Topics in Mathematics II - Actuarial Mathematics
Solutions to Exercises on Hand-Out 2
Frank Coolen (CM206 - Frank.Coolen@durham.ac.uk), February 2008
2-1. We have
10
q
45
= 1 −
10
p
45
= 1 −
45
55
=
10
55
, and
10|10
q
45
=
10
p
45
−
20
p
45
=
10
55
, both of which can also be
directly found by the fact that, for this model, T (45) is uniformly distributed on [0, 55]. This latter fact also makes
immediately clear that µ
45
=
1
55
, which can be verified by evaluating µ
45
= −
d
dt
ln
t
p
x
at t = 45 − x, where
−
d
dt
ln
100 − x − t
100 − x
=
1
100 − x − t
.
(Note: it is also fine just to use, for example, x = 0 here and evaluate the above at t = 45.)
2-2. In this problem, it is easiest to use
t
p
x
=
x+t
p
0
x
p
0
, leading to
17
p
19
=
8
9
,
15
q
36
= 1 −
15
p
36
=
1
8
, and
15|13
q
36
=
15
p
36
−
28
p
36
=
1
8
. To calculate µ
36
, use µ
x+t
=
g
x
(t)
t
p
x
, and set x = 0 and t = 36, in which case g
0
(t) = −
d
dt t
p
0
=
1
200
1 −
t
100
−
1
2
, leading to µ
36
=
1
200
[
1−
36
100
]
=
1
128
. The expected value of T (36) is easily calculated via E(T (x)) =
R
∞
0
t
p
x
dt, which here gives (be careful to use the correct interval on which
t
p
36
is positive for the integral!) E(T (36)) =
R
64
0
t+36
p
0
36
p
0
dt =
1
8
R
64
0
(64 − t)
1
2
=
128
3
.
2-3. In this problem, we have
Z
t
0
µ
x+s
ds = [− ln(85 − s) − 3 ln(105 − s)]
t
s=0
= ln
"
85
85 − t
105
105 − t
3
#
,
so
t
p
x
= e
−
R
t
0
µ
x+s
ds
=
85 − t
85
105 − t
105
3
.
This gives
20
p
x
= 0.4057, and hence
20
q
x
= 1 −
20
p
x
= 0.5943. Further,
20|10
q
x
=
30
q
x
−
20
q
x
=
20
p
x
−
30
p
x
= 0.4057 −
0.2358 = 0.1699,
20|
q
x
=
21
q
x
−
20
q
x
=
20
p
x
−
21
p
x
= 0.4057 − 0.3855 = 0.0202, and
5
p
x+20
=
25
p
x
20
p
x
=
0.3122
0.4057
= 0.7695.
2-4. E(T (45)) =
Z
∞
0
46
46 + t
3
dt = 46
3
−
1
2
(46 + t)
−2
∞
0
= 23.
2-5. Let µ
x+t
= µ for t ∈ [0, 1), then p
x
= e
−
R
1
0
µdt
= e
−µ
, and it is given that this is equal to 1 − q
x
= 0.84. So we
want t such that
t
p
x
= e
−
R
t
0
µ
x+u
du
= e
−tµ
= 0.84
t
= 0.95, hence t =
ln 0.95
ln 0.84
= 0.2942.
2-6.
2|2
q
20
=
2
p
20
−
4
p
20
= e
−0.002
− e
−0.004
= 0.001994.
2-7. (a) e
x
= E(K(x)) = 0 × P (K(x) = 0) + E(K(x)|K(x) ≥ 1) × P (K(x) ≥ 1) = p
x
(1 + e
x+1
), where we use
that E(K(x)|K(x) ≥ 1) = 1 + E(K(x + 1)). (Confirm this for yourself, it is similar to the logical relation that
[T (x)|T (x) ≥ 1] = 1 + T (x + 1).)
(b) From (a) we know that p
x
=
e
x
1+e
x+1
, so
2
p
75
=
1
p
751
p
76
=
e
75
1+e
76
×
e
76
1+e
77
=
10
11
= 0.909.
2-8. The relation
s+t
p
x
=
s
p
x
×
t
p
x+s
leads, with t = 1, x = 30, s = 20, to p
50
=
21
p
30
20
p
30
, so q
50
= 1 − p
50
=
111
6000
=
0.0185. The force of mortality at 30 + t is µ
30+t
= −
d
dt
ln
t
p
30
=
−1
t
p
30
×
d
dt
(
t
p
30
) =
70+2t
7800−70t−t
2
, and setting t = 20
gives µ
50
=
110
6000
= 0.0183.
2-9. (a) Let µ
x
be the constant value of µ
x+u
for 0 < u < 1, then
u
p
x
= e
−
R
u
0
µ
x
ds
= e
−uµ
x
= (p
x
)
u
.
(b) To calculate the central death rate m
x
, consider the denominator:
R
1
0 u
p
x
du =
R
1
0
(p
x
)
u
du =
R
1
0
e
u ln(p
x
)
du =
p
x
−1
ln p
x
=
q
x
µ
x
, using that, in this situation, µ
x
= − ln p
x
, which actually gives µ
x
= 0.5447 as p
x
= 0.580. Therefore, it
follows immediately that, in this situation, m
x
= µ
x
= 0.5447.
(c) We can write m
x
=
R
1
0
u
p
x
µ
x+u
du
R
1
0
u
p
x
du
, hence in general m
x
is a weighted average of the values µ
x+u
for u ∈ (0, 1).
But, of course it is a specially chosen weighted average, namely such that q
x
, with the actual µ
x+u
for u ∈ (0, 1),
is identical to the probability that T (x) ≤ 1, under constant force of mortality m
x
over the interval from time x to
x + 1. To show this, denote q
x
in the first case by P
µ
x+t
(T (x) ≤ 1), and in the second case by P
m
x
(T (x) ≤ 1). Then
P
m
x
(T (x) ≤ 1)
=
Z
1
0
P
x
(T (x) > u)m
x
du = m
x
Z
1
0
u
p
x
du =
R
1
0 u
p
x
µ
x+u
du
R
1
0 u
p
x
du
×
Z
1
0
u
p
x
du
=
Z
1
0
u
p
x
µ
x+u
du = P
µ
x+t
(T (x) ≤ 1).
2-10. As
10
p
35
= 0.81, and
10
p
35
= e
−
R
10
0
µ
35+u
du
= e
−k[35u+0.5u
2
]
10
u=0
= e
−400k
, we have e
−400k
= 0.81. Similarly, we
calculate
20
p
40
= e
−
R
20
0
µ
40+u
du
= e
−1000k
= e
−400k
5
2
= (0.81)
5
2
= 0.59049.
2-11. For this model, ages x ∈ [0, 105] are considered possible, so we restrict x to these values throughout this exercise.
The cumulative distribution function of the future lifetime of a life aged x is G
x
(t) =
t
105−x
for t ∈ [0, 105 − x), and
this function is of course equal to 1 for larger t. This straightforwardly leads to
t
p
x
=
105−t−x
105−x
for x ≤ 105 − t
0
for x > 105 − t
so
20
p
x
=
85−x
105−x
for x ≤ 85
0
for x > 85
q
x
=
1
105−x
for x ≤ 104
1
for x > 104
5|
q
x
=
5
p
x
−
6
p
x
=
1
105−x
for x ≤ 99
100−x
105−x
for 99 < x ≤ 100
0
for x > 100
2-12. We use
t
p
x
=
l
x+t
l
x
and
t
p
x+s
=
l
x+s+t
l
x+s
, the latter relation is, of course, particularly based on the pretty strong
assumption that the given time table is also suitable in s years time, for a life now aged x (note: this even plays a
role for all probabilities here, as future survival probabilities may well be updated with more information available
on a particular generation of the population - as discussed in the lectures). This leads to
(a)
55
p
10
= 0.7763,
45
p
10+10
= 0.7833,
35
p
10+20
= 0.7929,
25
p
10+30
= 0.8090,
15
p
10+40
= 0.8417,
5
p
10+50
= 0.9201.
(b)
55
q
10
= 0.2237.
(c) q
10
= 0.00085, q
50
= 0.00592, q
90
= 0.18877, these are also directly given in the provided table.
(d)
20|10
q
10
= 0.01939,
20|10
q
50
= 0.30184.
(e)
40|
q
10
= 0.00546,
80|
q
10
= 0.02059,
40|
q
50
= 0.02232.
All these probabilities clearly illustrate the effect of ageing. The probabilities in (e) are relatively small, which is of
course due to the fact that these probabilities are for events of surviving the first given number of years, and then
dying in the next year - for the first one, the probability of dying at age 50 is small, while for the last two surviving
to age 90 has pretty small probability.
2-13.
19|17
q
21
=
36
q
21
−
19
q
21
=
19
p
21
−
36
p
21
=
l
40
−l
57
l
21
= 0.1.
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