payable on death before age 60
...

(b) Assuming that the death in service benefit is $200 000, and assuming that
the death benefit is paid immediately on death, calculate the EPV at age 55
of the death in service benefit
...
At age 55 the member’s salary rate is $85 000 per year
...

Basis:
Service table from Table 9
...

Interest rate 6% per year effective
...
1; all salary increases occur half-way
through the year of age, on average
...
4 A new member aged 35 exact, expecting to earn $40 000 in the
next 12 months, has just joined a pension plan
...

There are no spousal benefits
...
Members contribute a percentage of salary, the rate depending on age
...

The employer contributes a constant multiple of members’ contributions
to meet exactly the expected cost of pension benefits
...
Assume all contributions are paid exactly half-way through the year of age in which they are paid
...
2
Survival after retirement: Standard Ultimate Survival Model
Interest:
4% per year effective
Exercise 9
...
Her starting
salary is $40 000 per year
...

At retirement she is to receive a pension payable monthly in advance,
guaranteed for 10 years
...
Using the

9
...

Assumptions:
Interest rate 7% per year effective before retirement, 5% per year
effective after retirement
Survival after retirement follows the Standard Ultimate Survival Model
(b) Now assume that this contribution rate is paid, but her salary increases at
a rate of 5% throughout her career, and interest is earned at 6% on her
contributions, rather than 7%
...
5% per year
...

Exercise 9
...
His salary in the year to the valuation date was $50 000
...

Calculate the value of the accrued death in service benefit and the normal
contribution rate for the death in service benefit
...
2
...

Salary scale follows Table 9
...

Projected unit credit funding method
...
7 A new company employee is 25 years old
...
All
contributions are paid by the employer, none by the employee
...
Salaries are assumed to increase at a
rate of 5% per year, increasing at each year end
...
6% for each year of service
...
The pension is payable monthly in advance
for life
...
The total accumulated
contribution is applied at the normal retirement age to purchase a monthly life
annuity-due
...


322

Pension mathematics

(b) Calculate the contribution, as a percentage of her starting salary, for the
retirement pension benefit for this life, for the year of age 25–26, using the
projected unit credit method
...
80
...
The annuity factor a65 is expected to be 11
...

¨
(c) Now assume that the employee joins the defined contribution plan
...
The annuity
(12)
factor a65 is expected to be 11
...
Assuming the employee stays in
¨
employment through to age 65, calculate (i) the projected fund at retirement
and (ii) her projected annual rate of pension, payable from age 65
...

(e) Explain briefly why the employer might prefer the defined contribution
plan even though the contribution rate is higher
...
8 In a pension plan, a member who retires before age 65 has their
pension reduced by an actuarial reduction factor
...

The plan sponsor wishes to calculate k such that the EPV at early retirement
of the reduced pension benefit is the same as the EPV of the accrued benefit
payable at age 65, assuming no exits from mortality or any other decrement
before age 65, and ignoring pay increases up to age 65
...

Calculate k for a person who entered the plan at age 25 and wishes to retire
at age (i) 55 and (ii) 60, using the following further assumptions:
Survival after retirement:
Interest rate:

Standard Ultimate Survival Model
6% per year effective

Exercise 9
...
The plan benefit is $350 per year
pension for each year of service, payable monthly in advance
...

Calculate the actuarial liability and the normal contribution for the age
retirement benefit for the member
...
2
...
Assume
6% per year interest and use the unit credit funding method
...
10 An employer offers a career average pension scheme, with
accrual rate 2
...
A plan member is aged 35 with five years past service,
and total past salary $175 000
...


9
...
2, calculate the actuarial liability and
the normal contribution for the age retirement benefit for the member
...
Post-retirement mortality follows
the Standard Ultimate Survival Model
...

Exercise 9
...
At the valuation date, 31 December

2008, she is exactly 45
...

• The final average salary is defined as the average salary in the two years
before exit
...
1
...
5% of final average salary for each year
of service
...
Allison has 15 years service at
the valuation date
...

• She could retire at 60
...
5% per month up to age 62
...
005)B where B would be the usual final salary benefit, calculated
as B = n SFin α, where n is the years of service from entry to age 62
...

• She could retire at age 65 with no actuarial reduction
...

(b) Calculate the EPV of Allison’s retirement pension for each of the possible retirement dates, assuming mortality is the only decrement
...

(c) Now assume Allison leaves the company and withdraws from active membership of the pension plan immediately after the valuation
...
She is entitled to a deferred pension
of 1
...
There is no COLA for the
benefit
...

Exercise 9
...

Benefit:
Normal retirement age:

$300 per year pension for each year of service
60

324

Pension mathematics

Survival model:
Interest:
Pension:
Pre-retirement exits:

Standard Ultimate Survival Model
6% per year effective
payable weekly, guaranteed for five years
mortality only

Active membership data at valuation
Service
Age for each employee
25
35
45
55

Number of employees

0
10
15
25

3
3
1
1

Inactive membership data at valuation
Age

Service

Number of employees

35
75

7
25

1 (deferred pensioner)
1 (pension in payment)

Exercise 9
...

You are given the following information
...
1
all increases occur on 31 December
each year
65
None
Standard Ultimate Survival Model

Membership

Name Age at entry
Giles
Faith

30
30

Age at
Salary at
1 January 2009 1 January 2008
35
60

38 000
47 000

Salary at
1 January 2009
40 000
50 000

9
...

(ii) Calculate the normal contribution rate in 2009 separately for Giles and
Faith, as a proportion of their 2009 salary, using the projected unit
credit funding method
...

(ii) Calculate the normal contribution rate in 2009 separately for Giles and
Faith, as a proportion of their 2009 salary, using the traditional unit
credit funding method
...

Answers to selected exercises
9
...
2 (a) $185 265
(b) $114 346
9
...
01171
(b) $2011
...
02
9
...
15
9
...
3%
(b) 56
...
6 Accrued death benefit value: $2 351
...
31
9
...
87%
(c) (i) $3 052 123, (ii) $277 466
9
...
448%
(b) 0
...
9 Actuarial liability: $1 842
...
45
9
...
65
Normal contribution: $1052
...
11 (a) 44
...
6%,
52
...
12 Total actuarial liability: $197 691
Total normal contribution: $8619
9
...
5%,
Faith: 25
...
1%,
Faith: 66
...
1 Summary
In this chapter we consider the effect on annuity and insurance valuation of
interest rates varying with the duration of investment, as summarized by a
yield curve, and of uncertainty over future interest rates, which we will model
using stochastic interest rates
...
In the final section we demonstrate the use of
Monte Carlo methods to explore distributions of uncertain cash flows and loss
random variables through simulation of both future lifetimes and future interest
rates
...
2 The yield curve
In practice, at any given time interest rates vary with the duration of the investment; that is, a sum invested for a period of, say, five years, would typically
earn a different rate of interest than a sum invested for a period of 15 years or
a sum invested for a period of six months
...
Note that, at least in principle, there is no uncertainty
over the value of v(t) although this value can change at any time as a result of
trading in the market
...


(10
...

326

10
...
1–10
...
The UK issues longer term bonds than most other countries, so the UK
yield curve is longer
...
Figure 10
...
0

4
...
0

2
...
0

0
...
1 Canadian government bond yield curve (spot rates), May 2007
...
0

Spot Rate of Interest (%)

5
...
0
3
...
0
1
...
0
0

5

10

15

20
25
Term (Years)

30

35

40

45

Figure 10
...


328

Interest rate risk
6
...
0
4
...
0
2
...
0
0
...
3 US government bond yield curve (spot rates), January 2002
...
0

4
...
0

2
...
0

0
...
4 US government bond yield curve (spot rates), November 2008
...
Figure 10
...
Both of these shapes are
relatively uncommon; the most common shape is that shown in Figures 10
...
4, a rising yield curve, flattening out after 10–15 years, with spot rates
increasing at a decreasing rate
...
This assumption has allowed us to use v t or e−δ t as discount functions for any term t,

10
...
When we relax this assumption, and allow interest rates to vary by term, the v t discount function is no longer appropriate
...
3 shows that the rate of interest on a one year US government bond
in January 2002 was 1
...
6%
...
The present value of a 20-year annuity-due of $1 per year
payable in advance, valued at 1
...
27; valued at 5
...
51
...
When we have a term structure this means we should discount each future payment using the spot interest rate appropriate to the term
until that payment is due
...

Since an investment now of amount v(t) in a t year zero-coupon bond will
accumulate to 1 in t years, v(t) can be interpreted as a discount function which
generalizes v t
...
The spot rates underlying the yield curve in Figure 10
...
63 for the 20-year annuity-due, closer to, but significantly
higher than the cost using the long term rate of 5
...

At any given time the market will determine the price of zero-coupon bonds
and this will determine the yield curve
...
Let f (t, t + k) denote the forward rate, contracted
at time zero, effective from time t to t + k, expressed as an effective annual rate
...
To determine forward rates in terms of spot rates of
interest, consider two different ways of investing $1 for t + k years
...
On the other hand, if the unit sum is invested
first for t years at the t year spot rate, then reinvested for k years at the k year
forward rate starting at time t, the accumulation will be (1+yt )t (1+f (t, t+k))k
...
That is
(1 + f (t, t + k))k =

(1 + yt+k )t+k
v(t)

...
The no arbitrage assumption is
discussed further in Chapter 13
...
3 Valuation of insurances and life annuities
The present value random variable for a life annuity-due with annual payments,
issued to a life aged x, given a yield curve {yt }, is
Kx

Y =

v(k)

(10
...
The expected present value of the annuity, denoted
a(x)y , can be found using the payment-by-payment (or indicator function)
¨
approach, so that
∞

a(x)y =
¨

k px v(k)
...
3)

k=0

Similarly, the present value random variable for a whole life insurance for (x),
payable immediately on death, is
Z = v(Tx )

(10
...


(10
...

By allowing for a non-flat yield curve we lose many of the relationships that
we have developed for flat interest rates, such as the equation linking ax and
¨
Ax
...
1 You are given the following spot rates of interest per year
...
032 0
...
038 0
...
043 0
...
046 0
...
048 0
...
3 Valuation of insurances and life annuities

331

Table 10
...
Calculations for
Example 10
...
0000
0
...
9335
0
...
8515
0
...
7679
0
...
6925
0
...
6257

t

0
...
88061
0
...
86337
0
...
84387
0
...
82188
0
...
79718
0
...
00000
0
...
78237
0
...
58919
0
...
42456
0
...
29073
0
...
18770

(a) Calculate the discount function v(t) for t = 1, 2,
...

(b) A survival model follows Makeham’s law with A = 0
...
00035
and c = 1
...
Calculate the net level annual premium for a 10-year term
insurance policy, with sum insured $100 000 payable at the end of the year
of death, issued to a life aged 80:
(i) using the spot rates of interest in the table above, and,
(ii) using a level interest rate of 4
...

Solution 10
...
1) for the discount function values and
(2
...
Table 10
...

(b) (i) The expected present value for the 10-year life annuity-due is
9

a(80 : 10 ) =
¨

v(k)k p80 = 5
...

k=0

The expected present value for the term insurance benefit is
1

9

100 000 A(80 : 10 ) =

k p80 (1 − p80+k )v(k

+ 1) = 66 739
...
72
...
8% per year flat yield curve gives a premium of
$13 181
...

2
In general, life insurance contracts are relatively long term
...
It is common actuarial practice to use
the long-term rate in traditional actuarial calculations, and in many cases, as
in the example above, the answer will be close
...
Overstating the interest results in a premium that is lower than
the true premium
...
With a rising yield curve, if a level interest rate is assumed, it should be
a little less than the long-term rate
...
3
...
1
...
We show that, if we take the premium
and the cash flow brought forward each year and invest them at the forward
rate, then there is exactly enough to fund the sums insured, provided that the
mortality and survival experience follows the assumptions
...

We will assume each year end cash flow for the policy is invested at the oneyear forward rate applying at the year end
...
The cash flows (in $000s) for a portfolio
of N = 100 000 contracts are given in Table 10
...
The entries in the table
are calculated as follows, where P is the premium of $13 213
...

• The premium income at time k, denoted Pk , is k p80 NP because at time k we

have k p80 N survivors with our deterministic model for mortality
...

• The net cashfow carried forward at time k + 1, denoted CFk+1 , is
CFk+1 = (Pk + CFk )(1 + f (k, k + 1)) − Ck+1
...
Over the course of the year interest is earned at rate
f (0, 1)
...


10
...
2
...
1; 100 000 contracts, in $000s
...
0320
0
...
0440
0
...
0510
0
...
0520
0
...
0560
0
...
At the inception of
the contract, we can lock in forward rates that will exactly replicate the required
cash flows
...
To get the policy value per surviving policyholder at k,
divide the net cash flow at k by the assumed number of survivors, 100 000 k px
...
83
...

First, we know that mortality is uncertain, so that the mortality related cash
flows are not certain
...
In Section
10
...
1 we discuss in detail what we mean by diversifiability, and under what
conditions it might be a reasonable assumption for mortality
...
Interest rate risk is inherently
non-diversifiable, as we shall discuss in Section 10
...
2
...
4 Diversifiable and non-diversifiable risk

Consider a portfolio consisting of N life insurance policies
...
, N , many quantities of interest for the ith policy
in this portfolio
...
In this case, N Xi represents the number of survivors after 10 years
...

i=1
Suppose for convenience that the Xi s are identically distributed with common
mean µ and standard deviation σ
...
Then
N

N

Xi = N µ

E

and

Xi = N σ 2 + N (N − 1)ρσ 2
...
Then
N

Xi = N σ 2

V
i=1

and the central limit theorem (which is described in Appendix A) tells us that,
provided N is reasonably large,
N

Xi ∼ N (N µ, N σ ) ⇒
2

N
i=1 Xi

√

i=1

− Nµ

Nσ

∼ N (0, 1)
...
More precisely, for any k > 0
N

N

Xi /N − µ ≥ k = Pr

Pr
i=1

Xi − N µ ≥ kN
i=1

= Pr

N
i=1 Xi

√

− Nµ

Nσ

√
k N
≥
σ


...
4 Diversifiable and non-diversifiable risk

335

be written as
√
k N
lim Pr |Z| ≥
N →∞
σ

= lim 2
N →∞

√
k N
−
σ

= 0,

where Z ∼ N (0, 1)
...
In this case we can reduce
i=1
the risk measured by Xi , relative to its mean value, by increasing the size of
the portfolio
...

So, we say that the risk within our portfolio, as measured by the random
variable Xi , is said to be diversifiable if the following condition holds
V[
lim

N →∞

N
i=1 Xi ]

N

= 0
...
In simple terms, a risk
is diversifiable if we can eliminate it (relative to its expectation) by increasing
the number of policies in the portfolio
...
Diversifiable risks are generally easier to deal with
than those which are not
...
4
...
2 we employed the no arbitrage principle to argue that the value
of a deterministic payment stream should be the same as the price of the zerocoupon bonds that replicate that payment stream
...
3
...
To do this we need to assume that the mortality risk
associated with a portfolio is diversifiable and we discuss conditions for this to
be a reasonable assumption
...
We will make the following two assumptions
throughout the remainder of this chapter, except where otherwise stated
...

(ii) The survival model for each of the N lives is known
...


336

Interest rate risk

The cash flow at any future time t for this group of policyholders will depend
on how many are still alive at time t and on the times of death for those not
still alive
...
However, with the two assumptions
above the mortality risk is diversifiable
...
This is demonstrated in the following example
...
2 For 0 ≤ t ≤ t + s, let Nt,s denote the number of deaths between
ages x + t and x + t + s from N lives aged x
...

N

Solution 10
...
Hence
V[Nt,s ] = N t px (1 − s px+t ) (1 − t px (1 − s px+t ))
⇒

V[Nt,s ]
=
N

⇒ lim

N →∞

t px (1 − s px+t )(1 − t px (1 − s px+t ))

N

V[Nt,s ]
= 0
...
There are exceptions; for example, for very old age mortality,
where the number of policyholders tends to be small, or where an insurance has
a very high sum at risk, in which case the outcome of that particular contract
may have a significant effect on the portfolio as a whole, or where the survival
model for the policyholders cannot be predicted with confidence
...
In the following section
we use this deterministic assumption for mortality to look at the replication of
the term insurance cash flows in Example 10
...


10
...
2 Non-diversifiable risk
In practice, many insurers do not replicate with forward rates or zero-coupon
bonds either because they choose not to or because there are practical difficulties
in trying to do so
...
4 Diversifiable and non-diversifiable risk

337

insurer can remove (much of) the investment risk, as shown in Table 10
...

However, while this removes the risk of losses, it also removes the possibility of
profits
...
For example, in some countries
it may not be possible to find risk free investments for terms longer than around
20 years, which is often not long enough
...
The rate of interest that would
be appropriate for an investment to be made over 20 years ahead could be very
difficult to predict
...
The risk that interest rates are lower than those expected in
the premium calculation is an example of non-diversifiable risk
...
The insurer decides
to invest all premiums in 10-year bonds, reinvesting when the bonds mature
...
If the insurer determines the premium assuming a fixed interest rate
of 6% per year, and the actual interest rate earned is 5% per year, then the
portfolio will make a substantial loss, and in fact each individual contract is
expected to make a loss
...
The key point here is
that the policies are not independent of each other with respect to the interest
rate risk
...
4
...
However, non-diversifiable risk is, arguably, even more
important
...
Note also that not all mortality
risk is diversifiable
...
4 below, we look at a situation where the
mortality risk is not fully diversifiable
...
3 we look at an
example of non-diversifiable interest rate risk
...
3 An insurer issues a whole life insurance policy to (40), with
level premiums payable continuously throughout the term of the policy, and
with sum insured $50 000 payable immediately on death
...
0001, B = 0
...
075
...
Show that the annual premium rate is P = $1 010
...


338

Interest rate risk

(b) Now suppose that the effective annual interest rate is modelled as a random
variable, denoted i, with the following distribution
...
25,
⎨
i = 5% with probability 0
...
25
...
Assume that the future lifetime is
independent of the interest rate
...
3 (a) At 5% we have
a40 = 14
...
29287

giving a premium of
P = 50 000

¯
A40
= $1 010
...

a40
¯

(b) Let S = 50 000, P = 1 010
...
The present value of the
future loss on the policy, L0 , is given by
¯
L0 = S viT − P aT i
...
As
¯
L0 |i = S viT − P aT i ,
¯
¯
E[L0 |i] = (S A40 − P a40 )|i
⎧
⎪ 1 587
...
25 (i = 4%),
⎨
=
0 with probability 0
...
49 with probability 0
...
6)

(10
...
25 (1 587
...
5 (0) + 0
...
49)
= $128
...


(10
...


(10
...
4 Diversifiable and non-diversifiable risk

339

We can interpret the first term as the risk due to uncertainty over the future
lifetime and the second term as the risk due to the uncertain interest rate
...
25
with probability 0
...
25

(i = 6%)
...

Also, from equation (10
...
432 ) 0
...
5 + (−1 071
...
25 − 128
...
882
...


(10
...

• The fixed interest assumption, 5% in this example, is what is often called

the ‘best estimate’ assumption – it is the expected value, as well as the most
likely value, of the future interest rate
...
In this example, using a 5% per year interest
assumption to price the policy leads to an expected loss of $128
...

• Breaking the variance down into two terms separates the diversifiable risk
from the non-diversifiable risk
...


340

Interest rate risk

Let L0,j denote the present value of the loss at inception on the jth policy
and let
N

L=

L0,j
j=1

so that L denotes the total future loss random variable
...
9), and noting that, given our assumptions at the start
of this section, the random variables {L0,j }N are independent and identically
j=1
distributed, we can write
V[L] = E[V[L|i]] + V[E[L|i]]
⎡ ⎡
⎤⎤
= E ⎣V ⎣

N

⎡ ⎡

L0,j |i⎦⎦ + V ⎣E ⎣

j=1

N

⎤⎤
L0,j |i⎦⎦

j=1

= E[N V[L0 |i]] + V[N E[L0 |i]]
= 196 364 762 N + 900 371 N 2
...
The first term
represents diversifiable risk since it is a multiple of N and the second term
represents non-diversifiable risk since it is a multiple of N 2
...
But, for a large portfolio, the contribution of
the interest uncertainty to the total standard deviation is far more important
than the future lifetime uncertainty
...
The following example shows that if these conditions do not hold,
mortality risk can be non-diversifiable
...
4 A portfolio consists of N identical one-year term insurance policies issued simultaneously
...
The insurer uses an effective interest rate
of 5% for all calculations but is unsure about the mortality of this group of
policyholders over the term of the policies
...
4 Diversifiable and non-diversifiable risk

341

distribution

q70

⎧
⎪0
...
25,
⎨
= 0
...
5,
⎪
⎩
0
...
25
...

(a) Let D(N ) denote the number of deaths during the one year term
...

lim
N →∞
N
(b) Let L(N ) denote the present value of the loss from the whole portfolio
...

lim
N →∞
N
Solution 10
...

Now
V[E[D(N )|q70 ]] = 0
...
022 − 0
...
25((0
...
025)N )2
= 4
...
25 × 0
...
022)N
+ 0
...
025(1 − 0
...
25 × 0
...
028)N
= 0
...

Hence
V[D(N )] = 4
...
0243705N
and so
√
V[D(N )]
= 0
...

N →∞
N
lim

342

Interest rate risk

(b) The arguments are as in part (a)
...

As
L(N ) = 50 000vD(N ) − 1 300N ,
we have
V[L(N )|q70 ] = (50 000v)2 V[D(N )|q70 ]
= (50 000v)2 N q70 (1 − q70 )
and
E[L(N )|q70 ] = 50 000vN q70 − 1 300N
...
025 − 0
...
5 × 10−6
...
02
...
5 Monte Carlo simulation
Suppose we wish to explore a more complex example of interest rate variation than in Example 10
...
If the problem is too complicated, for example if
we want to consider both lifetime variation and the interest rate uncertainty,
then the numerical methods used in previous chapters may be too unwieldy
...
Using Monte Carlo
techniques allows us to explore the distributions of present values for highly

10
...

If the sample is large enough, we can get good estimates of the moments of the
distribution, and, even more interesting, the full picture of a loss distribution
...

In this section we demonstrate the use of Monte Carlo methods to simulate
future lifetimes and future rates of interest, using a series of examples based on
the following deferred annuity policy issued to a life aged 50
...

• Level premiums of amount P = $4447 per year are payable continuously
throughout the period of deferment
...

• Basis for all calculations:
• The survival model follows Gompertz’ law with parameters B = 0
...
07
...

• The force of interest applying at age 65 is denoted r
...
In the
final example we will assume that r has a fixed but unknown value
...
5 Assume the force of interest from age 65 is 6% per year, so that
r = 0
...

(a) Calculate the EPV of the loss on the contract
...

Solution 10
...
This
gives the expected present value of the loss as
10 000 × 0
...
51058 + 4 447 × 1
...
49338
= −$6735
...


344

Interest rate risk

(b) The present value of the loss, L, can be written in terms of the expected
future lifetime, T50 , as follows
¯
P T50 v T50 − P aT50
L=
∗
v 15 − P a15
¯
10 000 a
¯
T50 −15

if T50 ≤ 15,
if T50 > 15
...
05) a15 5% (0
...
06
10

= Pr[T50 > 30
...
109 p50 = 0
...


2

Example 10
...
Assume that the force of interest from age 65 is 6% per year:
u1 = 0
...
51720,

u3 = 0
...


Solution 10
...
Each simulated uj
generates a simulated future lifetime tj through the inverse transform method,
where
uj = FT (tj )
...
Hence
u = FT (t)
= 1 − e−(B/ log(c))c

50 (ct −1)

−1
⇒ t = FT (u)

=

1
log(c)(log(1 − u))
log 1 −
log(c)
B c50

So
−1
t1 = FT (0
...
266,
−1
t2 = FT (0
...
314,
−1
t3 = FT (0
...
969
...


(10
...
5 Monte Carlo simulation

345

These simulated lifetimes can be checked by noting in each case that
tj q50 = uj
...
If (50) dies after
exactly 10
...
266P, the present value of the premiums paid is P a10
...
266 P e−10
...
266

δ

= −$8383
...


Similarly, the other two simulated future lifetimes give the following losses
¯
L0 = 10 000e−15δ a9
...
969

− P a15 δ = −$13 223
...
36
...


2

Example 10
...
6, generating 5000 values of the present
value of future loss random variable
...

(b) Calculate a 95% confidence interval for the expected value of the present
value of the loss
...

(d) Calculate a 95% confidence interval for the probability that the contract
generates a loss
...
7 Use an appropriate random number generator to produce a
sequence of 5000 U (0, 1) random numbers, {uj }
...
11) to generate corresponding values of the future lifetime, {tj }, and the present value of
the future loss for a life with future lifetime tj , say {L0,j }, as in Example 10
...

The result is a sample of 5000 independent values of the future loss random
variable
...

l
(a) The precise answers will depend on the random number generator (and
seed value) used
...
74;
l

sl = $15 733
...


(b) Let µ and σ denote the (true) mean and standard deviation of the present
value of the future loss on a single policy
...

j=1

Hence
⎡
σ

1
Pr ⎣µ − 1
...
96 √
j=1

σ
5000

⎦ = 0
...


Since ¯ and sl are estimates of µ and σ , respectively, a 95% confidence
l
interval for the mean loss is
¯ − 1
...
96 √ sl
l
l

...
86, − 6156
...

(c) Let L− denote the number of simulations which produce a loss, that is, the
number for which L0,i is positive
...
Then
L− ∼ B(5000, p)
and our estimate of p, denoted p, is given by
ˆ
p=
ˆ

l−
5000

where l − is the simulated realization of L− , that is, the number of losses
which are positive out of the full set of 5000 simulated losses
...
96
ˆ

p(1 − p)
ˆ
ˆ
p(1 − p)
ˆ
ˆ
, p + 1
...
Our calculations gave a total
ˆ
of 1563 simulations with a positive value for the expected present value of
the future loss
...
3126
ˆ

10
...
2998, 0
...

Different sets of random numbers would result in different values for each
of these quantities
...

As we have seen in Example 10
...
38 and 0
...
The 95% confidence intervals calculated
in Example 10
...
We used
simulation in this example to illustrate the method and to show how accurate
we can be with 5000 simulations
...
The next example demonstrates this in the
case where the force of interest from age 65 is fixed but unknown
...
8 Repeat Example 10
...
06, 0
...
Assume the random variables T50
and r are independent
...
8 For each of the 5000 simulations generate both a value for T50 ,
as in the previous example, and also a value of r from the N (0
...
0152 )
...
The simulated value of the present value of the loss for this
simulation, L0,j , is
L0,j =

P ti v tj − P atj
¯

10 000 a∗
¯
tj −15

if tj ≤ 15,
v 15

− P a15
¯

if tj > 15
...
The
remaining steps in the solution are as in Example 10
...

Our simulation gave the following results
...
5; sl = $16 903
...

l
Hence, an approximate 95% confidence interval for the mean loss is
(−6689, −5752)
...
3004,
ˆ

348

Interest rate risk

with an approximate 95% confidence interval for this probability of
(0
...
3131)
...
The probability of loss is not
significantly different from the fixed interest case
...
6 Notes and further reading
The simple interest rate models we have used in this chapter are useful for
illustrating the possible impact of interest rate uncertainty, but developing more
realistic interest rate models is a major topic in its own right, beyond the scope of
this text
...

We have shown in this chapter that uncertainty in the mortality experience is a
source of non-diversifiable risk
...
See, for example, Willets et al
...
In these circumstances,
the assumptions about the survival model in Section 10
...
1 may not be reasonable and so a significant aspect of mortality risk is non-diversifiable
...
6–10
...
Monte Carlo methods could be used to model uncertainty about the survival model; for example,
by assuming that the two parameters in the Gompertz formula were unknown
but could be modelled as random variables with specified distributions
...
A general
introduction is presented in e
...
Ross (2006), and Glasserman (2004) offers
a text more focused on financial modelling
...
(2007)
...
7 Exercises
Exercise 10
...
35
89
...
45
79
...
79

10
...

(b) Calculate the one-year forward rates, at t = 0, 1, 2, 3, 4
...
99
...
2 Consider an endowment insurance with sum insured $100 000
issued to a life aged 45 with term 15 years under which the death benefit is
payable at the end of the year of death
...
035 +

t

...
77
...

(c) Calculate the net policy value for a policy still in force three years after
issue, using the rates implied by the original yield curve, using the premium
basis
...
77
...

Exercise 10
...
One-half of all the policies have a sum
insured of $10 000, and the other half have a sum insured of $100 000
...

The insurer wishes to measure the uncertainty in the total present value of
claims in the portfolio
...

(a) Calculate the standard deviation of the present value of the benefit for an
individual policy, chosen at random
...

(c) By comparing the portfolio of 100 policies with a portfolio of 100 000
policies, demonstrate that the mortality risk is diversifiable
...
4 (a) The coefficient of variation for a random variable X is
defined as the ratio of the standard deviation of X to the mean of X
...


350

Interest rate risk

(b) An insurer issues a portfolio of identical 15-year term insurance policies to
independent lives aged 65
...

The mortality for the portfolio is assumed to follow Makeham’s law
with A = 0
...
7 × 10−6
...
124, as in the Standard
Ultimate Survival Model, or 1
...
The insurer models this uncertainty
assuming that there is a 75% probability that c = 1
...
114
...
The effective rate of interest is assumed to be 6% per year
...

(ii) Calculate the coefficient of variation of the total present value of
benefits for the portfolio assuming that 10 000 policies are issued
...

Exercise 10
...
The insurer calculates the premium assuming an interest rate of 7%
per year effective, and using the Standard Ultimate Survival Model
...

(b) Suppose that the effective annual interest rate is a random variable, i, with
the following distribution:
⎧
⎪5% with probability 0
...
25,
⎪
⎩
11% with probability 0
...

Write down the EPV of the net future loss on the policy using the mean
interest rate, and the premium calculated in part (a)
...

(d) Calculate the EPV and the standard deviation of the present value of the
net future loss on the policy
...

(e) Comment on the results
...
6 An insurer issues 15-year term insurance policies to lives aged 50
...
Level premiums

10
...

The insurer assumes the lives are subject to Gompertz’ law of mortality with
B = 3 × 10−6 and c = 1
...

(a) Generate 1000 simulations of the future loss
...

(c) Calculate a 90% confidence interval for the mean future loss
...
Does it lie in your
confidence interval in (c)?
(e) Repeat the 1000 simulations 20 times
...
The insurer models the interest
rate on all policies, I , as a lognormal random variable, such that
1 + I ∼ LN (0
...
02412 )
...
Comment on the effect of
interest rate uncertainty
...
7 An actuary is concerned about the possible effect of pandemic
risk on the term insurance portfolio of her insurer
...

(a) State, with explanation, whether pandemic risk is diversifiable or nondiversifiable
...


Answers to selected exercises
10
...
2 (b)
(c)
(d)

(0
...
05881, 0
...
05754, 0
...
05988, 0
...
05625, 0
...
05489)
$4395
...
50
$13 548
We show the first three rows of the cash flow table
...
77
4204
...
99

Forward rate
f (k, k + 1)

Expected
claims
outgo Ck+1

Net cash flow
carried forward
CFk+1

1
...
0441
1
...
11
83
...
47

4298
...
01
13 513
...
3 (a) $19 784
(b) $193 054
10
...
2337
(ii) 0
...
2192
10
...
13
(b) $0
(c) $7325
...
80, $8489
...
6 (d) −$184
...

(g) Term insurance is not very sensitive to interest rate uncertainty, as the
standard deviation of outcomes with interest rate uncertainty is very
similar to that without interest rate uncertainty
...
1 Summary
In this chapter we introduce emerging costs, or cash flow analysis for traditional
life insurance contracts
...

Traditional actuarial analysis focuses on determining the EPV of a cash flow
series, usually under a constant interest rate assumption
...
Using cash flow projections to model
risk offers much more flexibility than the EPV approach and provides actuaries
with a better understanding of the liabilities under their management and the
relationship between the liabilities and the corresponding assets
...
First we consider only those cash
flows generated by the policy, then we introduce reserves to complete the cash
flow analysis
...
We show how cash flow analysis can be used
to set premiums to meet a given measure of profit
...


11
...
2
...
The details of the policy are as follows
...
Level annual
premiums, denoted P, of amount $1500 are payable throughout the term
...
It would be very common to choose one month as the
interval since in practice premiums are often paid monthly
...

The purpose of a profit test is to identify the profit which the insurer can claim
from the contract at the end of each time period, in this case at the end of each
year
...
For ease of presentation, we ignore the
possibility of lapsing in this example
...

For this example, we use the following profit test basis
...
5% per year effective on all cash flows
...

3
...

q60+t = 0
...
001 t for t = 0, 1,
...


The initial expenses represent the acquisition costs for the policy
...
For each year that the policy is still in force, cash flows contributing to
the surplus emerging at the end of that year are the premium less any renewal
expense, interest earned on this amount and the expected cost of a claim at the
end of the year
...
1
...
2 P)
...

For the first policy year there is a premium payable at time 0, but no expenses
since these are included in the row for t = 0
...
5% and the
expected death claims, payable at time 1, are q60 S = 0
...

Hence the emerging surplus, or net cash flow, at time 1 is
1500 + 82
...
5
...
For example, the net cash flow at

11
...
1
...
2
...
00
0
...
50
52
...
50
52
...
50
52
...
50
52
...
50

82
...
61
79
...
61
79
...
61
79
...
61
79
...
61

1000
1100
1200
1300
1400
1500
1600
1700
1800
1900

−700
...
50
427
...
11
227
...
11
27
...
89
−172
...
89
−372
...
035 × 1500 + 0
...
035 × 1500)
− 100 000 × (0
...
001) = −72
...

In Table 11
...
, 10, Et denotes the renewal expenses incurred at the start of the
year from t − 1 to t
...
2
...
1 reveals a typical feature of net cash flows: several of the net cash
flows in later years are negative
...
The expected cash flow values in the final column of Table 11
...
1 and 6
...

In Chapter 7 we explained why the insurer needed to set aside assets to cover
negative expected future cash flows
...
In modelling cash flows, we use
reserves rather than policy values
...
The reserve may be equal to the
policy value, or may be some different amount
...
Usually, though, for traditional insurance, the policy value
calculation will be used to set reserves, perhaps using a conservative basis
...
1 does not require a reserve
since it will have been paid as soon as the policy was issued
...
In practice the reserve basis is likely to
be more conservative than the profit test basis
...

Interest:
Survival model:

4% per year effective on all cash flows
...
011 + 0
...
, 9
...
e
...
63,

and all functions are calculated using the reserve basis
...
2
...
2 are amounts that the insurer needs to assign
from its assets to support the policy
...
To see how to do this, consider, for example,
the reserve required at time 1, 1 V = 410
...
This amount is required for every
policy still in force at time 1
...
05 × (1 − 0
...
95
...

The expected proportion of policyholders surviving to the start of the following
time period, i
...
to age 61, is p60
...
In general, the cost at the end of the year from t − 1 to t of setting up a
reserve of amount t V at time t for each policy still in force at time t is t V p60+t−1
...
2 Profit testing for traditional life insurance

357

Table 11
...
Reserves for the 10-year
term insurance in Section 11
...

t

tV

t

tV

0
1
2
3
4

0
...
05
740
...
90
1150
...
94
1193
...
74
827
...
45

Table 11
...
Emerging surplus, per policy in force at start of year, for the
10-year term insurance in Section 11
...

t
0
1
2
3
4
5
6
7
8
9
10

t−1 V

P

Et

It

S q60+t−1

t V p60+t−1

Prt

0
...
05
740
...
90
1150
...
94
1193
...
74
827
...
45

1500
1500
1500
1500
1500
1500
1500
1500
1500
1500

700
...
0
52
...
50
52
...
50
52
...
50
52
...
50
52
...
50
102
...
36
134
...
87
146
...
25
138
...
14
105
...
59
732
...
04
1135
...
86
1175
...
70
813
...
89
0
...
00
176
...
99
131
...
26
137
...
68
138
...
72
133
...
71

The profit test calculations, including reserves, are set out in Table 11
...
Here
It denotes the interest earned in the year from t − 1 to t, E0 denotes the initial
expenses incurred at time 0 and for t = 1, 2,
...
Note that the cost of
setting up a reserve, t V p60+t−1 , is a cost to the insurer at the end of the year
from t − 1 to t, whereas the reserve t V is a positive asset at the start of the
following year
...
37 − 0
...
055(0
...
37)
− 100 000 × 0
...
74 × 0
...
41
...
, 10, the calculation of Pr t in Table 11
...

Many actuaries prefer to write this in the equivalent form
Pr t = (P − Et )(1 + i) +
where

tV

tV

− Sq60+t−1 ,

is called the change in reserve in year t and is defined as
tV

= (1 + i) t−1 V − t V p60+t−1
...
The incoming and outgoing reserves each year are not real
income and outgo in the same way as premiums, claims and expenses, but
accounting transfers
...
, Pr10 ) is called the profit vector for the contract
...
Multiplying Prt by t−1 p60 gives
a vector each of whose elements is the expected profit at the end of each year
given only that the contract was in force at age 60
...
, 10
...
1)

, where
0,

1,
...
, 9 p60 Pr10 )
(11
...
The profit signature is the key to
assessing the profitability of the contract
...
55, 125
...
96, 130
...
39, 130
...
35, 124
...
75, 113
...


11
...
There are a number of ways to measure profit, all based on
the profit signature
...
Given a profit signature
( 0 , 1 ,
...


(11
...
3 Profit measures

359

For the example in Section 11
...
24%
...

One problem with the internal rate of return is that there may be no real
solution to equation (11
...
However, we can still use
the risk discount rate to calculate the expected present value of future profit
(EPVFP), also called the net present value (NPV) of the contract
...
Then the NPV is the present value, at rate r, of the projected
profit signature cash flows, so that
n

NPV =

t
t vr
...
2, suppose the insurer uses a risk discount rate
of 10% per year
...
48
...
For a contract with level premiums
of P per year payable mthly throughout an n year contract issued to a life aged
x, the profit margin is
Profit Margin =

NPV
(m)

P ax:n
¨

(11
...

For the example in Section 11
...
48
NPV
=
= 1
...

P a60:10
¨
9 684
Another profit measure is the NPV as a proportion of the acquisition costs
...
2, the acquisition costs are $700, so the NPV is 17
...

Our final profit measure is the discounted payback period (DPP), also
known as the break-even period
...


t=0

The DPP represents the time until the insurer starts to make a profit on the
contract
...
2, the DPP is eight years
...
Most of the inputs we have used in the emerging
surplus calculation are in practice uncertain, for example we do not usually
know what interest rates and mortality rates will be
...

These measures of profitability can be used to calculate a premium
...
2
...
45, which gives the revised profit signature equal to
(−732
...
99, 290
...
88, 291
...
27, 287
...
65, 276
...
01, 259
...


This gives an internal rate of return of 40
...
97, profit margin = 10%, NPV as a percentage of acquisition
costs = 146
...

It is interesting to see how the reserve basis affects the profitability of the
contract
...
This will have the effect of increasing
the size of the reserves required, so that, for example 3 V = 1001
...
13 rather than 988
...
74, respectively
...
48 to $122
...
On the other
hand, weakening the reserve basis by using an interest rate of 5% gives a higher
NPV of $126
...
By increasing the size of the reserves, the insurer is being
required to assign more of its assets to the policy
...
5% per year in
our example
...
The intuition is that the reserve is assumed
to be invested conservatively, so higher reserves mean tying up more assets in
conservative investments, reducing the profitability
...
4 A further example of a profit test
The term insurance example used throughout Section 11
...
The policy itself was relatively uncomplicated – term insurance, level annual premiums, sum insured payable at the end
of the year of death – and we assessed its profitability assuming no allowance
for withdrawals and by calculating cash flows at annual intervals
...
However, the basic principles
are unchanged
...
4 A further example of a profit test

361

Example 11
...
The benefits under the policy are
• $50 000 if at the end of a month the life is disabled, having been healthy at

the start of the month,
• $100 000 if at the end of a month the life is dead, having been healthy at the
start of the month,
• $50 000 if at the end of a month the life is dead, having been disabled at the
start of the month,
• $50 000 if the life survives as healthy to the end of the term
...

The survival model used for profit testing is shown in Figure 11
...
The transition intensities µ01 , µ02 , µ03 and µ12 are constant for all ages x with values
x
x
x
x
per year as follows:
µ01 = 0
...
015,
x

µ03 = 0
...
03
...

• Interest: 7% per year
...

• The benefit on withdrawal is payable at the end of the month of withdrawal

and is equal to 80% of the sum of the reserve held at the start of the month
and the premium paid at the start of the month
...

• The gross premium and net premium policy values are calculated using the
same survival model as for profit testing except that withdrawals are ignored,
so that µ03 = 0 for all x
...
1 Multiple state model for Example 11
...


362

Emerging costs for traditional life insurance

• The net premium policy values are calculated using an interest rate of 5%

per year
...
25% per year
...

(a) Calculate the monthly premium on the net premium policy value basis
...

(c) Calculate the monthly gross premium
...

(e) Calculate the internal rate of return
...

Before solving this example, we remark that in practice it would be very unlikely
that a policyholder would withdraw late into the term of a policy such as this
one
...
This assumption
simplifies the formulae we require to conduct the profit test
...
1 We have a survival model, as shown in Figure 11
...
The difference between the parameterizations is that
the former allows for withdrawals, whereas the latter does not
...
The
following probabilities are useful in our calculations
...
035t},
µ01 + µ02 ds = exp{−0
...
03t} = t px ,

11
...
035s} 0
...
03(t − s)} ds

0

= 2(exp{−0
...
035t}),
01
t p55

t

=
0

t

=

00 01
11
s p55 µ55+s t−s p55+s ds

exp{−0
...
01 exp{−0
...
025t} − exp{−0
...
03t} = t px
...
03/12} =

1
12

12
px ,

5
(1 − exp{−0
...
03/12} − exp{−0
...
025/12} + 2 exp{−0
...
035/12}),
7
= 2(exp{−0
...
035/12}),

03∗
px =
01∗
px

1
12

01
px = 2(exp{−0
...
03/12})
...
Then
119

P

t

t
12

00
00
p55 v 12 = 50 000 10 p55 v 10

t=0
119

+ 50 000

t
12

00
p55

t=0

1
12

01
p55+

t
12

+

119

+ 100 000

t
12

00
p55

t=0

1
12

02
p55+

t
12

v

t
12

t+1
12

01
p55

1
12

12
p55+


...
13

t
12

v

t+1
12

364

Emerging costs for traditional life insurance

119

t

and

t
12

01
p55 v 12 = 3
...
00
...
If
t is an exact number of months, then the policy value is calculated before
payment of a premium and after payment of any benefits due at that time
...
, 12 , 0 from these starting values using the formulae
1

(t V (0) + P ) 1
...
05 12 =

1
12

11
p55+t

1
t+ 12 V

(1)

+ 50 000

1
12

12
p55+t
...
4
...
Then, using the equivalence
principle,
119

0
...
25% per year
...
26

11
...
4
...
1
...

11
3 12

4
1
4 12

tV

(0)

tV

(1)

t years

tV

(0)

tV

(1)

0
...
32

−
10 301
...
19
36 761
...
14
2769
...
40

...
19

...


37 273
...


2663
...


15 237
...
67

9 10
12

48 818
...
86

7157
...
16

11
9 12

49 407
...
00

124
...
00

15 613
...
75

10

and
119

t

t
12

01
p55 v 12 = 3
...
27
...
Parts of the calculation
are shown in Tables 11
...
6
...
5 is as follows
...

(2) denotes the reserve held at the start of the time interval for a life who
is healthy at that time, t− 1 V (0)
...

(3) denotes the gross premium, $484
...

(4) denotes the expenses payable in the time interval
...
21 = 1000 + 0
...
05 × 484
...

1
The entry for t = 12 is zero since all the initial expenses have been
assigned to the row for t = 0
...
05P,
which is assumed to be incurred at the start of the month
...
07 12 − 1)(t− 1 V (0) + P − Et )
...
5
...
1 assuming the life is healthy
at the start of the month
...


Dis
...
Death With’l
ben
...
ben
...

(6) (7)
(8)
(9)

Healthy
res
...
00 484
...
21
0
...
74 41
...
56 124
...
32

278
...
21
33
...
32 484
...


...
21

...
18 41
...
51 124
...


...


0
...


558
...


9
...


0
1
12
2
12

It
(5)

9 10
12

48 233
...
27

24
...
32 41
...
21 124
...
43 48 676
...
24

11
9 12

48 818
...
27

24
...
63 41
...
10 124
...
82 49 263
...
27

10

49 407
...
27

24
...
96 41
...
00 124
...
21 49 854
...
30

Table 11
...
Emerging surplus for Example 11
...

t
years
(1)

1
t− 12

1
12
2
12
3
12

0
...
00

0
...
00

0
...
49

58
...
84

10 218
...
28

10 244
...


57
...


124
...


10 161
...


16
...


9 10
12

370
...
10

124
...
24

0
...
86

1
...
84

124
...
39

10

124
...
70

124
...
00

0
...


V (0)

(2)

It
(3)

Death
ben
...

(5)

(1)

Prt
(6)

(6) is the expected disability benefit payable at the end of the month,
01∗
50 000 1 p55+t , in respect of a life who was healthy at the start of the
12
month but disabled at the end of the month
...
This expected cost is 1 p55+t t V (1)
...
This expected cost is
02∗
100 000 1 p55+t
...
4 A further example of a profit test

367

(9) is the expected cost of the withdrawal benefit
...
8
...
This expected cost is 1 p55+t t V (0)
...


The key to the columns in Table 11
...

1
(1) denotes the time interval from t − 12 to t, measured in years
...
No cash flows are included for the
12

(3)
(4)

(5)

(6)

1
first month, corresponding to t = 12 , since the life is healthy when the
policy is issued
...

is the expected death benefit payable at the end of the month for a
life who was disabled at the start of the month
...

12
denotes the expected cost of setting up the reserve required at the start
of the following month for a life who remains disabled throughout the
11∗
month
...

12
denotes the expected surplus emerging at the end of the month in respect
of a policyholder who was disabled at the start of the month, so that
(1)

Prt

(e) The profit signature vector,
lated as

= (2) + (3) – (4) – (5)
...
,

11
9 12 ,

(0)

= Pr0

1 2
12 , 12 ,
...


10 )

, is calcu-

368

Emerging costs for traditional life insurance
Table 11
...
Calculation of the profit signature for Example 11
...


t years
0
1
12
2
12

(0)

Prt

−1024
...
15

00∗
1 p
t− 12 55

(1)

01∗
1 p
t− 12 55

Prt

−
0
...
00
1
...
00
0
...
21
33
...

11
3 12

9
...

34
...
9971

...
8744

16
...

11
...
0008

...
0338

9
...

30
...

11
7 12

35
...
13

...
38

0
...
8694

...
7602

11
...
31

...
71

0
...
0351

...
0607

31
...
81

...
55

8
1
8 12

...
27
73
...

93
...
7580
0
...

0
...
54
4
...

0
...
0612
0
...

0
...
06
55
...

66
...
27

0
...
39

0
...
85

10

95
...
7067

0
...
0719

67
...
7
...


This gives an internal rate of return of 32
...

(f) The net present value, NPV, is given by
120

k

NPV =

k
12

k=0

(1 + 0
...
29
...
e
...
15)− 12

= 3
...
6 Exercises

369

and the NPV as a percentage of the acquisition costs is
NPV/ (0
...
0%
...


This gives a discounted payback period of five years and five months
...
5 Notes and further reading
In Example 11
...
This is a common feature of
profit tests in practice
...
In Example 11
...

For each of the policies considered in this chapter, benefits are payable at the
end of a time period
...
For example, for the term insurance policy considered in Section 11
...
If, instead, the death benefit had been payable immediately
on death, then we could allow for this in our profit test by assuming all deaths
occurred in the middle of the year
...
1 would all be adjusted by multiplying by a factor of 1
...

Throughout this chapter we have used deterministic assumptions for all the
factors
...
In Chapter 12 we describe how we might use stochastic scenarios for
emerging cost analysis for equity-linked contracts
...


11
...
1 A five year policy with annual cash flows issued to a life (x)
produces the profit vector
Pr = (−360
...
66, 14
...
19, 388
...
00),

370

Emerging costs for traditional life insurance

where Pr 0 is the profit at time 0 and Pr t (t = 1, 2,
...

The survival model used in the profit test is given by qx+t = 0
...
0005t
...

Calculate the NPV for this policy using a risk discount rate of 10% per year
...

Comment briefly on the difference between your answers to parts (b)
and (c)
...


Exercise 11
...
Initial expenses are $500 plus 50% of the first
monthly premium; renewal expenses are 5% of each monthly premium after
the first
...
00022, B = 2
...
124
...

Exercise 11
...
The sum insured, $100 000, is payable at the end of the year of death
...
The reserve at each year
end is 30% of the gross premium
...
008, q61 = 0
...
010, q63 = 0
...

(b) Calculate the profit signature for the contract
...

(d) Calculate the profit margin for the contract using a risk discount rate of
12% per year
...
6 Exercises

371

(e) Calculate the discounted payback period using a risk discount rate of 12%
per year
...

(g) If the insurer has a hurdle rate of 15% per year, is this contract satisfactory?
Exercise 11
...
The sum insured is $100 000, payable at the end of the year of death
or on survival to age 75
...
The insurer assumes that initial expenses will be $300, and renewal
expenses, which are incurred at the beginning of the second and subsequent
years in which a premium is payable, will be 2
...

The funds invested for the policy are expected to earn interest at 7
...
The insurer holds net premium reserves, using an interest rate of 6% per
year
...
00022, B = 2
...
124
...

Calculate the gross annual premium
...
5 Repeat Exercise 11
...
5% of each monthly premium after
the first
...
6 A life insurance company issues a special 10-year term insurance
policy to two lives aged 50 at the issue date, in return for the payment of a single
premium
...

• In the event of either of the lives dying within 10 years, a sum insured of

$100 000 is payable at the year end
...
(If both lives die within 10 years and in
the same year, a total of $300 000 is paid at the end of the year of death
...

Survival model: Assume the two lives are independent with respect to survival
and the model for each follows Makeham’s law with parameters A = 0
...
7 × 10−6 and c = 1
...

Interest: 4% per year
Expenses: 3% of the single premium at the start of each year that the contract
is in force
...

(b) Calculate the reserves on the premium basis assuming that
(i) only one life is alive, and
(ii) both lives are still alive
...
5% of the premium at issue, increasing at 4% per year
Exercise 11
...
The annuity commences
at the end of the year of death of the wife and is payable subsequently while the
husband is alive, for a maximum period of 20 years after the commencement
date of the policy
...
The
premium for the policy is payable annually while the wife and husband are
both alive and for a maximum of five years
...

Survival model: Assume the two lives are independent with respect to survival
and the model for each follows Makeham’s law with parameters A = 0
...
7 × 10−6 and c = 1
...

Interest: 4% per year
Expenses: Initial expense of $300 and an expense of 2% of each annuity
payment whenever an annuity payment is made
...

(b) Calculate the NPV for the policy assuming:
a risk discount rate of 15% per year,
expenses and the survival model are as in the premium basis, and
interest is earned at 6% per year on cash flows
...
8 A life aged 60 purchases a deferred life annuity, with a five year
deferred period
...
All payments
are contingent on survival
...

If the policyholder dies before the first annuity payment, the insurer returns
her gross premium, with interest of 5% per year, at the end of the year of her
death
...
9(0
...
7 × 10−6 × 1
...
6 Exercises

373

(b) Gross premium reserves are calculated using the premium basis
...

(c) The insurer conducts a profit test of the contract assuming the following
basis:
Survival model: µx = 0
...
7 × 10−6 × 1
...

(ii) Calculate the profit margin for the contract using a risk discount rate
of 10% per year
...
1 (a) (−360
...
66, 14
...
43, 377
...
29)
(b) $487
...
69
(e) 42
...
2 Selected values are Pr 30 = 54
...
75, measuring time in
months
11
...
00, 60
...
23, 195
...
08)
(b) (−300
...
16, 292
...
19, 313
...
19
(d) 8
...
4 $4553
...
5 $394
...
6 (a) $4 180
...
04, and (ii) 4 V = $3146
...
71, 5 = $177
...
42
11
...
79
(b) $779
...
8 (a) $192 805
...
51 and 10 V = $226 245
...
90 and 10 = $2 429
...
8%

12
Emerging costs for equity-linked insurance

12
...
We explore
deterministic emerging costs techniques with examples, and demonstrate that
deterministic profit testing cannot adequately model these contracts
...

Finally we discuss the use of quantile and conditional tail expectation reserves
for equity-linked insurance
...
2 Equity-linked insurance
In Chapter 1 we described some modern insurance contracts where the main
purpose of the contract is investment
...

These contracts are called unit-linked insurance in the UK and parts of
Europe, variable annuities in the USA (though there is often no actual annuity component) and segregated funds in Canada
...
The basic premise of these contracts is that a
policyholder pays a single or regular premium which, after deducting expenses,
is invested on the policyholder’s behalf
...
Regular management charges are deducted from the
fund by the insurer and paid into the insurer’s fund to cover expenses and
insurance charges
...

374

12
...
There may also be a guaranteed
minimum death benefit (GMDB)
...
One example is the bid-offer spread
...
There may also be an allocation percentage; if 101% of the premium
is allocated to units at the offer price, and there is a 5% bid-offer spread, then
101% of 95% of the premium (that is 95
...
The bid-offer spread mirrors the practice in unitized investment funds that are major competitors for policyholders’
investments
...
3 Deterministic profit testing for equity-linked insurance
Equity-linked insurance policies are usually analysed using emerging surplus
techniques
...
It is the insurer’s
cash flows that are important in pricing and reserving, but since the insurer’s
income and outgo depend on how much is in the policyholder’s fund, we must
first project the cash flows for the policyholder’s fund and then use these to
project the cash flows for the insurer’s fund
...

The following two examples illustrate these calculations
...
1 A 10-year equity-linked contract is issued to a life aged 55 with
the following terms
...
The insurer deducts
a 5% expense allowance from the first premium and a 1% allowance from
subsequent premiums
...

At the end of each year a management charge of 0
...


376

Emerging costs for equity-linked insurance

If the policyholder dies during the contract term, a benefit of 110% of
the value of the policyholder’s year end fund (after management charge
deductions) is paid at the end of the year of death
...

If the policyholder surrenders the contract, he receives the value of the
policyholder’s fund at the year end, after management charge deductions
...

If the policyholder holds the contract to the maturity date, he receives the
greater of the value of the policyholder’s fund and the total of the premiums
paid
...

(a) Assume the policyholder’s fund earns interest at 9% per year
...

(b) Calculate the profit vector for the contract using the following basis
...
005
...
All surrenders occur at the
end of a year immediately after the management charge deduction
...

Renewal expenses: 0
...

Interest: The insurer’s funds earn interest at 6% per year
...

(c) Calculate the profit signature for the contract
...

Solution 12
...
1
...
1 is as follows
...

(2) This shows the allocated premium invested in the policyholder’s fund
at time t − 1
...

(4) This shows the interest income on the combined premium and fund
brought forward at the rate assumed for the policyholder’s fund, 9%
per year
...

(6) This shows the management charge, at 0
...

(7) This shows the remaining fund, which is carried forward to the
next year
...
2
...
3 Deterministic profit testing

377

Table 12
...
Projection of policyholder’s fund for Example 12
...

t
(1)

Allocated
premium
(2)

Fund
b/f
(3)

Interest
(4)

Fund
at t −
(5)

Management
charge
(6)

Fund
c//f
(7)

1
2
3
4
5
6
7
8
9
10

4750
4950
4950
4950
4950
4750
4950
4950
4950
4950

0
...
67
10 914
...
26
23 921
...
01
39 144
...
83
56 961
...
02

427
...
98
1427
...
10
2598
...
56
3968
...
75
5572
...
43

5 177
...
65
17 291
...
36
31 470
...
58
48 063
...
58
67 483
...
45

38
...
47
129
...
77
236
...
80
360
...
44
506
...
00

5 138
...
17
17 162
...
60
31 234
...
77
47 702
...
14
66 977
...
45

Table 12
...
Emerging surplus for Example 12
...

t
(1)

Unallocated
premium
(2)

Expenses
(3)

Interest
(4)

Management
charge
(5)

Expected
death benefit
(6)

Prt
(7)

0
1
2
3
4
5
6
7
8
9
10

0
...
00
50
...
00
50
...
00
50
...
00
50
...
00
50
...
00
0
...
00
25
...
00
25
...
00
25
...
00
25
...
00

0
...
00
1
...
50
1
...
50
1
...
50
1
...
50
1
...
00
38
...
47
129
...
77
236
...
80
360
...
44
506
...
00

0
...
57
5
...
58
11
...
62
19
...
85
28
...
49
38
...
00
301
...
52
147
...
31
246
...
73
363
...
46
499
...
60

The key to the information in Table 12
...

(1) The entries for t are the years of the contract, from time t − 1 to time
t, except for t = 0 which represents the issue date
...

(3) This shows the insurer’s expenses at the start of the year
...
This is consistent with our calculation of
emerging costs for traditional policies in Chapter 11
...

(5) This shows the management charge, which is taken directly from
Table 12
...

(6) This shows the expected death benefit, which is 10% of the year end
fund value from Table 12
...
We
need only 10% of the fund as the rest is paid from the policyholder’s
fund – in the insurer’s cash flows we consider only income and outgo
that are not covered by the policyholder’s fund
...


Note that there is no projected cost in Table 12
...
45, is greater than the guarantee, 10×5000 =
$50 000
...
, 10
...
) The values
are shown in Table 12
...

(d) The NPV is calculated by discounting the profit signature at the risk discount
rate of interest, r = 15%, so that
10

NPV =

t

(1 + r)−t = $531
...


t=0

2
Example 12
...

The policyholder pays a single premium of $10 000, of which 3% is taken
by the insurer for expenses and the remainder, the allocated premium, is
invested in suitable assets
...
06% of the policyholder’s fund is transferred to the insurer’s fund
...
In addition, the insurer guarantees a minimum benefit
...
05t−1 ),
where t = 1, 2,
...


12
...
3
...
1
...
00000
1
...
89550
0
...
84224
0
...
00
301
...
70
124
...
50
206
...
83384
0
...
82552
0
...
81729

t

252
...
27
353
...
99
470
...
In the second year a
surrendered contract pays 95% of the policyholder’s fund
...

If the policyholder holds the contract to the maturity date, she receives
the money in the policyholder’s fund with a guarantee that the payout will
not be less than $10 000
...

Survival model: The force of mortality is constant for all ages and equal to
0
...

Death benefit: This is paid at the end of the month in which death occurs
...
The probability
of surrendering at the end of any particular month is 0
...
002 in the second year and 0
...

Interest: The policyholder’s fund earns interest at 8% per year effective
...

Initial expenses: 1% of the single premium plus $150
...
008% of the single premium plus 0
...
Renewal expenses are
payable at the start of each month after the first
...

(b) Construct a table showing the projected policyholder’s fund assuming the
policy remains in force throughout the term
...

(d) Calculate the NPV for the contract using a risk discount rate of 12% per year
...
2 (a) The probability of not dying in any month is
exp{−0
...
9995
...
004) exp{−0
...
9955

in the first year,

(1 − 0
...
006/12} = 0
...
001) exp{−0
...
9985

in subsequent years
...
4 shows the projected policyholder’s fund at selected durations
assuming the policy remains in force throughout the five years
...
The minimum death benefit is also given in the table –
in the first year this is the full premium and it increases by 5% at the start
of each year
...
4 show the cash flows for the
1
policyholder’s fund in the month from time t − 12 to time t
...
5
...
These are the cash flows for
the time period starting two months after the issue of the policy
...

The amount of the management charge, $5
...
4 and is paid into the insurer’s fund at the start of the period
...
78, are calculated as 0
...
0001 ×
9 819
...

The interest is calculated as (1
...
89 − 1
...
02
...
The insurer’s fund has to pay
the extra 1%, so the expected cost, $0
...
006/12})×0
...
57
...
03, is paid by the insurer’s
fund at the end of the month and is calculated as (1 − exp{−0
...
01 × 9 876
...
The expected cost of this GMDB is zero
after three months since, using the assumptions in the profit testing basis,
101% of the policyholder’s fund is greater than the minimum death benefit
thereafter
...
3 Deterministic profit testing

381

Table 12
...
Deterministic projection of the policyholder’s fund
for Example 12
...

Allocated
premium

t
1
12
2
12
3
12
4
12
5
12
6
12

Management
charge

Interest

Fund
c/f

Minimum
DB

9 700

0
...
00

62
...
41

10 000
...
41

5
...
77

9 819
...
00

0

9 819
...
89

63
...
57

10 000
...
57

5
...
51

9 934
...
00

0

9 934
...
96

63
...
07

10 000
...
07

...


...
00

...


...
25

...


...
33

...


...
00

...


...
74

6
...
53

10 407
...
00

0

10 407
...


...


6
...


...


66
...


...


10 467
...


...


10 500
...


...


0

11 094
...
66

71
...
97

10 500
...
97

...


...
70

...


...
75

...


...
03

...


...
00

...


...
85

7
...
49

11 965
...
00

0

11 965
...


...


7
...


...


76
...


...


12 034
...


...


11 576
...


...


0


...


...
32

7
...
02

12 829
...
25

0

12 829
...


...


7
...


...


82
...


...


12 904
...


...


12 155
...


...


0

13 676
...
21

87
...
62

12 155
...


...


1
1
1 12


...


...


...


2
1
2 12


...


...


...


3
1
3 12


...


...


...


4
1
4 12


...


...


...


Lapses in the first two years are a source of income for the insurer’s fund
since, on surrendering her policy, the policyholder receives less than the
full amount of the policyholder’s fund
...
95, is calculated as exp{−0
...
004 ×
0
...
57
...


382

Emerging costs for equity-linked insurance

Table 12
...
Deterministic projection of the insurer’s fund for Example 12
...

Unallocated Management
Minimum
premium
charge
Expenses Interest 1% DB
DB
Lapses

t
0
1
12
2
12
3
12
4
12
5
12
6
12


...


...


...

2
1
2 12

...


...


...

4
1
4 12

...


...
00
0
...
00
0
...
00
1
...
00
0
...
00
0
...
00
3
...
00
304
...
86

1
...
02

0
...
04

3
...
93

0

5
...
78

0
...
05

0
...
95

8
...
93

1
...
02

0
...
00

3
...
08

0

5
...
79

0
...
05

0
...
99

8
...
00

...


...
80

...


...
02

...


...
05

...


...
00

...


...
02

...


...
21
6
...


...


1
...
84

...


...
02
0
...


...


0
...
05

...


...
00
0
...


...


4
...
05

...


...
18

...


...
50
5
...


...


0
0

6
...
70

...


...
91
1
...


...


0
...
02

...


...
06
0
...


...


0
...
00

...


...
12
0
...


...


5
...
74

...


...
14
7
...


...


1
...
00

...


...
02
0
...


...


0
...
06

...


...
00
0
...


...


0
...
00

...


...
11
5
...


...


0
0

7
...
70

...


...
08
2
...


...


0
...
02

...


...
06
0
...


...


0
...
00

...


...
00
0
...


...


5
...
57

...


...
21

2
...
02

0
...
00

0
...
99


...


...


...



...


...


...



...


...
6 shows for selected durations the expected profit at the end of each
month per policy in force at the start of the tth month (Pr t ), the probability
that the policy is in force at the start of the month (given only that it was in
force at time 0) and the profit signature, t , which is the product of these
two elements
...

Hence
60

NPV =
k=0

k

k
12

(1 + r)− 12 = $302
...

2

12
...
6
...
2
...
00
304
...
0000
1
...
00
304
...
93

0
...
90

8
...
9910

7
...
08

0
...
97

8
...
9821

7
...
9777

...


...


...


8
...


...

8
...
42

...


...
9516
0
...


...


8
...


...

8
...
14

...


...


...


5
...
74

...


...
9235
0
...


...


5
...
37

...


...


...


5
...
14

...


...
9070
0
...


...


4
...
66

...


...


...


5
...
57

...


...
8908
0
...


...


4
...
96

...


...
99

0
...
24

t
0
1
12
2
12
3
12
4
12
5
12
6
12


...


...
In the
first example the guarantee had no effect at all on the calculations, and in the
second the effect was negligible
...
In practice, even though the policyholder’s fund may earn on average
a return of 9% or more, the return could be very volatile
...
We can explore the
sensitivity of the emerging profit to adverse scenarios by using stress testing
...
1 there is a GMMB – the final payout is guaranteed to be at
least the total amount invested, $50 000
...
The result is that
the GMMB still has no effect, and the NPV changes from $531
...
45
...
However, under the deterministic model there is no way to turn this
analysis into a price for the guarantee
...
The
insurer’s cash flows depend on the policyholder’s fund, and the policyholder’s
fund depends on market conditions
...

The investment risks in equity-linked insurance cannot be treated deterministically
...
In the next section we
develop the methodology introduced in this section to allow appropriately for
uncertainty
...
4 Stochastic profit testing
For traditional insurance policies we often assume that the demographic uncertainty dominates the investment uncertainty – which may be a reasonable
assumption if the underlying assets are invested in low risk fixed interest securities of appropriate duration
...
The uncertainty involved in equity-linked insurance is very
different
...
The uncertainty in the investment performance is a far more important
element, and it is not diversifiable
...

Using a deterministic profit test does not reflect the reality of the situation
adequately in most cases
...
The profit measure for an equity-linked
contract is modelled more appropriately as a random variable rather than a
single number
...

The good news is that we have done much of the work for stochastic profit
testing in the deterministic profit testing of the previous section
...
In this section we replace the deterministic investment scenarios with
stochastic scenarios
...
5, and used already for
this purpose with interest rates in Chapter 10
...
4 Stochastic profit testing

385

Using Monte Carlo simulation, we generate a large number of outcomes for
the investment return on the policyholder’s fund
...
The profit test proceeds exactly as described in the deterministic approach,
except that we repeat the test for each simulated investment return outcome, so
we generate a random sample of outcomes for the contract, which we can use
to determine the probability distribution for each profit measure for a contract
...
The policyholder may have a choice of funds available,
involving greater or lesser amounts of uncertainty
...
This assumption, which is very important in financial
modelling, can be expressed as follows
...
be a sequence of random
variables, where Rt represents the accumulation at time t of a unit amount
invested in an equity fund at time t −1, so that Rt −1 is the rate of interest earned
in the year
...
Note
that if Rt has a lognormal distribution with parameters µt and σt2 , then
log Rt ∼ N (µt , σt2 )
...

We demonstrate stochastic profit testing for equity-linked insurance by considering further the 10-year policy discussed in Example 12
...
In the discussion
of Example 12
...
3 we assumed a rate of return of 9% per year
on the policyholder’s fund
...
We
now assume that the accumulation factor for the policyholder’s fund over
the tth policy year is Rt , where the sequence {Rt }10 satisfies the indepent=1
dent lognormal assumption
...
074928 and σ 2 = 0
...
Note that the expected accumulation
factor each year is
E[Rt ] = eµ+σ

2 /2

= 1
...
3
...
7 shows the results of a single simulation of the investment returns
on the policyholder’s fund for the policy in Example 12
...

The values in column (2), labelled z1 ,
...
These values are converted to simulated values from the
specified lognormal distribution using rt = exp{0
...
15zt }, giving

386

Emerging costs for equity-linked insurance
Table 12
...
A single simulation of the profit test
...
95518
−2
...
23376
0
...
62022
0
...
22754
0
...
61893
−0
...
24384
0
...
89571
1
...
98206
1
...
89655
1
...
98225
1
...
31
60
...
07
144
...
57
230
...
33
298
...
38
375
...
94
8 010
...
61
19 159
...
89
30 494
...
16
39 490
...
89
49 717
...
00
306
...
03
107
...
70
192
...
69
249
...
17
332
...
71

−650
...
38
74
...
80
135
...
37
200
...
61
250
...
52
78
...
The values {rt }10 are
t=1
a single simulation of the random variables {Rt }10
...
09 used in the calculation of Table 12
...
The values in columns (4) and (5) are calculated in the
same way as those in columns (6) and (7) in Table 12
...
09
...

The values in column (6) are calculated in the same way as those in column (7)
in Table 12
...
This deduction was not needed in our calculations in Section 12
...
45, was greater than the GMMB
...
995 × (50 000 − 49 717
...
64
...
3
...
7 should be compared with the corresponding
values in Tables 12
...
3, respectively
...
4 Stochastic profit testing

387

per year, the NPV using this single simulation of the investment returns on the
policyholder’s fund is $232
...

To measure the effect of the uncertainty in rates of return, we generate a large
number, N , of sets of rates of return and for each set carry out a profit test as
above
...
, N
...
From this sample we can estimate the
mean, standard deviation and percentiles of this distribution
...

Let m and s be the estimates of the mean and standard deviation of NPV
...
96 √ , m + 1
...


It is important whenever reporting summary results from a stochastic simulation to give some measure of the variability of the results, such as a standard
deviation or a confidence interval
...
8
...
Let {NPV(i) }1 000
i=1
denote the simulated values for NPV arranged in ascending order
...
This would be
true for any value lying between NPV(500) and NPV(501) , and taking the
mid-point is a conventional approach
...

The results in Table 12
...
Under the deterministic analysis, the profit test showed no liability
for the guaranteed minimum maturity benefit, and the contract appeared to be
profitable overall – the net present value was $531
...
Under the stochastic
analysis, the GMMB plays a very important role
...

However, it does have a significant effect on the mean, which is considerably
lower than the median
...
Note also that an estimate of the probability

388

Emerging costs for equity-linked insurance
Table 12
...
Results from 1000 simulations
of the net present value
...
91
600
...
28, 417
...
82
498
...
51
87
897

that the net present value is negative, calculated using a risk discount rate of
15% per year, is
N − /N = 0
...

This profit test reveals what we are really doing with the deterministic test,
which is, approximately at least, projecting the median result
...


12
...
In fact, the expectation is usually taken over the future lifetime
uncertainty (given fixed values for the mortality rates), not the uncertainty
in investment returns or non-diversifiable mortality risk
...

The example studied in Section 12
...
The risk cannot be quantified deterministically
...

For this reason it is not advisable to use the equivalence premium principle
when there is significant non-diversifiable risk
...


12
...
8
...
We can extend this principle
to the pricing of equity-linked policies
...

The example studied throughout Section 12
...
82 and the expected net present
value, $380
...
6% of the acquisition costs, $650
...
However, we can investigate the effects of changing
the structure of the policy
...
4, Table 12
...
8 for four changes to the policy
structure
...

(1) Increasing the premium from $5 000 to $5 500, and hence increasing the
GMMB to $55 000 and the acquisition costs to $700
...
75% to 1
...

(3) Increasing the expense deductions from the premiums from 5% to 6% in
the first year and from 1% to 2% in subsequent years
...

In each of the four cases, the remaining features of the policy are as described
in Example 12
...

Increasing the premium, change (1), makes little difference in terms of our
chosen profit criterion
...

The premium for an equity-linked contract is not like a premium for a traditional contract, since most of it is unavailable to the insurer
...

Increasing the management charge, change (2), or the expense loadings,
change (3), does increase the expected net present value to the required level
but the probability of a loss is still greater than 5%
...
This is a demonstration of the
important principle that risk management begins with the design of the benefits
...
9
...

Change
(1)

(2)

(3)

(4)

E[NPV ]
433
...
60
594
...
33
SD[NPV ]
660
...
97
619
...
96
95% CI for E[NPV ] (392
...
51) (894
...
60) (556
...
09) (436
...
19)
5%-ile
−930
...
22
−724
...
29
Median of NPV
562
...
66
721
...
00
95%-ile
929
...
44
1 051
...
51
86
78
80
46
N−
897
882
894
939
N∗

An alternative, and in many ways more attractive, method of setting a premium for such a contract is to use modern financial mathematics to both price
the contract and reduce the risk of making a loss
...


12
...
6
...
This, like the use of the
equivalence principle to calculate a premium, is an example of the application of
the expected value principle
...
When the risk is non-diversifiable,
which is usually the case for equity-linked insurance, the expected value principle is inadequate both for pricing, as discussed in Section 12
...

Consider the further discussion of Example 12
...
4
...
82, in present value terms calculated using the
risk discount rate of 15% per year, on each policy issued
...

Calculating reserves for policies with significant non-diversifiable risk
requires a methodology that takes account of more than just the expected value

12
...
Such methodologies are called risk measures
...

There are two common risk measures used to calculate reserves for nondiversifiable risks: the quantile reserve and the conditional tail expectation
reserve
...
6
...
Suppose we have a random loss L
...

If L has a continuous distribution function, FL , the α-quantile reserve is Qα ,
where
Pr [L ≤ Qα ] = α,

(12
...


If FL is not continuous, so that L has a discrete or a mixed distribution, Qα
needs to be defined more carefully
...

To see how to apply this in practice, consider again Example 12
...
4
...
In other words, after paying the acquisition costs the insurer wishes
to set aside an amount of money, 0 V , so that, with probability 0
...

We need some notation
...
In practice, j will be a conservative rate of interest,
00
probably much lower than the risk discount rate
...
This is consistent with our notation
from Chapter 8 since our underlying model for the policy contains three states
– in force (which we denote by 0), lapsed and dead
...
4, and for each
of these we calculate Pr t,i , the profit emerging at time t, t = 1, 2,
...
For simulation i we calculate

392

Emerging costs for equity-linked insurance

the EPV of the future loss, say Li , as
10

Li = −
t=1

00
t−1 p55

Pr t,i

...
2)

Note that in the definition of Li we are considering future profits at times
t = 1, 2,
...

Then 0 V is set equal to the upper 95th percentile point of the empirical
distribution of L obtained from our simulations, provided that the upper 95th
percentile is positive, so that the reserve is positive
...

Calculations by the authors, with N = 1000 and j = 0
...
56
...
56 for each policy issued, it will be able to meet its
future liabilities with probability 0
...
These assumptions relate to
expenses,
lapse rates,
the survival model, and, in particular, the diversification of the mortality
risk,
the interest rate earned on the insurer’s fund,
the interest rate earned on the reserve,
the interest rate model for the policyholder’s fund,
the accuracy of our estimate of the upper 95th percentile point of the loss
distribution
...
In practice, the insurer
will review its reserves at regular intervals, possibly annually, during the term
of the policy and adjust the reserve if necessary
...
On the other hand, if the experience in the first year has
been favourable, the insurer may be able to reduce the reserve
...

In our example, the initial reserve, 0 V = $1259
...
This amount is expected to earn interest at a rate,
6%, considerably less than the insurer’s risk discount rate, 15%
...
6 Stochastic reserving

393

substantial reserves, which may not be needed when the policy matures, will
have a serious effect on the profitability of the policy
...
6
...
One problem with the quantile approach
is that it does not take into consideration what the loss will be if that 1 − α worst
case event actually occurs
...
The Conditional Tail Expectation
(or CTE) was developed to address some of the problems associated with the
quantile risk measure
...

As for the quantile reserve, the CTE is defined using some confidence level
α, where 0 ≤ α ≤ 1, which is typically 90%, 95% or 99% for reserving
...
The worst 1−α part of the loss distribution
is the part above the α-quantile, Qα
...


(12
...
If Qα falls in a probability mass, that is, if there is some > 0
such that Qα+ = Qα , then, if we consider only losses strictly greater than Qα ,
we are using less than the worst 1 − α of the distribution; if we consider losses
greater than or equal to Qα , we may be using more than the worst 1 − α of
the distribution
...
3) as follows
...
Then
CTEα =

(β − α)Qα + (1 − β ) E[L|L > Qα ]

...
4)

It is worth noting that, given that the CTEα is the mean loss given that the
loss lies above the VaR at level α (at least when the VaR does not lie in a
probability mass) then CTEα is always greater than or equal to Qα , and usually
strictly greater
...

Suppose the insurer wishes to set a CTE0
...
1 and throughout

394

Emerging costs for equity-linked insurance

Sections 12
...
5 and 12
...
2
...
2), with the rate of
interest j per year we expect to earn on reserves, exactly as in Section 12
...
2
...
6
...
06, the 50
worst losses, that is, the 50 highest values of Li , ranged in value from $1260
...
41, and the average of these 50 values is $3603
...
Hence we set the
CTE0
...
11
...

(1) The CTE reserve in our example has been estimated using simulations
based only on information available at the start of the policy
...
If the returns are good in the early years
of the contract, then it is possible that the probability that the guarantee will
cost anything reduces, and part of the reserves can be released back to the
insurer before the end of the term
...


12
...
4 Comments on reserving
The examples in this chapter illustrate an important general point
...
Several major life insurance companies have found their solvency at risk through issuing guarantees that were
not adequately understood at the policy design stage, and were not adequately
reserved for thereafter
...
This is a
passive approach to managing the risk and is usually not the best way to manage
solvency or profitability
...

The active approach to risk mitigation and management comes from option
pricing theory
...
There is an extensive

12
...
In
Chapter 13 we review the science of option risk management, at an introductory
level, and in Chapter 14 we apply the science to equity-linked insurance
...
7 Notes and further reading
A practical feature of equity-linked contracts in the UK which complicates the
analysis a little is capital and accumulation units
...
This contract design has been developed to
defray the insurer’s acquisition costs at an early stage
...
We would
generally simulate values for the interest earned on assets, and we might also
simulate expenses and withdrawal rates
...
2 demonstrates this
...
The major risk for such insurance is misestimation of the underlying mortality rates
...
It is therefore useful with term insurance to treat the force of mortality as a stochastic
input
...
It is
intuitive, easy to understand and to apply with simulation output
...
The CTE is used
for stochastic reserving and solvency testing for Canadian and US equity-linked
life insurance
...
In particular, she gives full definitions of quantile and CTE reserves, and shows how to simulate the emerging
costs and calculate profit measures when stochastic reserving is used
...
8 Exercises
Exercise 12
...
The policyholder deposits $100, and the insurer deducts 3% for expenses and profit
...
5% of the premium
...
At the end
of one year the policyholder receives the fund proceeds; if the proceeds are less
than the initial $100 investment the insurer pays the difference
...
09, 0
...


396

Emerging costs for equity-linked insurance

Let F1 denote the fund value at the year end
...
e
...
05 − (3 − 2
...

(a)
(b)
(c)
(d)
(e)

Calculate Pr[F1 < 100]
...

Show that the fifth percentile of the distribution of R is 0
...

Hence, or otherwise, calculate Q0
...

Let f be the probability density function of a lognormal random variable
with parameters µ and σ 2
...
95 (L0 )
...
Compare the results
of your simulations with the accurate values calculated in (a)–(e)
...
2 A life insurer issues a special five-year endowment insurance
policy to a life aged 50
...
The maturity
benefit on survival to age 55 is $20 000
...

Reserves are required at integer durations for each policy in force, are
independent of the premium, and are as follows:
0V

= 0, 1 V = 3 000, 2 V = 6 500, 3 V = 10 500, 4 V = 15 000, 5 V = 0
...
The profit objective is that the
EPV of future profit must be 1/3 of the gross annual premium, using a risk
discount rate of 10% per year
...
8 Exercises

397

(a) Calculate the annual premium
...
07, 0
...

(i) Estimate the probability that the policy will make a loss in the final
year, and calculate a 95% confidence interval for this probability
...

Compare this with the 95% confidence interval for the probability
determined from your simulations
...

Exercise 12
...
The policyholder is aged 60 and pays an annual premium
of $100
...

The death benefit is the greater of $500 and the amount of the fund, payable
at the end of the year of death
...

Mortality rates assumed are: q60 = 0
...
0028, q62 = 0
...
0037 and q64 = 0
...
There are no lapses
...

(b) Assume that the insurer’s fund earns interest of 6% per year
...
Calculate the profit signature for the contract assuming that no
reserves are held
...

(d) Explain how you would estimate the 99% quantile reserve and the 99%
CTE reserve for this contract
...
Immediately before the final premium
payment the policyholder’s fund is $485
...
09 and σ 2 = 0
...


398

Emerging costs for equity-linked insurance
Let L4 represent the present value of future loss random variable at time 4,
using an effective rate of interest of 6% per year
...

(ii) Calculate Q99% (L4 ) assuming that insurer’s funds earn 6% per year as
before
...
4 An insurer used 1 000 simulations to estimate the present value
of future loss distribution for a segregated fund contract
...
10 shows the
largest 100 simulated values of L0
...
10
...

6
...
865
7
...
416
9
...
284
11
...
322
15
...
357

(a)
(b)
(c)
(d)

6
...
918
7
...
508
9
...
814
11
...
327
15
...
774

6
...
949
7
...
583
9
...
998
11
...
404
15
...
998

6
...
042
7
...
739
9
...
170
12
...
415
15
...
200

6
...
106
7
...
895
9
...
287
12
...
625
16
...
944

6
...
152
7
...
920
9
...
314
12
...
733
16
...
957

6
...
337
8
...
981
10
...
392
13
...
925
17
...
309

6
...
379
8
...
183
10
...
546
13
...
076
17
...
226

6
...
413
8
...
335
10
...
558
13
...
091
17
...
709

6
...
430
8
...
455
10
...
647
14
...
343
17
...
140

Estimate Pr[L0 > 10]
...

Estimate Q0
...

Estimate CTE0
...


Exercise 12
...

In the first policy year, 25% of the premium is allocated to the policyholder’s
fund, followed by 102
...
The units are
subject to a bid-offer spread of 5% and an annual management charge of 1% of
the bid value of units is deducted at the end of each policy year
...

If the policyholder dies during the term of the policy, the death benefit of
$3000 or the bid value of the units, whichever is higher, is payable at the end
of the policy year of death
...
8 Exercises

399

the end of each policy year
...
On maturity, 110% of the bid value of the
units is payable
...
5% per year
5
...
5% of the second and subsequent years’
premiums
8
...


(a) Calculate the profit margin for the policy on the assumption that the
company does not hold reserves
...

(ii) Calculate the effect on the profit margin of a reserve requirement of
$400 at the start of the second, third and fourth years, and $375 at the
start of the fifth year
...

(c) An actuary has suggested the profit test should be stochastic, and has generated a set of random accumulation factors for the policyholder’s funds
...
07 and σ 2 = 0
...
Using
the random standard normal deviates given below, conduct the profit test
using your simulated accumulation factors, and hence calculate the profit
margin, allowing for the reserves as in (b):
−0
...
09365,

0
...
67706,

Answers to selected exercises
12
...
37040
(b) $107
...
10300
...
2

12
...
4

12
...
54
(e) (i) $3
...
83
(a) $3739
...
528, (0
...
572)
(ii) 0
...
488, (0
...
532)
(a) (3
...
14, 9
...
88, 16
...
27, 1
...
77, 4
...
76)
(e) (i) 0
...
50
(a) 0
...
036, 0
...
30
(d) $21
...
56%
(b) (ii) Reduces to 0
...
43%

13
Option pricing

13
...

First, we discuss the no arbitrage assumption, which is the foundation for all
modern financial mathematics
...

We discuss the Black–Scholes–Merton option pricing formula, and, in particular, demonstrate how it may be used both for pricing and risk management
...
2 Introduction
In Section 12
...
A methodology for managing this
risk, stochastic pricing and reserving, was set out in Sections 12
...
6
...
At the end of the contract, the capital may not be needed, but having to maintain large reserves is
expensive for the insurer
...

Since the non-diversifiable risks in equity-linked contracts and some pension
plans typically arise from financial guarantees on maturity or death, and since
these guarantees are very similar to the guarantees in exchange traded financial
options, we can use the Black–Scholes–Merton theory of option pricing to
price and actively manage these risks
...

There are several reasons why it is very helpful for an insurance company
to understand option pricing and financial engineering techniques
...
Also, by understanding financial engineering
methods an insurer can make better risk management decisions
...

There are many different types of financial guarantees in insurance contracts
...


13
...
The assumption is more colloquially known as the ‘no
free lunch’ assumption, and states quite simply that you cannot get something
for nothing
...

If we assume that there are no arbitrage opportunities in a market, then
it follows that any two securities or combinations of securities that give
exactly the same payments must have the same price
...
If
A = B, then an investor could buy the asset with the lower price and sell the
more expensive one
...

The no arbitrage assumption is very simple and very powerful
...

Replication is a crucial part of the framework
...

For example, suppose an insurer incurs a liability, under which it must deliver
the price of one share in Superior Life Insurance Company in one year’s time,
and the insurer wishes to value this liability
...
4 Options

403

the current value is $400 and the insurer assumes the share price in one year’s
time will follow a lognormal distribution, with parameters µ = 6
...
162
...
25
...
45
...
45 in a bond, which in one year will pay $438
...
It will
almost surely be either too much or not enough
...

In this simple case, that means holding a replicating portfolio of one share in
Superior Life Insurance Company
...
In one year,
the portfolio is exactly sufficient to pay the creditor, whatever the outcome
...
It cannot be worth $413
...
45, and replicate it for $400, giving a risk free profit (or
arbitrage) of $13
...

Replication does not require a model; we have eliminated the uncertainty in
the payoff, and we implicitly have a risk management strategy – buy the share
and hold it until the liability falls due
...

In practice, in most securities markets, arbitrage opportunities arise from
time to time and are very quickly eliminated as investors spot them and trade
on them
...


13
...
In this section we introduce the language of
options and explain how some option contracts operate
...

The holder of a European call option on a stock has the right (but not the
obligation) to buy an agreed quantity of that stock at a fixed price, known as
the strike price, at a fixed date, known as the expiry or maturity date of the
contract
...
The holder of a European call
option on this stock with strike price K and maturity date T would exercise the
option only if ST > K, in which case the option is worth ST − K to the option
holder at the maturity date
...
Thus, the payoff at time T under the option is
(ST − K)+ = max(ST − K, 0)
...
The holder of a European put option would
exercise the option only if ST < K, since the holder of the option could sell the
stock at time T for K then buy the stock at the lower price of ST in the market
and hence make a profit of K − ST
...
The option would not be exercised at the
maturity date in the case when ST > K, since the option holder would then be
selling stock at a lower price than could be obtained by selling it in the market
...

In making all of the above statements, we are assuming that people act rationally
when they exercise options
...
For example, a call option guarantees that the holder of the option
pays no more than the strike price to buy the underlying stock at the maturity
date
...
The names
‘European’ and ‘American’ are historical conventions, and do not signify where
these options are sold – both European and American options are sold worldwide
...
Many of the
options embedded in life insurance contracts are European-style
...
When K = St , or
even when K is close to St , we say the option is ‘at-the-money’
...


13
...
5 The binomial option pricing model
13
...
1 Assumptions
Throughout Section 13
...

Although the binomial model is simple, and not very realistic, it is useful
because the techniques we describe below carry through to more complicated
models for a stock price process
...

• There is a frictionless financial market in which there exists a risk free asset

•

•

•
•

(such as a zero-coupon bond) and a risky asset, which we assume here to be
a stock
...

The financial market is modelled in discrete time
...
Changes in asset prices and the exercise date for an option
can occur only at these same dates
...
This means there are
just two possible states one period later if we start at a given time and price
...
These trades do not impact the
prices
...
This is achieved by selling an asset they do not own, so the investor
‘owes’ the asset to the lender
...


We start by considering the pricing of an option over a single time period
...


13
...
2 Pricing over a single time period
To illustrate ideas numerically, consider a stock whose current price is $100
and whose price at time t = 1 will be either $105 or $90
...
03 per unit of time
...
To see this, suppose 100er > 105
...
At

406

Option pricing

time t = 1 the investor would then have $100er , part of which would be used
to buy one unit of stock in the market to wipe out the negative holding, leaving
a profit of either $(100er − 105) or $(100er − 90), both of which are positive
...

Now, consider a put option on this stock which matures at time t = 1 with
a strike price of K = $100
...
As we are assuming that
there are no trading costs in buying and selling stocks, the option holder could
use the sale price of $100 to buy stock at $90 at time t = 1 and make a profit
of $10
...
However, if the stock price falls, the seller of the put option
has a liability of $10
...
This means that we
look for a portfolio of assets at time t = 0 that will exactly match the payoff
under the put option at time t = 1
...
Then at time t = 1, the portfolio is worth
aer + 105b
if the stock price goes up, and is worth
aer + 90b
if the stock price goes down
...
To achieve this we
require that
aer + 105b = 0,
aer + 90b = 10
...
9312
...
9312 of the risk free asset and a short holding
of −2/3 units of stock exactly matches the payoff under the put option at time
t = 1, regardless of the stock price at time t = 1
...


13
...
26
...
We note here that under the no arbitrage assumption, we must have
dS0 < S0 er < uS0
...

The hedge portfolio consists of $a in the risk free asset and $bS0 in stock
...

S0 (u − d )

The option price at time 0 is a + bS0 , the value of the hedge portfolio, which
we can write as
e−r q (K − dS0 )

(13
...

u−d

(13
...

An interesting feature of expression (13
...
1)
could be thought of as the discounted value of the expected payoff under the

408

Option pricing

option
...
If q were the probability of a downward movement in the
stock price, then qe−r (K − dS0 ) would be the EPV of the option payoff
...
In
fact, nowhere in our determination of the price of the put option have we needed
to know the probabilities of the stock price moving up or down
...
It is important to remember
though that we have not used a probabilistic argument here, we have used
instead a replication argument
...
The following example demonstrates this for a general payoff
...
1 Consider an option over one time period which has a payoff Cu
if the stock price at the end of the period is uS0 , and has a payoff Cd if the stock
price at the end of the period is dS0
...
2)
...
1 We construct the replicating portfolio which consists of $a in the
risk-free asset and $bS0 in stock so that
aer + buS0 = Cu ,
aer + bdS0 = Cd ,
giving
b=

Cu − Cd
(u − d )S0

and
a = e−r Cu − u
= e−r

Cu − Cd
u−d

u
d
Cd −
Cu
...
5 The binomial option pricing model

409

Hence the option price is
a + bS0 = e−r
= Cu

u
C u − Cd
d
Cd −
Cu +
u−d
u−d
u−d
1 − de−r
u−d

= e−r Cu

+ Cd

er − d
u−d

ue−r − 1
u−d

+ Cd

u − er
u−d

= e−r (Cu (1 − q) + Cd q)
...
Note that q has not
been defined as the probability that the stock price is equal to dS0 at time t = 1,
and, in general, will not be equal to this probability
...

Under the binomial framework that we use here, there is some real probability that the stock price moves down or up
...
The true distribution is referred to by different names, the physical measure, the real world measure, the subjective measure or nature’s measure
...
The artificial
distribution that arises in our pricing of options is called the risk neutral measure, and in the language of probability theory is called the Q-measure
...
In what follows, we use EQ to denote expectation with respect to the
Q-measure
...
We know that in the real world investors
require extra expected return for extra risk
...

Example 13
...

Solution 13
...


410

Option pricing

Then
EQ [e−r S1 ] = e−r ((1 − q)uS0 + qdS0 )
= e−r

er − d
u−d

uS0 +

u − er
u−d

= S0
...
2 shows that under the risk neutral measure, the stock
price at time t = 0 is the EPV under the Q-measure of the stock price at time
t = 1
...
Under
the P-measure we expect the accumulation factor to exceed er on average, as a
reward for the extra risk
...
5
...
We now extend this
idea to pricing an option over two time periods
...

Let us now assume that in each of our two time periods, the stock price can
either increase by 5% of its value at the start of the time period, or decrease by
10% of its value
...

As before, we consider a put option with strike price $100, but this time the
exercise date is at the end of the second time period
...
1,
the stock price at time t = 2 is $110
...
50 if the stock price moves up once and down once, and $81
...
This means that the put option
will be exercised if at time t = 2 the stock price is $94
...
00
...
Suppose
first that at time t = 1 the stock price is $105
...
Suppose
this portfolio contains $au of the risk free asset and bu units of stock, so that the
replicating portfolio is worth $(au + 105bu )
...
50 if the stock
price moves down in the second time period since the option will be exercised

13
...
25

¨
¨
¨
¨ 105 rr
r
¨¨

¨
¨¨
r
rr

S0 = 100

rr
r

rr
r
¨
¨¨

¨ Pu
¨
¨
¨

¨¨
r
rr

rr
r

Stock price

¨

90

¨¨
rr

rr
rr

P0

94
...
00

0

5
...
1 Two-period binomial model
...
The equations that determine au and bu are
au er + 110
...
5bu = 5
...
3492 and au = 37
...
This shows that the replicating
portfolio at time t = 1, if the stock price at that time is 105, has value Pu =
$0
...

Similarly, if at time t = 1 the stock price is $90, we can find the replicating
portfolio whose value at time t = 1 is $(ad + 90bd ), where the equations that

412

Option pricing

determine ad and bd are
ad er + 94
...
50,
ad er + 81bd = 19,
since if the stock price rises to $94
...
50,
and if the stock price falls to $81, the payoff under the option is $19
...
0446
...
04
...
At this time point we want to find a
portfolio that replicates the possible amounts required at time t = 1, namely
$0
...
04 if it
goes down to $90
...
70,
aer + 90b = 7
...
4233 and a = 43
...
The replicating portfolio has value P0
at time t = 0, where
P0 = a + 100b = $1
...

There are two important points to note about the above analysis
...
The second point is that the replicating
portfolio is self-financing
...
80 in the risk free asset
and a short holding of −0
...
The replicating portfolio at time t = 1 then matches
exactly the option payoff at time t = 2
...

What we have done in this process is an example of dynamic hedging
...
This process
works for any number of steps, but if there is a large number of time periods it

13
...
However, if all we want to work out is the option price,
the result we saw for a single time period, that the option price is the discounted
value of the expected payoff at the expiry date under the Q-measure, also holds
when we are dealing with multiple time periods
...
05, d = 0
...
03
...
2), the probability of a downward movement in the stock price under the
Q-measure is
q=

1
...
03
= 0
...
05 − 0
...
5 × 2(1 − q)q = $1
...

This gives the option price as
1
...
06 = $1
...


13
...
4 Summary of the binomial model option pricing technique
• We use the principle of replication; we construct a portfolio that replicates the

option’s payoff at maturity
...

• We use dynamic hedging – replication requires us to rebalance the portfolio
at each time step according to the movement in the stock price in the previous
time step
...
However, there are important links
between the real world (P-measure) model and the risk neutral (Q-measure)
model
...
From this we showed that in
the risk neutral world the stock price after a single time period also has a
two-point distribution with the same possible values, uS0 and dS0 , but the
probabilities of moving up or down are not linked to those of the real world
model
...
This
artificial distribution is called the risk neutral measure because the mean accumulation of a unit of stock under this distribution is exactly the accumulated
value of a unit investment in the risk free asset
...

The binomial model option pricing framework is clearly not very realistic, but we can make it more flexible by increasing the number of steps in
a unit of time, as discussed below
...


13
...
6
...

• The market consists of zero-coupon bonds (the risk free asset) and stocks
•

•

•
•
•

•

(the risky asset)
...
This assumption simplifies the
presentation but can easily be relaxed if necessary
...
In the two-period binomial example we showed
how the replicating portfolio was rebalanced (costlessly) after the first time
unit
...

There are no transactions costs associated with trading the stocks and bonds
...

Stocks and bonds can be bought or sold in any quantities, positive or negative;
we are not restricted to integer units of stock, for example
...

In the real world, the stock price, denoted St at time t, follows a continuous time lognormal process with some parameters µ and σ 2
...


Clearly these are not realistic assumptions
...
We also know
that yield curves are rarely flat
...
6 The Black–Scholes–Merton model

415

model works remarkably well, both for determining the price of options
and for determining risk management strategies
...

A lognormal stochastic process with parameters µ and σ has the following
characteristics
...
3)

which implies that
log

St+τ
∼P N (µτ , σ 2 τ )
...
Our choice of parameters µ and σ 2 here
uses the standard statistical parameterizations
...
It is important to check what µ represents
when it is used as a parameter of a lognormal distribution
...
The
parameter µ is the mean log-return over a unit of time, and σ is the standard
deviation of the log-return over a unit of time
...
Some information on the lognormal distribution
is given in Appendix A
...
(This is the same as in the binomial model, where
the stock price movement in any time interval is independent of the movement
in any other time interval
...

The lognormal process assumed in the Black–Scholes–Merton model can be
derived as the continuous time limit, as the number of steps increases, of
the binomial model of the previous sections
...
7
for interested readers
...
6
...

• There is a unique risk neutral distribution, or Q-measure, for the stock price

process, under which the stock price process, {St }t≥0 , is a lognormal process
with parameters r − σ 2 /2 and σ 2
...
4)

Q

where Et denotes expectation using the risk neutral (or Q) measure, using all
the information available up to time t
...

Important points to note about this result are:
• Over any fixed time interval, say (t, t + τ ) where τ > 0, the stock price

accumulation factor, St+τ /St , has a lognormal distribution in the risk neutral
world with parameters (r − σ 2 /2)τ and σ 2 τ , so that
St+τ
∼Q LN ((r − σ 2 /2)τ , σ 2 τ ),
St

(13
...

St

We have added the subscript Q as a reminder that these statements refer to
the risk neutral, or Q-measure model
...
This follows from the previous point
since
Et [St+τ /St ] = exp((r − σ 2 /2)τ + τ σ 2 /2) = er τ
...
2 for the binomial model
...

•

13
...

• The volatility parameter, σ , is the same for both measures
...
5, the corresponding risk
neutral model is also binomial; the limit as the number of steps increases in
the (risk neutral) binomial model is then also a lognormal process
...
Note that the parameter µ,
the mean log-return per unit time for the P-measure, does not appear in the
specification of the Q-measure
...

• Formula (13
...
1) and the two-period binomial
model (Section 13
...
3)
...

• A mathematical derivation of the Q-measure and of formula (13
...
Interested readers should consult the references in
Section 13
...

Now consider the particular case of a European call option with strike price K
...


(13
...

Q

Now note that, under the Q-measure,
ST /St ∼ LN ((r − σ 2 /2)(T − t), σ 2 (T − t))
...

St

(13
...
Since the mean of this random variable
is
∞

x f (x)dx = exp{µ + σ 2 /2},

0

we have
∞

x f (x)dx = exp{µ + σ 2 /2} 1 −

a

= exp{µ + σ 2 /2}

log a − µ − σ 2
σ

− log a + µ + σ 2
σ


...
7) for c(t) gives
c(t) = e−r(T −t) St er(T −t)
− e−r(T −t) K 1 −
= St

− log(K/St ) + (r − σ 2 /2)(T − t) + σ 2 (T − t)
√
σ T −t
log(K/St ) − (r − σ 2 /2)(T − t)
√
σ T −t

log(St /K) + (r + σ 2 /2)(T − t)
√
σ T −t

− e−r(T −t) K

log(St /K) + (r − σ 2 /2)(T − t)
,
√
σ T −t

which we usually write as
c(t) = St

(d1 (t)) − Ke−r(T −t)

(d2 (t)) ,

(13
...

(13
...
8) and r appears only in the second term, this formula suggests that the
replicating portfolio at time t for the call option comprises

13
...


Indeed, this is the self-financing replicating portfolio required at time t
...

If the strike price is very small relative to the stock price we see that (d1 (t))
tends to one and (d2 (t)) tends to zero
...

For a European put option, with strike price K, the option price at time t is
p(t), where
Q

p(t) = Et

e−r(T −t) (K − ST )+ ,

which, after working through the integration, becomes
p(t) = Ke−r(T −t)

(−d2 (t)) − St

(−d1 (t)) ,

(13
...

The replicating portfolio for the put option comprises
•

(−d2 (t)) units of zero-coupon bonds with face value K, maturing at time
T , with value at time t
Ke−r(T −t)

plus
• a short holding of

(−d2 (t)) ,

(−d1 (t)) units of the stock, with total value at time t
−St

(−d1 (t))
...


You are asked to prove the first of these formulae as Exercise 13
...
These two
formulae show that, for these options, the replicating portfolio has a portion

420

Option pricing

St d v(t)/dSt invested in the stock, and hence a portion v(t) − St d v(t)/dSt
invested in the bond, where v(t) is the value of the option at time t
...
The quantity d v(t)/dSt is known as the delta of the option
at time t
...

Example 13
...

(a) Use formulae (13
...
10) to show that, using the Black–Scholes–
Merton framework,
c(t) + K e−r(T −t) = p(t) + St
...
11)

(b) Use a no-arbitrage argument to show that formula (13
...

Solution 13
...
8) and (13
...
11)
...
The first comprises the call option
plus a zero-coupon bond with face value K maturing at time T ; the second
comprises the put option plus one unit of the stock
...
At time T the first portfolio will be worth K if ST ≤ K, since
the call option will then be worthless and the bond will pay K, and it will be
worth ST if ST > K, since then the call option would be exercised and the
proceeds from the bond would be used to purchase one unit of stock
...
This will be worth K if ST ≤ K,
since the put option would be exercised and the stock would be sold at
the exercise price, K, and it will be worth ST if ST > K, since the put
option will then be worthless and the stock will be worth ST
...
6 The Black–Scholes–Merton model

421

portfolios have the same payoff at time T under all circumstances, they
must have the same value at all other times, in particular at time t
...
11)
...

2
Example 13
...
The insurer keeps 3% of
the premium to cover all expenses, then invests the remainder in a mutual fund
...
In the UK similar products are called unit trusts or investment
trusts
...
085 and σ 2 = 0
...
The mutual
fund does not pay out dividends; any dividends received from the underlying
portfolio are reinvested
...
The insurer guarantees that the payout at the maturity date will
not be less than the original $1000 investment
...

(b) Calculate the real world probability that the guarantee applies at the
maturity date
...

Solution 13
...
This will accumulate over the two years of the contract to
some random amount, S2 , say
...
In other words, the
policyholder has the right at the maturity date to receive a price of $1000
from the insurer for the mutual fund stocks
...

If the mutual fund stocks are worth more than $1000, then the policyholder
just takes the proceeds and the insurer has no further liability
...
So the price of the put option at inception is
p(0) = Ke−rT

(−d2 (0)) − S0

(−d1 (0))

422

Option pricing
where
log(S0 /K) + (r + σ 2 /2)T
= 0
...
3493,
√
σ T
√
d2 (0) = d1 (0) − σ T = 0
...
4584,

d1 (0) =

giving
p(0) = 414
...
794 = $75
...

So the 3% expense charge, $30, is insufficient to fund the guarantee cost
...
599% of the initial investment
...
599% as the expense loading, the price of
the guarantee would be even greater, as we would invest less money in the
mutual fund at inception whilst keeping the same strike price
...
This means that
log(S2 /S0 ) ∼ N (2µ, 2σ 2 ),
which implies that
log S2 ∼ N (log S0 + 2µ, 2σ 2 )
...
311
...
311
...
Figure 13
...
Where this line crosses the line x = y (shown as a dotted
line) we have a solution
...
72% (i
...
the expense loading is around $107
...
Alternatively, Excel
Solver gives the solution that an expense loading of 10
...

2

13
...
2 Expense loading plotted against option cost for Example 13
...


Finding the price is only the first step in the process
...
In part (a)
of Example 13
...
79
in two-year zero-coupon bonds and short selling $338
...
99
...

In practice, continuous rebalancing is impossible
...

Example 13
...
4 above, where an insurer has issued
a guarantee which matures in two years
...

In Table 13
...

Assume, as in Example 13
...
Determine the cash flows arising assuming that the insurer
(a) invests the entire option cost in the risk-free asset,
(b) invests the entire option cost in the mutual fund asset,

424

Option pricing
Table 13
...
Table of mutual fund
stock prices for Example 13
...

Time, t
(months)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

St
$
970
...
07
959
...
93
921
...
25
1045
...
59
945
...
77
932
...
11
906
...
86
831
...
99
785
...
36
707
...
87
715
...
74
675
...
71
766
...
79 to zerocoupon bonds and −$338
...

Solution 13
...
66) = $233
...


13
...
99e2r = $83
...
This leaves a shortfall at maturity of
$(233
...
98) = $149
...

(b) If the option cost is invested in the mutual fund asset, it will accumulate to
75
...
66/970) = $60
...
28
...
41;
• the stock part of the hedge accumulates in proportion to the mutual
fund share price, with final value
−338
...
66/970) = −$267
...
41 − 267
...
64, which means that the insurer is liable
for an additional cash flow at maturity of $42
...
In this case the
total cost of the guarantee is the initial hedge cost of $75
...
70
...
79er = $436
...

Mutual fund: −338
...
11/970) = −$316
...

So the value of the hedge portfolio immediately before rebalancing
is $119
...

The rebalanced hedge is found from formula (13
...
26 − 504
...
70
...
57 − 98
...
87 back
to the insurer, as the value of the initial hedge more than pays for the
rebalanced hedge
...

Bonds: 603
...
19
...
56 × (766
...
11) = −$426
...

Total hedge portfolio value: $207
...

We need $233
...
06
...

Time Value of hedge
Cost of
Final
Net cash flow
(years) brought forward new hedge guarantee cost
$
0
1
2

0
119
...
28

75
...
70
–

–
–
233
...
99
−20
...
06

(iii) Here, we repeat the exercise in (b) but we now accumulate and rebalance each month
...
2
...
In
the final month, the total reflects the cost of the guarantee payoff
...

We see how the rebalancing frequency affects the cash flows; with
a monthly rebalancing frequency, all the cash flows required are relatively small, after the initial hedge cost
...
However, the amounts are small, demonstrating
that if the insurer follows this rebalancing strategy, there is little additional cost involved after the initial hedge cost, even though the final
guarantee payout is substantial
...
26 in this case
...

2
This example demonstrates that in this case, where the option matures in-themoney, the dynamic hedge is remarkably efficient at converging to the payoff
with only small adjustments required each month
...
In practice,
many hedge portfolios are rebalanced daily or even several times a day
...
The
hedge is a form of insurance, and, as with all insurance, there is a cost even
when there is no claim
...
7 Notes and further reading

427

Table 13
...
Cash flow calculations for Example 13
...

Time
(months)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

New hedge portfolio
Bonds

Mutual Fund

Total

Old hedge
brought forward

Net cash flow
$

414
...
09
437
...
69
505
...
15
332
...
22
492
...
18
531
...
60
603
...
54
776
...
22
882
...
97
965
...
54
981
...
44
991
...
84

−338
...
17
−358
...
83
−411
...
59
−283
...
43
−411
...
25
−445
...
81
−504
...
58
−628
...
33
−693
...
74
−697
...
77
−712
...
59
−675
...
71

75
...
92
78
...
86
94
...
56
48
...
79
81
...
94
85
...
78
98
...
96
148
...
88
188
...
24
268
...
76
268
...
84
315
...
13
233
...
00
84
...
99
87
...
93
75
...
88
60
...
56
97
...
56
79
...
18
146
...
52
176
...
62
246
...
58
266
...
57
296
...
90
296
...
34

75
...
76
−2
...
88
−1
...
96
1
...
33
0
...
48
−2
...
76
−0
...
50
−2
...
55
−1
...
03
−0
...
27
−0
...
01
0
...
00
0
...
7 Notes and further reading
This chapter offers a very brief introduction to an important and exciting area
...
For a description of the history of options and option pricing,
see Boyle and Boyle (2001)
...
The original proof is given in
Cox et al
...

We assumed from Section 13
...
1 onwards that the stock did not pay any dividends
...


428

Option pricing
13
...
1 Let c(t) denote the price of a call option on a non-dividend paying
stock, using the Black–Scholes equation (13
...
Show that
dc(t)
=
dSt

(d1 (t))
...

Exercise 13
...
5,
E Q [Sn ] = S0 ern
...

Exercise 13
...
1St
0
...


Zero-coupon bonds are available for all integer durations, with a risk-free rate
of interest of 6% per time period compounded continuously
...

(a) Find the price and the replicating portfolio for the option assuming it is
issued at t = 0 and matures at t = 1
...
Find the
price and the replicating portfolio at t = 0 and at t = 1
...
4 Consider a two-period binomial model for a non-dividend paying
security with price St at time t, where S0 = 1
...
2St
0
...


At time t = 2 option A pays $3 if the stock price has risen twice, $2 if it has
risen once and fallen once and $1 if it has fallen twice
...


13
...
879% per period
...
5
...

(b) Calculate the price of option A and show that it is different to the price of
option B
...

Exercise 13
...
The price of a six-month
European call option with a strike price of $420 is $41
...

Assume the Black–Scholes pricing formula applies
...
State the assumptions you make in the calculation
...

(c) Calculate the delta of the option
...

Exercise 13
...
25 St or 0
...
The
risk-free rate of interest is 10% per time unit effective
...

The value of S0 is 100
...
25,
D2 = 2 if S2 = 100,
⎩
0 if S2 = 64
...

(c) Derive the corresponding hedging strategy, i
...
the combination of the
underlying security and the risk-free asset required to hedge an investment
in the derivative security
...

Exercise 13
...
00
...
$1 held in cash between times t and t + 1 receives interest to become
$1
...
The stock price after t time units is denoted by St
...

(b) Calculate the price (at time t = 0) of a derivative contract written on the
stock with expiry date t = 2 which pays $10
...
00 (and otherwise pays 0)
...
3 (a)
(b)
13
...
5 (a)
(b)
(c)
(d)
13
...
7 (a)
(b)

$15
...
61
$1
...
633,
Option B: $1
...
55
38
...
42%
Long 5 342
...
1019
0
...
4667 (decrease)
$4
...
1 Summary
In this chapter we describe financial options embedded in insurance contracts,
focusing in particular on the most straightforward options which appear as
guaranteed minimum death and maturity options in equity-linked life insurance
policies effected by a single premium
...


14
...
Nevertheless, these guarantees are not
negligible – failure to manage the risk from apparently innocuous guarantees
has led to significant financial problems for some insurers
...
However,
in the case when the equity-linked contract incorporates financial guarantees
that are essentially the same as the financial options discussed in Chapter 13,
we can use the more sophisticated techniques of Chapter 13 to price and manage the risks associated with the guarantees
...

To show how the guarantees can be viewed as options, recall Example 12
...
Consider, for
now, the GMMB only
...
The fund
value is variable, moving up and down with the underlying assets
...

Let Ft denote the value of the policyholder’s fund at time t
...
2, the benefit for policies still in force at the maturity date, say
at time n, (the term is n = 5 years in Example 12
...
As the policyholder’s fund contributes the
amount Fn , the insurer’s additional liability is h(n), where
h(n) = max(P − Fn , 0)
...

Recognizing that the fund value process {Ft }t≥0 may be considered analogous to
a stock price process, and that P is a fixed, known amount, the guarantee payoff
h(n) is identical to the payoff under an n-year European put option with strike
price $P, as described in Section 13
...
So, while in Chapter 12 we modelled
this contract with cash flow projection, we have a more appropriate technique
for pricing and valuation from Chapter 13, using the Black–Scholes–Merton
framework
...

There are a few differences between the options embedded in equity-linked
contracts and standard options traded in markets
...

(1) The options embedded in equity-linked contracts have random terms to
maturity
...
The GMDB expires on the
death of the policyholder, if that occurs during the term of the contract
...
The underlying risky asset
process represents the value of a traded stock or stock index
...


14
...

Throughout this chapter we consider equity-linked contracts paid for by a
single premium, P, which, after the deduction of any initial charges, is invested
in the policyholder’s fund
...
We make all the assumptions in Section 13
...
1 relating to the
Black–Scholes–Merton framework
...


14
...
3
...

Suppose a GMMB under a single premium contract guarantees that the payout
at maturity, n years after the issue date of the contract, will be at least equal to
the single premium, P
...
This payoff is conditional on the policy remaining in
force until the maturity date
...
For
simplicity here we ignore surrenders and assume all policyholders are aged x
at the commencement of their policies, and are all subject to the same survival
model
...

Consider the situation at the issue of the contract
...
If the policyholder does survive n
years, the GMMB does apply at time n, and we know that the amount required
at the issue of the contract to fund this guarantee is
E0 e−rn (P − Fn )+
...

Q

434

Embedded options

Note that we are adopting a mixture of two different methodologies here
...

Suppose that the total initial expenses are a proportion e of the single premium, and the management charge is a proportion m of the policyholder’s fund,
deducted at the start of each year after the first
...

S0

Since we are interested in the relative increase in St , we can assume S0 = 1
without any loss of generality
...
) Then
Fn = P(1 − e)(1 − m)n−1 Sn
...
We can now apply
formula (13
...
Then the price at the issue date of a GMMB, guaranteeing a
return of at least the premium P, is
π(0) = P n px ξ ξ −1 e−rn (−d2 (0)) −

(−d1 (0))

= P n px e−rn (−d2 (0)) − ξ (−d1 (0))
where
d1 (0) =

log(ξ ) + (r + σ 2 /2)n
√
σ n

and
√
d2 (0) = d1 (0) − σ n
...
1)

14
...
Any guarantee can be viewed as a financial option
...
In equation (14
...
In
other cases when the only random quantity in the payoff function is the fund
value at maturity, we can use exactly the same approach as in equation (14
...

Q

Example 14
...
After a deduction of 3% for initial
expenses, the premium is invested in an equity fund
...
5% is deducted from the fund at the start of every year except the
first
...

The risk free rate of interest is 5% per year, continuously compounded, and
stock price volatility is 25% per year
...
0001, B = 0
...
075
...
Calculate the revised cost at issue of the GMMB as a percentage of the single
premium, commenting on any additional assumptions required
...
1 (a) With n = 10 we have
ξ = (1 − 0
...
005)9 = 0
...
932148,
√
σ n
√
d2 (0) = d1 (0) − σ n = 0
...
106275 P
Q

and
10 p60

= 0
...
0716P
...
16% of the single premium
...
55 E0 e−10r h(10) = 0
...
106275P = 5
...

The assumption that 55% of policies reach maturity is reasonable if we
assume that survival, allowing for mortality and lapses, is a diversifiable
risk which is independent of the stock price process
...

2

14
...
2 Reserving
We have already defined the reserve for an insurance contract as the capital set
aside during the term of a policy to meet future obligations under the policy
...

Using the Black–Scholes–Merton approach, the value of the guarantee is
interpreted as the value of the portfolio of assets that hedges, or replicates, the
payoff under the guarantee
...
If the mortality
and lapse experiences follow the basis assumptions, the payoffs from the options
will be precisely the amounts required for the guarantee payments
...

Increasingly, insurers are hedging their own guarantees
...
When the insurer
retains the risk, the contribution to the policy reserve for the guarantee will be
the cost of maintaining the replicating portfolio
...

Suppose we consider the GMMB from Section 14
...
1, where the guarantee
liability for the insurer at maturity, time n, is (P − Fn )+ , and where the issue
price was π(0) from equation (14
...
The contribution to the reserve at time

14
...


Note here that the expense factor ξ = (1−e)(1−m)n−1 does not depend on t,
but the reserve at time t does depend on the stock price at time t, St
...
In
Q
particular, Et assumes knowledge of the stock price process at t, St
...
In practice though, it is not possible to
hedge the guarantee perfectly, as the assumptions of the Black–Scholes–Merton
formula do not apply exactly
...

Determining an appropriate reserve for the unhedgeable risk is beyond the
scope of this book, but could be based on the stochastic methodology described
in Chapter 12
...
2 Assume that the policy in Example 14
...
Assuming there are no lapses, calculate the
contribution to the reserve from the GMMB at this time given that, since the
policy was purchased, the value of the stock has
(a) increased by 45%, and
(b) increased by 5%
...
2 (a) Recall that in the option valuation we have assumed that the
return on the fund, before management charge deductions, is modelled by
the index {St }t≥0 , where S0 = 1
...
45
...
927213 as in Example 14
...
So
d1 (6) = 1
...
741983,
4 p66

= 0
...
035892P = $358
...

(b) For S6 = 1
...
39
...

2

14
...
4
...
The most
common guarantees on death are a fixed or an increasing minimum death benefit
...
In the USA, the guaranteed minimum
payout on death might be the accumulation at some fixed rate of interest of all
premiums paid
...

We approach GMDBs in the same way as we approached GMMBs
...
If the insurer knew at the issue
of the policy that the life would die at age x + t, the insurer could cover the
guarantee by setting aside
v(0, t) = E0 e−rt h(t)
Q

at the issue date, where Q is again the risk neutral measure for the stock price
process that underlies the policyholder’s fund
...
4 Guaranteed minimum death benefit

439

be set aside to cover the GMDB, denoted π(0), is found by averaging over the
possible ages at death, x + t, so that
n

π(0) =

v(0, t)t px µx+t dt
...
2)

0

If the death benefit is payable at the end of the month of death rather than
immediately, the value of the guarantee becomes
12n

π(0) =

v (0, j/12)
j=1

| qx
...
3)

Notice that (14
...
3) are similar to formulae we have met in earlier
chapters
...


(14
...
2) and (14
...
In each expression we are finding the expected amount required at time 0 to provide a death
benefit (and in each case we require 0 at time n with probability n px )
...
4) the amount required if death occurs at time t is the present value of
the payment at time t, namely Sv t , whereas in expression (14
...

Example 14
...
A single premium of P = $10 000 is invested in an equity fund
...
25% are deducted at the start of each month
...

Calculate the value of the guarantee on the following basis
...
0001, B = 0
...
075
Risk free rate of interest: 5% per year, continuously compounded
Volatility: 25% per year
Solution 14
...
0025 denote the monthly management
1
charge
...
, 12 , is
h(k) = max(Pe0
...
05k
− Sk , 0
...
, 60 ), the payoff is a multiple of the payoff
12
under a put option with strike price e0
...
Before applying formula
(13
...
9) to include the maturity date, so we now write these as
d1 (t, T ) and d2 (t, T ) where T is the maturity date
...
10) with strike price e0
...
05, to obtain the first term in formula
(13
...
Thus, if v(0, k) denotes the value at time
0 of the guarantee at time k, then

v(0, k) = P(1 − m)12k
=P

(−d2 (0, k))
− S0 (−d1 (0, k))
(1 − m)12k

(−d2 (0, k)) − (1 − m)12k (−d1 (0, k))

where, from (13
...
05k ) + (r + σ 2 /2) k
√
σ k

and d2 (0, k) = d1 (0, k) − σ

√
k,

with σ = 0
...

Table 14
...

Using these values in formula (14
...
7838% of
the single premium, or $278
...

2

14
...
2 Reserving
We now apply the approach of the previous section to reserving for a GMDB on
the assumption that the insurer is internally hedging
...
Suppose
that the payoff function under the guarantee at time s is h(s)
...
4 Guaranteed minimum death benefit

441

Table 14
...
Spreadsheet excerpt for the GMDB in Example 14
...

d1 (0, k)

d2 (0, k)

v(0, k)

1/12
2/12
3/12
4/12
5/12
6/12
7/12

...


...
001400
0
...
002425
0
...
003130
0
...
003704

...


...
070769
−0
...
122575
−0
...
158244
−0
...
187237

...


...
16
431
...
79
623
...
20
776
...
99

...


...
002248
0
...
002265
0
...
002282
0
...
002299

...


...
010477
0
...
010662
0
...
010844

−0
...
534293
−0
...
543585
−0
...
30
2735
...
88
2789
...
63

0
...
002709
0
...
002725
0
...
Then
Q

v(t, s) = Et

e−r(s−t) h(s)
...

Example 14
...
3 is still in force three
years and six months after the issue date
...
5 = 1
...
5 = 1
...


442

Embedded options

Solution 14
...
3, the strike price for an
option expiring at time s is e0
...
Since we are valuing the option
at time t < s, the time to expiry is now s − t
...
10)
we have

v(t, s) = P(1 − m)12s
= P e0
...
05s e−0
...
05s ) + (r + σ 2 /2)(s − t)
√
σ s−t

and
d2 (t, s) = d1 (t, s) − σ

√
s − t
...
5, we calculate v(3
...
, 5 and multiply each value by the mortality
probability, s−t− 1 | 1 q63
...
The resulting valuation is
12 12

(a) $30
...
5 = 1
...
05 when S3
...


2

Example 14
...
An initial expense deduction of 4% of the premium is made, and the
remainder of the premium is invested in an equity fund
...
6% per year
...
There is no guaranteed minimum maturity benefit
...

(b) Calculate the value of the excess amount of the death benefit over the fund
value at the date of death six years after the issue date, as a percentage of
the policyholder’s fund at that date
...

Basis:
Survival model: Makeham’s law, with A = 0
...
00035 and
c = 1
...
4 Guaranteed minimum death benefit

443

Risk free rate of interest: 5% per year, continuously compounded
Volatility: 25% per year
Solution 14
...
96e−0
...

Second, we note that the excess amount of the death benefit over the fund
value at the date of death can be viewed as a GMDB equal to 10% of the
fund value at the date of death
...
1 Fs = 0
...
006s Ss
...
096 e−0
...

Q

Q

In the previous chapter we saw that under the risk neutral measure the EPV
of a stock price at a future point in time is the stock price now
...

Q

Since S0 = 1, we have
v(0, s) = S0 × 0
...
006s = 0
...
006s
...
096

e−0
...
096 A 1

55:10 δ=0
...
5)

= 0
...

So the value of the GMDB at the inception of the policy is 2
...

(b) The value at time t < s of the option that would be needed to fund the
GMDB if the policyholder were to die at time s, given that the policy is in

444

Embedded options
force at t, is, for a unit premium,
Q

v(t, s) = Et

e−r(s−t) h(s) = 0
...
96St e−0
...


The total contribution to the reserve for the GMDB for a policy still in force
at time t, with original premium P, is then
10−t

π(t) = P
0

v(t, w + t) w p55+t µ55+t+w d w
10−t

= 0
...
006(w+t) w p55+t µ55+t+w d w
10−t

= 0
...
006t
0

e−0
...
096 P St e−0
...

55+t:10−t δ=0
...
096 P S6 e−0
...
6%

= 0
...
036

× 0
...


The fund value at time t = 6 is
F6 = 0
...
036 ,
and so the reserve, as a proportion of the fund value, is
¯
0
...
036 A 1

61:4 δ=0
...
96 P S6 e−0
...
1A 1

61:4 δ=0
...
0124
...
24% of the policyholder’s fund
value
...
5 Pricing methods for embedded options
In discussing pricing above, we have expressed the price of a GMMB and a
GMDB as a percentage of the initial premium
...
That
is, the price of the option would come from the initial deduction of eP in the
notation of Section 14
...
1 above
...


14
...
A more common
expense loading in North America is a management charge, applied as a regular
percentage deduction from the policyholder’s fund
...
3
...
4
...

Consider a single premium equity-linked policy with a term of n years issued
to a life aged x
...
3
...
Also, we assume that
mortality is a diversifiable risk which is independent of the stock price process
...
3
...
4
...
Suppose these guarantees consist of
a payment of amount h(t) if the life dies at time t (< n) and a payment h(n) if
the life survives to the end of the term
...
Allowing for the probabilities of death
and survivorship, we have
n

π(0) =
0

E0 [h(t) e−rt ] t px µx+t dt + n px E0 [h(n) e−rn ]
...

Let c denote the component of the management charge that is required to
fund the guarantees from a total (fixed) management charge of m (> c) per
year
...

Assume that the management charge is deducted daily, which we treat as a
continuous deduction
...

Hence, the risk premium received in the time interval t to t + dt for a policy
still in force is (loosely) c P St e−mt dt
...
This cost is c P e−mt dt

446

Embedded options

since an investment of this amount at time 0 in the stock will accumulate to
c P St e−mt dt at time t (recall that S0 = 1)
...

¯

The risk premium c is chosen so that the value to the insurer of the risk premiums
to be received is equal to the cost at time 0 of setting up the replicating portfolios
to pay the guarantees, so that
c = π(0)/(P ax:n δ=m )
...
The risk premium
c is a component of the total management charge m, but we need to know
m to calculate the right-hand side of this equation for c
...
In some cases there may be no solution
...

If the management charge is deducted less frequently, say annually in
advance, we can use the same principles as above to derive the value of the
risk premiums
...
Ignoring survivorship, the
amount of the risk premium to be received at time t (t = 0, 1,
...
Allowing
for survivorship, this value is c P (1 − m)t t px and so the value at time 0 of all
the risk premiums to be received is
n−1
t px

c P (1 − m)t = c P ax:n i∗
¨

t=0

where
i∗ = m/(1 − m) so that 1/(1 + i∗ ) = 1 − m
...
6 In Example 14
...
25%
of the fund value and the GMDB option price was determined to be 2
...


14
...
20% per month is allocated to commission and administrative expenses
...
05% per month is
sufficient to cover the risk premium for the option
...
3
...
6 The risk neutral value of the risk premium of c per month is
E0 cF0 + cF1/12 e−r/12 1/12 p60 + · · · + c F59/12 e−59r/12 59/12 p60
Q

= c P S0 1 + (1 − m) 1/12 p60 + (1 − m)2 2/12 p60 + · · · + (1 − m)59 59/12 p60
= 12 c P S0 a(12)
¨

60:5

where the annuity interest rate is i such that
1/12

vi

= (1 − m) ⇒ i = (1 − m)−12 − 1 = 3
...


The annuity value is 4
...
05% per
month is $259
...

The value of the guarantee at the inception date, from Example 14
...
38 so the risk premium of 0
...
The insurer needs to revise the pricing structure for this product
...
6 Risk management
The option prices derived in this chapter are the cost of either buying the appropriate options in the market, or internally hedging the options
...
It would be inappropriate to charge an option
premium using the Black–Scholes–Merton framework, and then invest the premium in bonds or stocks with no consideration of the dynamic hedging implicit
in the calculation of the cost
...

Under the assumptions of the Black–Scholes–Merton model, and provided
the mortality and lapse experience is as assumed, the hedge portfolio will mature
to the precise cost of the guarantee
...
So hedging is a form of risk mitigation
...
Generally, if the risk is not hedged, the reserves required using
the stochastic techniques of Chapter 12 will be considerably greater than the
hedge costs
...
In reality, the rebalancing will
be less frequent
...

If the hedge portfolio is rebalanced at discrete points in time (e
...
monthly),
there will be small costs (positive or negative) incurred as the previous hedge
portfolio is adjusted to create the new hedge portfolio
...
5
...
3
...

The hedge portfolio is invested partly in zero-coupon bonds, maturing at time
n, and partly (in fact, a negative amount, i
...
a short sale) in stocks
...

dSt

For a GMDB, the approach is identical, but the option value is a weighted
average of options of all possible maturity dates, so the hedge portfolio is a
mixture of zero coupon bonds of all possible maturity dates, and (short positions
in) stocks
...


The stock part of the hedge portfolio has value
n−t

St
0

d
v(t, w + t)
dSt

w px+t

µx+t+w d w
...
7 Emerging costs

449

zero-coupon bond at time t is (loosely)
v(t, t + w) − St

d
v(t, t + w)
dSt

w px+t

µx+t+w d w
...
More
complex strategies are also possible, bringing options and futures into the hedge,
but these are beyond the scope of this book
...
Many jurisdictions are moving towards market consistent
valuation for accounting purposes, even where the insurers do not use hedging
...
7 Emerging costs
Whether the insurer is hedging internally or buying the options to hedge, the
profit testing of an equity-linked policy proceeds as described in Chapter 12
...
In this section, we first explore deterministic
profit testing, and then discuss how to make the profit test stochastic
...
Suppose
we are projecting the emerging cash flows for a single premium equity-linked
policy with a term of n years and with a GMDB and/or a GMMB, for a given
stock price scenario
...

Assuming the insurer hedges the options internally, the income to and outgo
from the insurer’s fund for this contract arise as follows:
Income: +
+
+
Outgo: −
−
−

Initial front-end-load expense deduction
...

Investment return on income over the 1/m year period
...

Initial hedge cost, at t = 0
...

− If the policyholder dies, there may be a GMDB liability
...


450

Embedded options

The part of this that differs from Chapter 12 is the cost of rebalancing the hedge
portfolio
...
5, for a standard put option, we looked at calculating
rebalancing errors for a hedge portfolio adjusted monthly
...
As in Example 13
...
If rebalancing is continuous
(in practice, one or more times daily), then the hedge adjustment will be (in
practice, close to) zero, and the emerging guarantee cost will be zero given that
the experience in terms of stock price movements and survival is in accordance
with the models used
...
Also, if the hedge cost is used to buy
options in the market, there will be no hedge adjustment cost and no guarantee
cost once the options are purchased
...
We break the hedge portfolio down into the stock part,
assumed to be invested in the underlying index {St }t≥0 , and the bond part,
invested in a portfolio of zero-coupon bonds
...


Then 1/m years later, the bond part of the hedge portfolio has appreciated by
a factor er/m and the stock part by a factor St+1/m /St
...


1
The rebalanced hedge portfolio required at time t + 1/m has value π(t + m ),
but is required only if the policyholder survives
...
So the total cost at time t + 1/m of rebalancing
the hedge, given that the policy was in force at time t, is

π(t +

1
m)

1
m

px+t − π bf (t +

1
m)

and the cost of the GMDB is
h(t +

1
m)

1
m

qx+t
...

m
m
If lapses are explicitly allowed for, then the mortality probability would be
replaced by an in-force survival probability
...
7 Emerging costs

451

In the following example, all of the concepts introduced in this chapter are
illustrated as we work through the process of pricing and profit-testing an equitylinked contract with both a GMDB and a GMMB
...
7 An insurer issues a five-year equity-linked policy to a life aged
60
...
The benefit on maturity or death is a
return of the policyholder’s fund, subject to a minimum of the initial premium
...

Management charges of 0
...

(a) Calculate the monthly risk premium (as part of the overall management
charge) required to fund the guarantees, assuming
(i) volatility is 25% per year, and
(ii) volatility is 20% per year
...
0001, B = 0
...
075
Lapses: None
Risk-free rate of interest: 5% per year, continuously compounded
(b) The insurer is considering purchasing the options for the guarantees in
the market; in this case the price for the options would be based on the
25% volatility assumption
...
Use the basis from part (a), assuming,
additionally, that expenses incurred at the start of each month are 0
...
The two stock
price scenarios are
(i) stock prices in the policyholder’s fund increase each month by 0
...
05%
...
Calculate all
1
2
the cash flows to and from the insurer’s fund at times 0, 12 and 12 per
policy issued for the following stock price scenarios
(i) stock prices in the policyholder’s fund increase each month by 0
...
05%,
and
(iii) S 1 = 1
...
9995
...
025% of the fund,
after deducting the management charge, and
the insurer holds no additional reserves apart from the hedge portfolio
for the options
...
7 (a) The payoff function, h(t), for t =

1
2
12 , 12 ,


...
003
...
Then
Q

e−r(s−t) h(s)

Q

e−r(s−t) P − P Ss (1 − m)12s

v(t, s) = Et
= Et

+

= P e−r(s−t) (−d2 (t, s)) − St (1 − m)12s (−d1 (t, s))
where
log(St (1 − m)12s ) + (r + σ 2 /2)(s − t)
√
σ s−t

d1 (t, s) =
and

d2 (t, s) = d1 (t, s) − σ

√
s − t
...
145977 P for σ = 0
...
112710 P for σ = 0
...


60
12

px
...
7 Emerging costs

453

Next, we convert the premium to a regular charge on the fund, c, using
(12)
60:5

π(0) = 12 c P a
¨

where the interest rate for the annuity is i = (1 − m)−12 − 1 = 3
...
26658
...
00285 for σ = 0
...
00220 for σ = 0
...

(b) Following the convention of Chapter 12, we use the stock price scenarios
to project the policyholder’s fund value assuming that the policy stays in
force throughout the five-year term of the contract
...
Outgo at the start of the month comprises the risk premium
for the option (which is paid to the option provider), and the expenses
...
At time t = k/12, where
k = 0, 1,
...
003)k
and g is the rate of growth of the stock price
...
003 Ft
...
0001 (1 − 0
...

• The risk premium is

0
...
003) Ft
...
003 − (1 − 0
...
0001 + 0
...

• The profit to the insurer, allowing for survivorship to time t, is
t

= t p60 ((0
...
003)(0
...
00285)) Ft − 20)
...
2
...
7 part (b), first stock price scenario
...
00
3010
...
92

...


...
78
3682
...
70
120
...
40

...


...
96
142
...
63
2852
...
45

...


...
27
3489
...
67
37
...
08

...


...
55
50
...
99775
0
...


...

0
...
85309

37
...
79
37
...


...

58
59

43
...
33

• The net present value of the profit using a risk discount rate of 10% per

year is
59

NPV =
k=0

k

k
12

1
...


Because the insurer is buying the options, there is no outgo for the insurer
in respect of the guarantees on death or maturity – the purchased options
are assumed to cover any liability
...
There are no
end-of-month cash flows, so we calculate the profit vector using cash flows
at the start of the month
...

Some of the calculations for the scenario where the stock price grows at
0
...
2
...
11
...
05% each
month, gives a NPV of $1463
...

1
2
(c) The items of cash flow for the insurer’s fund at times 0, 12 and 12 , per
policy issued, are shown in Table 14
...
The individual items are as follows:
Income: the management charge (1)
...
7 Emerging costs

455

Table 14
...
Cash flows for Example 14
...

Time,
t
Scenario

Management
GMDB and
charge
Expenses GMMB
(1)
(2)
(3)

Cost of
hedge
(4)

Net cash
flow
(5)

112 709
...
79
112 709
...
79
112 709
...
79

0

(i)
(ii)
(iii)

3000
3000
3000

249
...
25
249
...
67
2982
...
67

249
...
82
249
...
87
0

−1380
...
40
−1380
...
96
4107
...
96

2
12

(i)
(ii)
(iii)

3007
...
63
2967
...
86
246
...
52

0
15
...
64

−1388
...
21
−1 352
...
92
4097
...
20

the net cash flow (5), calculated as
(5) = (1) − (2) − (3) − (4)
...

(1) Management charge
t p60 P St

× 0
...
003
...
99712t+1 × 0
...


(3) Death benefit (for t > 0)
( t− 1 p60 − t p60 ) P(1 − St × 0
...

12

(4) The cost of setting up the hedge portfolio at time 0 is the same for each
1
stock price scenario and is equal to 106 π(0)
...
05/12 +

0S

1
12

)
...
Hence, the net cost of rebalancing the hedge portfolio

456

Embedded options
Table 14
...
Hedge portfolios for Example 14
...

Investment scenario

Time
t

(i)

(ii)

(iii)

0

π(t)
ϒt
t St

112 710
417 174
−304 465

112 710
417 174
−304 465

112 710
417 174
−304 465

1
12

π(t)
ϒt
t St

111 342
415 700
−304 358

113 478
421 369
−307 891

111 342
415 700
−304 358

2
12

π(t)
ϒt
t St

109 956
414 172
−304 216

114 253
425 626
−311 373

114 097
425 216
−311 119

at this time per policy originally issued is
1
12

p60 π(1/12) − (ϒ0 e0
...


1
12

Similarly, the net cost of rebalancing the hedge portfolio at time
policy originally issued is
2
12

p60 π(2/12) −

The values of π(t), ϒt and

1
12

t

p60 (ϒ 1 e0
...


are shown in Table 14
...


12

2

We note several important points about this example
...
This is particularly true for the
internal hedging case, part (c)
...
If the lognormal model
for stock prices is appropriate, then the expected present value (under the
P-measure) of the hedge rebalancing costs will be close to zero
...
7 the present value is significant
and negative, meaning that the hedge portfolio value brought forward each
month is more than sufficient to pay for the guarantee and new hedge
portfolio at the month end
...


14
...
In the table below we show some
summary statistics for 500 simulations of the NPV for part (c), again calculated using a risk discount rate of 10% per year
...
20 per year
...

This is because this scenario is highly unrepresentative of the true stock
price process
...

(3) If we run a stochastic profit test under part (b), where the option is purchased
in the market, the variability of simulated NPVs is very small
...
The mean NPV
over 500 simulations is approximately $2137, and the standard deviation
of the NPV is approximately $766, assuming the same parameters for the
stock price process as for (c) above
...
Using the first scenario (increasing
prices) generates a NPV of $137 053 and using the second gives $2381
...


14
...

Hardy (2003) gives some examples and information on practical ways to manage the risks
...
Much
more convoluted options are sold, particularly in association with variable annuity policies
...
Also, the guarantee may specify that after an
introductory period, the policyholder could withdraw 5% of the initial premium

458

Embedded options

per year for 20 years
...
New variants are being created regularly, reflecting
the strong interest in these products in the market
...
2 we noted three differences between options embedded in
insurance policies and standard options commonly traded in financial markets
...
Both
of these issues have been addressed in this chapter
...
One
of the implications is that the standard models for short-term options may not be
appropriate over longer terms
...
There is considerable empirical evidence that
the lognormal model is not a good fit for stock prices in the long run
...
Sources for further information
include Hardy (2003) and Møller (1998)
...

In some countries annual premium equity-linked contracts are common
...

Bacinello (2003) discusses an Italian style annual premium policy
...
(2008) give an introduction to some of the issues around equitylinked insurance, including a discussion of a guaranteed minimum income
benefit, another more complex embedded option
...
9 Exercises
Exercise 14
...

(a) Calculate the value of the GMMB at the issue date for a single premium
of $100
...

(c) Calculate the value of the GMMB two years after issue, assuming that the
policy is still in force, and that the underlying stock prices have decreased
by 5% since inception
...
9 Exercises
Basis and policy information:
Age at issue:
Front end expense loading:
Annual management charge:
Survival model:
Lapses:
Risk free rate:
Volatility:

459

60
2%
2% at each year end (including the first)
Standard Ultimate Survival Model
5% at each year end except the final year
4% per year, continuously
compounded
20% per year

Exercise 14
...
A single premium of $10 000 is invested in an equity fund
...
At the end of
the month of death before age 70, the death benefit is 105% of the policyholder’s
fund subject to a minimum of the initial premium
...

(b) Express the cost of the death benefit as a continuous charge on the fund
...
3 An insurer issues a range of 10-year variable annuity guarantees
...
The policy carries
a guaranteed minimum maturity benefit of 100% of the premium
...
e
...

(b) Calculate the probability that the guaranteed minimum maturity benefit will
mature in-the-money under the Q-measure
...

(d) Calculate the price of the option
...
Explain why this value would not be suitable
...
We simulate the payoff under the risk

460

Embedded options
neutral measure, discount at the risk-free rate and take the mean value
to estimate the Q-measure expectation
...


Basis:
Survival model:
Stock price appreciation:
Risk free rate of interest:
Management charges:

No mortality
Lognormally distributed, with µ = 0
...
25 per year
4% per year, continuously compounded
3% of the fund per year, in advance

Exercise 14
...
There is a guaranteed minimum maturity benefit equal to
the initial premium of $100
...
The insurer offers the policyholder a reset option, under which
the policyholder may reset the guarantee to the current fund level, in which
case the remaining term of the policy will be increased to 10 years
...

(b) Determine the threshold value for F5 (i
...
the fund at time 5) at which the
option to reset becomes more valuable than the original option
...
18 per year
5% per year, continuously compounded
1% of the fund per year, in advance
3%

Exercise 14
...
The premium is $100 000
...
25% of the
fund are deducted at the start of each month
...
19%
of the fund
...
The
actuary first works out the hedge rebalancing cost each month then inserts
that into the profit test
...
9 Exercises

461

Table 14
...
Single scenario of stock prices for stochastic profit test for
Exercise 14
...

t
0
1
2
3
4
5
6
7
8
9
10
11
12

St

t

St

t

St

t

St

t

St

1
...
95449
0
...
97371
1
...
01181
0
...
98733
0
...
91293
0
...
88248
0
...
92420
0
...
02563
1
...
25234
1
...
10038
0
...
04213
1
...
14174
1
...
09292
1
...
27355
1
...
31999
1
...
20481
1
...
23876
1
...
09478
1
...
09203
1
...
05115
1
...
18018
1
...
34264
1
...
39327
1
...
41652
1
...
34578
1
...
50309
1
...
45134
1
...
40476
1
...
39672
1
...
25762
1
...
6
...
5, in $1 000s
...


...

59
60

St
1
...
95449
0
...
25762
1
...
540
11
...
592

...


...
200
0
...
585
−29
...
528

...


...
658
–

38
...
668
41
...


...

7
...
955
11
...


...

0
...
619

–
−0
...
109

...


...
326
−0
...
5 represent one randomly generated
scenario
...

(i) Table 14
...
Use the stock price scenario in Table 14
...

Calculate the present value of the hedge rebalance costs at an effective
rate of interest of 5% per year
...
7 shows the first two rows of the profit test for this scenario
...


462

Embedded options
Table 14
...
Profit test table for Exercise 14
...


Time, t
(months)

Management
costs

Expenses

Hedge costs

100 000
...
38
96 261
...


...


Ft

250
...
03
240
...
00
61
...
57

10 540
...
99
−109
...
21
200
...
24

Complete the profit test and determine the profit margin (NPV as a
percentage of the single premium) for this scenario
...

Basis for hedging and profit test calculations:
Survival:
Standard Ultimate Survival Model
Lapses:
None
Risk-free rate:
5% per year, continuously
compounded
Volatility:
20% per year
Incurred expenses – initial:
1% of the premium
Incurred expenses – renewal: 0
...
1 (a)
(b)
(c)
14
...
3 (a)
(b)
(c)
(d)
14
...
61
0
...
08
$107
...
13%
0
...
60819
$6033
$18 429
The original option value is $4
...
46
...
4 both options have value $6
...


14
...
5 (b)

463

(i) The PV of rebalancing costs is −$1092
...
23%
(iii) We note that the initial hedge cost converts to a monthly outgo
of 0
...
255%, compared with income of 0
...
Overall we would not expect this contract to be profitable
on these terms
...
1 Probability distributions

In this appendix we give a very brief description of the probability distributions used in this book
...


A
...
1 Binomial distribution

If a random variable X has a binomial distribution with parameters n and p,
where n is a positive integer and 0 < p < 1, then its probability function is
Pr[X = x] =

n x
p (1 − p)n−x
x

for x = 0, 1, 2,
...
This distribution has mean np and variance np(1 − p),
and we write X ∼ B(n, p)
...


(A
...
1
...
1 Probability distributions

465

for a ≤ x ≤ b, and has probability density function
f (x) =

1
b−a

for a < x < b
...

A
...
3 Normal distribution

If a random variable X has a normal distribution with parameters µ and σ 2
then its probability density function is
f (x) =

−(x − µ)2
1
√ exp
2σ 2
σ 2π

for −∞ < x < ∞, where −∞ < µ < ∞ and σ > 0
...

The random variable Z defined by the transformation Z = (X −µ)/σ has
mean 0 and variance 1 and is said to have a standard normal distribution
...

The traditional approach to computing probabilities for a normal random
variable is to use the relationship
Pr[X ≤ x] = Pr[Z ≤ (x − µ)/σ ]
and to find the right-hand side from tables of the standard normal
distribution, or from an approximation such as
(x) ≈ 1 −

1
2

1 + a 1 x + a 2 x 2 + a3 x 3 + a4 x 4 + a5 x 5 + a6 x 6

−16

for x ≥ 0 where
a1 = 0
...
0000380036,

a2 = 0
...
0000488906,

a3 = 0
...
0000053830
...
5 × 10−7
...
For example, in Excel we can find Pr[X ≤ x] from

466

Appendix A
...
(Changing this parameter to FALSE gives
the value of the probability density function at x
...
Suppose we want to find the value xp such that
Pr[Z > xp ] = p where Z ∼ N (0, 1) and 0 < p ≤ 0
...
We can find this
approximately as
xp = t −
where t =

a0 + a1 t + a2 t 2
1 + d 1 t + d 2 t 2 + d3 t 3

log(1/p2 ) and
a0 = 2
...
432788,

a1 = 0
...
189269,

a2 = 0
...
001308
...
5×10−4
...
5, but in practical actuarial applications this case rarely arises
...
Specifically, we can find x such that Pr[X ≤ x] = p using
= NORMINV(p, µ, σ )
...
1
...
This distribution has mean
exp{µ + σ 2 /2} and variance exp{2µ + σ 2 }(exp{σ 2 } − 1), and we write
X ∼ LN (µ, σ 2 )
...
1 Probability distributions

467

We can calculate probabilities as follows
...

√ exp
2σ 2
yσ 2π

Now substitute z = log y, so that the range of the integral changes to
(−∞, log x), with dz = dy/y
...
Thus, we can compute probabilities for a lognormally distributed random variable from the standard
normal distribution
...

In Chapters 12 and 13 we used the result that if X ∼ LN (µ, σ 2 ) then
a

x f(x)dx = exp{µ + σ 2 /2}

0

log a − µ − σ 2
σ


...

√ exp
2σ 2
σ 2π

(A
...
Probability theory

Combining the exponential terms, the exponent becomes
z−

(z − µ)2
−1
=
2σ 2
2σ 2
−1
=
2σ 2
−1
=
2σ 2
−1
=
2σ 2

z 2 − 2µz + µ2 − 2σ 2 z
z 2 − 2(µ + σ 2 )z + µ2
z 2 − 2(µ + σ 2 )z + (µ + σ 2 )2 + µ2 − (µ + σ 2 )2
(z − (µ + σ 2 ))2 − 2µσ 2 − σ 4

− z − (µ + σ 2 )
=
2σ 2

2

+µ+

σ2

...
We can now
write
a

x f(x)dx =

0

log a
−∞

− z − (µ + σ 2 )
1
√ exp
2σ 2
σ 2π

= exp µ +

σ2
2

log a
−∞

2

exp µ +

σ2
dz
2

− z − (µ + σ 2 )
1
√ exp
2σ 2
σ 2π

2

dz
...
2)
...
2) we see that the mean of the lognormal
distribution with parameters µ and σ 2 is
exp µ +

σ2

...
3 Functions of a random variable

469

A
...

Suppose that X1 , X2 , X3 ,
...
Now
define the sum Sn = n Xi so that E[Sn ] = nµ and V[Sn ] = nσ 2
...

The central limit theorem can be used to justify approximating the
distribution of a (finite) sum of independent and identically distributed random variables by a normal distribution
...
e
...
Then using
moment generating functions we see that the distribution of Sn is B(n, p)
since
E[exp{tSn }] = E[exp{t(X1 + X2 + · · · + Xn )}]
n

E[exp{tXi }]

=
i=1
n

(pet + 1 − p)

=
i=1

= (pet + 1 − p)n
...
1) with n = 1
...
Thus we can think of a binomial random variable as the sum of Bernoulli random variables, and, provided the number
of variables being summed is large, we can approximate the distribution of
this sum by a normal distribution
...
3 Functions of a random variable

In many places in this book we have considered functions of a random variable
...
We have also evaluated the expected
value and higher moments of functions of a random variable
...
We
quote results only, giving references for these results in Section A
...


470

Appendix A
...
3
...
Let g be a function and let Y = g(X ), so that
the possible values for Y are g(0), g(1), g(2),
...
,
Y takes the value g(x) if X takes the value x
...


(A
...
Higher
moments are similarly computed
...
we have
∞

E[Y ] =

g(x)r Pr[X = x]
...
, and define Y = v X where 0 < v < 1
...

1 − qv r

A
...
2 Continuous random variables

We next consider the situation of a continuous random variable, X , distributed on (0, ∞) with probability density function f (x) for x > 0
...
Then we can compute the expected value of Y as
∞

E[Y ] = E[g(X )] =

g(x)f (x)dx
...
4)

0

As in the case of discrete random variables, the expected value of Y can be
found without explicitly stating the distribution of Y , and higher moments
can be found similarly
...
3) – probability
function has been replaced by probability density function, and summation
by integration
...
3 Functions of a random variable

471

It can be shown that Y has a probability density function, which we denote
h, given by
h(y) = f g −1 (y)

d −1
g (y)
dy

(A
...
However, formula (A
...

For example, suppose that X has an exponential distribution with parameter λ
...
Then by formula (A
...

λ+δ

The alternative (and more complicated) approach to finding E[Y ] is to first
identify the distribution of Y , then find its mean
...

dy
δy
By formula (A
...

δ

Thus
1

E[Y ] =
0

yh(y)dy =

λ
δ

1
0

yλ/δ dy =

λ y(λ/δ)+1
δ (λ/δ) + 1

1

=
0

λ

...
In any event, the key point is that
a function of a random variable is itself a random variable with its own
distribution, but because of formula (A
...

A
...
3 Mixed random variables
Most of the mixed random variables we have encountered in this book have
a probability density function over an interval and a mass of probability

472

Appendix A
...
For example, under an endowment insurance with term
n years, there is probability density associated with payment of the sum
insured at time t for 0 < t < n, and a mass of probability associated with
payment at time n
...
4
...


More generally, suppose that X is a random variable with probability density
function f over some interval (or possibly intervals) which we denote by
I , and has masses of probability, Pr[X = xi ], at points x1 , x2 , x3 ,
...

i

For example, suppose that Pr[X ≤ x] = 1 − e−λx for 0 < x < n, and
Pr[X = n] = e−λn
...
If we
define Y = e−δX , then
n

E[Y ] =

e−δx λe−λx + e−δn e−λn

0

λ
1 − e−(λ+δ)n + e−(λ+δ)n
λ+δ
1
λ + δe−(λ+δ)n
...
4 Conditional expectation and conditional variance

Consider two random variables X and Y whose first two moments exist
...
In particular,
E[Y ] = E [E[Y |X ]]

(A
...


(A
...
Consider

A
...
6)
...
8)

y

(this is just the bivariate version of formula (A
...
By the rules of conditional
probability,
Pr[X = x, Y = y] = Pr[Y = y|X = x] Pr[X = x]
...
9)

Then setting g(X , Y ) = Y and using (A
...
9) we obtain
y Pr[Y = y|X = x] Pr[X = x]

E[Y ] =
x

y

Pr[X = x]

=
x

y Pr[Y = y|X = x]
y

Pr[X = x]E[Y |X = x]

=
x

= E [E[Y |X ]]
...
7) we have
V[Y ] = E[Y 2 ] − E[Y ]2
= E[E[Y 2 |X ]] − E[Y ]2
= E V[Y |X ] + E[Y |X ]2 − E[Y ]2
= E [V[Y |X ]] + E E[Y |X ]2 − E [E[Y |X ]]2
= E [V[Y |X ]] + V [E[Y |X ]]
...
5 Notes and further reading
Further details on the probability theory contained in this appendix can
be found in texts such as Grimmett and Welsh (1986) and Hogg and
Tanis (2005)
...


Appendix B
Numerical techniques

B
...
The first,
the trapezium rule has the advantage of simplicity, but its main disadvantage
is the amount of computation involved for the method to be very accurate
...
We now outline each method, and give numerical illustrations of both
...
3
...

B
...
1 The trapezium rule

Under the trapezium rule, the interval (a, b) is split into n intervals, each of
length h = (b − a)/n
...
+

a+h
a+(j+1)h

a+nh
a+(n−1)h

f (x)dx
...
1 Numerical integration

475

Table B
...
Values of I ∗
under the trapezium rule
...
64504
12
...
64258
12
...
64242

We obtain the value of I under the trapezium rule by assuming that f is a
linear function in each interval so that under this assumption
a+(j+1)h

f (x)dx =

a+jh

h
2

(f (a + jh) + f (a + (j + 1)h)) ,

and hence
I =h

1
2 f (a) + f (a

⎛

= h ⎝ 1 f (a) +

+ h) + f (a + 2h) +
...

2

j=1

To illustrate the application of the trapezium rule, consider
20

I∗ =

e−0
...


0

We have chosen this integral as we can evaluate it exactly as
I∗ =

1
1 − e−0
...
64241,
0
...
We have a = 0 and b = 20, and for our numerical illustration we have
set n = 20, 40, 80, 160 and 320, so that the values of h are 1, 0
...
25, 0
...
0625
...
1 shows the results
...


476

Appendix B
...
2
...

n

I∗

10
20
40

12
...
6424112
12
...
1
...


h
3

This approximation arises by approximating the function f by a quadratic
function that goes through the three points (a, f (a)), (a + h, f (a + h)) and
(a + 2h, f (a + 2h))
...

Let us again consider
20

I∗ =

e−0
...


0

To seven decimal places, I ∗ = 12
...
2 shows numerical
values for I ∗ when n = 10, 20 and 40
...
2 that the calculations are considerably more accurate than under the trapezium rule
...


B
...
3
...

m

Im

60
70
80
90
100

34
...
75059
34
...
75155
34
...
1
...
For example, we saw in Chapter 2 that
the complete expectation of life is given by
∞

◦

ex =

t px dt
...
For example, looking at the integrand in the above expression,
we might say that the probability of a life aged x surviving a further 120 − x
years is very small, and so we might replace the upper limit of integration by
120 − x, and perform numerical integration over the finite interval (0, 120 −
x)
...

To illustrate this idea, consider the following integral from Section 2
...
2
◦
where we computed ex for a range of values for x in Table 2
...
Table B
...
These values have been calculated using repeated
Simpson’s rule
...

For example, with m = 70, setting n = 140 results in h = 0
...
This maintains
consistency between successive calculations of Im values
...
From this table our conclusion is that, to five decimal
◦
places, e40 = 34
...


478

Appendix B
...
2 Woolhouse’s formula

Woolhouse’s formula was used in Chapter 5
...
We use the Euler–Maclaurin formula which is
concerned with numerical integration
...
For a function f , the Euler–Maclaurin formula can
be written as
b

N

f (x)dx = h

a

f (a + ih) −
i=0

+

h2
12

1
(f (a) + f (b))
2

f (a) − f (b) −

h4
720

f (a) − f (b) + · · · , (B
...
We shall apply this formula twice, in each
case ignoring second and higher order derivatives of f
...
1) is
n
1
f (i) − 2 (f (0) + f (n)) +

1
12

f (0) − f (n)
...
2)

i=0

Second, setting a = 0, b = n and N = mn for some integer m > 1 (so that
h = 1/m), the left-hand side of (B
...


(B
...
2) and (B
...
(B
...
4) gives the first three terms of Woolhouse’s
formula, and in actuarial applications it usually suffices to apply only these
terms
...
3 Notes and further reading

479

B
...
Details of the derivation of the trapezium rule and repeated
Simpson’s rule can be found in standard texts on numerical methods such
as Burden et al (1978) and Ralston and Rabinowitz (1978)
...
1 The inverse transform method

The inverse transform method allows us to simulate observations of a random variable, X , when we have a uniform U (0, 1) random number generator
available
...

The result follows for the following reason: if U ∼ U (0, 1), then F −1 (U )
has the same distribution as X
...
First, we note that as the distribution function F is continuous, it is a monotonic increasing function
...

˜
Now let X = F −1 (U )
...
2 Simulation from a normal distribution

481

since F is a monotonic increasing function
...

Example C
...
1254,

u2 = 0
...
7548,

from the U (0, 1) distribution
...
1 Let F denote the distribution function of an exponentially
distributed random variable with mean 100, so that
F(x) = 1 − exp{−x/100}
...
8746 = 13
...
5471 = 60
...
2452 = 140
...


2

C
...
In many situations, for example if we wish to create a large
number of simulations of an insurance portfolio over a long time period,
it is much more effective in terms of computing time to use a programming language rather than a spreadsheet
...

Without going into details, we now state the two most common
approaches to simulating values from a standard normal distribution
...
3
...
Simulation
C
...
1 The Box–Muller method

The Box–Muller method is to first simulate two values, u1 and u2 , from a
U (0, 1) distribution, then to compute the pair
x=

−2 log u1 cos(2π u2 )

y=

−2 log u1 sin(2π u2 )

which are random drawings from the standard normal distribution
...
643 and u2 = 0
...
1703
and y = 0
...


C
...
2 The polar method
From a computational point of view, the weakness of the Box–Muller
method is that we have to compute trigonometric functions to apply it
...


If s < 1, we compute
x = v1

−2 log s
,
s

y = v2

−2 log s
s

which are random drawings from the standard normal distribution
...

For example, if u1 = 0
...
279, we find that v1 = 0
...
4420 and hence s = 0
...
As the value of s is less than 1, we
proceed to compute x = 0
...
3450
...
3 Notes and further reading
Details of all the above methods can be found in standard texts on simulation,
e
...
Ross (2006), or on probability theory, e
...
Borovkov (2003)
...
(2004)
...
National Vital Statistics Reports; Vol 53, No
...
Hyattsville, Maryland: National Center for
Health Statistics
...
and Stegun, I
...
(1965)
...
New York: Dover
...
Australian Life Tables 2000–
02
...

[4] Bacinello,A
...
(2003)
...
North American
Actuarial Journal 7, 3, 1–17
...
(2006)
...
Chichester: John Wiley & Sons
...
A
...
Elements of Stochastic Modelling
...

[7] Bowers, N
...
, Gerber, H
...
, Hickman, J
...
, Jones, D
...
and Nesbitt,
C
...
(1997)
...
Itasca: Society of
Actuaries
...
P
...
P
...
Derivatives: The Tools that
Changed Finance
...

[9] Boyle, P
...
and Schwartz, E
...
(1977)
...
Journal of Risk and Insurance 44,
639–66
...
J
...
S
...
The pricing of equitylinked life insurance policies with an asset value guarantee
...

[11] Burden, R
...
and Faires, J
...
(2001)
...

Pacific Grove: Brooks/Cole
...
J
...
(2004)
...
New Jersey: Princeton
University Press
...
Continuous Mortality
Investigation Report, Number 12
...

[14] Continuous Mortality Investigation (2006)
...
www
...
org
...
pdf
...
A
...
(1992)
...
Oxford: Oxford University Press
...
R
...
D
...
The Theory of Stochastic
Processes
...

[17] Cox, J
...
, Ross, S
...
and Rubinstein, M
...
Options pricing: a
simplified approach
...

[18] Dickson, D
...
M
...
Premiums and reserves for life insurance
products
...

[19] Forfar, D
...
, McCutcheon, J
...
and Wilkie, A
...
(1988)
...
Transactions of the Faculty of Actuaries
41, 97–269
...
(2004)
...
New York: Springer-Verlag
...
(1825)
...
Philosophical Transactions of the Royal Society
of London 115, 513–85
...
L
...
E
...
(1994)
...
Upper Saddle River: Addison-Wesley
...
and Welsh, D
...
A
...

Oxford: Oxford University Press
...
R
...
Investment Guarantees: Modeling and Risk Management for Equity-Linked Life Insurance
...

[25] Hoem, J
...
(1983)
...
H
...
Opperman, T
...
Thiele
and J
...
Gram
...

[26] Hoem, J
...
(1988)
...
Transactions of the XXIII International
Congress of Actuaries, Volume R, pp
...

[27] Hogg, R
...
and Tanis, E
...
(2005)
...
Upper Saddle River: Prentice Hall
...
-C
...
On binomial option pricing
...

[29] Hull, J
...
(2005)
...

Upper Saddle River: Prentice Hall
...
C
...
P
...
S
...
J
...
and
Wilson, D
...
E
...
Variable Annuities
...

[31] Macdonald, A
...
(1996)
...
I: multiple state, binomial and Poisson
models
...

[32] Macdonald, A
...
, Waters, H
...
and Wekwete, C
...
(2003a)
...

Scandinavian Actuarial Journal, 1–27
...
S
...
R
...
T
...
The
genetics of breast and ovarian cancer II: A model of critical illness
insurance
...

[34] Makeham, W
...
(1860)
...
Journal of the Institute of Actuaries 8, 301–10
...
C
...
Derivatives Markets, 2nd edition
...

[36] McGill, D
...
, Brown, K
...
, Haley, J
...
and Schieber, S
...
(2005)
...
New York: Oxford
University Press
...
(1998)
...
ASTIN Bulletin 28, 17–47
...
(1977)
...
London: Heinemann
...
(1995)
...
Insurance: Mathematics & Economics 17, 171–80
...
English Life Tables No
...

London: The Stationery Office
...
H
...
A
...
T
...
P
...
Numerical Recipes, 3rd edition
...

[42] Ralston, A
...
(1978)
...
New York: McGraw-Hill
...
F
...
Life, Death and Money
...

[44] Rolski, T
...
, Schmidt, V
...
(1999)
...
Chichester: John Wiley &
Sons
...
M
...
Stochastic Processes, 2nd edition
...

[46] Ross, S
...
(2006)
...
Burlington: Elsevier
Academic Press
...
(1965)
...
Scandinavian Actuarial Journal, 184–211
...
R
...
An approach to the study of multiple state
models
...

[49] Waters, H
...
and Wilkie,A
...
(1988)
...
Journal of the Institute
of Actuaries 114, 569–80
...
C
...
P
...
A
...
L
...
,
Macdonald, A
...
, Miller, K
...
, Richards, S
...
, Robjohns, N
...
P
...
R
...
Longevity in the 21st century
...

[51] Woolhouse, W
...
B
...
On an improved theory of annuities and
assurances
...


Author index

Abramovitz and Stegun 473, 479
Arias 67
Australian Government Actuary 67
Bacinello 458
Blake 319
Borovkov 482
Bowers et al
...
427
Dickson 279
Forfar et al
...
137
Grimmett and Welsh 473
Hardy 395, 457, 458

Hoem 220, 278
Hogg and Tanis 473
Hsia 427
Hull 427
Ledlie et al
...
278, 348
Makeham 35
McDonald 427
McGill et al
...
348
Ralston and Rabinovitz 479
Renn 15
Rolski et al
...
348
Woolhouse 137

487

Index

Accident hump 50
Accidental death model 232
Accrual rate 13, 307
Accrued benefit 307
Accumulation units 395
Acquisition cost 354
Actuarial liability 314
Actuarial notation 26, 330
Actuarial reduction factor 322
Actuarial value 76
Adverse selection 10, 215
Age rating 165
Age retirement pension 13, 307
Age-at-death random variable 17
Aggregate survival models 56
Alive-dead model 230-231
Allocated premium 375
Allocation percentage 375
American options 404
Analysis of surplus 200
Annuitant 108
Annuity 4, 11–12, 107–141
Certain 108
Comparison by payment frequency 121
Deferred 123–125
Equity-indexed 7
Guaranteed 125–126
Increasing arithmetically 127–129
Increasing geometrically 129–130
Joint life 11, 234, 264
Last survivor 12, 234, 263
Life 4, 107–137
Payable 1/mthly 118–121
Payable annually 108–115
Payable continuously 115–117
Regular premium deferred annuity 11, 187

Reversionary annuity 12, 264
Single premium deferred annuity 11
Single premium immediate annuity 11
Term Life annuity 11, 108
Variable annuity 7, 374
Whole Life annuity 11, 108
Anti-selection 215
Assessmentism 2
Asset shares 200–203
Assurance (also, see insurance) 4
At-the-money 404
Australian Government Actuary 51
Australian Life Tables 2000–02 50, 51, 57
Basis 74
Bernoulli distribution 469
Bid-offer spread 375
Binomial distribution 42, 43, 464
Binomial option pricing model 405–414
Black-Scholes-Merton model 401, 414–427,
432, 447
Bonus 5, 100, 158
Box-Muller method 482
Breslau mortality 2
Call option 403, 417–418
Canada Life Tables 2000–02 70
Capital units 395
Career Average Earnings plan 13, 312–314
Cash refund payout option 172
Cash value 213
Central limit theorem 334, 469
Claims acceleration approach 94
Coefficient of variation 349
Commission 8, 150
Common shock model 281
Complete expectation of life 29, 117

488

Index
Compound reversionary bonus 5, 100, 158
Conditional expectation 472
Conditional Tail Expectation (CTE) 393–4
Conditional variance 472
Confidence interval 345, 387
Constant addition to force of mortality 165
Constant force of mortality 48
Contingent insurance 263
Continuous Mortality Investigation (CMI) 53,
67, 102
Table A14 53, 54, 55, 58
Table A2 54, 58
Table A21 69
Table A23 68
Table A5 64
Continuous time stochastic process 231
Cost of living adjustment (COLA) 309
Critical illness Insurance 12, 279
CTE (conditional tail expectation) 393–4
Curtate expectation of life 32
Curtate future lifetime 33, 78
De Witt, Johan 15
Death in service benefit 298
Death strain at risk 195, 199
Deferred acquisition cost 154
Deferred insurance benefits 91–93
Deferred life annuity 123–125
Deferred mortality probablity 27
Defined benefit pension (DB) 13–14, 291, 297
Defined contribution pension (DC) 13, 291,
294–297
Delta (of an option) 420
Delta hedge 449
Demutualization 14
Dependent probabilities for a multiple
decrement model 260
Direct marketing 8
Disability income insurance model 233–234,
361
Disability retirement 298
Discounted payback period 359, 360, 369
Diversifiable mortality risk 335
Diversifiable risk 333–342
Diversification 156, 333
Dividends (stock) 414, 427
Dividends (with profit insurance)
see ‘bonus’
Dynamic hedging 410, 412, 426
Effective rate of discount 75
Embedded option 401, 431–463

489

Emerging costs for equity-linked insurance
374–400, 449–457
Emerging costs for traditional insurance
353–373
Emerging surplus 357, 365, 366
Employer sponsored pension plan 12, 290–325
Endowment insurance 5, 89
English Life Table 15 50, 53, 67
Equity-Indexed Annuity 7
Equity-Linked Insurance 7–8, 374–400,
431–463
Segregated Fund 7, 374
Unit Linked 7, 374
Variable Annuity 7, 374
Equivalence principle 146, 388
Euler Maclaurin formula 132, 478
Euler, Leonhard 137
Euler’s method 212, 245
European options 403
Expectation of life – complete 29
Expectation of life – curtate 33
Expected present value (EPV) 76
Expected present value of future profit 359,
360
Expected shortfall 393
Expenses 150, 434
Expiry date 403
Extra risks 165–169
Failure rate 36
Family income benefit 138
Final average salary 292
Final salary pension plan 13, 306–312
Force of interest 75
Force of mortality 21–26
Force of transition 237
Forward rate of interest 329
Fractional age assumptions 44
Constant force of mortality 48
Uniform distribution of deaths 44, 49
Front-End Load 8, 445
Functions of a random variable 469
Funding for pension benefits 314–318
Future lifetime random variable 17, 76, 230
Future loss: gross 145
Future loss: net 145
Geometric Brownian motion 414
Gompertz’ law of mortality 24, 31, 32, 34, 38,
39, 52
Gompertz-Makeham formula, GM(r,s) 35, 102
Graunt, John 2
Gross future loss 145

490

Index

Gross premium 150–154
Gross premium policy value 183
Group life insurance 2
Guaranteed life annuity 125–126
Guaranteed minimum death benefit (GMDB)
375, 378, 401, 431
Pricing 438–440
Reserving 440–444
Guaranteed minimum maturity benefit
(GMMB) 374, 376, 401, 431
Pricing 433–436, 444–447
Reserving 436–438
Guaranteed minimum withdrawal benefit 457
Halley, Edmund 2
Hazard rate 36
Hedge portfolio 406–407, 448, 450
Hedging 402, 410, 436
Hurdle rate 359, 360
Ill-health retirement 298
Immediate annuity: term 114
Immediate annuity: whole life 113
Income protection Insurance 12
see also ‘disability income insurance’
Increasing annuities 127–130
Increasing insurance 98
Independent probabilities for a multiple
decrement model 261
Indicator random variables 96
Initial expenses 150, 434
Insurable interest 2
Insurance 3–14
Critical Illness 12
Disability income 233
Endowment 5, 89–91
Equity-Linked 7, 374–400, 431–463
Income protection 12, 233
Joint life 234, 264
Long term care 12
Pure endowment 88–89
Term 4, 86–88
Unitized with-profit 7
Universal Life 6
Whole Life 5, 14, 78–86
With-Profits 5, 100
Insurer’s fund 374, 375
Integrals over an infinte interval 477
Interest rate risk 326–352
Internal rate of return (IRR) 359, 360, 368
International actuarial notation 26, 330
In-the-money 404
Inverse transform method 344

Jensen’s inequality 141
Joint life and last survivor model 234–235,
261–270
Actuarial notation 263
Contingent insurance 263
Independent survival models 270
Last survivor annuity 263
Last survivor insurance 263
Joint life annuity 11, 234, 264
Joint life insurance 234, 264
Kolmogorov’s forward equations 242–3
Kologorov, Andrei Nikolaevich 278
Lapse 213
Lapse and re-entry option 220
Lapse supported insurance 259
Last survivor annuity 12, 234, 263
Last survivor insurance 263
Life annuity 107
Life insurance underwriting 55
Life table 41
Lifetime distribution 17, 42
Limiting age 21, 84
Liquidity risk 215
Lognormal distribution 385, 415, 466
Lognormal process 414, 433
Log-return 415
Long in an asset 405
Long Term Care Insurance 12
Maclaurin, Colin 137
Makeham’s law of mortality 35, 38, 65, 74,
101, 283
Management charge 374, 375, 378, 432, 445
Margins 155
Market consistent valuation 449
Markov property 235
Markov, Andrei Andreyevich 278
Maturity date 403
Mixed random variables 471
Monte Carlo simulation 342–348, 384–388
Morbidity risk 12
Mortality improvement 348
Mortality profit 199
Mortality rate 27
Multiple decrement models 256–261
Dependent probabilities 260
Independent probabilities 261
Multiple state models 230–289
Accidental death model 232
Alive-dead model 230–231

Index
Disability income insurance model
233–234, 251–255
Joint life and last survivor model 234–235,
261–273
Kolmogorov’s forward equation 242–3
Multiple decrement models 256–261
Permanent dsability model 232–233
Policy values 250–256
Premiums 247–250
Thiele’s differential equation 250–256
Transitions at specified ages 274–278
Mutual insurance 14
National life tables 49–52
Australian Life Tables 2000–02 50, 57, 67
Canada Life Tables 2000–02 70
English Life Table 15 1990–92 (ELT 15) 50,
53, 67
US Life Table 2002 50, 62, 67, 70
Nature’s measure 409
Negative policy values 220
Net amount at risk 195
Net cash flow 354
Net future loss 145
Net premium 42, 146–150
Net premium policy value 183
Net present value 359, 360, 368
New business strain 154
No arbitrage 330, 402–403
Nominal rate of discount 75
Nominal rate of interest 75
Non-diversifiable risk 332–342, 401
Non-forfeiture law 214
Non-participating (non-par) business 5
Normal approximation 140
Normal contribution 314–315
Normal cost 319
Normal distribution 162, 465
Normal Lives 9
Numerical integration 85, 474–477
Repeated Simpson’s rule 476–477
Trapezium rule 474–475
Office premium 142
Option pricing 401–430
Options 403–404
Out-of-the-money 404
Paid-up sum insured 214
Pandemic risk 351
Participating (par) business 5
Past service benefits 307
Pension Benefits 12–13

491

Pension Plan Valuation 290–325
Actuarial liability 314
Actuarial reduction factor 322
Age retirement 298
Cost of living adjustment (COLA) 309
Current Unit Credit 312
Death-in-service 298, 319
Deferred pension 309
Disability retirement 298
Funding plans 314–318
Ill-health retirement 298
Normal contribution 314–315
Normal cost 318
Projected unit credit (PUC) 312, 315–7
Service table 304
Traditional unit credit (TUC) 312, 315–7
Withdrawal 298, 309–310
Perinatal mortality 50
Permanent dsability model 232–233
Physical measure 409
P-measure 409
Polar method 482
Policy alterations 213–219
Policy values 176–229
Between premium dates 205–207
Definition 183
For policies with annual cash flows 182–196
Gross premium policy value 183
Multiple state models 250–256
Negative policy values 220
Net premium policy value 183
Recursive formulae for policy values
191–196, 204–5, 207
Retrospective policy value 219
Thiele’s differential equation 207–213,
250–256
With discrete cash flows other than annually
203
Policyholder’s fund 374, 375, 432
Pop-up benefit 284
Portfolio percentile premium principle
162–165
Preferred Lives 9, 56
Preferred mortality 56
Premium 1, 10–11, 142–175
Gross premium 142, 150–4
Level premium 2
Mathematical premium 142
Net premium 142, 146–150
Office premium 142
Premium principles 146
Risk premium 142, 445
Single premium 10, 142

492

Index

Premium principles 146
Equivalence principle 146, 388
Portfolio percentile principle 162–165
Pre-need insurance 8
Present value of future loss 145–6, 176–7
Probability distributions 464–469
Profit 154–161, 196–200
Profit margin 359, 360, 368
Profit measures 358–360
Discounted payback period 359, 360, 369,
375
Expected present value of future profit 359,
360
Hurdle rate 359, 360
Internal rate of return (IRR) 359, 360, 368,
375
Net present value 359, 360, 368, 375, 376,
379
Profit margin 359, 360, 368, 375
Risk discount rate 359, 360, 376, 379
Profit signature 358, 368, 375, 376, 378–9
Profit test basis 354
Profit testing 353–400
Deterministic 375–377, 457
Equity-linked life insurance 374–400,
449–457
Profit signature 358, 368, 375, 376, 378–9
Profit vector 358, 368, 375, 376, 378
Stochastic 384–388, 457
Traditional life insurance 353–373
Profit vector 358
Projected unit credit funding (PUC) 312
Proportionate paid-up sum insured 224
Proprietary insurer 14
Prospective reserve 183
Pure endowment insurance 88
Put option 403, 418–420
Put-call parity 421
Q-measure 409, 416
Quantile premium principle 389
Quantile reserve 391
Radix 41
Rated Lives 9
Rating Factors 8
Real world measure 409
Rebalancing errors 423–425, 450
Rebalancing frequency 426
Recursions 81, 130, 204, 207
Recursive calculation of policy values
191–195
Regular premium deferred annuity 11, 148

Related single decrement model 260
Renewal expenses 150
Repeated Simpson’s rule 476–477
Replicating cash flows 332–334, 402
Replicating portfolio 402, 406–407, 414
Reserve 183, 355–358
GMDB 440–444
GMMB 436–438
Reserve basis 356
Retirement benefit 12, 291, 306
Retrospective policy value 219
Reversionary annuity 12, 234, 264
Reversionary Bonus 5, 100, 158
Risk discount rate 359, 360
Risk free rate of interest 414, 433
Risk management 389, 394, 402,
447–449
Risk measures 391
Risk neutral measure 409
Risk premium 142, 445
Runge-Kutta method 220
Salary scale 291–2
Segregated funds 7, 374
Select and ultimate survival model 56
Select life table 59–66
Select lives 56, 101, 136
Select period 57
Selection 56, 101, 136
Self-financing hedge portfolio 412
Semi-Markov process 278
Service table 297–306
Short selling 405
Simpson’s Rule 476
Repeated Simpson’s rule 476
Simulation 342–348, 384–388, 480–482
Normal distribution 481–2
Inverse transform method 480–481
Single decrement model 260
Single premium 10, 142
Single premium deferred annuity 11
Single premium immediate annuity 11
Spot rate of interest 326
Standard normal distribution 465
Standard Select Survival Model 144–145
Table of annuity values 144
Standard Ultimate Survival Model 74
Table of Ax 83
Status for multiple lives 262
Stochastic interest rate 338, 343, 348
Stochastic pricing 388–390
Stochastic process 231
Stochastic reserving 390–395

Index
Stochastic simulation 343
see also ‘Monte Carlo simulation’
Stock price proces 414, 433
Strike price 403, 434
Subjective measure 409
Sum at risk 195
Sum insured 1
Surrender value 213
Survival function
Conditions for valid distribution 19–20
Survival function 18
Tail conditional expectation 393
Tail value at risk 393
Target replacement ratio 294
Term annuity due 112
Term continuous annuity 117
Term immediate annuity 114
Term Insurance Convertible Term 5
Term Insurance Yearly Renewable Term
(YRT) 4
Term Insurance 4–5, 52, 86
The 1/mthly case 87
The annual case 86
The continuous case 86
Term Life Annuity 11, 108
Term life annuity due 112
Term structure of interest rates 326
Terminal Bonus 5
Termination expenses 150
Thiele, Thorvald 220
Thiele’s differential equation 207–213
Total pensionable earnings 312
Traditional unit credit funding (TUC) 312
Transactions costs 414
Transition intensity 237
Trapezium rule 474–475

493

UK Government Actuary 67
Ultimate mortality models 56
Unallocated premium 375
Underwriting 8–10, 55
Uniform distribution 44, 464
Uniform distribution of deaths (UDD) 44, 49,
93, 131, 150
Uninsurable lives 9
Unit Linked Insurance 7, 374
Unitized with-profit insurance 7
Universal Life Insurance 6
US Life Tables 2002 50, 62, 67, 70
Utmost good faith 10
Valuation 183
Value-at-Risk 391–2
Variable Annuity 7, 374
Variable insurance benefits 96–101
Volatility 415
Whole Life Annuity 11, 108
Whole life annuity due 109
Whole life continuous annuity 115
Whole life immediate annuity 113
Whole Life Insurance 5, 14, 75
The 1/mthly case 79
The annual case 78
The continuous case 75
Withdrawal benefit 13, 309
With-Profits 5
Woolhouse’s formula 132–135, 150, 478
Yearly Renewable Term (YRT) 4
Yield curve 326
Zero coupon bonds 326, 405, 414



---