Exercises_-_Chapter_7_-_Solutions - ACTSCI 331 Life...

ACTSCI 331 - Life Contingencies I
Solutions (Chapter 7)
1. (a) The bene°t premium is 2000
20
P
°
A
1
35: 30 j
±
which is obtained by
2000
20
P
°
A
1
35: 30 j
±
= 2000
A
1
35: 30 j
a
35: 20 j
=
2000 (0 : 06873)
11 : 39534
= 12 : 063
(b) The prospective loss at time 10 is de°ned as
10
L =
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
2000 v T ° 10 ° 2000
20
P
°
A
1
35: 30 j
±
a
T ° 10 j
, 10 < T < 20
2000 v T ° 10 ° 2000
20
P
°
A
1
35: 30 j
±
a
10 j
, 20 < T < 30
0 ° 2000
20
P
°
A
1
35: 30 j
±
a
10 j
, T > 30
,
while the prospective loss at time 25 is de°ned as
25
L =
²
2000 v T ° 25 , 25 < T < 30
0 , T > 30
.
(c) To obtain the values of
10
L at T = 12 : 3 , we simply replace T by 12.3 in the de°nition of
10
L by
choosing the expression in the appropriate domain of de°nition. It follows that
(
10
L j T = 12 : 3) = 2000 v
12 : 3 ° 10
° 2000
20
P
°
A
1
35: 30 j
±
a
12 : 3 ° 10 j
= 2000 v
2 : 3
° 12 : 063 a
2 : 3 j
= 2000 e
° 0 : 06(2 : 3)
° 12 : 063
°
1 ° e
° 0 : 06(2 : 3)
0 : 06
±
= 1716 : 3
Identical technique for
25
L at T = 27 : 4
(
25
L j T = 27 : 4) = 2000 v
27 : 4 ° 25
= 2000 v
2 : 4
= 2000 e
° 0 : 06(2 : 4)
= 1731 : 8
(d) The bene°t reserve
10
V is given by
10
V = 2000 A
1
45: 20 j
° 2000
20
P
°
A
1
35: 30 j
±
a
45: 10 j
= 2000 (0 : 09710) ° 12 : 063 (7 : 35175)
= 105 : 52
1

while
25
V = 2000 A
1
60: 5 j
= 2000 (0 : 07952)
= 159 : 04
(e) From the de°nition of
10
L , one identi°es that
10
L is a strictly decreasing function of the time until
death T for which (
10
L j T = 10) = 2000 . Thus,
Pr (
10
L ± 1800 j T > 10) = Pr ( T > t
±
j T > 10) , (1)
for which t ± is such that
10
L is taken the value 1800 , i.e.
(
10
L j T = t
±
) = 1800 .
To do so, we shall identify in which brackets of the de°nition of
10
L the value t
±
belongs ( (10 ; 20) ,
(20 ; 30) or (30 ; 1 ) ). Note that (
10
L j T = 10) = 2000 and
(
10
L j T = 20) = 2000 v
10
° 2000
20
P
°
A
1
35: 30 j
±
a
10 j
= 2000 e
° 0 : 06(10)
° (12 : 063)
°
1 ° e
° 0 : 06(10)
0 : 06
±
= 1006 : 9 .
Given that
10
L is a decreasing function of T , it follows that t
±
2 (10 ; 20) . Then, t
±
is the unique
solution of
2000 v
t
°
° 10
° 2000
20
P
°
A
1
35: 30 j
±
a
t ° ° 10 j
= 1800 ,
which is equivalent to
2000 v
t
°
° 10
° 12 : 063
°
1 ° v
t
°
° 10
0 : 06
±
= 1800
°
2000 +
12 : 063
0 : 06
±
v
t
°
° 10
= 1800 +
12 : 063
0 : 06
.
It follows that
v
t
°
° 10
=
1800 +
12 : 063
0 : 06
2000 +
12 : 063
0 : 06
,
or
t
±
= 10 °
1
0 : 06
ln
°
1800 +
12 : 063
0 : 06
2000 +
12 : 063
0 : 06
±
= 11 : 5877 .
Substituting back t ± in (1), one °nds
Pr (
10
L ± 1800 j T > 10) = Pr ( T > 11 : 5877 j T > 10)
=
1 : 5877
p
45
= exp
³
° 0 : 0005246 ² 1 : 1
45
´
1 : 1
1 : 5877
° 1
µ¶
= 0 : 99377 ,
2

2. (a) The bene°t premium 4000 P
´
20 j
A
45
µ
is calculated as follows:
4000 P
´
20 j
A
45
µ
= 4000
20 j
A
45
a
45: 20 j
=
4000 (0 : 11630)
11 : 01577
= 42 : 23
(b) The prospective loss at time 10 is de°ned as
10
L =
(
0 ° 4000 P
´
20 j
A
45
µ
a
T ° 10 j
, 10 < T < 20
4000 v
T ° 10
° 4000 P
´
20 j
A
45
µ
a
10 j
, T > 20
,
while the prospective loss at time 25 is de°ned as
25
L =
³
4000 v T ° 25 , T > 25 .
(c) To obtain the values of
10
L at T = 12 : 3 , we simply replace T by 12.3 in the de°nition of
10
L by
choosing the expression in the appropriate domain of de°nition. It follows that
(
10
L j T = 12 : 3) = 0 ° 4000 P
´
20 j
A
45
µ
a
12 : 3 ° 10 j
= ° 4000 P
´
20 j
A
45
µ
a
2 : 3 j
= ° 42 : 23
°
1 ° e
° 0 : 06(2 : 3)
0 : 06
±
= ° 90 : 725
Identical technique for
25
L at T = 27 : 4
(
25
L j T = 27 : 4) = 4000 v
27 : 4 ° 25
= 4000 v
2 : 4
= 4000 e
° 0 : 06(2 : 4)
= 3463 : 6
(d) The bene°t reserve
10
V is given by
10
V = 4000
´
10 j
A
55
µ
° 4000 P
´
20 j
A
45
µ
a
55: 10 j
= 4000 (0 : 22524) ° 42 : 23 (7 : 09581)
= 601 : 3
while
25
V = 4000 A
70
= 4000 (0 : 56246)
= 2249 : 8
(e) From the de°nition of
10
L , one notices that (
10
L j T = 10) = 0 . Also,
10
L is a strictly decreasing
function in (10 ; 20) . At T = 20 ,
10
L experiences a jump
´
10
L j T = 20
°
µ
= 0 ° 4000 P
´
20 j
A
45
µ
a
10 j
´
10
L j T = 20
+
µ
= 4000 v
10
° 4000 P
´
20 j
A
45
µ
a
10