SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE ...

152 Pages·2013·0.55 MB·English

SOCIETY OF ACTUARIES

EXAM MLC Models for Life Contingencies

EXAM MLC SAMPLE SOLUTIONS

The following questions or solutions have been modified since this document was
prepared to use with the syllabus effective spring 2012,

Prior to March 1, 2012:
Questions: 151, 181, 289, 300
Solutions: 2, 284, 289, 290, 295, 300

Changed on March 19, 2012: Questions: 20, 158, 199 (all are minor edits)

Changed on April 24, 2012: Solution: 292

Changed on August 20, 2012: Questions and Solutions 38, 54, 89, 180, 217 and 218
were restored from MLC-09-08 and reworded to conform to the current presentation.
Question 288 was reworded to resolve an ambiguity. A typo in Question 122 was
corrected. Questions and Solutions 301-309 are new questions

Changed on August 23, 2012: Solution 47, initial formula corrected; Solution 72,
minus signs added in the first integral

Changed on January 7, 2013: Question 300 modified and Solutions to 300 and 305
modified

Changed on April 5, 2013: Solution to 22 modified

Changed on September 9, 2013: Typo in Question 32 fixed.

Changed on September 30, 2013: Error in Solution to Question 80 fixed.

Copyright 2011 by the Society of Actuaries

MLC-09-11 PRINTED IN U.S.A.
Question #1
Answer: 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

MLC‐09‐11 1
Question #2
Answer: 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
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.


For example A1  1 E 
x:10  10 x
E e10
10 x

A 
x


With those relationships, the solution becomes
1000A 1000A1  E A 
x  x:10 10 x x10
1000 0.06 1e0.060.0410e0.060.0410 0.07 
   
0.060.04 0.070.05
10000.601e10.5833e1
 
593.86

MLC‐09‐11 2
Question #3
Answer: D

  1
EZ bvt p  dt  e0.06te0.08te0.05t dt
t t x xt
0 0 20

1 100  5
 e0.07t 
  
20 7  0 7
EZ2  bvt2 p  dt e0.12te0.16te0.05t 1 dt  1 e0.09tdt
  t t x xt
0 0 20 20 0

1 100  5
 e0.09t 
  
20 9  0 9
2
5 5
VarZ  0.04535
 
9 7

Question #4
Answer: C

Let ns = nonsmoker and s = smoker

b g b g b g
k = q ns pns q s ps
xk xk xk xk
0 .05 0.95 0.10 0.90
1 .10 0.90 0.20 0.80
2 .15 0.85 0.30 0.70

A1ns v qns v2 pns qns
x:2 x x x1
1 1
0.05 0.950.10 0.1403
1.02 1.022

A1s v qs  v2 ps qs
x:2 x x x1
1 1
0.10 0.900.20 0.2710
1.02 1.022

A1  weighted average = (0.75)(0.1403) + (0.25)(0.2710)
x:2
= 0.1730

MLC‐09‐11 3
Question #5
Answer: B

 1 2 3
    0.0001045
x x x x

p e0.0001045t
t x

APV Benefitset1,000,000 p 1dt
t x x
0
et500,000 p 2dt
t x x
0
e200,000 p 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
Answer: B


EPV Benefits1000A1   E 1000vq
40:20 k 40 40k
k20

EPV Premiumsa   E 1000vq
40:20 k 40 40k
k20
Benefit premiums Equivalence principle 
 
1000A1   E 1000vq a  E 1000vq
40:20 k 40 40k 40:20 k 40 40k
k20 20
1000A1 /a
40:20 40:20
161.320.27414369.13

14.81660.2741411.1454
5.11

While this solution above recognized that 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:

MLC‐09‐11 4
EPV Benefits 1000A 161.32
40

EPV Premiums =a  E  E 1000vq
40:20 20 40 k 60 60k
k0
a  E 1000A
40:20 20 40 60
14.81660.2741411.14540.27414369.13
 
11.7612101.19
11.7612101.19161.32
161.32101.19
 5.11
11.7612

Question #7
Answer: C

ln1.06

A  A  0.530.5147
70 i 70
0.06
1 A 10.5147
a  70  8.5736
70 d 0.06/1.06
a 1vp a 10.978.57368.8457
 
69 69 70
1.06
a2 2a 21.000218.84570.25739
69 69
8.5902
m1
Note that the approximation am a  works well (is closest to the exact
x x
2m
answer, only off by less than 0.01). Since m = 2, this estimate becomes
1
8.8457 8.5957
4

Question #8 - Removed

Question #9 - Removed

MLC‐09‐11 5
Question #10
Answer: E

d = 0.05  v = 0.95

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 1 A /d 10.35076/0.0512.9848
40 40
1000A 350.76
so P  40  27.013
40 a 12.9848
40

E L K 10 1000ARevised P aRevised 549.1827.0139.0164305.62
10 40 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Revised 1 ARevised/d 10.54918/0.059.0164
50 50

Question #11
Answer: 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 

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
MLC‐09‐11 6
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
MLC‐09‐11 7
1vT
Alternatively, here is a solution based on a 
T 
 1 vT 
 
Var a Var  
T  
 
vT 
Var  since VarX constantVarX

 

VarvT
 since VarconstantXconstant2VarX
2
2A A 2
x x
 which is Bowers formula 5.2.9
2

This could be transformed into 2A 2Vara  A2, which we will use to get
x T x
2A NSand 2A S.
x x

2A  Ev2T
x  
 Ev2T NSProbNSEv2T SProbS
   
 2Vara NSA NS2ProbNS
 T x 
2Vara SA S2ProbS
 T x 
 0.018.5030.28620.70
 
0.018.8180.44420.30
 
0.166830.700.285320.30
0.20238

A  EvT
x  
 EvT NSProbNSEvT SProbS
   
0.2860.700.4440.30
0.3334
MLC‐09‐11 8
2A A 2
  x x
Var a 
T 2

0.202380.33342
 9.12
0.01

Question #12 - Removed

Question #13
Answer: D

Let NS denote non-smokers, S denote smokers.

ProbT tProbT t NSProbNSProbT t SProbS
1e0.1t0.71e0.2t0.3
10.7e0.1t 0.3e0.2t

S (t)0.3e0.2t 0.7e0.1t
0
Want tˆ such that 0.751S tˆ or 0.25 S tˆ
0 0
0.250.3e0.2tˆ 0.7e0.1tˆ 0.3e0.1tˆ2 0.7e0.1tˆ

Substitute: let xe0.1tˆ
0.3x20.7x0.250

0.7 0.490.30.254
This is quadratic, so x
20.3

x0.3147

e0.1tˆ 0.3147 so tˆ11.56

MLC‐09‐11 9