Exam M Actuarial Models - Sample Solutions - The Online Test Page

100 Pages·2005·0.37 MB·English

SOCIETY OF ACTUARIES

EXAM M ACTUARIAL MODELS

EXAM M SAMPLE SOLUTIONS

Copyright 2005 by the Society of Actuaries

Some of the questions in this study note are taken from past SOA examinations.

M-09-05 PRINTED IN U.S.A.
Question #1
Key: E

q = p − p
2 30:34 2 30:34 3 30:34
p =(0.9)(0.8)=0.72
2 30

p =(0.5)(0.4)=0.20
2 34
p =(0.72)(0.20)=0.144
2 30:34
p =0.72+0.20−0.144=0.776
2 30:34
p =(0.72)(0.7)=0.504
3 30
p =(0.20)(0.3)=0.06
3 34
p =(0.504)(0.06)=0.03024
3 30:34
p =0.504+0.06−0.03024
3 30:34
=0.53376

q =0.776−0.53376
2 30:34
=0.24224

Alternatively,

q = q + q − q
2 30:34 2 30 2 34 2 30:34
b g
= p q + p q − p 1− p
2 30 32 2 34 36 2 30:34 32:36
= (0.9)(0.8)(0.3) + (0.5)(0.4)(0.7) – (0.9)(0.8)(0.5)(0.4) [1-(0.7)(0.3)]
= 0.216 + 0.140 – 0.144(0.79)
= 0.24224

Alternatively,

q = q × q − q × q
2 30:34 3 30 3 34 2 30 2 34
=(1− p )(1− p )−(1− p )(1− p )
3 30 3 34 2 30 2 34
=(1−0.504)(1−0.06)−(1−0.72)(1−0.20)
=0.24224

(see first solution for p , p , p , p )
2 30 2 34 3 30 3 34
Question #2
Key: E

1000A =1000⎡A1 + A ⎤
x ⎣ x:10 10 x⎦
=1000⎡∫10e−0.04te−0.06t(0.06)dt+e−0.4e−0.6∫∞e−0.05te−0.07t(0.07)dt⎤
⎢⎣ 0 0 ⎥⎦
=1000⎡0.06∫10e−0.1tdt+e−1(0.07)∫∞e−0.12tdt⎤
⎢⎣ 0 0 ⎥⎦
⎡ 10 ∞⎤
=1000 0.06⎡−e−0.10t⎤ +e−1(0.07)⎡−e−0.12t⎤
⎢ ⎣ 0.10 ⎦ ⎣ 0.12 ⎦ ⎥
⎣ 0 0 ⎦
=1000⎡0.06⎡1−e−1⎤+ 0.07e−1⎡1−e−1.2⎤⎤
⎣0.10⎣ ⎦ 0.12 ⎣ ⎦⎦
=1000(0.37927+0.21460)=593.87

Because this is a timed exam, many candidates will know common results for constant
force
and constant interest without integration.

µ
For example A1 = (1− E )
x:10 µ+δ 10 x
E =e−10(µ+δ)
10 x
µ
A =
x
µ+δ

With those relationships, the solution becomes
1000A =1000⎡A1 + E A ⎤
x ⎣ x:10 10 x x+10⎦
⎡⎛ 0.06 ⎞( −(0.06+0.04)10) −(0.06+0.04)10⎛ 0.07 ⎞⎤
=1000 1−e +e
⎢⎜ ⎟ ⎜ ⎟⎥
⎣⎝0.06+0.04⎠ ⎝0.07+0.05⎠⎦
=1000⎡(0.60)(1−e−1)+0.5833e−1⎤
⎣ ⎥⎦
=593.86
Question #3
Key: A

⎧c(400−x) x<400
B=⎨
⎩ 0 x≥400
100= E(B)=ci400−cE(X ∧400)
⎛ 300 ⎞
=ci400−ci300 1−
⎜ ⎟
⎝ 300+400⎠
⎛ 4⎞
=c 400−300i
⎜ ⎟
⎝ 7⎠
100
c= =0.44
228.6

Question #4
Key: C

Let N =# of computers in department
Let X = cost of a maintenance call
Let S = aggregate cost

Var(X)= ⎡Standard Deviation (X)⎤2 = 2002 = 40,000
⎣ ⎦
E(X2)= Var(X)+⎡E(X)⎤2
⎣ ⎦
=40,000+802 =46,400
E(S)= N×λ×E(X)= N×3×80= 240N
Var(S)= N×λ×E(X2)= N×3×46,400=139,200N

We want 0.1≥Pr(S >1.2E(S))
⎛ S−E(S) 0.2E(S) ⎞ 0.2×240N
≥Pr⎜ > ⎟⇒ ≥1.282=Φ(0.9)
139,200N 139,200N 373.1 N
⎝ ⎠
2
⎛1.282×373.1⎞
N ≥ =99.3
⎜ ⎟
⎝ 48 ⎠
Question #5
Key: B

(τ) (1) (2) (3)
µ =µ +µ +µ =0.0001045
x x x x
p(τ) =e−0.0001045t
t x

APV Benefits= ∫∞e−δt1,000,000 p (τ)µ(1)dt
t x x
0
+∫∞e−δt500,000 p (τ)µ(2)dt
t x x
0
+∫∞e−δτ200,000 p (τ)µ(3)dt
t x x
0
1,000,000 ∞ 500,000 ∞ 250,000 ∞
= ∫ e−0.0601045tdt+ ∫ e−0.0601045tdt+ ∫ e−0.0601045tdt
2,000,000 0 250,000 0 10,000 0
=27.5(16.6377)=457.54

Question #6
Key: B

∞
APV Benefits=1000A1 + ∑ E 1000vq
40:20 k 40 40+k
k=20
∞
APV Premiums=πa(cid:5)(cid:5) + ∑ E 1000vq
40:20 k 40 40+k
k=20
Benefit premiums ⇒Equivalence principle ⇒
∞ ∞
1000A1 + ∑ E 1000vq =πa(cid:5)(cid:5) +∑ E 1000vq
40:20 k 40 40+k 40:20 k 40 40+k
k=20 20
π=1000A1 /a(cid:5)(cid:5)
40:20 40:20
161.32−(0.27414)(369.13)
=
14.8166−(0.27414)(11.1454)
=5.11

While this solution above recognized that π=1000P1 and was structured to take
40:20
advantage of that, it wasn’t necessary, nor would it save much time. Instead, you could
do:

APV Benefits =1000A =161.32
40
∞
APV Premiums =πa(cid:5)(cid:5) + E ∑ E 1000vq
40:20 20 40 k 60 60+k
k=0
=πa(cid:5)(cid:5) + E 1000A
40:20 20 40 60
=π⎡14.8166−(0.27414)(11.1454)⎤+(0.27414)(369.13)
⎣ ⎦
=11.7612π+101.19
11.7612π+101.19=161.32
161.32−101.19
π= =5.11
11.7612

Question #7
Key: C

ln(1.06)
δ
A = A = (0.53)=0.5147
70 i 70 0.06
1− A 1−0.5147
a(cid:5)(cid:5) = 70 = =8.5736
70 d 0.06/1.06
a(cid:5)(cid:5) =1+vp a(cid:5)(cid:5) =1+⎛0.97⎞(8.5736)=8.8457
⎜ ⎟
69 69 70 ⎝1.06⎠
a(cid:5)(cid:5)(2) =α(2)a(cid:5)(cid:5) −β(2)=(1.00021)(8.8457)−0.25739
69 69
=8.5902
(m−1)
Note that the approximation a(cid:5)(cid:5)(m) ≅ a(cid:5)(cid:5) − works well (is closest to the exact answer,
x x
2m
1
only off by less than 0.01). Since m = 2, this estimate becomes 8.8457− =8.5957
4

Question #8
Key: C

The following steps would do in this multiple-choice context:

1. From the answer choices, this is a recursion for an insurance or pure
endowment.
2. Only C and E would satisfy u(70) = 1.0.
1+i
3. It is not E. The recursion for a pure endowment is simpler: u(k)= u(k−1)
p
k−1
4. Thus, it must be C.

More rigorously, transform the recursion to its backward equivalent,
u(k−1) in terms of u(k):
⎛ q ⎞ ⎛1+i ⎞
u(k)=−⎜ k−1 ⎟+⎜ ⎟ u(k−1)
p p
⎝ k−1⎠ ⎝ k−1⎠
p u(k)=−q +(1+i)u(k−1)
k−1 k−1
u(k−1)=vq +vp u(k)
k−1 k−1

This is the form of (a), (b) and (c) on page 119 of Bowers with x=k−1. Thus, the
recursion could be:

A =vq +vp A
x x x x+1
or A1 =vq +vp A 1
x:y−x x x x+1:y−x−1
or A =vq +vp A
x:y−x x x x+1:y−x−1

Condition (iii) forces it to be answer choice C

u(k−1)= A fails at x=69 since it is not true that
x
A =vq +(vp )(1)
69 69 69
u(k−1)= A1 fails at x=69 since it is not true that
x:y−x
A1 =vq +(vp )(1)
69:1 69 69
u(k−1)= A is OK at x=69 since
x:y−x
A =vq +(vp )(1)
69:1 69 69

Note: While writing recursion in backward form gave us something exactly like page
119 of Bowers, in its original forward form it is comparable to problem 8.7 on page 251.
Reasoning from that formula, with π =0 and b =1, should also lead to the correct
h h+1
answer.
Question #9
Key: A

You arrive first if both (A) the first train to arrive is a local and (B) no express arrives in
the 12 minutes after the local arrives.
P(A)=0.75
Expresses arrive at Poisson rate of (0.25)(20)=5 per hour, hence 1 per 12 minutes.

e−110
f (0)= =0.368
0!
A and B are independent, so
P(A and B)=(0.75)(0.368)=0.276

Question #10
Key: E

d = 0.05 → v = 0.095

At issue
49
A = ∑vk+1 q =0.02(v1+...+v50)=0.02v(1−v50)/d =0.35076
40 k 40
k=0
and a(cid:5)(cid:5) =(1− A )/d =(1−0.35076)/0.05=12.9848
40 40
1000A 350.76
so P = 40 = =27.013
40 a(cid:5)(cid:5) 12.9848
40

E( L K(40)≥10)=1000ARevised −P a(cid:5)(cid:5)Revised =549.18−(27.013)(9.0164)=305.62
10 50 40 50

where
24
ARevised = ∑vk+1 qRevised =0.04(v1+...+v25)=0.04v(1−v25)/d =0.54918
50 k 50

k=0
and a(cid:5)(cid:5)Revised =(1− ARevised)/d =(1−0.54918)/0.05=9.0164
50 50
Question #11
Key: E

Let NS denote non-smokers and S denote smokers.

The shortest solution is based on the conditional variance formula

Var(X)= E(Var(X Y))+Var(E(X Y))
Let Y = 1 if smoker; Y = 0 if non-smoker
1− AS
E(a Y =1)= aS = x
T x δ

1−0.444
= =5.56
0.1
1−0.286
( )
Similarly E a Y =0 = =7.14
T 0.1
E(E(a Y))= E(E(a 0))×Prob(Y=0)+E(E(a 1))×Prob(Y=1)
T T T
=(7.14)(0.70)+(5.56)(0.30)
=6.67
E⎡(E(a Y))2⎤ =(7.142)(0.70)+(5.562)(0.30)
⎢⎣ T ⎥⎦
= 44.96
Var(E(a Y))=44.96−6.672 =0.47
T
E(Var(a Y))=(8.503)(0.70)+(8.818)(0.30)
T
=8.60
( )
Var a =8.60+0.47=9.07
T

Alternatively, here is a solution based on
Var(Y)= E(Y2)−⎡E(Y)⎤2, a formula for the variance of any random variable. This
⎣ ⎦
can be
transformed into E(Y2)= Var(Y)+⎡E(Y)⎤2 which we will use in its conditional form
⎣ ⎦
E((a )2 NS)=Var(a NS)+⎡E(a NS)⎤2
T T ⎣ T ⎦

⎡( )2⎤ ( )2
Var⎡a ⎤ = E a − E⎡a ⎤
⎣ T ⎦ ⎢⎣ T ⎥⎦ ⎣ T ⎦
E⎡a ⎤ = E⎡a S⎤×Prob[S]+E⎡a NS⎤×Prob[NS]
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
T T T
=0.30a S +0.70a NS
x x
0.30(1− AS) 0.70(1− ANS)
x x
= +
0.1 0.1
0.30(1−0.444)+0.70(1−0.286)
= =(0.30)(5.56)+(0.70)(7.14)
0.1
=1.67+5.00=6.67

E⎡(a )2⎤ = E⎡a 2 S⎤×Prob[S]+E⎡a 2 NS⎤×Prob[NS]
⎢⎣ T ⎥⎦ ⎣ T ⎦ ⎣ T ⎦
( ( ) ( )2⎞
=0.30 Var a S + E⎡⎣a S⎤⎦ ⎟
T T ⎠

( )
( ) ( )2
+0.70 Var a NS +E a NS
T T
=0.30⎡8.818+(5.56)2⎤+0.70⎡8.503+(7.14)2⎤
⎣ ⎦ ⎣ ⎦
11.919 + 41.638 = 53.557

Var⎡a ⎤ =53.557−(6.67)2 =9.1
⎣ ⎦
T

1−vT
Alternatively, here is a solution based on a =
T δ
⎛ 1 vT ⎞
( )
Var a =Var⎜ − ⎟
T δ δ
⎝ ⎠
⎛−vT ⎞
= Var⎜ ⎟ since Var(X +constant)= Var(X)
δ
⎝ ⎠

Var(vT)
= since Var(constant×X)=constant2×Var(X)
δ2
2A −(A )2
x x
= which is Bowers formula 5.2.9
δ2

This could be transformed into 2A =δ2Var(a )+ A2, which we will use to get
x T x
2A NS and 2A S.
x x