Math 630, Problem Set 1
1.
Given: l1 = 9700, q1 = q2 = .02, q4 = .026, and d3 = 232. Determine the number of
survivors to age 5.
2.
Given t| qx = .10 for t = 0, 1, . . . , 9, calculate 2 px+5 .
3.
Given S0 (x) =
10000−x2
10000
for 0 ≤ x ≤ 100, calculate q32 .
(
4.
A person age 70 is subject to the force of mortality µ70+t =
Calculate
.01, t ≤ 5
.02, t > 5
.
20 p70 .
2x
10000−x2
for 0 ≤ x ≤ 100, determine qx .
5.
Given µx =
6.
For a standard insured, the force of mortality µ30+t for 0 ≤ t ≤ 1 and p30 = .95. For a
preferred insured, the force of mortality is µ30+t − c for 0 ≤ t ≤ 1. Determine c such
that qx is reduced by 25%.
7.
A life table for severely disabled is created by modifying an existing life table by doubling
the force of mortality at all ages. In the original table, q75 = .12. Calculate q75 in the
modified table.
8.
In DML with ω = 105, calculate
9.
In a DML model, q10 = 1/45. Determine µ10 .
10|20 q25 .
10. A life is subject to constant force of mortality µ. Given that e50 = 24, determine µ.
◦
11. Given that mortality follows DML and e30 = 30, find q30 .
◦
12. Mortality follows DML. Var(T15 ) = 675. Calculate e25 .
◦
13. Given that S0 (x) = e−0.05x , find e30 .
14. Hens lay an average of 30 eggs each month until death. The survival function for
m
where m is in months. 100 hens have survived to age 12
hens is S0 (m) = 1 − 72
months. Calculate the expected total number of eggs to be laid by these 100 hens in
their remaining lifetimes.
15. Given: e35 = 49 and p35 = .995. If µx is doubled for 35 ≤ x ≤ 36, what is the revised
value of e35 ?
16. Given that 1 qx+ 3 =
4
4
3
31
and mortality is UDD from age x to x + 1, calculate qx .
17. Deaths are UDD between integral ages, qx = 0.10, and qx+1 = 0.15. Calculate 0.3|0.5 qx+0.4 .
18. A mortality study is conducted for the age interval [x, x + 1]. If a constant force of
mortality applies over the interval, 1 qx+ 1 = 0.05. Calculate 0.25 qx+0.1 assuming UDD
4
10
over the interval.
19. Given
x
qx
90
0.10
91
0.12
92
0.13
q[x] = 0.5qx ,
93
0.15
94
0.16
and
q[x]+1 = 0.75qx+1 ,
l[91] = 10000,
find l[90] if the select period is 2 years.
20. Given a 2-year select period and the table
x 1000q[x]
40
.438
41
.453
42
.477
43
.510
44
.551
1000q[x]+1
.574
.599
.634
.680
.737
1000qx+2
.699
.738
.790
.856
.937
calculate 10001|2 q[41] .
21. Given a 2-year select period
x l[x]
24 —
25 —
26 —
l[x]+1
—
—
—
lx+2
x+2
42,683
26
35,000
27
26,000
28
and q[x]+1 = 1.5q[x+1] , q[x]+2 = 1.2q[x+1]+1 , calculate l[26] .
22. Mortality follows Gompertz Law with B = 0.002 and c = 1.03, find t such that t p45 = 0.6.
Answers
2x+1
6. 0.013072;
1. 8848;
2. 3/5;
3. 0.007242;
4. 0.70469;
5. 10000−x
2;
7. 0.2256;
8. 1/4;
9. 1/45;
10. .040822;
11. 1/60;
12. 40;
13. 20;
14. 90000; 15. 48.755; 16. 0.3; 17. 0.059375;
18. 0.04725; 19. 11279.53;
20. 1.336;
21. 36944;
22. 37.12674.
Solutions to Selected Problems
4.
20 p70
Z
= exp −
20
Z
µ70+t dt = exp −
Z
= exp −
5
Z
20
0.01 dt −
0
◦
20
µ70+t dt −
µ70+t dt =
5
0.02 dt = e−0.35 .
5
ω − 30
1
1
= 30 ⇒ ω = 90. So q30 =
= .
2
ω − 30
60
◦
13. S0 (x) = e−0.05x ⇒ CFM with µ = 0.05. So e30 =
18. First note that
Under CFM,
Z
0
0
11. e30 = 30 ⇒
5
0.35 px
0.25 qx+0.1
= 0.1 px · 0.25 px+0.1
= 0.05 ⇒
0.25 px+0.1
1
1
=
= 20.
µ
0.05
(1)
= 0.95 ⇒ e0.25µ = 0.95 ⇒ eµ = 0.954 = px
Under UDD, (1) becomes 1 − (0.35)qx = (1 − 0.1qx )(0.25 px+0.1 ).
Using qx = 1 − px = 1 − 0.954 in the last equation, we get 0.25 px+0.1 = 0.04725.
20. We have 1000 · 1|2 q[41] = 1000 · p[41] · 2 q[41]+1 . Further more, p[41] = 1 − q[41]
2 q[41]+1
and
= 1 − 2 p[41]+1 = 1 − p[41]+1 · p41+2 = 1 − (1 − q[41]+1 )(1 − q41+2 ) =
= q[41]+1 + q41+2 − q[41]+1 · q41+2 .
Then use the values given in the table.
21. (Sketch)
• First note that q[x]+2 = qx+2 , so we are actually given that qx+2 = 1.2q[x+1]+1 .
• Using the given l26 , l27 and l28 , we can find p26 and p27 .
• Using p26 and p27 with qx+2 = 1.2q[x+1]+1 , we can find p[25]+1 and p[26]+1 .
• Using p[25]+1 and p[26]+1 with q[x]+1 = 1.5q[x+1] , we can find p[26] and p[27] .
• Using the given l28 = l26 + 2, p[26]+1 , and p[26] , we can find l[26] .
22.
Under Gompertz’s law, µx = Bcx = (0.002)(1.03)x . So,
Z t
0.003
45+s
45
t
· 1.03 · (1.03 − 1) = 0.6
(0.002)(1.03)
ds = exp −
t p45 = exp −
ln 1.03
0
Solving for t in the last equation.