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Title: associated single decrement tables
Description: This note is for master's or bachelor's students to learn about associated single decrement table calculations. It is as straightforward as possible.
Description: This note is for master's or bachelor's students to learn about associated single decrement table calculations. It is as straightforward as possible.
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Associated Single Decrement Tables
𝑡
(𝜏)
𝑡 𝑝𝑥
(𝜏)
= 𝑒 − ∫0 𝜇𝑥+𝑠𝑑𝑠
In this model, the multiple decrement model with
dependent rates, the transition probabilities depend
on all the forces
...
′(𝑗)
𝑡 𝑞𝑥
𝑡 (𝑗)
′(𝑗)
= 1 − 𝑡 𝑝𝑥
𝑡
=∫
0
𝑛
′(𝑗)
∏ 𝑡𝑝𝑥
′(𝑗)
𝑠𝑝𝑥
(𝑗)
𝜇𝑥+𝑠 𝑑𝑠
(𝜏)
= 𝑡𝑝𝑥
𝑗=1
To go from multiple-decrement to single-decrement probabilities:
(𝜏)
1
...
2
...
Calculate 𝑡 𝑝𝑥′(𝑗) using 𝑡𝑝𝑥′(𝑗) = 𝑒 − ∫0 𝜇𝑥+𝑠𝑑𝑠
(𝜏)
4
...
To go from associated single-decrement probabilities to multiple-decrement probabilities:
1
...
Calculate 𝑡 𝑝𝑥 by multiplying the
exponentiating
...
Calculate 𝑡 𝑞𝑥 using 𝑡𝑞𝑥 = ∫0 𝑡 𝑝𝑥 𝜇𝑥+𝑡 𝑑𝑠
-
(𝑗)
(𝑗)
the 𝜇𝑥 ’s, integrating, and
1
If decrement j is uniform in the single-decrement table, then 𝑠𝑝𝑥′(𝑗) 𝜇𝑥+𝑠 = 𝑤−𝑥
Constant force of decrement
(𝜏)
(𝑗)
If the total force of decrement 𝜇𝑥+𝑠 and one of the forces of decrement 𝜇𝑥+𝑠 are constant between integral ages,
(𝑗)
𝑞𝑥
then apply for this formula:
′(𝑗)
𝑠𝑝𝑥
=(
(𝜏) 𝑞(𝜏)
𝑠𝑝𝑥 ) 𝑥
0≤𝑠≤1
Uniform in the multiple-decrement tables
′(𝑗)
𝑠𝑝𝑥
(𝜏)
= ( 𝑠𝑝𝑥 )
𝑙𝑛 𝑠𝑝𝑥′(𝑗)
(𝜏)
𝑙𝑛 𝑠𝑝𝑥
(𝑗)
(𝜏)
Let 𝜇𝑥 and 𝜇𝑥 be the constant forces of decrement:
(𝜏)
1
...
(𝑗)
𝑞𝑥
(𝜏)
𝑞𝑥
0≤𝑠≤1
(𝑗)
=
𝑞𝑥
(𝜏)
𝑞𝑥
(𝑗)
′(𝑗)
= 𝑒 −𝑠 𝜇𝑥
𝑠𝑝𝑥
(𝜏)
(𝜏)
−𝑠 𝜇𝑥
𝑠𝑝𝑥 = 𝑒
(𝑗)
𝜇𝑥
(𝜏)
(𝜏)
′(𝑗)
= ( 𝑠𝑝𝑥 )𝜇𝑥
𝑠𝑝𝑥
2
...
1
(𝑗)
𝑞𝑥 = ∫
0
1
(𝜏)
𝑞𝑥 = ∫
3
...
0
(𝜏) (𝑗)
𝑡𝑝𝑥 𝜇𝑥 𝑑𝑠
(𝜏) (𝜏)
𝑡𝑝𝑥 𝜇𝑥
(𝜏)
𝑞𝑥
1
(𝜏)
𝑠𝑝𝑥 𝑑𝑠
(𝜏)
0
1
(𝜏)
𝑠𝑝𝑥 𝑑𝑠
𝑑𝑠 = 𝜇𝑥 ∫
0
(𝑗)
𝑞𝑥
(𝑗)
= 𝜇𝑥 ∫
(𝑗)
𝜇𝑥
=
(𝜏)
𝜇𝑥
4
...
(𝑗)
(𝑗)
𝜇𝑥
′(𝑗)
𝑠𝑝𝑥
(𝜏)
𝑞𝑥
(𝜏)
(𝜏)
(𝜏)
= ( 𝑠𝑝𝑥 )𝜇𝑥 = ( 𝑠𝑝𝑥 )𝑞𝑥
′(𝑗)
(𝜏) 𝑙𝑛𝑝𝑥
(𝑗)
•
𝑞𝑥 = 𝑞𝑥 (
•
When 𝑡𝑞𝑥 = 𝑡𝑞𝑥 and 𝑡 𝑞𝑥 = 𝑡𝑞𝑥 for 𝑡 ≤ 1
...
(𝑗)
The trick is to use the 𝑡 𝑝𝑥′(𝑗) 𝜇𝑥+𝑡 = 𝑞𝑥′(𝑗)
• If there are two decrements, then
𝑡
(1)
(𝜏) (1)
𝑡𝑞𝑥 = ∫0 𝑠𝑝𝑥 𝜇𝑥+𝑠 𝑑𝑠
𝑡
(1)
= ∫0 𝑠𝑝𝑥′(1) 𝑠𝑝𝑥′(2) 𝜇𝑥+𝑠 𝑑𝑠
𝑡
(𝜏)
𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑠𝑝𝑥 =
′(1)
′(2)
𝑠𝑝𝑥
𝑠𝑝𝑥
(1)
= ∫0 ( 𝑠𝑝𝑥′(1) 𝜇𝑥+𝑠 ) 𝑠 𝑝𝑥′(2) 𝑑𝑠
𝑡
′(1)
= ∫0 𝑞𝑥
=
=
′(2)
(1 − 𝑠𝑞𝑥 ) 𝑑𝑠
𝑡
𝑞𝑥′(1) ∫0 (1 − 𝑠𝑞𝑥′(2) ) 𝑑𝑠
′(2)
𝑡2𝑞
𝑞𝑥′(1) (𝑡 − 2𝑥 )
′(1) (1)
′(1)
𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑠𝑝𝑥 𝜇𝑥+𝑠 = 𝑞𝑥
′(2)
& 𝑠𝑝𝑥
′(2)
= 1 − 𝑠𝑞𝑥
𝑢𝑛𝑑𝑒𝑟 𝑈𝐷𝐷
= 𝑞𝑥′(1) (1 −
•
′(2)
𝑞𝑥
2
)
𝑤ℎ𝑒𝑛 𝑡 = 1
If there are three decrements, then
(1)
𝑡𝑞𝑥
=
′(1)
(1
𝑡𝑞𝑥
−
𝑡 (𝑞𝑥′(2) + 𝑞𝑥′(3) )
2
+
𝑡 2 𝑞𝑥′(2) 𝑞𝑥′(3)
)
3
Summary: 0 ≤ 𝑠 + 𝑡 ≤ 1
Assumptions
Formula
(𝑗)
𝑞𝑥
′(𝑗)
𝑠 𝑝𝑥+𝑡
Decrements are uniformly distributed in the multipledecrement table, or are constant
(𝑗)
𝑞𝑥
=
(𝜏)
(𝜏)
= ( 𝑠𝑝𝑥+𝑡 )𝑞𝑥
(𝜏)
𝑞𝑥
(
𝑙𝑛𝑝𝑥′(𝑗)
(𝜏)
𝑙𝑛𝑝𝑥
)
𝑞𝑥′(2)
)
2
𝑠 ′(2)
(1)
′(1)
(1 − ( ) 𝑞𝑥 )
𝑠𝑞𝑥 = 𝑠𝑞𝑥
2
′(2)
′(3)
𝑞𝑥 + 𝑞𝑥
𝑞𝑥′(2) 𝑞𝑥′(3)
′(1)
= 𝑞𝑥 (1 −
+
)
2
3
(1)
𝑞𝑥 = 𝑞𝑥′(1) (1 −
There are 2 decrements, and the decrements are
uniformly distributed in the associated singledecrement tables
(1)
𝑞𝑥
There are 3 decrements and the decrements are
uniformly distributed in the associated singledecrement tables
(1)
𝑠 𝑞𝑥
=
𝑞𝑥′(1)
(𝑠 −
𝑠 2 (𝑞𝑥′(2) + 𝑞𝑥′(3) )
2
𝑠 3 𝑞𝑥′(2) 𝑞𝑥′(3)
)
+
3
Continuous insurances
(𝑗)
1
...
Then the net single premium for the insurance is
∞
𝑣𝑡
𝐴=∫
0
𝑛
(𝜏)
(𝑗)
𝑡𝑝𝑥 ∑ 𝜇𝑥+𝑡
𝑗=1
(𝑗)
𝑏𝑡 𝑑𝑡
2
Title: associated single decrement tables
Description: This note is for master's or bachelor's students to learn about associated single decrement table calculations. It is as straightforward as possible.
Description: This note is for master's or bachelor's students to learn about associated single decrement table calculations. It is as straightforward as possible.