EXERCISE 1
ACTUARIAL MATHEMATICS
SURVIVAL MODELS
1. Given: 8 1=6
< t
S0 (t) = 1 , 0 t 120.
: 120
0, otherwise.
Determine:
(a) the probability that (0) survives to age 30 .
(b) the probability that (30) dies by age 50.
(c) the probability that (40) survives at least 25 years.
Answer
1=6 1=6
30 3
(a) We want Pr (T0 > 30) = S0 (30) = 1 = = 0:9532.
120 4
50 1=6
S0 (30 + 20) 1 120
(b) We want Pr (T30 20) = F30 (20) = 1 S30 (20) = 1 =1 1=6
=
S0 (30) 1 12030
0:0410 (low, as expected).
S0 (40 + 25)
(c) We want Pr (T40 > 25) = S40 (25) = = 0:9395 (high, as expected).
S0 (40)
1
2. Given: 8 1=6
< t
S0 (t) = 1 , 0 t 120.
: 120
0, otherwise.
Check all six conditions on S0 (t).
Answer
S0 (0) = (1 0)1=6 = (1)1=6 = 1
lim S0 (t) = 0
t!1
This is true as S0 (t) = 0 for t > 120.
" #
5=6
1 t 1
S00 (t) = 1
6 120 120
S00 (t) is negative.
S0 (t) is a non-increasing function of t.
S0 (t) is di¤erentiable.
t S0 (t) and t2 S0 (t) are 0 for t > 120, so lim t S0 (t) = 0 and lim t2 S0 (t) = 0.
t!1 t!1
In this model, the age 120 is a limiting age. This is the maximum age that anyone can
live to.
If a survival model has a limiting age, then lim S0 (t) = 0, lim t S0 (t) = 0, and lim t2 S0 (t) =
t!1 t!1 t!1
0 are automatically satis…ed.
2
3. Given: 8 1=6
< t
S0 (t) = 1 , 0 t 120.
: 120
0, otherwise.
Derive a formula for S30 (t), the survival function for (30).
Answer
We want S30 (t). By de…nition,
1=6 1=6
30 + t 120 30 t
1 1=6
S0 (30 + t) 120 120 120 30 t
S30 (t) = = 1=6
= 1=6
=
S0 (30) 30 120 30 120 30
1
120 120
1=6 1=6
90 t t
= = 1
90 90
8 1=6
< t
S30 (t) = 1 , 0 t 90.
: 90
0, otherwise.
In this model, the limiting age is 120. It "embedded" in the survival function for (30).
Our person is already 30 years old. There are a max of 90 years of future lifetime available
for (30). Limiting age is still 120.
3
4. Given the following survival model:
x 1=6
S0 (x) = 1 , 0 x 120; 0, otherwise.
120
This is a Generalized De Moivre model. In this case, we have = 1=6. The De Moivre
model (UDD) is a special case with = 1. Determine x .
Answer
f0 (x)
x = .
S0 (x)
In this case,
1 x 5=6
f0 (x) = 1 .
720 120
So,
5=6
x 5=6 1 120 x
1
720
1 720
120
= 120 = .
x
x 1=6 120 x
1=6
1
120 120
Simplifying this, we get
1
1 120 x
x = .
720 120
We can rewrite this as
1 120 1 1 1=6
x = = = .
720 120 x 6 120 x 120 x
In general, we can write x as
x = .
! x
4
5. Under the Gompertz law of mortality,
B cx
Sx (t) = exp ct 1 .
log (c)
Suppose that B = 0:0001 and c = 1:05. Find the probability that (25) survives to age 35.
Answer
We want S25 (10).
In our case,
!
0:0001 (1:05)25
S25 (10) = exp 1:0510 1 = 0:9956 (high, as expected):
log (1:05)
5
6. Given:
Survival model: Generalized De Moivre with = 1=6 and ! = 120.
Find ex for x = 0, 50, and 100.
Answer
R1
We know E (Tx ) = ex = 0 t px dt.
In our case, we have a limiting age, so
Z 120 x Z 120 x 1=6
t
ex = t px dt = 1 dt
0 0 120 x
" #120 x
7=6
6 t
= (120 x) 1
7 120 x
0
6 6
= (120 x) [0 1] = (120 x) .
7 7
6
e0 = (120) = 102:86 (expected age at death of (0) ).
7
6
e50 = (120 50) = 60 (expected age at death of (50) ).
7
6
e100 = (120 100) = 17:14 (expected age at death of (100) ).
7
In this problem, we ended up with a nice closed form expression for ex . It is possible to
end up with a result that need to be approximated using numerical methods. On a test,
you don’t need to worry about a case like this. I’d give you enough information to get an
answer (i.e., tables).
6
7. Given:
Mortality: Generalized De Moivre with ! = 120 and = 1=6.
Find V ar (Tx ) for x = 0, 50, and 100.
Answer
We showed that
t
t px =1 , 0 t 120 x.
120 x
V ar (Tx ) = E (Tx2 ) [E (Tx )]2 .
Z 120 x Z 120 x 1=6
t
E Tx2 =2 t t px dt = 2 t 1 dt
0 0 120 x
Substitution:
t
Let y = 1
120 x
) t = (120 x) (1 y)
) dt = (120 x) dy
t 2 (0; 120 x) ) y 2 (1; 0)
Z 0
E Tx2 = 2 (120 x) (1 y) y 1=6 (120 x) dy
1
Z 1
2
= 2 (120 x) (1 y) y 1=6 dy
0
1
6 7=6 6 13=6
= 2 (120 x)2 y y
7 13 0
6 6
= 2 (120 x)2 .
7 13
So,
2
2 6 6 6
V ar (Tx ) = 2 (120 x) (120 x)
7 13 7
= 0:056514914 (120 x)2 .
V ar (T0 ) = 813:8148.
V ar (T50 ) = 276:9231.
V ar (T100 ) = 22:606.
We found that V ar (Tx ) # as x ".
