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Assignment 2 Solution
1 of 5
- You are given that a) Deaths follow the constant force assumption over each year of age, b) qx = 0.06, c) qx+ 1 = 0.07, d) 0. 054 : 1
x
A
Find : 2
2 1 x
A.
- Assume that the mortality follows the DeMoivre’s Law where w = 100. Let Z be the present value of the payment of a 10-year term insurance payable upon death for (20). Find a) the Z- density, fZ(z) b) the distribution function of Z , FZ(z) c) E [ Zk ], k = 1, 2, 3, … d) the probability that the premium is not sufficient.
Assignment 2 Solution
Due: Friday, October 20, 2006 by 3pm
2 of 5
10 80
ln 10 80
70
10 (^7080)
z
v z
z v
z
z δ
δ
- An annual decreasing 10-year term insurance pay (60) benefit of $(10 – k ), k = 0, 1, 2, …, 9, if (60) dies in year k + 1. You are given that a) A 60 : 1 = 0. 96 ,
b) The actuarial present value (premium) for this insurance is 5 if q 60 = 0.1. Find the actuarial present value (premium) if q 60 = 0.15 and if qx is unchanged for all other ages. HINT: Use (^ )^ (^ )^ (^11) : 1
1 DA (^) x : n |= nvqx + vpx DAx + n −
- Assume there is constant force of mortality μ and constant force of interest δ, let Z be the present value of one unit of whole life insurance that is payable at the end of the year of death bought by a life aged x. a) Find the mean of Z in terms of μ and δ. b) Find the variance of Z in terms of μ and δ.
Constant force of mortality: (^) k px = e −^ μ k , q (^) x + k = 1 − e −^ μ
( )
( ) ( )
( ) ( ) 0
( ) 0
( )
0
( 1 ) 0
1
[ ] 1
δ μ
δ δ
δ μ
μ δ μ
δ δ μ δ μ μ δ
δ μ μ
− +
− −
− +
− − +
− ∞
=
− + − ∞
=
− − +
∞
=
− + − − ∞
=
∑ ∑
∑ ∑
e
e e
e
e
e
e e e e
A EZ v p q e e e
k
k k
k
k
k k k
k x x k k x
Assignment 2 Solution
Due: Friday, October 20, 2006 by 3pm
4 of 5
( )
- 218865987
- 15
30
- 15 1 30
110801
0
1 (^80) =
= (^) ∑
− −
=
A a k
k
ii) (^) ∫
∞ (^) −
0 A (^) x E [ Z ] e δ ttpx μ( x t ) dt
( )
ln
[ ]
30
- (^150)
1
30 0
11080 0 80
^ =
^ =
= =∫ ∫
−
t
t t
A EZ dt dt
b)
( )
- 0795 ln
1
- 15
1 90
1
- 15
1 90
1 90
1
- 15
[ ]^1 90
- (^150) 1
90 0
90 0
^ =
=
=
= (^) ∫ ∫
t
t t EZ dt dt
c)
( )
- 0221064 ln
1
- 15
1 80
1
- 15
1 80
1 80
1
- 15
1 [ ]
80
- (^1510) 1
80 10
80 10
^ =
=
=
= (^) ∫ ∫
t
t t EZ dt dt
- A life insurance policy pays $1000 at the moment of death before age 65 and $2000 at the moment of death after age 65. Calculate the mean and variance of the present value of the benefits at age 50 under this policy if δ = 0.08 and μ ( x ) = 0.02 for all x.
T T
v T Z v T
[ ] (^1) 15| :
(^1000 2000) x x
E Z = A + A
[ 1.^5 ]
15 0
(^150). 08 0. 02 : (^15 )
A^1 x^ = (^) ∫ vttpx μ( x + t ) dt =∫ e −^ te − t 0. 02 dt = − e −
[ 1.^5 ] 15
- 08 0. 02 15 15 |
∞ ∞ (^) − − A x = (^) ∫ vttpx μ x + tdt =∫ e te t dt = e
[ ] (^1) 15| :
(^1000 2000) x x
E Z = A + A = 244.
Assignment 2 Solution
Due: Friday, October 20, 2006 by 3pm
5 of 5
E^ [ Z^2 ]^ = 10002 2 Ax : 15 + 20002152 | Ax
[ 2.^7 ]
15 0
(^1520). 08 0. 02 0
2 : 15
2
A^1 x^ = ∫ vttpx μ( x + t ) dt =∫ e −^ ⋅ te − t 0. 02 dt = − e −
[ 2.^7 ]
15
- 08 0. 02 15
(^215) | 2
∞ ∞ (^) −⋅ −
A x = ∫ vttpx μ x + tdt =∫ e te t dt = e
[ ] 1000 2000 133512. 9487
2 15 |
2 : 15
2 22 EZ = Ax + Ax =
[ ] [ ] ( [ ]) 133512. 9487 ( 244. 626 ) 73671. 05
2 2 2
VarZ = EZ − EZ = − =
