Suggested solutions to DHW textbook exercises
Exercise 2.5
Clearly, F0 (t) is the cdf of an Exponential with mean 1/. So T0 has an Exponential distribution.
(a) Since S0 (t) = et , we have
Sx (t) =
e(x+t)
S0 (x + t)
=
= et .
S0 (x)
ex
Thus, we see that Tx
Suggested solutions to DHW textbook exercises
Exercise 2.4
(a) To show S0 is a legitimate survival function, we show 3 conditions:
(i) S0 (0) = 1: trivial
(ii) lim S0 (x) = 0: Since all parameters A, B , C and D are all positive, then the term
x
Ax + 1 Bx
Suggested solutions to DHW textbook exercises
Exercise 2.3
We are given
(100 x)1/2
1
100 x =
, for 0 x 100.
10
10
The probability that a newborn will die between ages 19 and 36 is given by
S0 (x) =
19|17 q0
= Pr[19 < T0 36] = S0 (19) S0 (36)
=
Prepared by
Suggested solutions to DHW textbook exercises
Exercise 2.2
(a) The implied limiting age is the solution to G( ) = 0 which leads us to
18000 110 2 = ( 90)( + 200) = 0.
Thus, = 90 since the limiting age cannot be negative.
(b) For G to be a legitimate survi
Suggested solutions to DHW textbook exercises
Exercise 2.1
(a) The probability that a newborn life dies before age 60 is given by
Pr[T0 60] = F0 (60) = 1 (1 60/105)1/5 = 1 (45/105)1/5 = 1 (3/7)1/5 = 0.1558791.
(b) The probability that (30) survives to at
Michigan State University
STT 455 - Actuarial Models I
Class Test 1
Monday, 7 October 2013
Total Marks: 100 points
Please write your name and student number at the spaces provided:
Name:
Section No.:
There are ve (5) multiple choice (MC) and one (1) writ
Suggested solutions to DHW textbook exercises
Exercise 2.11
(a) It is not dicult to show that under Makehams law, we have
x
(A + Bcz )dz
S0 (x) = exp
= exp Ax +
0
B
(cx 1)
log(c)
.
It follows therefore that
tpx
= Sx (t) =
S0 (x + t)
S0 (x)
exp A(x + t) +
Suggested solutions to DHW textbook exercises
Exercise 2.12
(a) For Makehams law, it can easily be veried that
px = exp A +
Bx
c (c 1)
log(c)
.
The following R code produces a table of px for x = 0 to x = 130:
A <- .0001
B <- .00035
c <- 1.075
px <- funct
Suggested solutions to DHW textbook exercises
Exercise 2.13
(a) We are given = 2x where refers to smokers and unstarred, non-smokers. It is easy
x
to verify that
t
tpx
t
+s ds
x
= exp
= exp 2
0
2
t
x+s ds
= exp
0
x+s ds
= ( tpx )2 .
0
Note that because
Life Tables and Selection
Lecture: Weeks 4-5
Lecture: Weeks 4-5 (STT 455)
Life Tables and Selection
Fall 2013 - Valdez
1 / 28
Chapter summary
Chapter summary
What is a life table?
also called a mortality table
tabulation of basic mortality functions
deriv
Insurance Benets
Lecture: Weeks 6-8
Lecture: Weeks 6-8 (STT 455)
Insurance Benets
Fall 2013 - Valdez
1 / 36
An introduction
An introduction
Central theme: to quantify the value today of a (random) amount to
be paid at a random time in the future.
main app
Suggested solutions to DHW textbook exercises
Exercise 3.9
(a) Let the constant force between ages [x + k, x + k + 1] be denoted by +k so that
x
px+k = ex+k ,
from which it follows that +k = log px+k . Therefore, we have
x
Pr[Rx s, Kx = k ]
Pr[k < Tx k +
Suggested solutions to DHW textbook exercises
Exercise 3.8
(a) Starting with p =
x
x+1 / x ,
we note that
x
Because
25
= 98363 =
26 ,
24
Starting with px]+2 =
[
x+1
p
x
=
=
22
x+3 / [x]+2 ,
x+3
px]+2
[
[x]+2
=
x+2 .
[21]+2
[x]+2 / [x]+1 ,
=
[20]+1
=
19
Fi
Suggested solutions to DHW textbook exercises
Exercise 3.2
When
x s
are given, it is better to use the direct linear interpolation formula for UDD
= (1 t)
x+t
x
+t
x+1
and the direct exponential interpolation formula for constant force
x+t
=
1t
x
t
x+1 .
Suggested solutions to DHW textbook exercises
Exercise 3.1
The gures below are based on the US Life Table, 2004 prepared by the Center for Disease
Control and Prevention (CDC). The table typies pattern of human population mortality.
0.1
0.01
0.001
100
90
Suggested solutions to DHW textbook exercises
Exercise 2.15
(a) We know that
spx ds
x =
e
=
0
0
S0 (x + s)
1
ds =
S0 (x)
S0 (x)
S0 (x + s)ds.
0
Using a change of variable of integration t = x + s, we nd that
x =
e
1
S0 (x)
S0 (x + t)dt =
0
1
S0 (x)
S0 (t)
Suggested solutions to DHW textbook exercises
Exercise 2.14
(a) Starting with
x =
e
1
tpx dt =
0
0
1+
tpx dt +
px t1px+1 dt
tpx dt = 1 +
1
1
1+
tpx dt
1
t1px+1 dt
spx+1 ds
=1+
0
1
= 1 + x+1
e
The inequalities hold because we know that tpx 1 for all x an