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Multiple Decrements
Insurance and annuities
(j)
1
...

Then the net single premium for the insurance is
𝑛

∞

A=∫

vt

0

(τ)
t px

(𝑗)

(j)

∑ 𝜇𝑥+𝑡 bt dt
𝑗=1

2
...
In any multiple decrement situation, each of the decrement is mutually exclusive, Z is the benefit random
(j)
variable for an insurance paying bt at time t upon occurrence of decrement j, then
𝑚
(τ)
(𝑗)
(𝑗)
E[Z] = ∫
𝑡 𝑝𝑥 ∑ 𝜇𝑥+𝑡 𝑏𝑡 dt
0
𝑗=1
𝑚
∞
(τ)
(𝑗)
(𝑗) 2
E[Z 2 ] = ∫ 𝑣 2𝑡 𝑡 𝑝𝑥 ∑ 𝜇𝑥+𝑡 (𝑏𝑡 ) dt
0
𝑗=1
(2)
(1)
∗
∞

𝑣𝑡

-

Special formula for additional EPV of additional benefit 𝑏 = 𝑏
secondary decrement with constant force 𝜇 (2) : 𝑏 ∗ 𝜇 (2) 𝑎𝑥: 𝑛

−𝑏

paid for the first n years on

Examples
1
...

q′(1)
= 0
...

x
′(2)
II
...
12 and is uniformly distributed with each year of age
...

q′(3)
= 0
...
5 in each year
...
5)(0
...
25(0
...
5)(0
...
2)(1 − 0
...
12))
= 0
...
0182 = 0
...
In a double-decrement table:
′(1)
I
...
1
II
...

1 − 0
...

={
𝑡𝑝𝑥
0
...

𝑙𝑥 = 1000
(2)
Calculate 𝑑𝑥
...
8, but it then drops instantaneously to 0
...
3 at 1; 30% of the population immediately before t=1 is decremented by
it
...
2 by the average surviving from decrement 1, which is the midpoint,
or 1 − 0
...
5) = 0
...
The multiply the point mass of 0
...
1 = 0
...

(2)
𝑞𝑥 = 0
...
95) + 0
...
9) = 0
...
46) = 460
3
...
Benefits are paid at the moment of
death
...

You are given:
I
...
001
II
...
02
III
...
06
Using the normal approximation, calculate the size of the fund needed at issue so that the probability that
there will be enough funds to pay all benefits is 95%
...

(𝜏)
−𝜇 (𝜏) 𝑡 = 𝑒 −0
...
6 )
𝐸[𝑍] = ∫ 𝑒 −0
...
02𝑡 0
...
06𝑡 𝑒 −0
...
001(9000)𝑑𝑡 =
+
0
...
08
0
0
= 339
...

Let 𝑍2 be the present value for the benefit received for a non-accidental death in the first 20 years
...

𝑍1 , 𝑍2 , 𝑍3 are mutually exclusive
...
12𝑡 𝑒 −0
...
001(100002 )𝑑𝑡
0

20

= 100,000 ∫ 𝑒 −0
...
8 )
=
0
...
96
𝜇 (2) = 𝜇 (𝜏) − 𝜇 (1)
20

𝐸[𝑍22 ] = ∫ 𝑒 −0
...
02𝑡 (0
...
001)(10002 )𝑑𝑡
0

20

= 19,000 ∫ 𝑒 −0
...
8 )
=
0
...
49
∞

𝐸[𝑍32 ] = ∫ 𝑒 −0
...
02𝑡 (0
...
14𝑡 𝑑𝑡
20

20,000 −2
...
14
= 8687
...
96 + 127,461
...
16 = 806,998
...
60 − 339
...
64
The funding needed for 95% probability of having enough fund is
500(339
...
645√500(691,543
...
00
4
...
The pure endowment benefit is 1
...
You are given:
I
...
001
...

𝐴45: 20 = 0
...

𝛿 = 0
...

(1)
- Let the decrement for accidental death be (1)
...
001

20

(𝜏)

(𝜏)

(1)

(𝜏)

𝐴 = ∫ 𝑣 𝑡 𝑡 𝑝45 (𝜇45+𝑡 + 𝜇45+𝑡 ) 𝑑𝑡 + 𝑣 20 20𝑝45
0

20

(𝜏) (𝜏)

(𝜏)

20

(𝜏)

= (∫ 𝑣 𝑡 𝑡𝑝45 𝜇45+𝑡 𝑑𝑡 + 𝑣 20 20𝑝45 ) + 𝜇 (1) ∫ 𝑣 𝑡 𝑡𝑝45 𝑑𝑡
0

0

= 𝐴45: 20 + 0
...
3 + 0
...
3 + 0
...
04
0
...
3)
= 0
...
04
= 0

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