THE UNIVERSITY OF HONG KONG
DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE
STAT3901 Life Contingencies I
Assignment 1 Solution
1. S0 (x):
Z
S0 (x) = exp
0
x
1
dt
10 + t
= exp(−[ln(10 + t)]x0 )
10
= exp ln
10 + x
=
10
10 + x
f0 (x):
f0 (x) = −
d
S0 (x)
dt
= 10(10 + x)−2
10
(10 + x)2
=
Alternate Solution:
f0 (x) = S0 (x)µ(x)
=
10
1
10 + x 10 + x
=
10
(10 + x)2
Sx (t):
Sx (t) =
S0 (x + t)
S0 (x)
=
10
10 + x
10 + x + t 10
=
10 + x
10 + x + t
p30 :
p30 =
=
10 + 30
10 + 30 + 1
40
41
1
5|15 q25 :
5|15 q25
= 15 q 30 5 p25
= 5 p25 (1 − 15 p30 )
=
35
40
(1 − )
40
55
=
21
88
2a.
f0 (x) = −
=
d
S0 (x)
dx
2(a − x)
a2
2b.
µ(x) =
f0 (x)
S0 (x)
=
2(a − x) a2
a2
(a − x)2
=
2
a−x
3. Step 1:
3 p20
= p202 p21
0.9 = 0.95p20
p20 =
18
19
Step 2:
4 p21
Z
= exp −
25
µx dx
21
= exp(−0.1)
Combining the results:
5 p20
= p204 p21
=
18
exp(−0.1)
19
= 0.857214396
2
4a.
t p20
Z
= exp −
20+t
20
1
100 − x
= exp([ln(100 − x)]20+t
20 )
80 − t
= exp ln
80
80 − t
0 ≤ t < 80
80
Z ∞
=
t p20 dt
=
e̊20
0
80
Z
80 − t
dt
80
=
0
= 80 −
802
2(80)
= 80 − 40
= 40
Alternate Solution:
Observe that T20 ∼ U (0, 80), therefore the average time of death is given by
80
2
4b. Since T20 follows the uniform distribution, the following relationship holds:
e20 +
1
= e̊20
2
e20 = 39.5
4c.
2
E(K20
)
=
79
X
k 2 k| q20
k=1
79
1 X 2
k
=
80 k=1
=
1 (79)(80)(159)
80
6
= 2093.5
2
V ar(K20 ) = E(K20
) − e220
= 533.25
3
= 40.
5. Life expectancy before adjustment:
Z 10 1
t
dt + 1010 p20
e̊20:10| =
60
0
1 102
50
=
+ 10
60 2
60
=
55
6
Life expectancy after adjustment:
Z 10 1
e̊20:10| =
t
dt + 1010 p20
60
0
Z 10
Z 1
tt−1 p21 p20 µ20+t dt + 109 p21 p20
tt p20 µ20+t dt +
=
1
0
Z
=
1
Z
tt p20 µ20+t dt + p20
9
(t + 1)t p21 µ21+t dt + 109 p21
0
0
= 8.474045371
Decrease in life expectancy =
55
6
− 8.474045371 = 0.692621296
Remarks: The calculations in this problem can be greatly simplified and reduced
by applying the following formula:
Z n
e̊x:n| =
t px dt
0
which can be proved using integration by parts.
6.
ex:n| =
n−1
X
kk| qx + nn px
k=1
=
n−1
X
k(k px − k+1 px ) + nn px
k=1
=
=
n−1
X
kk px −
n−1
X
k=1
k=1
n−1
X
n
X
kk px −
k=1
(k + 1 − 1)k+1 px + nn px
kk px +
k=2
= px − nn px +
n
X
k=2
=
n
X
k px
k=1
4
n
X
k px
k=2
k px
+ nn px
+ nn px
7a.
50 p20
70
Z
= exp −
20
1
100 − x
= exp([ln(100 − x)]70
20 )
30
= exp ln
80
=
3
8
7b. We are given the following:
Bc30 = 0.001
Bc80 = 0.15
Therefore:
c50 = 150
B=
50 p20
0.001
3
(150) 5
= 0.00004947
Z 70
x
= exp −
Bc dx
20
Bcx
= exp −
ln c
70 !
20
B(c70 − c20 )
= exp −
ln c
= 0.579372007
7c. We are given the following:
A + Bc10 = 0.0014
A + Bc20 = 0.0024
A + Bc30 = 0.0042
5
Therefore:
Bc10 (c10 − 1) = 0.001
Bc20 (c10 − 1) = 0.0018
c10 = 1.8
B = 0.000694444
A = 0.00015
Z 70
x
(A + Bc )dx
50 p20 = exp
20
= exp −50A − B
c70 − c20
ln c
= 0.500307798
7d.
50 p20
Z
= exp −
70
0.001xdx
20
2
70 − 202
= exp −0.001
2
= exp(−2.25)
= 0.105399225
8a.
1−t
1
=
+
x+t p0
x p0
x p0
x+t p0
=1−t+
1
x+1 p0
tx p0
x+1 p0
(1 − t)px + t
1
=
px
t px
t px
=
px
px + t(1 − px )
=
px
px + tqx
6
8b.
u qx+tt px
= t|u qx
= t px − t+u px
u qx+t
=1−
=1−
t+u px
t px
px
px + tqx
px + (t + u)qx px
=
px + (t + u)qx − (px + tqx )
px + (t + u)qx
=
uqx
px + (t + u)qx
8c.
µx+t =
− dtd t px
t px
=
px qx
px + tqx
px (px + tqx )2
=
qx
px + tqx
9.
Z
1
t px dt
e̊x =
+ px e̊x+1
0
Z
1
(1 − tqx )dt + 62(1 − qx )
56 =
0
1
56 = 1 − qx + 62 − 62qx
2
62.5qx = 7
qx = 0.112
10a.
1|1 q[60]+1
= p[60]+1 q60+2
= (1 − 0.11)(0.13)
= 0.1157
7
10b.
0.2 q[60]+1
=
=
0.2 q[60]
1 − 0.7q[60]
0.09(0.2)
1 − 0.7(0.09)
= 0.019210245
10c.
0.7 q[61]+0.7
= 1 −0.7 p[61]+0.7
= 1 − 0.4 p[61]+10.3 p[61]+0.7
= 1 − (p[61]+1 )0.4 (p[61] )0.3
= 1 − (0.88)0.4 (0.9)0.3
= 0.079410914
10d.
0.8 q[62]+0.7
= 1 −0.3 p[62]+0.70.5 p[62]+1
0.3q[62]
p[62]+1
=1− 1−
p[62] + q[62]
p[62]+1 + 0.5q[62]+1
0.87
= 1 − (1 − 0.3(0.11))
0.87 + 0.5(0.13)
= 0.100224599
8