ACTL 90006 Life Insurance Models I
SEMESTER 1, 2014 MID-SEMESTER EXAM
2 May 2014, 3:15–4:15pm
Time allowed: 55 minutes.
Reading time: 5 minutes.
Total number of pages 2.
This exam contributes 20% to the assessment in this subject.
Instructions to invigilators:
No books, lecture notes or other personal materials such as handbooks for calculators
are allowed in the examination room.
Instructions to candidates:
Attempt all Four questions. The number of marks allocated is shown at the end of
each question. The total marks is 40.
Electronic calculators may be used. Only one model of calculator may be used which
must be one of the following:
Casio FX82 (with or without any suffix)
Casio FX83 (with or without any suffix)
Casio FX85 (with or without any suffix)
Sharp EL531 (with or without any suffix)
Texas Instruments BA II Plus (with or without any suffix)
Texas Instruments TI-30 (with or without any suffix).
Write the make and model number of your calculator in the form below.
Student number
Calculator make
Calculator model
1
1. In a certain life table q75 = .05, q76 = .08, q77 = .12, q78 = .20. Assume l75 =
10, 000. Let random variable N be the number of lives surviving to age 77 out of
l75 = 10, 000 lives and let D be the number of lives died before reaching age 77 so
that D +N = l75 = 10, 000. Calculate E(N ), Var(N ), Cov(N,D). [7 marks]
————————————————————————————————————
2. In a certain life table q75 = .05, q76 = .08, q77 = .12, q78 = .20.
(a) Compute f75(1.5) to 4 significant digits, if the uniform distribution of deaths
assumption applies, where fx(t) is the density of Tx at time t.
(b) Compute 0.5|1.5q75.5 to 4 significant digits, if the constant force mortality
assumption applies.
(c) Let Tx be the future lifetime of a life aged x and let Kx = [Tx] be the
curtate future lifetime. Define Rx = Tx Kx to be the fractional part of
Tx. Calculate P
(
R75 ≤ 0.5|K75 = 2
)
to 4 significant digits, if the Balducci
assumption applies. [5+3+5=13 marks]
———————————————————————————————————–
3. Let i.i.d random sample X1, X2, . . . , Xn follow an unknown distribution and let
x1, x2, . . . , xn be their observed values.
If the unknown distribution is U( θ, θ), where θ > 0 is unknown. Find the
maximum likelihood estimate (MLE) of θ.
If the unknown distribution is U(a, b), where 0 < a < b are unknown. Find
the MLE of a and b, respectively. [5+5=10 marks]
———————————————————————————————————–
4. T60 is assumed to follow an unknown distribution function F . A mortality study
of 4 individuals aged 60 resulted in two lifetimes (in years): 15, 30; the remaining
two lives withdrew at ages 70 and 80. The likelihood function
L = [1 F (10)][F (15) F (15 )][1 F (20)][F (30) F (30 )].
Find F (10), F (15 ) and F (30) to maximize L.
Let F (t) be the K-M estimate of F (t). Find F (15).
Plot F (t). [4+4+2=10 marks]
———————————————————————————————————–
END OF EXAM
2
Solutions to the mid-semester exam
1. (a) N ～ B(l0, 2p75), D ～ B(l0, 2q75), E[N ] = l0 2p75 = l0p75p76 = 8740
Var[N ] = l0 2p75(1 2p75) = l0p75p76(1 p75p76) = 1101.24 = Var(D).
(b) 0 = Var(l0) = Var(N + D) = Var(D) + Var(N ) + 2Cov(N,D) this gives
2Cov(N,D) = Var(D) Var(N ) = 2l0l0p75p76(1 p75p76) and Cov(N,D) =
Var(N ) = 1101.24.
2. (a) f75(1.5) = 1.5p75μ75+1.5 = p75 0.5p76μ76+0.5 = p75q76 = 0.076.
(b) 0.5|1.5q75.5 = 0.5p75.5 2p75.5 = p
0.5
75 p
0.5
75 p76p
0.5
77 = 0.1335.
(c)
P
(
R75 ≤ 0.5|K75 = 2
)
=
P
(
R75 ≤ 0.5, K75 = 2
)
P(K75 = 2)
=
2p75 0.5q77
2p75q77
=
0.5q77
q77
=
0.5× q77
1 0.5× q77
1
q77
= 0.5319 .
3. (a) f(xi) =
1
2θ
, θ ≤ xi ≤ θ, i = 1, 2, . . . , n. This shows that |xi| ≤ θ for
i = 1, 2, . . . , n, alternatively, θ ≥ max{|x1|, |x2|, . . . , |xn|}. The likelihood
function is
L =
n∏
i=1
f(xi) =
1
(2θ)n
, θ ≥ max{|x1|, |x2|, . . . , |xn|}.
As L is decreasing of θ, the MLE of θ is θ = max{|x1|, |x2|, . . . , |xn|}.
(b) f(xi) =
1
b a , a ≤ xi ≤ b, i = 1, 2, . . . , n. This shows that
a ≤ min{x1, x2, . . . , xn}, b ≥ max{x1, x2, . . . , xn}.
The likelihood function is
L =
n∏
i=1
f(xi) =
1
(b a)n
, a ≤ min{x1, x2, . . . , xn}, b ≥ max{x1, x2, . . . , xn}.
The MLE of a is
a = min{x1, x2, . . . , xn},
and the MLE of b is
b = max{x1, x2, . . . , xn}.
4. In order to maximize L, we require that
F (10) = F (15 ) = 0, F (30 ) = F (20) = F (15) (to be determined in what follows) , F (30) = 1.
so that
L = F (15)[1 F (15)]2 = a(1 a)2
which is maximized by a = F (15) = 1/3, by setting the derivative to be zero.
The plot is omitted.
3
