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Assignment 1 Solution
1 of 7
- Suppose that 20000
2 x x s x
= for x ≥ 0
a) Find w (the oldest age that a life can live)
Since s ( w )= 0 ,we have 0 200 100 20000
2
= ⇒ =− =
= w or w
w w sw
Therefore, w = 100 as w =− 200 is impossible.
b) Calculate the probability of surviving to age 20.
2
=
s =
c) Calculate the probability that a life aged 20 will survive to age 40.
2
=
s =
20 20 =^ = =
s
s p
d) Find the probability that a life aged 20 will die between 30 and 40.
s
s s q
e) Find the probability that a life aged 19 will die before age 36.
36 − 19 19 = 17 19 =^1 − = − =
s
s q q
- The life table is given as follow:
x qx
i) Find the probability that (26.5) will survive to age 28.25 under
NOTE: In general: (^28). 25 − 26. 5 p 26 (^). 5 = 1. 75 p 26. 5 = 0. 5 p 26. 5 ⋅ p 27 ⋅ 0. 25 p 28
a) the UDD assumption
26
26
- 5 26. 5 0. 5 26. 5 = − ⋅
q
q p q
p 27 = 1 − q 27 = 1 − 0. 0232 = 0. 9768
25 p 28 =^1 −^0. 25 q 28 =^1 −^0.^25 q 28 =^1 −^0.^25 ⋅^0.^0254 =^0.^99365
75 p 26. 5 =^0. 5 p 26. 5 ⋅ p 27 ⋅ 0. 25 p 28 =^0.^960149
Assignment 1 Solution
Due: Friday, October 6, 2006 by 3pm
2 of 7
b) the constant-force assumption
1 1 ( 1 (^ ) )^ (^ )^ ( 1 )^ ( 1 0. 0213 )^0. 989293
5 0. 5 26
5 26
5
5 p 26. 5 =^ − 0. 5 q 26. 5 = − − p 26 = p = − q = − =
p 27 = 1 − q 27 = 1 − 0. 0232 = 0. 9768
( 1 ) ( 1 0. 0254 ) 0. 993589
25 0. 25
25 p 28 =^ − q 28 = − =
75 p 26. 5 =^0. 5 p 26. 5 ⋅ p 27 ⋅ 0. 25 p 28 =^0.^960145
c) the Balducci’s assumption
( )
26
26
- 5 26. 5 0. 5 26. 5 = − ⋅ = − ⋅ = − − −
= − = q q
q p q
p 27 = 1 − q 27 = 1 − 0. 0232 = 0. 9768
( ) ( )
28
- 25 28 = − −
q
p p
- 75 p 26. 5 =^0. 5 p 26. 5 ⋅ p 27 ⋅ 0. 25 p 28 =^0.^960141
ii) Find (^26) : 1. 5
o e under
a) the constant-force assumption
( )
ln 0. 9768
ln 0. 9787
ln 0. 9787
- 5
0
- 5
0
ln 26
1
0
- 5
0 26 27
1
0
ln
- 5
1 26 1 27
1
0 26
- 5
1 26
1
0 26
- 5
0
26 : 1. (^526)
27
26
∫
∫ ∫
∫ ∫
∫ ∫
∫
−
t
t t p
t
t p
t
t
t t
t
p e dt
e dt p p dt
p dt p p dt
p dt p dt
e p dt
o
b) the Balducci’s assumption
Assignment 1 Solution
Due: Friday, October 6, 2006 by 3pm
4 of 7
- 3 3 0. 3 0. 2 5 0. 3 0. 20 0. 10 0. 51
1 3 5
[ ( 96 ) ] ( 2 1 )
[ ( ) ] ( 2 1 )
1 96 2 96 3 96
9996
1
96
2
1
2
= + × × + × × × =
= + +
= −
= −
−
=
∞
=
p p p
EK k p
EK x k p
k
k
k
k x
[ ]
[ ( 96 )] [ ( 96 ) ] [ ( 96 )] 0. 51 0. 366 0. 376044
( ) [ ( ) ] [ ( )]
2 2 2
2 2
= − = − =
= −
VarK EK E K
VarK x EK x EK x
- If x
x −
μ ( ) for 0 < x < 50 ,
a) find s(x).
2 50
(^0 )
( ) exp () exp
x dt t
s x t dt
x
μ for 0 < x < 50
b) Calculate the exact value of 20
o e.
30
0
2
30
0 2
2
30
0
5020
0
(^2020)
^ =
−
dt
t
dt
t
dt
s
s t
e tp dt
o
c) Calculate the exact value of e 20. (HINT:
1
=
nn n x
n
x
- 5056
900 60 900
1
30
30
29
1
2
29
1
50201 2
1
20 96
=
= − +
−
= =
− −
k = k k
k k k
k e p
d) Calculate the exact value of 20 : 5
e.
Assignment 1 Solution
Due: Friday, October 6, 2006 by 3pm
5 of 7
( ) ( )
180
11 4
900 60 900
1
30
30
5
1
2
5
1
5 2
1
20 : 5 20
=
= − +
− = (^) ∑ =∑ ∑
k = k = k =
k k k
k e p
- Assume that the force of mortality follows the DeMoivre’s law, where x
x −
for 0 ≤ x < 80 with l 0 = 800. Find
x s x = −
x
x t t px −
lx = l 0 s ( x )= 800 − 10 x
l 20 = l 0 s ( 20 )= 800 − 10 ⋅ 20 = 600
l 21 = l 0 s ( 21 )= 800 − 10 ⋅ 21 = 590
∫
1
0
L (^) x lxex : 1 lx tpxdt
o
1
0
1 20 20 0 20
∫ ∫ dt
t L l tp dt
a) 1 m 20
20
20 21 1 20 =
L
l l m
b) 10 L 20
10
0
10
0
∫ ∫ dt
t L l tp dt
c) a(20)
20 21
20 21
−
l l
L l a
d) (^20)
o e
0
60
0 20
∫ ∫
dt
t
e p dt
t
o
e) T 20
20 =^020 =^800 ⋅^30 =^24000
o
T l e
Assignment 1 Solution
Due: Friday, October 6, 2006 by 3pm
7 of 7
[ ] [ ] [ ]
( (^) [ ]) (^) [ ] [ ]
[ ]
1
0
1
0 55 1
1
0
2
1 55 1 55 1
1
0 55
2
1 55
1
0 55
2
0 55
∫
∫ ∫
∫ ∫
∫ ∫ ∫
− +
t dt
t dt p dt
tq dt p p dt
p dt p dt p dt
t
t
t t t
h) (^) [ ] ∫
∞
2 55 tp dt
HINT:
∫ ∫ ∫
∞ ∞ = + a
a f ( t ) dt f ( t ) dt f ( t ) dt 0 0
and recognize [ ] (^) [ ] ∫
∞
0
e x tpxdt
o
[ ] [ ] [ ] [^ ]^ [ ]
2
0
(^5555)
2
0 55 0 55 2 55
∫ ∫ ∫ ∫
∞ ∞ tp^ dt tp dt tp dt e tp^ dt
o
i) 57
o e
HINT: Use result from (h).
[ ] [ ] [ ]
57
57 0 57 0 2 55 55 2 2 55
∫ ∫ ∫
∞ ∞
∞
o
o
e
tp^ dt p tp dt tp dt e
j) (^) [ 56 ]
o e
HINT: Use result from (i).
[ ] (^) [ ] [ ] [ ] (^ [ ])^5 850
0 57
1
0 56 1 56 1 57
1
0 56 0
56 =^ ∫ 56 =∫ + ∫ =∫ − + ∫ =
∞ ∞ −
∞ e (^) tp dt tp dt p t p dt tq dt tp dt
o
