1. No category

SOCIETY OF ACTUARIES
EXAM MLC Models for Life Contingencies
EXAM MLC SAMPLE SOLUTIONS
Copyright 2013 by the Society of Actuaries
The solutions in this study note were previously presented in study note MLC-09-08
and MLC-09-11. They have been edited for use under the 2014 learning objectives
and textbook. Most solutions are mathematically the same as those in MLC-09-11.
Solutions whose wording has changed are identified with a * before the question
number. The only question which has changed mathematically is 300. Questions
310-318 are new.
Some of the questions are taken from past SOA examinations. No questions from
published exams after 2005 are included except 310-313, which come from exams of
2012 or 2013. The November 2006 Exam M and May 2007, May 2012, November
2012, May 2013 and November 2013 Exam MLC are available at
http://www.soa.org/education/exam-req/syllabus-study-materials/edu-multiple-choiceexam.aspx.
The average time allotted per multiple choice question will be shorter beginning with
the Spring 2014 examination. Some of the questions here would be too long for the
new format. However, the calculations, principles, and concepts they use are still
covered by the learning objectives. They could appear in shorter multiple choice
questions, perhaps with intermediate results given, or in written answer questions.
Some of the questions here would still be appropriate as multiple choice questions in
the new format.
The weight of topics in these sample questions is not representative of the weight of
topics on the exam. The syllabus indicates the exam weights by topic.
April 2014 changes: In Solutions 122B and 194, instances of subscripts of the form u:v
were changed to be just u or v as appropriate.
Further April 2014 changes: Solution to 175 has a typo corrected. Solution to 248 is
the negative of the previous solution.
MLC-09-13
PRINTED IN U.S.A.
Question #1
Answer: E
2
q30:34  2 p30:34  3 p30:34
2
p30   0.9  0.8   0.72
2
p34   0.5  0.4   0.20
2
p30:34   0.72  0.20   0.144
2
p30:34  0.72  0.20  0.144  0.776
3
p30   0.72  0.7   0.504
3
p34   0.20  0.3  0.06
3
p30:34   0.504  0.06   0.03024
3
p30:34  0.504  0.06  0.03024
 0.53376
2 q30:34
 0.776  0.53376
 0.24224
Alternatively,
2 q30:34  2 q30  2 q34  2 q30:34
b
g
 2 p30q32  2 p34q36  2 p30:34 1  p32:36
= (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,
2
q30:34  3 q30  3 q34  2 q30  2 q34
 1  3 p30 1  3 p34   1  2 p30 1  2 p34 
 1  0.504 1  0.06   1  0.72 1  0.20 
 0.24224
(see first solution for
MLC-09-13
2
p30 ,
2
p34 ,
3
p30 ,
3 p34
2
)
Question #2
Answer: E
1000 Ax  1000  Ax1:10  10 Ax 


10

 1000   e0.04t e0.06t (0.06)dt  e 0.4 e 0.6  e 0.05t e 0.07 t (0.07)dt 
0
 0

10

 1000 0.06 e0.1t dt  e1 (0.07)  e0.12t dt 
0
0


0.10 t 10
0.12 t  

 1000 0.06   e0.10   e1 (0.07)   e0.12  

0

0 

1 
1  e1   0.07
 1000  0.06
 0.12 e 
 0.10 
 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 Ax1:10 
10 E x
Ax 

 
e
1  10 Ex 
10    

 
With those relationships, the solution becomes
1000 Ax  1000  Ax1:10  10 Ex Ax 10 



0.06 
0.07  
 0.06 0.04 10
 0.06  0.04 10 
 1000 
e 
 1 e


 0.07  0.05  
 0.06  0.04 


 1000  0.60  1  e1  0.5833 e1 


 593.86
MLC-09-13
3
Question #3
Answer: D


1
dt
20
1  100 
5
0.07 t  
 

  e

0
20  7 
7
2

1
1 
bt vt t px  x t dt   e0.12t e 0.16t e 0.05t
dt   e 0.09t dt
0
20
20 0
1  100  0.09t  5
 
 
 e
0 9
20  9  
E  Z    bt v t t px  x t dt   e0.06 t e 0.08 t e 0.05 t
0
0
E  Z 2   

0
 
2
5 5
Var  Z       0.04535
9 7
Question #4
Answer: C
Let ns = nonsmoker and s = smoker
k=
q xb  kg
ns
pxb  kg
q xb gk
s
px 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
1 ns
A x:2 
ns
v qx  
1
 0.05
1.02
A x:2 
1 s
v
1
1.02
qx 
s
ns
px 
v2
1
1.022
ns
ns
0.95  0.10  0.1403
 v2
 0.10  
qx 1
1
1.02 2
px 
s
qx 1
s
0.90  0.20  0.2710
A1x:2  weighted average = (0.75)(0.1403) + (0.25)(0.2710)
= 0.1730
MLC-09-13
4
s
Question #5
Answer: B
 x    x1   x 2   x 3  0.0001045
t

px   e0.0001045t


1
APV Benefits   e t 1,000,000 t px   x  dt
0


2
  e t 500,000 t px    x   dt
0


3
  e 200,000 t px    x   dt
0

1,000,000  0.0601045t
500,000  0.0601045t
250,000  0.0601045t
e
dt 
e
dt 
e
dt


2,000,000 0
250,000 0
10,000 0
 27.5 16.6377   457.54
*Question #6
Answer: B
1
EPV Benefits  1000 A40:20

EPV Premiums   a40:20 

 k E401000vq40k
k  20

 k E401000vq40k
k  20
Net premiums  Equivalence principle 
1
1000 A40:20



 k E401000vq40k   a40:20   k E401000vq40k
k  20
20

1
 1000 A40:20

/ a40:20
161.32   0.27414  369.13
14.8166   0.27414 11.1454 
 5.11
1
While this solution above recognized that   1000P40:20
and was structured to take
advantage of that, it wasn’t necessary, nor would it save much time. Instead, you
could do:
MLC-09-13
5
EPV Benefits  1000 A40  161.32
EPV Premiums = a40:20 
  a40:20 

20 E40  k E60 1000vq60  k
k 0
20 E40 1000 A60
  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 
 0.53  0.5147
0.06
1  A70 1  0.5147
a70 

 8.5736
d
0.06 /1.06
 0.97 
a69  1  vp69a70  1  
  8.5736   8.8457
 1.06 
A70   A70 
i
 
a69
   2  a69    2   1.00021 8.8457   0.25739
2
 8.5902
 m  1 works well (is closest to the exact
m
Note that the approximation ax   ax 
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-09-13
6
Question #10
Answer: E
d = 0.05  v = 0.95
At issue

49



A40   v k 1 k q40  0.02 v1  ...  v50  0.02v 1  v50 / d  0.35076
k 0
and a40  1  A40  / d  1  0.35076  / 0.05  12.9848
so P40 
1000 A40 350.76

 27.013
a40
12.9848
E  10 L K40  10   1000 A50
Revised
 P40 a50
 549.18   27.013 9.0164   305.62


Revised
where
Revised
A50
and
24
  v k 1 k q50
Revised
k 0
Revised
a50

1  A
Revised
50


 0.04 v1  ...  v 25  0.04v 1  v 25 / d  0.54918
 / d  1  0.54918 / 0.05  9.0164
*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  AxS
E aT Y  1  axS 



1  0.444
 5.56
0.1
1  0.286
Similarly E aT Y  0 
 7.14
0.1



E  E  aT Y    E  E  aT 0    Prob  Y=0   E  E  aT 1   Prob  Y=1
  7.14  0.70    5.56  0.30 
 6.67
MLC-09-13
7
 
E  E aT Y

     7.142   0.70   5.562   0.30
2
 44.96
 
  44.96  6.672  0.47
E  Var  aT Y     8.503 0.70    8.818  0.30 
Var E aT Y
 8.60
 
Var aT  8.60  0.47  9.07
Alternatively, here is a solution based on
 
Var(Y )  E Y 2   E Y   , a formula for the variance of any random variable.
This can be
2
 
transformed into E Y 2  Var Y    E Y   which we will use in its conditional
form
 
E aT
2

2




2
NS  Var aT NS   E aT NS 


 
Var  aT   E  aT

2

 E  aT 


2
E  aT   E  aT S  Prob S  E  aT NS  Prob  NS
 0.30axS  0.70axNS


0.30 1  AxS
  0.70 1  A 
NS
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  aT

2

 E  aT 2 S  Prob S  E  aT 2 NS  Prob  NS
 
 

2
 0.30 Var aT S  E  aT S 





0.70 Var aT NS  E aT NS

2

2
2
 0.30 8.818   5.56    0.70 8.503   7.14  




11.919 + 41.638 = 53.557
Var  aT   53.557   6.67   9.1
2
MLC-09-13
8
Alternatively, here is a solution based on aT 
1  vT

1 v 
Var aT  Var   
  
  vT 
 Var 
 since Var  X  constant   Var  X 
  
T
 
  since Var  constant  X   constant
Var vT


2

2
Ax   Ax 
2
 Var  X 
2
2
This could be transformed into 2Ax   2 Var  aT   Ax2 , which we will use to get
2
Ax NS and 2AxS .
Ax  E  v 2T 
2
 E  v 2T NS  Prob  NS  E  v 2T S  Prob  S 

 

  

2
  2Var aT NS  Ax NS   Prob  NS 


2
  2 Var aT S  AxS   Prob  S 


  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
Ax  E  vT 
 E  vT NS  Prob  NS  E  vT S  Prob  S
  0.286  0.70    0.444  0.30 
 0.3334
Ax   Ax 
2
 
Var aT 

MLC-09-13
2
2
0.20238  0.33342
 9.12
0.01
9
Question #12 - Removed
Question #13
Answer: D
Let NS denote non-smokers, S denote smokers.
Prob T  t   Prob T  t NS  Prob  NS  P rob T  t S  P rob  S




 1  e0.1t  0.7  1  e0.2t  0.3
 1  0.7e0.1t  0.3 e0.2t
S0 (t )  0.3e0.2 t  0.7e0.1t
Want tˆ such that 0.75  1  S0  tˆ  or 0.25  S0  tˆ 
ˆ

ˆ

ˆ 2
0.25  0.3e0.2t  0.7e0.1t  0.3 e0.1t
ˆ
 0.7e0.1t
ˆ
Substitute: let x  e0.1t
0.3x2  0.7 x  0.25  0
This is quadratic, so x 
0.7  0.49   0.3 0.25 4
2  0.3
x  0.3147
ˆ
e0.1t  0.3147
MLC-09-13
so tˆ  11.56
10
*Question #14
Answer: A
At a constant force of mortality, the net premium equals the force of mortality and
so   0.03 .

0.03
2
Ax  0.20 

2   2  0.03
   0.06
Var  0 L  
where A 
2
Ax   Ax 
 a 2

 

2

0.20   13 
 0.06
0.09 
0.03 1

0.09 3
2
2
a
 0.20
1
1

   0.09
Question #15 - Removed
*Question #16
Answer: A
1000 P40 
A40 161.32

 10.89
a40 14.8166
 a 
 11.1454 
1000 20V40  1000  1  60   1000  1 
  247.78
 14.8166 
 a40 
 20V  5000P40  (1  i)  5000q60
21V 
P60

 247.78  (5)(10.89)  1.06  5000  0.01376   255
1  0.01376
[Note: For this insurance, 20V  1000 20V40 because retrospectively, this is identical
to whole life]
Though it would have taken much longer, you can do this as a prospective
reserve. The prospective solution is included for educational purposes, not to
suggest it would be suitable under exam time constraints.
1000P40  10.89 as above
1
1000 A40  4000 20 E40 A60:5
 1000P40  5000P40  20 E40 a60:5  
1
 A60  5 E60 A65  0.06674
where A60:5
MLC-09-13
11
20 E40  5 E60
a65
a40:20  a40  20 E40 a60  11.7612
a60:5  a60  5 E60 a65  4.3407
1000  0.16132    4000  0.27414  0.06674  
 10.89 11.7612   510.89  0.27414  4.3407    0.27414 0.68756  9.8969 
161.32  73.18  128.08  64.79
1.86544
 22.32

Having struggled to solve for  , you could calculate
above)
calculate 21V recursively.
20V
20V
prospectively then (as
1
 4000 A60:5
 1000 A60  5000 P40 a60:5   5 E60 a65
  4000  0.06674   369.13   5000  0.01089  4.3407    22.32  0.68756  9.8969 
 247.86 (minor rounding difference from 1000 20V40 )
Or we can continue to
21V
21V
prospectively
1
 5000 A61:4
 1000 4 E61 A65  5000P40 a61:4   4 E61 a65
where 4 E61 
l65 4  7,533,964 
v 
  0.79209   0.73898
l61
 8,075, 403 
1
A61:4
 A61  4 E61 A65  0.38279  0.73898  0.43980
a61:4
 0.05779
 a61  4 E61 a65  10.9041  0.73898  9.8969
 3.5905
21V
  5000  0.05779   1000  0.73898 0.43980 
  5 10.89  3.5905   22.32  0.73898 9.8969 
 255
Finally. A moral victory. Under exam conditions since prospective net premium
reserves must equal retrospective net premium reserves, calculate whichever is
simpler.
MLC-09-13
12
Question #17
Answer: C
Var  Z   2 A41   A41 
2
A41  A40  0.00822  A41  (vq40  vp40 A41 )
 A41  (0.0028 /1.05   0.9972 /1.05  A41)
 A41  0.21650
2

A41  2 A40  0.00433  2 A41  v 2 q40  v 2 p40 2 A41


 2 A41  (0.0028 /1.052  0.9972 /1.052
2
A41  0.07193
Var  Z   0.07193  0.216502
 0.02544
Question #18 - Removed
Question #19 - Removed
MLC-09-13
13

2
A41 )
*Question #20
Answer: D
 x    x1t   x 2t
 0.2 xt   x 2t
  x 2t  0.8 xt
'1
qx 
k
3
 1
'1
px 
1
 1 e
  0.2 k t 2 dt
0
 1 e
0.2
k
3
 0.04
 ln 1  0.04  /  0.2   0.2041
k  0.6123
 2
2 qx

px   x 2t dt  0.8
2
0 t
2
0



 0.8 2 qx   0.8 1  2 px 
2
t
px   xt dt

( )
  xt d t
 
0
2 px  e
2
e
 k t2 d t
0

8 k
e 3

  8   0.6123
3
e
 0.19538
 2
2 qx
 0.8 1  0.19538  0.644
MLC-09-13
14
Question #21
Answer: A
k
min(k ,3)
f(k)
f  k    min(k ,3) 
0
1
2
3+
0
1
2
3
0.1
(0.9)(0.2) = 0.18
(0.72)(0.3) = 0.216
1-0.1-0.18-0.216 =
0.504
0
0.18
0.432
1.512
f  k    min  k ,3 
0
0.18
0.864
4.536
2.124
5.580
E  min( K ,3)  2.124

E  min  K ,3 
2
  5.580
Var  min( K ,3)  5.580  2.1242  1.07
Note that E  min( K ,3) is the temporary curtate life expectancy, ex:3 if the life is
age x.
Question #22
Answer: B
 0.1 60
 0.08 60
e    e   
2
 0.005354
S0 (60) 
 0.1 61
 0.08 61
e     e   
2
 0.00492
0.00492
q60  1 
 0.081
0.005354
S0 (61) 
MLC-09-13
15
2
*Question #23
Answer: D
Let q64 for Michel equal the standard q64 plus c. We need to solve for c.
Recursion formula for a standard insurance:
20V45
  19V45  P45  1.03  q64 1  20V45 
Recursion formula for Michel’s insurance
20V45
  19V45  P45  0.01 1.03   q64  c  (1  20V45 )
The values of 19V45 and 20V45 are the same in the two equations because we are
told
Michel’s net premium reserves are the same as for a standard insurance.
Subtract the second equation from the first to get:
0   1.03 (0.01)  c(1  20V45 )
c
(1.03)  0.01
1  20V45 
0.0103
1  0.427
 0.018

MLC-09-13
16
Question #24
Answer: B
K is the curtate future lifetime for one insured.
L is the loss random variable for one insurance.
LAGG is the aggregate loss random variables for the individual insurances.
 AGG is the standard deviation of LAGG .
M is the number of policies.
 
L  v K 1   aK 1   1   v K 1  
d
 d
1  Ax 
E  L    Ax   ax   Ax  
d
 0.75095 
 0.24905  0.025 
  0.082618
 0.056604 
 
Var  L   1  
 d
2

2
Ax 
Ax2

2

E  LAGG   M E  L   0.082618M
Var  LAGG   M Var  L   M (0.068034)   AGG  0.260833 M
L
 E  LAGG   E  LAGG  
Pr  LAGG  0   AGG


 AGG
 AGG 


0.082618M 
 Pr  N (0,1) 


M  0.260833 

0.082618 M
0.260833
 M  26.97
 1.645 
 minimum number needed = 27
MLC-09-13

0.025 
2

 1 
 0.09476   0.24905   0.068034
 0.056604 
17
Question #25
Answer: D
1  v K 1
for K  0,1, 2,...
d
Z 2  Bv K 1
for K  0,1,2,...
Z1  12,000
Annuity benefit:
Death benefit:
1  v K 1
 Bv K 1
d
12,000 
12,000  K 1

B
v
d
d 

Z  Z1  Z 2  12,000
New benefit:
2


12,000 

K 1
Var( Z )   B 
 Var v
d


12,000
Var  Z   0 if B 
 150,000 .
0.08
In the first formula for Var  Z  , we used the formula, valid for any constants a and
b and random variable X,
Var  a  bX   b2Var  X 
Question #26
Answer: A
 xt: y t   xt   y t  0.08  0.04  0.12
Ax   xt /   xt     0.5714


Ay   y t /  y t    0.4


Axy   xt: y t /  xt: y t    0.6667


axy  1/  xt: y t    5.556
Axy  Ax  Ay  Axy  0.5714  0.4  0.6667  0.3047
Premium = 0.304762/5.556 = 0.0549
MLC-09-13
18
*Question #27
Answer: B
P40  A40 / a40  0.16132/14.8166  0.0108878
P42  A42 / a42  0.17636/14.5510  0.0121201
a45  a45  1  13.1121
E  3 L K42  3  1000 A45  1000P40  1000P42 a45
 201.20  10.89  12.1213.1121
 31.39
Many similar formulas would work equally well. One possibility would be
1000 3V42  1000P42  1000P40  , because prospectively after duration 3, this differs
from the normal net premium reserve in that in the next year you collect 1000P40
instead of 1000P42 .
*Question #28
Answer: E
E  min T ,40    40  0.005  40   32
2
32   t f  t  dt   40 f  t  dt
40
w
0
40
  t f  t  dt   t f  t  dt  40 .6 
w
w
0
40
 86   tf  t  dt
w
40
 tf  t dt  54
w
40
  t  40  f  t  dt  54  40 .6   50
w
e 40 
40
MLC-09-13
S0  40 
.6
19
*Question #29
Answer: B
d  0.05  v  0.95
Step 1 Determine p x from Kevin’s work:
608  350vpx  1000vqx  1000v2 px  px1  qx1 
608  350  0.95 px  1000  0.95 1  px   1000  0.9025 px 1
608  332.5 px  950 1  px   902.5 px
px  342 / 380  0.9
Step 2 Calculate 1000 Px:2 , as Kira did:
608  350  0.95  0.9   1000 Px:2 1   0.95  0.9  
 299.25  608  489.08
1000 Px:2 
1.855
The first line of Kira’s solution is that the expected present value of Kevin’s net
premiums is equal to the expected present value of Kira’s, since each must equal
the expected present value of benefits. The expected present value of benefits
would also have been easy to calculate as
1000 0.95 0.1  1000 0.952  0.9   907.25


Question #30
Answer: E
Because no premiums are paid after year 10 for (x), 11Vx  Ax11
One of the recursive reserve formulas is
10V

h 1V

 hV   h  1  i   bh1qxh
 32,535  2,078  1.05  100,000  0.011  35,635.642
px  h
0.989
 35,635.642  0  1.05  100,000  0.012  36,657.31  A
11V 
x 11
0.988
MLC-09-13
20
Question #31
Answer: B
t 

The survival function is S0 (t )   1   :
 
Then,
x
t 


ex 
and t px   1 

2
  x
105  45

e45 
 30
2
105  65
e65 
 20
2

e45:65  

40
0 t
p45:65dt  
40 60  t
0
FG
H
60

40  t
dt
40
1
60  40 2 1 3
60  40  t 
t  t
60  40
2
3
IJ
K
40
0
 1556
.




e 45:65  e45  e 65  e 45:65
 30  20  1556
.  34
In the integral for e45:65 , the upper limit is 40 since 65 (and thus the joint status
also)
can survive a maximum of 40 years.
Question #32
Answer: E
4  S0 (4) / S0 (4)

d
i
  e4 / 100
1  e4 / 100

e4 / 100
1  e4 / 100

e4
 1.202553
100  e4
MLC-09-13
21
Question # 33
Answer: A
L ln px big OP  qb g LM ln e OP
qxbi g  qxb g M
M ln p b g P x M ln e-b g P
bi g
N
 qxb g 
Q
x
N

Q
 bi g
 b g
 x b g   x b1g   x b 2 g   x b 3g  15
.
qxb g  1  e  b g  1  e1.5
=0.7769
b0.7769gb g  b0.5gb0.7769g
2
qxb2 g 
 b g
15
.
 0.2590
Question # 34
Answer: D
22
A 60  v 3

B
B
 q 60  2

B
pay at end
live
of year 3
2 years in year 3
 v4

pay at end
of year 4

2 p 60
1
1.03
3
3p 60
then die

q60 3
live
then die
3 years
in year 4
1  0.09  1  0.11  0.13 
1
1.03
4
1  0.09  1  0.11 1  0.13  0.15
 0.19
MLC-09-13
22
Question # 35
Answer: B
ax  ax:5 5Ex ax 5
a x:5
5 Ex
1  e 0.07b5g

 4.219 , where 0.07     for t  5
0.07
 e 0.07b5g  0.705
a x5 
1
 12.5 , where 0.08     for t  5
0.08
b
gb g
 ax  4.219  0.705 12.5  13.03
Question #36
Answer: D

1
2
px   p' x  p' x  0.8  0.7   0.56
1
qx 
 
 
 ln p ' 1 
x
 q  since UDD in double decrement table


 ln p   x
x


 ln  0.8  

 0.44
 ln  0.56  
 0.1693
1
0.3 qx  0.1

0.3qx 
1
 0.053

1  0.1qx 
To elaborate on the last step:
 Number dying from cause 


1 between x  0.1 and x  0.4 
1

0.3 qx  0.1 
Number alive at x  0.1
Since UDD in double decrement,


1
lx   0.3 qx 



lx  1  0.1qx 
MLC-09-13

23
*Question #37
Answer: E
The net premium is
oL
1
1
    0.04  0.04333
ax
12
 v T  (0.04333  0.0066)aT  0.02  0.003aT
 1 vT 
 v T  0.04693 
  0.02



 0.04693  0.04693
 v T 1 
 0.02





 
Var  o L   Var v
T
2
 0.04693 
1 
  0.1(4.7230)  0.4723



Question #38
Answer: D
 0.7 0.1 0.2 


T   0.3 0.6 0.1 
 0
0
1 

 0.52 0.13 0.35 


T   0.39 0.39 0.22 
 0
0
1 

2
Actuarial present value (APV) prem = 800(1 + (0.7 + 0.1) + (0.52 + 0.13)) = 1,960
APV claim = 500(1 + 0.7 + 0.52) + 3000(0 + 0.1 + 0.13) = 1800
Difference = 160
Question # 39 - Removed
MLC-09-13
24
Question # 40
Answer: D
Use Mod to designate values unique to this insured.
b
b
g
g b gb g
 1000b0.36933 / 111418
.
.
g  3315
a60  1  A60 / d  1  0.36933 / 0.06 / 106
.
 111418
.
1000 P60  1000 A60 / a60
d
i
Mod
Mod
Mod
A60
 v q60
 p60
A61 
d
b
b
i
gb
g
1
01376
.
 0.8624 0.383  0.44141
106
.
g
Mod
a Mod  1  A60
/ d  1  0.44141 / 0.06 / 106
.  9.8684
d
Mod
Mod
E 0 LMod  1000 A60
 P60a60
i
b
g
 1000 0.44141  0.03315 9.8684
 114.27
*Question # 41
Answer: D
The prospective reserve at age 60 per 1 of insurance is A60 , since there will be no
future premiums. Equating that to the retrospective reserve per 1 of coverage, we
have:
s40:10
A60  P40
 P50Mod s50:10 20 k40
10 E50
1
a40:10
a
A40
A40
:20
Mod 50:10
A60 

 P50

a40 10 E40 10E50
10 E50
20 E40
0.36913 
016132
.
7.70
7.57
0.06

 P50Mod

14.8166 0.53667 0.51081
0.51081 0.27414
b
gb
g
0.36913  0.30582  14.8196 P50Mod  0.21887
1000 P50Mod  19.04
MLC-09-13
25
Alternatively, you could equate the retrospective and prospective reserves at age
50. Your equation would be:
A50 
a50:10
P50Mod
1
A40 a40:10 A40:10



a40 10 E40 10 E40
1
where A40
 A40 10 E40 A50
:10
b
gb
g
 016132
.
 0.53667 0.24905
 0.02766
d
.
7.70
0.02766


ib g 14016132
.8166 0.53667 0.53667
.
b1000gb014437
g  19.07

0.24905  P50Mod 7.57 
1000 P50Mod
7.57
Alternatively, you could set the expected present value of benefits at age 40 to the
expected present value of net premiums. The change at age 50 did not change the
benefits, only the pattern of paying for them.
A40  P40 a40:10  P50Mod
016132
.

10 E40
a50:10
.
FG 016132
I
H 14.8166JK b7.70g  d P ib0.53667gb7.57g
b1000gb0.07748g  19.07

Mod
50
1000 P50Mod
4.0626
Question # 42
Answer: A
d xb2g  qxb2g  lxb g  400
b g
d xb1g  0.45 400  180
qx b2 g 
d xb2 g
400

 0.488
l b g  d b1g 1000  180
x
x
pxb2g  1  0.488  0.512
Note: The UDD assumption was not critical except to have all deaths during the
year so that 1000 - 180 lives are subject to decrement 2.
MLC-09-13
26
Question #43
Answer: D
Use “age” subscripts for years completed in program. E.g., p0 applies to a person
newly hired (“age” 0).
Let decrement 1 = fail, 2 = resign, 3 = other.
1
1
1
Then q0b g  14 , q1b g  15 , q2b g  13
q0b2g  15 , q1b2g  13 , q2b2g  18
q0b3g  110 , q1b3g  19 , q2b3g  14
b
gb
gb
g
This gives p0b g  1  1 / 4 1  1 / 5 1  1 / 10  0.54
b gb gb g
pb g  b1  1 / 3gb1  1 / 8gb1  1 / 4g  0.438
So 1b g  200, 1b g  200 b0.54g  108 , and 1b g  108 b0.474g  512
.
p1b g  1  1 / 5 1  1 / 3 1  1 / 9  0.474

2



0
1
q2b1g  log p2b1g / log p2b g q2b g
2
c h / logb0.438g 1  0.438
 b0.405 / 0.826gb0.562g
q2b1g  log
2
3
 0.276
d2b1g  l2b g q2b1g
 512
. 0.276  14
b gb
g
Question #44 - Removed
MLC-09-13
27
Question #45
Answer: E
ex 
For the given life table function:
x
2
1
k qx 
x
  x 1

Ax 
k b
1   x 1 k 1
v
  x k b
a  x
Ax 
ax 
v k 1 k qx 
x
1  Ax
d
e50  25    100 for typical annuitants
ey  15  y  Assumed age = 70
A70 
a30
 0.45883
30
a70  9.5607
500000  b a70  b  52, 297
Question #46
Answer: B
10 E30:40  10p30 10
d p v id p v ib1  i g
 b E gb E gb1  i g
 b0.54733gb0.53667 gb179085
.
g
p40 v10 
10
10
30
10
10
10
40
10
10
30
10
40
 0.52604
The above is only one of many possible ways to evaluate
which should give 0.52604
a30:40:10  a30:40 10 E30:40 a3010:4010
b
g b
gb
g
 b13.2068g  b0.52604gb114784
.
g
 a30:40  1  0.52604 a40:50  1
 7.1687
MLC-09-13
28
10
10 p30 10 p40 v ,
all of
*Question #47
Answer: A
Equivalence Principle, where  is annual net premium, gives
1000 A35  ( IA)35    a35

d
1000 A35
1000  0.42898

(1199143
.
 616761
.
)
a35  IA 35
b gi
428.98
582382
.
 73.66

We obtained a35 from
a35 
1  A35 1  0.42898

 1199143
.
d
0.047619
Question #48 - Removed
Question #49
Answer: C
 xy   x   y  014
.

0.07
Ax  Ay 

 0.5833
   0.07  0.05
 xy
014
.
014
.
1
1
Axy 


 0.7368 and axy 

 5.2632
 xy   014
.  0.05 019
.
 xy   014
.  0.05
P
Axy
axy

Ax  Ay  Axy
MLC-09-13
axy

b
g
2 0.5833  0.7368
 0.0817
5.2632
29
Question #50
Answer: E
 20V  P20 1  i   q40 1  21V   21V
b0.49  0.01gb1 ig  0.022b1 0.545g  0.545
b1  ig  b0.545gb1  0.022g  0.022
0.50
 111
.
 21V  P20 1  i   q41 1  22V   22V
. g  q b1  0.605g  0.605
b0.545.01gb111
41
q41 
0.61605  0.605
0.395
 0.028
Question #51
Answer: E
1000 P60  1000 A60 / a60
b
gb
g
 1000bq  p A g / b106
.  p a g
 c15  b0.985gb382.79gh / c106
.  b0.985gb10.9041gh  33.22
 1000 v q60  p60 A61 / 1  p60 v a61
60
60
61
60
61
Question #52 - Removed
MLC-09-13
30
Question #53
Answer: E
g   ln(0.96)  0.04082
 x02t: y t  0.04082  0.01  0.03082
h   ln(0.97)  0.03046
 x01t: y t  0.03046  0.01  0.02046
5


5
pxy  5 pxy00  exp    x01t: y t   x02t: y t   x03t: y t dt  e 5(0.06128)  0.736
0
Question #54
Answer: B
Transform these scenarios into a four-state Markov chain, as shown below.
State
from year t – 3
to year t – 2
from year t – 2
to year t – 1
0
Decrease (D)
Decrease (I)
Probability that year t
will decrease from year t
-1
0.8
1
Increase
Decrease
0.6
2
Decrease
Increase
0.75
3
Increase
Increase
0.9
0.80 0.00 0.20 0.00 
0.60 0.00 0.40 0.00 

The transition probability matrix is 
0.00 0.75 0.00 0.25


0.00 0.90 0.00 0.10 
Note for example that the value of 0.2 in the first row is a transition from DD to DI,
that is, the probability of an increase in year three after two successive years of
decrease.
The requested probability is that of starting in state 0 and then being in either state
0 or 1 after two years. That requires the first row of the square of the transition
probability matrix. It can be obtained by multiplying the first row of the matrix by
each column. The result is [0.64 0.15 0.16 0.05]. The two required probabilities are
0.64 and 0.15 for a total of 0.79. The last two values in the vector do not need to
be calculated, but doing so provides a check (in that the four values must sum to
one).
Alternatively, the required probabilities are for the successive values DDID and
DDDD. The first case is transitions from state 0 to 2 and then 2 to 1 for a
MLC-09-13
31
probability of 0.2(0.75) = 0.15. The second case is transitions from 0 to 0 and then
0 to 0 for a probability of 0.8(0.8) = 0.64 for a total of 0.79.
MLC-09-13
32
Question #55
Answer: B
lx    x  105  x
 t p45  l45t / l45  (60  t ) / 60
Let K be the curtate future lifetime of (45). Then the sum of the payments is 0 if
K  19 and is K – 19 if K  20 .
F 60  K IJ  1
60 K
b40  39...1g  b40gb41g  13.66

60
2b60g

20 a45 
 1  GH
60
K  20
Hence,
c
h
c
Prob K  19  13.66  Prob K  32.66
h
b g
 ProbbT  33g
 Prob K  33 since K is an integer
 33p45 
l78 27

l45 60
 0.450
MLC-09-13
33
Question #56
Answer: C
Ax 
2
Ax 
d IAi
z
E
0s x
zd

0
 0.4
 
x
 0.25    0.04
  2





z

0 s
Ax ds
Ax ds
ib g
e0.1s 0.4 ds
F e I
 b0.4gG
H 01. JK
0.1s


0
0.4
4
01
.
Alternatively, using a more fundamental formula but requiring more difficult
integration.
c IA h
x


z
z

0

0
bg
b0.04g e
t t px  x t e   t dt
t e 0.04 t
 0.04
z

0
0.06t
dt
t e 0.1t dt
(integration by parts, not shown)

t
1
 0.04

e 0.1 t
0
01
. 0.01
0.04

4
0.01
FG
H
MLC-09-13
IJ
K
34
Question #57
Answer: E
Subscripts A and B here distinguish between the tools and do not represent ages.

We have to find e AB

eA 

eB 

FG 1  t IJ dt  t  t
H 10K
20
z
z FGH
z FGH
10
0
IJ
K
t
t2
1  dt  t 
7
14
7
0
7
e AB 
2 10

1
t
7
7
0
 49 
0
49
 35
.
14
IJ FG1  t IJ dt  z FG1  t  t  t IJ dt
K H 10K
H 10 7 70K
2
7
0
t2 t2
t3
t  
20 14 210
7
0
 7
49 49 343


 2.683
20 14 210



 10  5  5

e AB  e A  e B  e AB
 5  35
.  2.683  5817
.
Question #58
Answer: A
xt  0.100  0.004  0.104
t
pxb g  e0.104t
Expected present value (EPV) = EPV for cause 1 + EPV for cause 2.
z
5
b
z
g
b
5
g
2000 e0.04t e0.104t 0100
.
dt  500,000 e0.04 t e0.104 t 0.400 dt
0
0
  2000  0.10   500, 000  0.004    e0.144t dt 
5
0
MLC-09-13
35


2200
0.144 5
1 e
 7841
0.144
Question #59
Answer: A
R  1  px  qx
S  1  px  e
k 
1
1
1
1
e 0
  xt dt   k dt
 xt  k  dt
0
since e 0
 e 0

 e 0
1
  xt dt  k dt
So S  0.75R  1  px  e k  0.75qx
1  0.75qx
px
px
1  qx
ek 

1  0.75qx 1  0.75qx
e k 
k  ln
LM 1  q OP
N1  0.75q Q
x
x
MLC-09-13
36
Question #60
Answer: C
A60  0.36913
2
d  0.05660
A60  0.17741
and
2
2
A60  A60
 0.202862



Expected Loss on one policy is E  L     100,000   A60 
d
d



2
Variance on one policy is Var  L    100,000   2 A60  A60
d

On the 10000 lives,
E  S   10,000E  L   and Var  S   10,000 Var  L  
2


The  is such that 0  E  S  / Var  S   2.326 since   2.326   0.99
 
 
10,000   100,000   A60 
d
d 
  2.326



2
100 100,000   2 A60  A60
d

 
 
100   100,000     0.36913
d 
d 
 2.326


100,000    0.202862 
d


0.63087  36913
d
 0.004719

100,000 
d


0.63087  36913  471.9  0.004719
d
d

36913  471.9

d 0.63087  0.004719
 59706
  59706  d  3379
MLC-09-13
37
Question #61
Answer: C
  0V    1  i   1000  1V  1V   q75
1V
 1.05  1000q75
Similarly,
2 V   1V     1.05  1000q76
3V
  2V    1.05  1000q77


1000 3V  1.053  1.052    1.05  1000  q75 1.052  1000 1.05  q76  1000  q77 *




1000  1000 1.052 q75  1.05q76  q77
1.05  1.05
3

2

 1.05
1000 x 1  1.052  0.05169  1.05  0.05647  0.06168

3.310125
1000 1.17796
 355.87
3.310125
* This equation is an algebraic manipulation of the three equations in three
unknowns  1V , 2V ,   . One method – usually effective in problems where benefit
= stated amount plus reserve, is to multiply the 1V equation by 1.052 , the 2V
equation by 1.05, and add those two to the 3V equation: in the result, you can
cancel out the 1V , and 2V terms. Or you can substitute the 1V equation into the
2 V equation, giving 2 V in terms of  , and then substitute that into the 3V
equation.
Question #62
Answer: D
A281:2 
z
2  t
e 1 72 dt
0
a28:2
3V
d
i
FG IJ
H K
b g
1
1  e2  0.02622 since   ln 106
.
 0.05827
72
71
 1 v
 19303
.
72

 500,000 A281:2  6643 a28:2 = 287
MLC-09-13
38
Question #63
Answer: D
Let Ax and a x be calculated with  x t and   0.06
Let Ax* and ax* be the corresponding values with  x t
increased by 0.03 and  decreased by 0.03
ax 
1  Ax


0.4
 6.667
0.06
ax  ax
*


    xs  0.03 ds 0.03t
*
0
e
dt
 Proof: ax  0 e

t


0
e 
t
  x s ds 0.03t 0.03t
0
e
e
t

 e
0
  x s ds
0
e 0.06 t dt
 ax 
Ax*  1  0.03 ax*  1  0.03 ax
b gb
 1  0.03 6.667
g
 0.8
MLC-09-13
39
dt
Question #64
Answer: A
bulb ages
Year
0
1
0
1
2
3
10000
0
1000
9000
100+2700
900
280+270+3150
2
3
0
0
6300
0
0
0
The diagonals represent bulbs that don’t burn out.
E.g., of the initial 10,000, (10,000) (1-0.1) = 9000 reach year 1.
(9000) (1-0.3) = 6300 of those reach year 2.
Replacement bulbs are new, so they start at age 0.
At the end of year 1, that’s (10,000) (0.1) = 1000
At the end of 2, it’s (9000) (0.3) + (1000) (0.1) = 2700 + 100
At the end of 3, it’s (2800) (0.1) + (900) (0.3) + (6300) (0.5) = 3700
1000 2800 3700


105
.
105
. 2 105
. 3
 6688
Expected present value 
Question #65
Answer: E

e25:25 
z
z
15
0 t
p25dt 15 p25
z
10
0t
p40 dt
F
IJ e
 e
dt  G e z
H
Kz
1
L1

1  e i  e M d1  e
d
.04
N.05
15

15 .04 t
0
.04 ds
0
0
.60
.60
 112797
.
 4.3187
 15.60
MLC-09-13
10 .05t
dt
.50
iOPQ
40
#
replaced
1000
2800
3700
Question #66
Answer: C
5
p 60 1 
e1  q je1  q jb1  q gb1  q gb1  q g
 b0.89gb0.87gb0.85gb0.84gb0.83g
60 1
60  2
63
64
65
 0.4589
Question # 67
Answer: E
1
     0.08      0.04
 
12.50  ax 
Ax 

 0.5
 

1
2
Ax 

  2 3
e j
Var aT 
2
Ax  Ax2
2
1

1
 3 4  52.083
0.0016
S.D.  52.083  7.217
MLC-09-13
41
*Question # 68
Answer: D
v  090
.  d  010
.
Ax  1  dax  1  010
. 5  0.5
b gb g
Net premium  
5000 Ax  5000vqx
ax

b5000gb0.5g  5000b0.90gb0.05g  455
5
Net premium reserve at the end of year 10  1 
0.2  1 
ax 10
ax
ax 10
 ax 10  4
5
b gb g
  a
 b5000gb0.6g  b455gb4g  1180
Ax 10  1  dax 10  1  010
. 4  0.6
10V
 5000 Ax 10
x 10
Question #69
Answer: D
v is the lowest premium to ensure a zero % chance of loss in year 1 (The present
value of the payment upon death is v, so you must collect at least v to avoid a loss
should death occur).
Thus v = 0.95.
2
E Z  vqx  v 2 px qx 1  0.95  0.25  0.95  0.75  0.2
bg
d i
b g
 0.3729
b g
b g
2
4
E Z  v qx  v px qx 1  0.95  0.25  0.95  0.75  0.2
2
2
4
 0.3478
b g d i c b gh
Var Z  E Z2  E Z
MLC-09-13
2
b
g
2
 0.3478  0.3729  0.21
42
Question #70
Answer: D
Expected present value (EPV) of future benefits =
  0.005  2000  0.04 1000  /1.06  1  0.005  0.04 0.008  2000  0.06 1000  /1.062
 47.17  64.60
 111.77
EPV of future premiums  1  1  0.005  0.04  /1.06 50
 1.9009  50
 95.05
E  1 L K55  1  111.77  95.05  16.72
Question #71 - Removed
Question #72
Answer: A
Let Z be the present value random variable for one life.
Let S be the present value random variable for the 100 lives.

E ( Z )  10 e  t e  t  dt
5
 10

 
e (   )5
 2.426
FG  IJ e b g
H 2   K
F 0.04IJ de i  11233
 10 G
.
H 0.16K
Var b Z g  E d Z i  c E b Z gh
d i
 2   5
E Z 2  102
0.8
2
2
2
 11233
.
 2.4262
 5.348
bg
bg
VarbSg  100 Varb Zg  534.8
E S  100 E Z  242.6
F  242.6
 1645
.
 F  281
534.8
MLC-09-13
43
Question #73
Answer: D
Prob{only 1 survives} = 1-Prob{both survive}-Prob{neither survives}
b
ge
 13p50 3p 50  13p50 13p 50
b
gb
gb
gb
j
gb
gb
g b
gb
g
 1  0.9713 0.9698 0.9682 0.9849 0.9819 0.9682  1  0.912012 1  0.93632
 
 0.912012
0.936320
 0.140461
Question # 74 - Removed
Question #75 - Removed
Question # 76
Answer: C
This solution applies the equivalence principle to each life. Applying the
equivalence principle to the 100 life group just multiplies both sides of the first
equation by 100, producing the same result for P.
EPV  Prems   P  EPV  Benefits   10q70v  10 p70q71v 2  Pp70 p71v 2
P
b10gb0.03318g  b10gb1  0.03318gb0.03626g  Pb1  0.03318gb1  0.03626g
108
.
108
. 2
 0.3072  0.3006  0.7988 P
P
0.6078
 3.02
0.2012
(EPV above means Expected Present Value).
MLC-09-13
44
108
. 2
*Question #77
Answer: E
Level net premiums can be split into two pieces: one piece to provide term
insurance for n years; one to fund the reserve for those who survive.
Then,
Px  Px1:n  Px:n1 nV
And plug in to get
b
gb
g
0.090  Px1:n  0.00864 0.563
Px1:n  0.0851
Another approach is to think in terms of retrospective reserves. Here is one such
solution:


V  Px  Px1:n sx:n
n

 Px  Px1:n
 aE
x:n
n

P
P  P 

P 
 Px  Px1:n
x
ax:n
1
x:n
ax:n
1
x
x:n
1
x:n
e
j
0.563  0.090  Px1:n / 0.00864
b
gb
g
Px1:n  0.090  0.00864 0.563
 0.0851
MLC-09-13
45
Question #78
Answer: A
b g
  ln 105
.  0.04879
Ax  
x
t
0

x
0

px  x t e  t dt
1  t
e dt for the given mortality function
x
1
a
  x x
From here, many formulas for the reserve could be used. One approach is:
Since
A50 
a50
50
 1  A50 
18.71
 0.3742 so a50  
  12.83
50
  

 1  A40 
19.40
 0.3233 so a40  
  13.87
60
60
  
0.3233
P  A40  
 0.02331
13.87
reserve   A50  P  A40  a50   0.3742   0.0233112.83   0.0751.
A40 
a60

Question #79
Answer: D
Ax  E vTx   E vTx NS   Prob  NS  E vTx S   Prob  S
0.03 
0.6




  0.70  
  0.30
 0.03  0.08 
 0.06  0.08 
 0.3195
Similarly, 2 Ax 
F
H
Var aT
I
b gK 
x
MLC-09-13
2
FG 0.03 IJ  0.70  FG 0.06 IJ  0.30  01923
.
.
H 0.03  016
H 0.06  016
. K
. K
Ax  Ax2
2

01923
.
 0.31952
 141
..
0.082
46
Question #80
Answer: B
2 q80:84
 2 q80  2 q84  2 q80:84
 0.5  0.4  1  0.6   0.2  0.15  1  0.1  0.5  0.4  0.2  0.15  (1  0.6  0.1)
= 0.10136
Using new p82 value of 0.3
 0.5  0.4  1  0.3  0.2  0.15  1  0.1  0.5  0.4  0.2  0.15  (1  0.3  0.1)
= 0.16118
Change = 0.16118 – 0.10136 = 0.06
Alternatively,
2 p80  0.5  0.4  0.20
3 p80  2 p80  0.6  0.12
2 p84  0.20  0.15  0.03
3 p84  2 p84  0.10  0.003
2 p80:84  2 p80  2 p84  2 p80 2 p84 since independent
3 p80:84
 0.20  0.03   0.20  0.03  0.224
 3 p80  3 p84  3 p80 3 p84
 0.12  0.003   0.12  0.003  0.12264
2 q 80:84  2 p80:84  3 p80:84
 0.224  0.12264  0.10136
Revised
3 p80  0.20  0.30  0.06
3
p 80:84  0.06  0.003   0.06  0.003
 0.06282
2 q 80:84  0.224  0.06282  0.16118
change  0.16118  0.10136  0.06
Question #81 - Removed
MLC-09-13
47
Question #82
Answer: A
5
b g  pb1g pb2g
p50
5 50 5 50
FG 100  55IJ e b gb g
H 100  50K
 b0.9gb0.7788g  0.7009
 0.05 5

Similarly
10
FG 100  60IJ e b g b g
H 100  50 K
 b0.8gb0.6065g  0.4852
b g 
p50
 0.05 10
b g  pb g  pb g  0.7009  0.4852
5 5 q50
5 50
10 50
 0.2157
Question #83
Answer: C
Only decrement 1 operates before t = 0.7
b g  b0.7g qb1g  b0.7gb010
. g  0.07
1

0.7 q40
40
Probability of reaching t = 0.7
since UDD
is
1-0.07 = 0.93
Decrement 2 operates only at t = 0.7, eliminating 0.125 of those who reached 0.7
b gb
g
b2g  0.93 0125
q40
.
 011625
.
MLC-09-13
48
Question #84
Answer: C
e
j
 1 2 p80v 2  1000 A80 
FG
H
 1
0.83910
106
. 2
vq80 v 3 2 p80q82

2
2
IJ  665.75  FG 0.08030  0.83910  0.09561IJ
K
. g
2b106
. g
H 2b106
K
3
b
g
b
b167524
.
g  665.75
g
 174680
.
 665.75   0.07156
  397.41
Where 2 p80 
3,284 ,542
 0.83910
3,914 ,365
b
gb
g
Or 2 p80  1  0.08030 1  0.08764  0.83910
*Question #85
Answer: E
At issue, expected present value (EPV) of benefits

  bt vt t p65 
0
65  t
dt
  1000  e0.04t  e 0.04t  t p65 65  t  dt

0
 1000

0
t
p65 65  t  dt  1000  q65  1000
EPV of premiums   a65  
FG 1 IJ  16.667
H 0.04  0.02K
Net premium   1000 / 16.667  60
2V


z
z

0

0
b g
b2u v u u p67  65 2  u du   a67
b g b gb
g
1000 e0.04b 2uge0.04u u p67  65 2  u du  60 16.667
 1000e0.08
z

0 u
b g
p67  65 2  u du  1000
 108329
.  q67  1000  108329
.  1000  8329
.
MLC-09-13
49
Question #86
Answer: B
(1)
ax:20  ax:20  120 Ex
(2)
ax:20 
(3)
Ax:20 
(4)
Ax  A1x:20  20 Ex Ax 20
1  Ax:20
d
A1x:20
 Ax:201
b gb g
0.28  A1x:20  0.25 0.40
A1x:20  018
.
Now plug into (3):
Ax:20  018
.  0.25  0.43
ax:20 
Now plug into (2):
Now plug into (1):
b
1  0.43
 1197
.
0.05 / 105
.
g
ax:20  1197
.  1  0.25  1122
.
Question #87 - Removed
Question #88
Answer: B
ex  px  pxex 1  px 
ex
8.83

 0.95048
1  ex 1 9.29
ax  1  vpx  v2 2 px  ....
a
x:2
 1  v  v 2 2 px  ...
ax:2  ax  vqx  5.6459  5.60  0.0459
v 1  0.95048  0.0459
v  0.9269
1
i   1  0.0789
v
MLC-09-13
50
Question #89
Answer: E
One approach is to enumerate the possible paths ending in F and add the
probabilities:
FFFF – 0.23 = 0.008
FFGF – 0.2(0.8)(0.5) = 0.080
FGFF – 0.8(0.5)(0.2) = 0.080
FGHF – 0.8(0.5)(0.75) = 0.300
The total is 0.468.
An alternative is to use matrix multiplication. The desired probability is the value in
the upper left corner of the transition probability matrix raised to the third power.
Only the first row needs to be evaluated.
The first row of the matrix squared is [0.44 0.16 0.40 0.00], obtained by multiplying
the first row by each column, in turn. The first row of the matrix cubed is obtained
by multiplying the first row of the squared matrix by each column. The result is
[0.468 0.352 0.080 0.100]. Note that only the first of the four calculations in
necessary, though doing the other three and observing that the sum is 1 provides
a check.
Either way, the required probability is 0.468 and the actuarial present value is
500v3 (0.468)  500(0.9)3 (0.468)  171 .
Question #90 – Removed
MLC-09-13
51
Question #91
Answer: E
1
1

75  60 15
1
1 3 1
60F 
  
   85
  60 15 5 25
60M 
t
10
t
 1
25
t
M
p65
 1
t
F
p60
Let x denote the male and y denote the female.
e x  5  mean for uniform distribution over  0,10  
e y  12.5  mean for uniform distribution over  0,25  
10 
t 
t 
e xy   1  1   dt
0
 10  25 
10 
7
t2 
  1 
t
 dt
0
 50 250 

 t 

7 2 t3 
t 

100
750 
10
0
 10 
7
1000
 100 
100
750
 10  7 




exy  ex  e y  exy  5 
MLC-09-13
4 13

3 3
25 13 30  75  26
 
 1317
.
2
3
6
52
*Question #92
Answer: B

1

 3

1
2
Ax 

  2 5
Ax 
Let P ( Ax ) denote the net premium.
c h
P Ax    0.04
F P c A hI A  A
Var b Lg  G 1 
j
H  JK e
F 0.04 IJ FG 1  FG 1IJ IJ
 G1 
H 0.08 K H 5 H 3K K
F 3I F 4 I
G J G J
H 2 K H 45K
2
x
2
2
x
x
2
2
2

1
5
*Question #93
Answer: A
Let  be the net premium
Let kV denote the net premium reserve at the end of year k.
For any n,  nV    1  i    q25n  n1V  p25n  n1V 
 n1V
Thus 1V   0 V    1  i 
2V
3V
  1V   1  i    1  i     1  i    s2


  2V   1  i    s2   1  i    s3
By induction (proof omitted)
nV   sn
For n  35, nV  a60 (expected present value of future benefits; there are no future
premiums)
a60   s35

a60
s35
For n  20,
MLC-09-13
20V
a 
  s20   60  s20
s 
 35 
53
Alternatively, as above
 nV    1  i   n1V
Write those equations, for n  0 to n  34
0 :  0V    1  i   1V
1:  1V   1  i   2V
2 :  2V   1  i   3V
34 :  34V    1  i   35V
34 k
Multiply equation k by 1  i 
and sum the results:
 0V    1  i 35   1V   1  i 34   2V   1  i 33    34V    1  i  
34
33
32
 34V 1  i   35V
1V 1  i   2V 1  i   3V 1  i  
For k  1, 2,
0V
35k
, 34, the k V 1  i 
1  i 35   1  i 35  1  i 34 
terms in both sides cancel, leaving
 1  i   35V

Since 0V  0
 s35  35V
 a60
(see above for remainder of solution)
MLC-09-13
54
*Question #94
Answer: B
Note: The symbol  x t: y t can be ambiguous. Here it means that both lives were
alive at ages x and y and the last survivor status is intact at time t (that is, at least
one of the two lives is alive). This meaning matches the wording of the question,
which does not use the symbol. Then,
t q y t p x  x  t  t q x t p y  y t
 x t: y t 
t q x  t p y  t px  t q y  t px  t p y
For (x) = (y) = (50)
5010.5:5010.5 
where
10.5
p50 
10.5 q50
10
2
 l60  l61   12 8,188,074  8,075, 403  0.90848
l50
8,950,901
 1  10.5 p50  0.09152
p50 
10.5
1
 10.5 q50  10 p50  q60  2
 0.09152  0.91478 0.01376  2 

 10.5 q50  10.5 p50   2   10.5 p50 2  0.09152  0.90848 2    0.908482
8,188,074
 0.91478
8,950,901
p50   50  10.5   10 p50  q60 since UDD
Alternatively, 10t  p50  10 p50 t p60
10t  p50:50   10 p50 
2
 t p60 
2
10t  p 50:50  2 10 p50 t p60   10 p50 
2
 t p60 
2
 2 10 p50 1  tq60    10 p50  1  tq60  since UDD
2
2
Derivative  2 10 p50 q60  2  10 p50  1  tq60  q60
2
Derivative at 10  t  10.5 is
2  0.91478 0.01376   0.91478 1   0.5 0.01376   0.01376  0.0023
2
10.5
p 50:50  2 10.5 p50   10.5 p50 
2
 2  0.90848    0.90848 
2
 0.99162
 (for any sort of lifetime) 
MLC-09-13

dp
dt    0.0023  0.0023
p
0.99162
55
 0.0023
Question #95
Answer: D
 xt   x1t   x 2t  0.01  2.29  2.30

P  P  vt t px   x 2t dt  50,000 vt t px   x1t dt  50,000 vt t px  xt dt
2
2
0
2 0.1t 2.3t
0
P  P e
e
0
2
2 0.1t 2.3t
 2.29dt  50,000 e
0
2 2.4 

1 e
P 1  2.29 
2.4

P = 11,194
e

 0.01dt  50,000 e0.1t e2.3t  2.3dt
2
2 2.4 


1 e
  50000  0.01
2.4


 2.3 
e
2 2.4 
*Question #96
Answer: B
ex  px  2 px  3px  ... 1105
.
b g
Annuity  v 3 3 p x 1000  v 4 4 p x  1000  104
.  ...


. g
1000b104
k 3 k
v
k px
k 3
 1000v 3

 k px
k 3
F 1 IJ
 1000v be  0.99  0.98g  1000G
H 104
. K
3
3
x
Let  = net premium.
e
j
 1  0.99v  0.98v 2  8072
2.8580  8072
  2824
MLC-09-13
56
 9.08  8072


2.4 
Question #97
Answer: B
b g
b ge
1
 a30:10  1000 A30  P IA 30
 10
:10

10
A30
j
1000 A30
b g
1
a30:10  IA 3010
 1010 A30
:

b
g
b
1000 0102
.
7.747  0.078  10 0.088
g
102
6.789
 15.024

Question #98
Answer: E
For the general survival function S0 (t )  1 

FG1  t IJ dt
H   30K
L t OP
 Mt 
N 2b  30g Q
e 30 
z
t

,0t  ,
  30
0
2
  30
0

  30
2
100  30
 35
2
Prior to medical breakthrough
  100  e 30 
After medical breakthrough
e 30  e30  4  39
so

e30
  39 

   30
2

    108
Question #99
Answer: A
L  100, 000v 2.5  4000a3
@5%
= 77,079
MLC-09-13
57
Question #100
Answer: D
  accid   0.001
 total   0.01
  other   0.01  0.001  0.009

Expected present value   500,000 e0.05t e0.01t  0.009  dt
0

10 50,000 e0.04t e0.05t e0.01t  0.001 dt
0
 0.009 0.001
 500, 000 

 100, 000
 0.06 0.02 
Question #101 Removed
Question #102
Answer: D
1000 20V  1000 Ax  20 

ax  20 
1000  19V  20 Px 1.06   qx 19 1000 
px 19
 342.03  13.72 1.06   0.01254 1000   369.18
0.98746
1  0.36918
 11.1445
 0.06 /1.06 
so 1000 Px  20  1000
MLC-09-13
Ax  20 369.18

 33.1
ax  20 11.1445
58
Question #103
Answer: B
k
 
1
k
 xt dt
 2  xt dt
 e 0
 e 0

 
k px
   x1t dt 
  e 0



k
2
  k px  where k px is from Illustrative Life Table, since    follows I.L.T.
2
10
6,616,155
 0.80802
8,188,074
6,396,609

 0.78121
8,188,074
p60 
11 p60
10
1
b  g  p b g  p b g
q60
10 60 11 60

b
10
p60
g b
2
11 p60
g
2
from I.L.T.
 0.80802 2  0.781212  0.0426
Question #104
Answer: C
Ps 
1
 d , where s can stand for any of the statuses under consideration.
a s
as 
1
Ps  d
1
 6.25
01
.  0.06
1
axy 
 8.333
0.06  0.06
ax  ay 
axy  axy  ax  ay
axy  6.25  6.25  8.333  4167
.
Pxy 
1
 0.06  018
.
4.167
MLC-09-13
59
Question #105
Answer: A
z
b
g
g j  48
d 0b g  1000 e b  0.04 gt   0.04 dt
1
0
 1000 1  e b  0.04
e
e b  0.04 g  0.952
b
  0.04   ln 0.952
g
 0.049
  0.009
z
b
g
d 3b1g  1000 e 0.049t 0.009 dt
4
3
 1000
0.009 b0.049 gb3g b0.049 gb4 g
e
e
 7.6
0.049
e
j
Question #106
Answer: B
This is a graph of lx  x .
b
g
 x would be increasing in the interval 80,100 .
The graphs of lx px , l x and lx2 would be decreasing everywhere.
Question #107
Answer: B
Variance  v30 15 px 15 qx
Expected value  v15 15 px
v30 15 px 15 qx  0.065 v15 15 px
v15 15qx  0.065  15 qx  0.3157
Since  is constant
15 q x
 px 
15

 1   px 
15

 0.6843
px  0.975
qx  0.025
MLC-09-13
60
*Question #108
Answer: E
1
 2
11V
I
11V

II
1   2


10V

11V
I
10V
I
0
II
 p
1 i
  II
 11V II 
qx 10
1000
px10

x 10
 p
1 i


x 10
10V
I
qx 10
1000
px 10
 10V II   II
 101.35  8.36 
 p
1 i
x 10
1.06 
1  0.004
 98.97
Question #109
Answer: A
2
EPV (x’s benefits)   v k 1bk 1 k px q x  k
k 0
b g
b gb g
b gb gb g
 1000 300v 0.02  350v 2 0.98 0.04  400v 3 0.98 0.96 0.06
 36,829
*Question #110
Answer: E
 denotes the net premium
V  EPV future benefits - EPV future premiums
1
0.6 
     0.326
108
.
.  q65 10
10V   108
11V 
p65
19
b

gb g b gb g
. g  b010
. gb10g
b5.0  0.326gb108
1  010
.
=5.28
MLC-09-13
61
Question #111
Answer: A
Expected present value Benefits 
 0.8 0.110,000    0.8 0.9  0.097 9,000 
 1, 239.75
1.062
1.063
  0.8   0.8  0.9  
1, 239.75  P 1 


1.062 
 1.06
 P  2.3955 
P  517.53  518
Question #112
Answer: A
1180  70a30  50a40  20a30:40
b gb g b gb g
1180  70 12  50 10  20a30:40
a30:40  8
a30:40  a30  a40  a30:40  12  10  8  14
100a30:40  1400
Question #113
Answer: B


0
0
a   at f (t )dt  

1  e0.05t 1
tet dt
0.05 (2)
1
0.05
zb


te
t
i
 te 1.05t dt
LM b g FG
IJ
H
K
N
1 L F 1 I O

.
M1  G . JK PP  185941
0.05 NM H 105
Q
1
t
1

 t  1 et 

e 1.05t
2
0.05
105
.
105
.
2
20,000  185941
.
 37,188
MLC-09-13
62
OP
Q

0
Question #114
Answer: C
Event
x0
Prob
 0.05
 0.95 0.10  0.095
 0.95 0.90  0.855
x 1
x2
Present Value
15
15  20 /1.06  33.87
15  20 /1.06  25/1.062  56.12
E  X    0.0515   0.095 33.87    0.855 56.12  51.95
E  X 2    0.0515   0.095 33.87    0.855 56.12  2813.01
2
 
2
2
Var  X   E X 2  E  X   2813.01   51.95  114.2
2
2
Question #115
Answer: B
Let K be the curtate future lifetime of (x + k)
kL
 1000v K 1  1000 Px:3  aK 1
When (as given in the problem), (x) dies in the second year from issue, the curtate
future lifetime of  x  1 is 0, so
1L
 1000v  1000 Px:3 a1
1000
 279.21
1.1
 629.88  630

The premium came from
A
Px:3  x:3
ax:3
Ax:3  1  d ax:3
Px:3  279.21 
MLC-09-13
1  d ax:3
ax:3

1
d
ax:3
63
Question #116
Answer: D
Let M = the force of mortality of an individual drawn at random; and T  future
lifetime of the individual.
Pr T  1
n
 E Pr T  1 M



z
zz
zd

0 0
0
bg
Pr T  1 M   f M  d
0
2 1
2
s
1
2
e   t dt d
1  e 
i 21 du  21 d2  e
2
 056767
.
Question #117
Answer: E
For this model:
1/ 60
1
(1)
40(1)t 

; 40
 20  1/ 40  0.025
1  t / 60 60  t
1/ 40
1
(1)
40(2)t 

; 40
 20  1/ 20  0.05
1  t / 40 40  t
 
40
 20  0.025  0.05  0.075
MLC-09-13
i 21 d1  e i
1 
64
2
*Question #118
Answer: D
Let   net premium
Expected present value of benefits =
 0.03 200,000 v  0.97 0.06 150,000 v 2  0.97 0.94 0.09 100,000 v 3
b gb
g b gb gb
g b gb gb gb
 5660.38  7769.67  6890.08
 20,32013
.
Expected present value of net premiums
 ax:3 
b gb g
 1  0.97v  0.97 0.94 v 2 
 2.7266 
20,32013
.
 
 7452.55
2.7266
1V

. g  b200,000gb0.03g
b7452.55gb106
1  0.03
 1958.46
Initial reserve, year 2 = 1V  
= 1958.56 + 7452.55
= 9411.01
Question #119
Answer: A
Let  denote the premium.
b g
L  bT v T   aT  1 i
T
 v T   aT
 1   aT
E L  1   ax  0
 
 L  1   aT  1 
b
g
aT
ax

1
ax
d
 ax  1  v T
i
 ax
v T  1   ax
v T  Ax


 ax
1  Ax
MLC-09-13
65
g
Question #120
Answer: D
(0, 1)
(1, 0.9)
(1.5, 0.8775)
(2, 0.885)
tp1
1
2
t
1
p1  (1  01
. )  0.9
2
p1  0.9 1  0.05  0.855
b gb
g
b
g
since uniform, 1.5 p1  0.9  0.855 / 2
 0.8775
e1:1.5  area between t  0 and t  1.5
 1  0.9 
 0.9  0.8775 

 (1)  
 (0.5)
2
 2 


 0.95  0.444  1.394
MLC-09-13
66
Alternatively,
e1:1.5  
1.5
t
0
1
0.5
0
0
p1dt   t p1dt  1 p1 
1
t
p2 dt
0.5
  (1  0.1t )dt  0.9 (1  0.05t )dt
0
0
1
0.5
 0.1t 2 
 0.05t 2 
 t 

0.9

t 

2 0
2 0


 0.95  0.444  1.394
Question #121
Answer: A
b g
10,000 A63 112
.  5233
A63  0.4672
Ax 1 
A64 
b g
Ax 1  i  qx
px
. g  0.01788
b0.4672gb105
1  0.01788
 0.4813
A65 
. g  0.01952
b0.4813gb105
1  0.01952
 0.4955
Single gross premium at 65 = (1.12) (10,000) (0.4955)
= 5550
b1  ig
MLC-09-13
2

5550
5233
i
5550
 1  0.02984
5233
67
Question #122A
Answer: C
Because your original survival function for (x) was correct, you must have
 x t  0.06   x02t: y t   x03t: y t   x02t: y t  0.02
 x02t: y t  0.04
Similarly, for (y)
 y t  0.06   x01t: y t   x03t: y t   x01t: y t  0.02
 x01t: y t  0.04
The first-to-die insurance pays as soon as State 0 is left, regardless of which state
is next. The force of transition from State 0 is
x01t: y t  x02t:y t  x03t:y t  0.04  0.04  0.02  0.10 .
With a constant force of transition, the expected present value is


0.10
 t
00
01
02
03
0.05t 0.10t
0 e t pxy (xt:y t  xt:y t  xt:y t )dt  0 e e (0.10)dt  0.15
MLC-09-13
68
Question #122B
Answer: E
Because (x) is to have a constant force of 0.06 regardless of (y)’s status (and vice23
versa) it must be that 13
x t   y t  0.06 .
There are three mutually exclusive ways in which both will be dead by the end of
year 3:
1: Transition from State 0 directly to State 3 within 3 years. The probability of this
is
3
0.02 0.10t
0.3
0 t p  dt  0 e 0.02dt   0.10 e 0  0.2(1  e )  0.0518
2: Transition from State 0 to State 1 and then to State 3 all within 3 years. The
probability of this is
3

3
0 t
00
xy
3
03
x t : y t
0.10t
3
0.10 t
pxy00  x01t: y t 3t p13
0.04(1  e 0.06(3t ) )dt
x  t dt   e
0
3
  0.04 e
0
 0.4(1  e
0.10 t
0.3
e
)e
0.18 0.04 t
0.18
e
(1  e
0.04 0.10t 0.04e0.18 0.04t
  
e

e
0.10
0.04
0.12
3
0
)  0.00922
3: Transition from State 0 to State 2 and then to State 3 all within 3 years. By
symmetry, this probability is 0.00922.
The answer is then 0.0518 + 2(0.00922) = 0.0702.
MLC-09-13
69
Question #122C
Answer: D
Because the original survival function continues to hold for the individual lives, with
a constant force of mortality of 0.06 and a constant force of interest of 0.05, the
expected present values of the individual insurances are
Ax  Ay 
0.06
 0.54545 ,
0.06  0.05
Then,
Axy  Ax  Ay  Axy  0.54545  0.54545  0.66667  0.42423
Alternatively, the answer can be obtained be using the three mutually exclusive
outcomes used in the solution to Question 122B.
1:


0

e0.05t t pxy00  x03t: y t dt   e0.05t e0.10t 0.02dt 
2 and 3:
0


0
0.02
 0.13333
0.15

13
e0.05t t pxy00  x01t: y t  e0.05 r r p11
x  t : y  t  x t  r : y t  r drdt
0


0
0
  e0.05t e0.10t 0.04 e0.05 r e0.06 r 0.06drdt 
0.04 0.06
 0.14545
0.15 0.11
The solution is 0.13333 + 2(0.14545) = 0.42423.
The fact that the double integral factors into two components is due to the
memoryless property of the exponential transition distributions.
MLC-09-13
70
Question #123
Answer: B
5
q35:45  5 q35  5 q45  5 q35:45
 5 p35q40  5 p45q50  5 p35:45q40:50
b
g
 p q  p q  p  p b1  p p g
 b0.9gb.03g  b0.8gb0.05g  b0.9gb0.8g 1  b0.97gb0.95g
 5 p35q40  5 p45q50  5 p35  5 p45 1  p40:50
5 35 40
5 45 50
5 35
5 45
40
50
 0.01048
Alternatively,
6
p35  5 p35  p40   0.90 1  0.03  0.873
6
p45  5 p45  p50   0.80 1  0.05  0.76
5 q35:45
 5 p35:45  6 p35:45
  5 p35  5 p45  5 p35:45    6 p35  6 p45  6 p35:45 
  5 p35  5 p45  5 p35  5 p45    6 p35  6 p45  6 p35  6 p45 
  0.90  0.80  0.90  0.80    0.873  0.76  0.873  0.76 
 0.98  0.96952
 0.01048
Question #124 – Removed
Question #125 - Removed
MLC-09-13
71
Question #126
Answer: E
Let
Y = present value random variable for payments on one life
S   Y  present value random variable for all payments
E Y  10a40  148166
.
dA
2
Var Y  10
2
40
d
2
 A40
d
i
2
ib
g
2
 100 0.04863  016132
.
106
. / 0.06
2
 70555
.
E S  100 E Y  14,816.6
Var S  100 Var Y  70,555
Standard deviation S  70,555  265.62
By normal approximation, need
E [S] + 1.645 Standard deviations = 14,816.6 + (1.645) (265.62)
= 15,254
*Question #127
Answer: B
Initial net premium 
Where
e
e
1
5 A30  4 A30
:20
j
5a30:35  4a30:20
b
g b
g b
g

5 010248
.
 4 0.02933
5 14.835  4 11959
.

0.5124  011732
.
0.39508

 0.015
74.175  47.836
26.339
b
g
j
1
A30
 A30:20  A30:201  0.32307  0.29374  0.02933
:20
and
a30:20 
1  A30:20
d

1  0.32307
 11959
.
0.06
106
.
FG IJ
H K
Comment: the numerator could equally well have been calculated as
A30  4 20E30 A50 = 0.10248 + (4) (0.29374) (0.24905) = 0.39510
MLC-09-13
72
Question #128
Answer: B
0.75
b gb g
px  1  0.75 0.05
 0.9625
0.75
b gb g
p y  1  0.75 .10
 0.925
0.75 qxy  1 0.75 pxy
b p gd p i since independent
= 1- b0.9625gb0.925g
 1
0.75
x
0.75
y
 01097
.
*Question #129
Answer: D
Let G be the gross premium.
Expected present value (EPV) of benefits = 100,000A35
EPV of premiums = Ga35
EPV of expenses = 0.1G  25   2.50 100  a35
Equivalence principle:
Ga35  100,000 A35   0.1G  25  250  a35
A35
 0.1G  275
a35
0.9G  100,000 P35  275
G  100,000
G
100 8.36   275
0.9
 1234
MLC-09-13
73
Question #130
Answer: A
The person receives K per year guaranteed for 10 years  Ka10  8.4353K
The person receives K per years alive starting 10 years from now 10 a40 K
b
g
Hence we have 10000  8.435310 E40a50 K
Derive
10 E40:
1
A40  A4010

:
10 E40
Derive a50 

b
10 E40
1
A40  A4010
:
A50

gA
50
0.30  0.09
 0.60
0.35
1  A50 1  0.35

 16.90
.04
d
104
.
Plug in values:
10,000  8.4353  0.60 16.90 K
c
b gb
gh
 18.5753K
K  538.35
MLC-09-13
74
Question #131
Answer: D
11
STANDARD: e25:11  
11
0
t 
t2

1

dt

t

 10.1933


2  75 0
 75 
e z
1
MODIFIED:
p25
 0.1ds
0
 e.1  0.90484
1
10 
t 
e25:11   t p25 dt  p25  1   dt
0
0
 74 
IJ
K
F t I
1 e

 e Gt 
01
.
H 2  74JK
 0.95163  0.90484b9.32432g  9.3886

z
1 0.1t
0
e
dt  e0.1
0.1
0.1
z FGH
10
0
1
t
dt
74
2
10
0
Difference
=0.8047
*Question #132
Answer: B
Comparing B & D: Prospectively at time 2, they have the same future benefits. At
issue, B has the lower net premium. Thus, B has the higher reserve.
Comparing A to B: Consider them retrospectively. At issue, B has the higher net
premium. Until time 2, they have had the same benefits, so B has the higher
reserve.
Comparing B to C: Visualize a graph C* that matches graph B on one side of t=2
and matches graph C on the other side. By using the logic of the two preceding
paragraphs, C’s reserve is lower than C*’s which is lower than B’s.
Comparing B to E: Reserves on E are constant at 0.
MLC-09-13
75
Question #133
Answer: C
Since only decrements (1) and (2) occur during the year, probability of reaching
the end of the year is
1
2
p60
b g  p60
b g  1  0.01 1  0.05  0.9405
b
gb
g
Probability of remaining through the year is
1
2
3
p60
.  0.84645
b g  p60
b g  p60
b g  1  0.01 1  0.05 1  010
b
gb
gb
g
Probability of exiting at the end of the year is
b3g  0.9405  0.84645  0.09405
q60
Question #134 - Removed
Question #135
Answer: D
EPV of regular death benefit 

zb

0
gd
ib
gd
zb

0
gd ib
gd i
100000 e- t 0.008 e-  t dt
i
100000 e-0.06t 0.008 e-0.008t dt
b
g
 100000 0.008 / 0.06  0.008  11,764.71
EPV of accidental death benefit 

zb
30
0
gd
ib
gd
i
zb
30
0
gd ib
100000 e-0.06t 0.001 e-0.008t dt
 100 1  e-2.04 / 0.068  1,279.37
Total EPV  11765  1279  13044
MLC-09-13
gd i
100000 e- t 0.001 e-  t dt
76
Question #136
Answer: B
b gb
g b gb
g
g b gb
g
l 60 .6  .6 79,954  .4 80,625
 80,222.4
b gb
l 60 1.5  .5 79,954  .5 78,839
 79,396.5
0.9 q 60 .6
80222.4  79,396.5
80,222.4
 0.0103

Question #137 - Removed
Question #138
Answer: A
b g  qb1g  qb2g  0.34
q40
40
40
1
2
 1  p40
b g p40
b g
0.34  1  0.75 p40
 b 2g
p40
 b 2g  0.88
q40
. y
 b 2g  012
q41
 b2g  2 y  0.24
b gb g
l b g  2000b1  0.34gb1  0.392g  803
b g  1  0.8 1  0.24  0.392
q41

42
MLC-09-13
77
Question #139
Answer: C
bg
Pr L  '  0  0.5
Pr 10,000v K 1   ' aK 1  0  0.5
From Illustrative Life Table,
47
p30  0.50816 and
48
p30 .47681
Since L is a decreasing function of K, to have
Pr L    0  0.5 means we must have L    0 for K  47.
b g
b g
b g
Highest value of L   for K  47 is at K = 47.
b g
L   at K  47  10,000 v 471    a471
b g
 609.98  16.589 
b
g
L    0  609.98  16.589    0
  
609.98
 36.77
16.589
Question #140
Answer: B
b g
Prb K  1g  p  p  0.9  0.81  0.09
Prb K  1g  p  0.81
. .81  2.72  2.4715
EbY g = .1  1.09  187
. .81  2.72  6.407
E dY i = .1  1 .09  187
VARbY g  6.407  2.4715  0.299
Pr K  0  1  px  01
.
1 x
2 x
2 x
2
2
2
2
2
MLC-09-13
78
Question #141
Answer: E
E  Z   b Ax
since constant force Ax   /(   )
E(Z) 
b  0.02 
b

 b/3
 
 0.06 
Var  Z   Var b v T   b 2Var v    b 2

2
Ax  Ax2

2
 
   

b 

   2      


 2 1
 4
 b2     b2  
10 9 
 45 
2
Var  Z   E  Z 
4 b
b2   
 45  3
4 1
b     b  3.75
 45  3
*Question #142
Answer: B
In general, for any premium P and corresponding loss at issue random variable L,
2
Var L  1  P 2 Ax  Ax2
b g b gc
Here P   
h
1
1
    0.08  0.12
ax
5
b g FGH ..1208IJK c A  A h .5625
F b.12gIJ c A  A h
and Varb L *g  G1 
H .08 K
b1  g b0.5625g .744
So Var b L *g 
b1  g
E L *  A .15a  1  a b .15g  1  5b.23g  .15
E L *  Varb L *g .7125
2
So Var L  1 
2
2
x
x
2
5
4
2
x
2
x
15 2
8
12 2
8
x
MLC-09-13
x
x
79
Question #143 - Removed
Question #144
Answer: B
Let l0b g  number of students entering year 1
superscript (f) denote academic failure
superscript (w) denote withdrawal
subscript is “age” at start of year; equals year - 1
p0b g  1  0.40  0.20  0.40
l2b g  10 l2b g q2b f g  q2b f g  01
.
b
g
q2b wg  q2b g  q2b f g  10
.  0.6  01
.  0.3
l1b gq1b f g  0.4 l1b g 1  q1b f g  q1b wg
e
j
q1b f g  0.4 1  q1b f g  0.3
e
q1b f g 
j
0.28
 0.2
14
.
p1b g  1  q1b f g  q1b wg  1  0.2  0.3  0.5
b g  q b w g  p b g q b w g  p b g p b  g q b w g
w
3 q0
0
0
1
0
1
2
b gb g b gb gb g
 0.2  0.4 0.3  0.4 0.5 0.3
 0.38
MLC-09-13
80
Question #145
Answer: D
e25  p25 (1  e26 )
N
M
e26
 e26
due to having the same 
1
N
M
M 0.05
p25
 exp    25
 0.1(1  t )dt   p25
e
 0 t

N
N
M
M
e25
 p25
(1  e26 )  e 0.05 p25
(1  e26 )  0.951e25
 9.51
Question #146
Answer: D
E[YAGG ]  100 E[Y ]  100(10, 000)ax
 100(10, 000)
1  Ax
 10, 000, 000

b10,000g 1 c A  A h
b10,000g b0.25g  b016

. g  50,000

 Y  Var Y 
2
2
2
b
x
2
x
g
 AGG  100 Y  10 50,000  500,000
0.90  Pr
LM F  E Y
N 
AGG
AGG
 1282
.

OP
Q
0
F  E YAGG
 AGG
F  1282
.  AGG  E YAGG
b
g
F  1282
.
500,000  10,000,000  10,641,000
MLC-09-13
81
Question #147
Answer: A
1
A30:3
 1000vq30  500v 2 1 q30  250v3 2 q30
2
3
 1  1.53 
 1 
 1.61 
 1 
 1.70 
 1000 

  500 
  0.99847  
  250 
  0.99847  0.99839  

 1.06  1000 
 1.06 
 1000 
 1.06 
 1000 
 1.4434  0.71535  0.35572  2.51447
1
1 1
 1 2
 0.00153 
1
a30:1  1 2  1 2 
 1  2 q30     0.97129  1 

2 2
2
 1.06 


1 1
   0.97129  0.999235
2 2
 0.985273
2.51447
Annualized premium 
0.985273
 2.552
 2
2.552
2
 1.28
Each semiannual premium 
Question: #148
Answer: E
b DAg
1
80:20
q80 .2
q80 .1
eb g j
1
 20vq80  vp80 DA 8119
:
bg
b g
b g
b g
b g b gb
2.9b12.225g

 12.267
20 .2
.8
1

DA 8119
:
106
.
106
.
13 106
. 4
1
 DA 8119

 12.225
:
.8
1
DA80
 20v .1  v .9 12.225
:20
13 
106
.
Question #149 - Removed
MLC-09-13
82
g
Question #150
Answer: A
t
1
100  x  t
t
 t

px  exp   
ds   exp ln(100  x  s) 0  


0 100  x  s
100  x


e50:60  e50  e60  e50:60
50
e50  
50
e60  
40
0
50  t
1 
t2 
dt   50t    25
50
50 
2 0
40
0
40  t
1 
t2 
dt   40t    20
40
40 
2 0
40
e50:60  
40
0
2
40 2000  90t  t
1 
t3 
 50  t  40  t 
2
dt

dt

2000
t

45
t


  14.67



0
2000
2000 
3 0
 50  40 
e50:60  25  20  14.67  30.33
Question #151
Answer: C
Ways to go 0  2 in 2 years
0  0  2; p   0.7  0.1  0.07
0  1  2; p   0.2  0.25   0.05
0  2  2; p   0.11  0.1
Total = 0.22
Binomial m = 100 q = 0.22
Var = (100) (0.22) (0.78) = 17
MLC-09-13
83
Question #152
Answer: A
For death occurring in year 2
0.3 1000
EPV 
 285.71
1.05
For death occurring in year 3, two cases:
(1) State 2  State 1  State 4: (0.2  0.1) = 0.02
(2) State 2  State 2  State 4: (0.5  0.3) = 0.15
0.17
Total
EPV =
0.17 1000
 154.20
1.052
Total. EPV = 285.71 + 154.20 = 439.91
Question #153 - Removed
*Question #154
Answer: C
Let  denote the single net premium.
1
  30 a35   A35
:30

30
a35
1
1  A35
:30
A
e

35:30
j
1
 A35
a65
:30
1
1  A35
:30


b.21.07g9.9
b1.07g
1.386
.93
 1.49

MLC-09-13
84
Question #155
Answer: E
0.4 p0
.5  e z

0.4
0
e F e2 x jdx
LM e22x OP.4
N Q0
e
0.8
.4 F F e 2 1I
H
K
e
.4 F 
.5  e.4 F .6128
bg
 ln .5  .4F .6128
 .6931  .4F .6128
 F  0.20
*Question #156
Answer: C
 9V  P  1.03  qx9b  1  qx9  10V
 qx 9  b  10V   10V
 3431.03  0.02904 872   10V
 10V  327.97
b   b  10V   10V  872  327.97  1199.97
 1

1
0.03 

P  b   d   1200 


 14.65976 1.03 
 ax

= 46.92
9V  net premium reserve at the start of year ten – P = 343 – 46.92 = 296.08
Question #157
Answer: B
d = 0.06  V = 0.94
Step 1 Determine p x
668  258vpx  1000vqx  1000v 2 px  px 1  qx 1 
668  258  0.94  px  1000  0.94  1  px   1000  0.8836  px 1
668  242.52 px  940 1  px   883.6 px
px  272/ 298.92  0.91
MLC-09-13
85
Step 2 Determine 1000 Px:2
668  258  0.94  0.91  1000 Px:2 1   0.94  0.91 
 220.69  668  479
1000 Px:2 
1.8554
Question #158
Answer: D
1
1
1
100,000  IA 40:10  100,000 v p40  IA 41:10  10 v10 9 p41 q50   A40:10
100,000 


see comment 


 8,950,901 
10 



0.99722
0.16736   9, 287, 264    0.00592  
 100,000
1.06 
1.0610





  0.02766 100,000 
=15,513
1
 A40  10 E40 A50
Where A40:10
 0.16132   0.53667  0.24905 
 0.02766
Comment: the first line comes from comparing the benefits of the two insurances.
At each of age 40, 41, 42,…,49  IA40:10 provides a death benefit 1 greater than
1
 IA141:10 . Hence the
1
A40:10
term. But  IA41:10 provides a death benefit at 50 of 10,
1
while  IA40:10 provides 0. Hence a term involving 9 q41  9 p41 q50 . The various v’s
1
and p’s just get all expected present values at age 40.
MLC-09-13
86
Question #159
Answer: A
10001V   1  i   qx 1000  10001V 
40  80 1.1  qx 1000  40 
qx 
88  40
 0.05
960
1 AS



G 
expenses 1  i   1000qx
px
100   0.4100  1.1  1000 0.05
1  0.05
60 1.1  50
 16.8
0.95
Question #160
Answer: C
1
At any age, px    e0.02  0.9802
1
1
qx    1  0.9802  0.0198 , which is also q x  , since decrement 2 occurs only at the
end of the year.
Expected present value (EPV) at the start of each year for that year’s death
benefits
= 10,000*0.0198 v = 188.1

px   0.9802*0.96  0.9410

Ex  px v  0.941 v  0.941*0.95  0.8940
EPV of death benefit for 3 years 188.1  E40 *188.1  E40 * E41 *188.1  506.60
MLC-09-13
87
Question #161
Answer: B
40
e30:40 
 t p30dt
0
  30  t
dt
  30
0
40


t
t2
2   30 
40
0
800
  30
 27.692
 40 
  95
Or, with a linear survival function, it may be simpler to draw a picture:
0
p30  1
40
30
70
e30:40  area =27.692  40
40
p30
p30  0.3846
1  40 p30 
2
  70
 0.3846
  30
  95
t
p30 
65  t
65
Var  E T    E T  
2
MLC-09-13
2
88
Using a common formula for the second moment:

Var T    2 t t px dt  ex2
0
 2
65
0
t 
 65 
t  

t 1   dt    1   dt 
0
 65 
  65  
2
 2*  2112.5  1408.333   65  65/ 2 
2
 1408.333  1056.25  352.08
Another way, easy to calculate for a linear survival function is

Var T    t 2 t px  x t dt 
0



0
t t px  x t dt
1
1 
 65
t  dt    t  dt 
0
65
65 

65 2
0
t3

3  65
65
0
 t2

 2  65
65 
0

2
2
2


 1408.33   32.5  352.08
2
With a linear survival function and a maximum future lifetime of 65 years, you
probably didn’t need to integrate to get E T30   e30  32.5
Likewise, if you realize (after getting   95 ) that T30 is uniformly distributed on (0,
65), its variance is just the variance of a continuous uniform random variable:
2
65  0 

Var 
 352.08
12
Question #162
Answer: E
218.15 1.06   10,000  0.02 
 31.88
1  0.02
 31.88  218.151.06    9,000  0.021  77.66
2V 
1  0.021
1V

MLC-09-13
89
Question #163
Answer: D

ex  ey   t px  0.95  0.952  ...
k 1
0.95
 19
1  0.95
exy  pxy  2 pxy  ...

 1.02  0.95 0.95  1.02  0.95  0.95  ...
2
1.02  0.95
 1.02 0.95  0.95  ... 
1  0.952
exy  ex  ey  exy  28.56
2
4
2
2
 9.44152
Question #164 - Removed
Question #165 - Removed
Question #166
Answer: E

ax   e0.08t dt  12.5
0

3
 0.375
8

3
2
Ax   e0.13t  0.03 dt   0.23077
0
13
2
1
2
 aT  Var aT   2  2 Ax   Ax    400 0.23077   0.375   6.0048







Ax   e0.08t  0.03 dt 
0
 
 
Pr aT  ax   aT   Pr aT  12.5  6.0048
1  vT

 Pr 
 6.4952  Pr 0.67524  e0.05T 
 0.05

 ln 0.67524 

 Pr T 
  Pr T  7.85374
0.05

 e0.037.85374  0.79
MLC-09-13
90
Question #167
Answer: A
5
 0.05 5
 
p50
 e     e0.25  0.7788
1
5 q55
5 1
 0.030.02 t
  55
dt    0.02 / 0.05 e0.05t
t  e
0

 0.4 1  e
0.25

= 0.0885
 

Probability of retiring before 60  5 p50
 5 q55
= 0.7788*0.0885
= 0.0689

MLC-09-13
1
91
5
0
Question #168
Answer: C
Complete the table:
l81  l80  d80  910
l82  l81  d81  830 (not really needed)
e x  ex 
1
2
1

 since UDD 
2

e x  e x  1 2
l  l  l   1

e x   81 82 83
l80

 2


e  1  l  l  l  Call this equation (A)
 80 2  80 81 82




e  1  l  l  Formula like (A), one age later. Call this (B)
 81 2  81 82


Subtract equation (B) from equation (A) to get




1
1
l81  e80   l80  e81   l81


2
2






910  8.5  0.51000  e81  0.5 920


e81 
8000  460  910
 8.21
920
MLC-09-13
92
Alternatively, and more straightforward,
910
 0.91
1000
830
p81 
 0.902
920
830
p81 
 0.912
910
p80 
e80  1 q 80  p 80 1  e81 
 
2  


1
where q80 contributes
since UDD
2
1


8.5  1  0.91   0.91 1  e81 
2


e81  8.291
1


e81  q81  p81 1  e82 
2


1


8.291  1  0.912   0.912 1  e82 
2


e82  8.043
1


e81  q81  p81 1  e82 
2


1
 1  0.902    0.902 1  8.043
2
 8.206
Or, do all the recursions in terms of e, not e , starting with e80  8.5  0.5  8.0 , then
final step e81  e81  0.5
MLC-09-13
93
Question #169
Answer: A
T
p x t
0
1
2
3
0.7
0.7


px
vt
vt t px
1
0.7
0.49
1
0.95238
0.90703
–
1
0.6667
0.4444
t

2
From above ax:3   vt t px  2.1111
t 0
 a
1000 2Vx:3  1000 1  x 2:1

ax:3


1 

  1000 1 
  526

 2.1111 

Alternatively,
Px:3 
1
 d  0.4261
ax:3

1000 2Vx:3  1000 v  Px:3

 1000  0.95238  0.4261
 526
You could also calculate Ax:3 and use it to calculate Px:3 .
MLC-09-13
94

Question #170
Answer: E
Let G be the gross premium.
Expected present value (EPV) of benefits = 1000A50 .
EPV of expenses, except claim expense = 15  1 a50
EPV of claim expense = 50A50 (50 is paid when the claim is paid)
EPV of premiums = G a50
Equivalence principle: Ga50  1000 A50  15  1 a50  50 A50
1050 A50  15  a50
G
a50
For the given survival function,
a50 1  1.0550
1 50 k
1 50 k
A50 
v
(
l

l
)

v
(2)


 0.36512
 50k 1 50k 100 
l50 k 1
50
0.05(50)
k 1
1  A50
a50 
 13.33248
d
Solving for G, G = 30.88
Question #171
Answer: A
4
 0.05 4
p50  e     0.8187
10
8
 0.05 10
p50  e     0.6065
 0.04 8
p60  e     0.7261
18
p50   10 p50  8 p60   0.6065  0.7261
 0.4404
414 q50  4 p50  18 p50  0.8187  0.4404  0.3783
MLC-09-13
95
Question #172
Answer: D
a40:5  a40  5 E40 a45
 14.8166   0.73529 14.1121
 4.4401
 a40:5  100,000 A45 v5 5 p40    IA40:5
1

  100,000 A45  5 E40 / a40:5   IA40:5
1

 100,000  0.20120  0.73529  /  4.4401  0.04042 
 3363
Question #173
Answer: B
Calculate the probability that both are alive or both are dead.
P(both alive) = k pxy  k px  k p y
P(both dead) = k qxy  k qx k q y
P(exactly one alive) = 1  k pxy  k qxy
Only have to do two year’s worth so have table
Pr(both alive)
Pr(both dead)
Pr(only one alive)
1
0
0
(0.91)(0.91) = 0.8281
(0.09)(0.09) = 0.0081
0.1638
(0.82)(0.82) = 0.6724
(0.18)(0.18) = 0.0324
0.2952
0.8281 0.6724 
0.1638 0.2952 
 1
 0
EPV Annuity  30, 000 


 20, 000 


  80, 431
0
1
2 
0
1.05
1.05 
1.051
1.052 
 1.05
 1.05
Alternatively,
0.8281 0.6724

 2.3986
1.05
1.052
0.91 0.82
ax:3  a y:3  1 

 2.6104
1.05 1.052
EPV  20,000 ax:3  20,000 a y:3 10,000 axy:3
axy:3  1 
(it pays 20,000 if x alive and 20,000 if y alive, but 10,000 less than that if
both are alive)
  20,000 2.6104   20,000  2.6104   10,000  2.3986  80, 430
MLC-09-13
96
Question #174
Answer: C
Let P denote the gross premium.


0
axIMP
0
P  ax   e t e t dt   e0.05t dt  20
E  L 
10
axIMP
 e
P
0.03t 0.02t
e
dt  e
0.0310  0.0210 
e
0

e
0.03t 0.01t
e
dt
0
l  e0.5 e0.5

 23
0.05
0.04
E  L  23  20  3

E  L 3

 15%
P
20
Question #175
Answer: C
1
A30:2
 1000vq30  500v 2 1 q 30
2
 1 
 1 
 1000 
  0.00153  500 
  0.99847  0.00161
 1.06 
 1.06 
 2.15875
Initial fund  2.15875 1000 participants = 2158.75
Let Fn denote the size of the fund at the end of year n.
F1  2158.75 1.07  1000  1309.86
F2  1309.86 1.069   500  900.24
Expected size of the fund at end of year 2 = 0 (since the amount paid was the
single net premium). Difference is 900.
MLC-09-13
97
Question #176
Answer: C
2
Var  Z   E  Z 2   E  Z 
E Z   


t
t t
0

v b 

0
E  Z 2   
px  xt dt   e0.08t e0.03t e0.02t  0.02  dt
0
0.02 2

 0.02  e0.07t dt 
7
0.07

0
2
 vt bt  t px xt dt  

0
e

  0.02 e0.12t  xt dt  2
0
Var  Z  
 
1 2

7
6
2

12

0.05t 2
1
e0.02t  0.02  dt
6
1 4

 0.08503
6 49
*Question #177
Answer: C
0.1
8  311
1.1
0.1
 1
 6   511
1.1
From Ax  1  d ax we have Ax  1 
Ax10
Ax  Ax  i
Ax 

3
0.1

 0.2861
11 ln 1.1
Ax 10 
5
0.1

 0.4769
11 ln 1.1
Let P be the net premium.
10V
 Ax10  P  ax10
 0.2861 
 0.4769  
6
 8 
= 0.2623
There are many other equivalent formulas that could be used.
MLC-09-13
98
Question #178
Answer: C

  100,000  e0.06t  e0.001t 0.001dt
Regular death benefit
0
0.001 

 100,000 

 0.06  0.001 
 1639.34
Accidental death
  100,000 e0.06t e0.001t  0.0002  dt
10
0
10
 20 e0.061t dt
0
1  e0.61 
 20 
  149.72
0.061


Expected Present Value  1639.34  149.72  1789.06
Question #179
Answer: C
00
p61
 560 / 800  0.70
01
p61
 160 / 800  0.20
10
p61
 0, once dead, stays dead
11
p61
 1, once dead by cause 1, stays dead by cause 1
00
01
11
p61
 p61
 p10
61  p61  0.70  0.20  0  1  1.90
MLC-09-13
99
Question #180
Answer: C
The solution requires determining the element in row 2 column 3 of the matrix
obtained by multiplying the first three transition probability matrices. Only the
second row needs to be calculated. After one period, the row is [0.0 0.7 0.3]. For
the second period, multiply this row by the columns of Q1 . The result is [0.07 0.49
0.44]. For the third period, multiply this row by the columns of Q2 . The result is
[0.1645 0.3465 0.4890] with the final entry providing the answer.
An alternative is to enumerate the transitions that result in extinction. Label the
states S (sustainable), E (endangered) and X (extinct). The possible paths are:
EX – 0.3
EEX – 0.7(0.2) = 0.14
EEEX – 0.7(0.7)(0.1) = 0.049
Total = 0.489
Note that if the species is not extinct by time 3, it will never become extinct.
Question #181
Answer: B
Pr(dies in year 1)  p 02  0.1
Pr(dies in year 2)  p 00 p 02  p 01 p12  0.8(0.1)  0.1(0.2)  0.10
Pr(dies in year 3)  p 00 p 00 p 02  p 00 p 01 p12  p 01 p11 p12  p 01 p10 p 02  0.095
EPV (benefits)  100, 000[0.9(0.1)  0.92 (0.10)  0.93 (0.095)]  24, 025.5
Pr(in State 0 at time 0)  1
Pr(in State 0 at time 1)  p 00  0.8
Pr(in State 0 at time 2)  p 00 p 00  p 01 p10  0.8(0.8)  0.1(0.1)  0.65
EPV ($1 of premium)  1  0.9(0.8)  0.92 (0.65)  2.2465
Net premium = 24,025.5/2.2465 = 10,695.
Question #182 - Removed
Question #183 - Removed
MLC-09-13
100
Question #184
Answer: B
1000P45a45:15   a60:15  15 E45  1000 A45
1000
A45
 a45  15 E45 a60     a60  15 E60 a75  15 E45   1000 A45
a45
201.20
14.1121   0.72988 0.5108111.1454
14.1121
 11.1454   0.68756 0.39994 7.2170    0.72988 0.51081  201.20
where
15 Ex
was evaluated as 5 Ex  10 Ex5
14.2573 9.9568    3.4154   201.20
  17.346
Question #185
Answer: A
1V
  0 V    1  i   1000  1V  1V  qx
2V
  1V   1  i    2000  2V  2V  qx1  2000
  1  i   1000q   1  i   2000q  2000
  1.08  1000  0.1    1.08  2000  0.1  2000
x
x 1
  1027.42
MLC-09-13
101
Question #186
Answer: A
Let Y be the present value of payments to 1 person.
Let S be the present value of the aggregate payments.
E Y   500 ax  500
 Y  Var Y  
1  Ax   5572.68
d
 5002
1
d2

2

Ax  Ax2  1791.96
S  Y1  Y2  ...  Y250
E  S   250 E Y   1,393,170
 S  250   Y  15.811388 Y  28,333
 S  1,393,170 F  1,393,170 
0.90  Pr  S  F   Pr 

28,333 
 28,333
F  1,393,170 

 Pr  N  0,1 
28,333 

0.90  Pr  N  0,1  1.28
F  1,393,170  1.28  28,333
=1.43 million
Question #187
Answer: A
bg
q41 1
q  b1g  1  p b1g  1  e p b g j
41
41
41
bg
q41 


1
2
l41   l40   d40   d40   1000  60  55  885
1

2

d41   l41   d41   l42   885  70  750  65
q41 
1
750

p41  
885
750 
 1  1  
q41

 885 
MLC-09-13
 
q41
65
135

65
135
 0.0766
102
Question #188
Answer: D

t 

S0 (t )  1  
 
d

t 
log  S0 (t )  
dt
 t
ex  

t 
x
1 
 dt 
 1
  x
 x 
0
1


e0new   old
 new
  new  2 old  1
2  1 
1
old
old
2  1 9 
 0new 
 
  old  4

4 
Question #189
Answer: C
Constant force implies exponential lifetime
Var T   E T    E T 
2
2
  0.1
2
1
1
 2     2  100
  
2

E  min T ,10     t  0.1e .1t dt   10  0.1e .1t dt
10
0
 te
10
.1t
 10e
1
.1t 10
0
1
 10e .1t
 10e  10e  10  10e
10
1
 10 1  e1   6.3
MLC-09-13

103
Question #190
Answer: A
% premium amount for 15 years


Gax:15  100,000 Ax  0.08G  0.02Gax:15    x  5  5ax 
Per policy for life
4669.95 11.35  51,481.97   0.08  4669.95   0.02 11.35 4669.95    x  5  5ax 
ax 
1  Ax 1  0.5148197

 16.66
d
0.02913
53,003.93  51,481.97  1433.67   x  5  83.30
4.99   x  5
x  9.99
The % of premium expenses could equally well have been expressed as
0.10G  0.02G ax:14 .
The per policy expenses could also be expressed in terms of an annuityimmediate.
MLC-09-13
104
Question #191
Answer: D
For the density where Tx  Ty ,


Pr Tx  Ty  
40

y
y 0 x 0
0.0005dxdy  

40
y 0
50
y
0.0005 x dy  
0
50
y 40
40
50
0
y 40
  0.0005 ydy  

40
x0 0.0005dxdy
y 40
0.0005 y 2
2
40
0
 0.02 y
0.0005 x
40
dy
0
0.02 dy
50
40
 0.40 + 0.20 = 0.60
For the overall density,
Pr Tx  Ty   0.4  0  0.6  0.6  0.36
where the first 0.4 is the probability that Tx  Ty and the first 0.6 is the probability
that Tx  Ty .
Question #192
Answer: B
The conditional expected value of the annuity, given  , is
1
.
0.01  
The unconditional expected value is
0.02
1
 0.01  0.02 
ax  100
d   100 ln 
  40.5
0.01 0.01  
 0.01  0.01 
100 is the constant density of  on the interval  0.01,0.02 . If the density were not
constant, it would have to go inside the integral.
MLC-09-13
105
Question #193
Answer: E
Recall ex 
x
2
ex:x  ex  ex  ex:x
ex:x  
t 
t 
 dt
1 
 1 
   x    y 
 x 
0
Performing the integration we obtain
x
ex:x 
3
2   x 
ex:x 
3
2   2a 
2   3a 
 3
 2  7a
3
3
2   3a 
2
  a   k 
3
3
3.5a  a  k  3.5a  3a 
(i)
(ii)
k 5
The solution assumes that all lifetimes are independent.
Question #194
Answer: B
Although simultaneous death is impossible, the times of death are dependent as
the force of mortality increases after the first death. There are two ways for the
benefit to be paid. The first is to have (x) die prior to (y). That is, the transitions are
State 0 to State 2 to State 3. The EPV can be written with a double integral


0

e0.04t t pxy00  x02t: y t  e0.04 r r p y22t  y23t  r drdt
0


0
0
  e0.04t e0.12t 0.06 e0.04 r e0.10 r 0.10drdt 
0.06 0.10
 0.26786
0.16 0.14
By symmetry, the second case (State 0 to State 1 to State 3) has the same EPV.
Thus the total EPV is 10,000(0.26786+0.26786) = 5,357.
MLC-09-13
106
Question #195
Answer: E
Let
k
p0  Probability someone answers the first k problems correctly.
2
p0   0.8  0.64
4
p0   0.8  0.41
2
p0:0   2 p0   0.642  0.41
4
p0:0   0.41  0.168
2
p0:0  2 p0  2 p0  2 p0:0  0.87
4
p0:0  0.41  0.41  0.168  0.652
2
4
2
2
Prob(second child loses in round 3 or 4) = 2 p0:0  4 p0:0
= 0.87-0.652
= 0.218
Prob(second loses in round 3 or 4 second loses after round 2) =
2
p0:0  4 p0:0
2
=
p0:0
0.218
 0.25
0.87
Question #196
Answer: E
If (40) dies before 70, he receives one payment of 10, and Y = 10. The probability
of this is (70 – 40)/(110 – 40) = 3/7
If (40) reaches 70 but dies before 100, he receives 2 payments.
Y  10  20v30  16.16637
The probability of this is also 3/7.
If (40) survives to 100, he receives 3 payments.
Y  10  20v30  30v60  19.01819
The probability of this is 1 – 3/7 – 3/7 = 1/7
E Y    3/ 7  10   3/ 7  16.16637  1/ 7  19.01819  13.93104
 
E Y 2   3/ 7  102   3/ 7  16.166372  1/ 7  19.018192  206.53515
 
Var Y   E Y 2   E Y   12.46
Since everyone receives the first payment of 10, you could have ignored it in the
calculation.
MLC-09-13
2
107
Question #197
Answer: C
2

E  Z    v k 1bk 1
k 0

k
px qx  k
 v  300   0.02  v 2  350  0.98 0.04   v3 400  0.98 0.96  0.06  
 36.8
 
2

E Z 2   v k 1bk 1
k 0

2
k
px qx  k
 v 2  300   0.02  v 4  350   0.98 0.04   v 6 4002  0.98 0.96  0.06
2
2
 11,773
 
Var  Z   E Z 2  E  Z 
2
 11,773  36.82
 10, 419
Question #198
Answer: A
Benefits +
3
0 Le  1000v 
Expenses
 0.20G  8   0.06G  2 v   0.06G  2 v2
at G  41.20 and i  0.05,
0L
 for K  2   770.59
MLC-09-13
108
– Premiums
 G a3
Question #199
Answer: D
P  1000P40
 235  P 1  i   0.015 1000  255  255
[A]
 255  P 1  i   0.020 1000  272  272
[B]
Subtract [A] from [B]
20 1  i   3.385  17
1 i 
Plug into [A]
20.385
 1.01925
20
 235  P 1.01925  0.015 1000  255  255
235  P 
255  11.175
1.01925
P  261.15  235  26.15
1000 25V 
 272  26.151.01925   0.0251000
1  0.025
 286
MLC-09-13
109
*Question #200
Answer: A
1.2
1
1
0.8
0.6
0.5
0.4
0.4
0.2
0
0
10
x
Give
n
0
s  x
1
55 q35
20 55 q15
20 55 q15
55 q35
 1
20
30
15
0.70
Linear
Interpolation
40
50
Give
n
25
60
35
0.50
0.48
Linear
Interpolation
S0  90 
0.16 32
 1

 0.6667
S0  35
0.48 48

S0  35  S0  90  0.48  0.16 32


 0.4571
S0 15
0.70
70

0.4571
 0.6856
0.6667
MLC-09-13
110
70
Give
n
75
0.4
80
90
0
100
Given
90
100
0.16
Linear
Interpolation
0
Alternatively,
20 55 q15
55 q35

20 p15  55 q35
55 q35

20 p15

S0  35 
S0 15 
0.48
0.70
 0.6856

*Question #201
Answer: A
S0 (80)  1 *  e ^  0.16*50   e ^  0.08*50    0.00932555
2
S0 (81)  1 *  e ^  0.16*51  e ^  0.08*51   0.008596664
2
p80  S0 81 / S0 80   0.008596664 / 0.00932555  0.9218
q80  1  0.9218  0.078
Alternatively (and equivalent to the above)
For non-smokers, px  e0.08  0.923116
50
px  0.018316
For smokers, px  e0.16  0.852144
50
px  0.000335
So the probability of dying at 80, weighted by the probability of surviving to 80, is
0.018316  1  0.923116   0.000335  1  0.852144 
= 0.078
0.018316  0.000335
MLC-09-13
111
Question #202
Answer: B
x
40
41
42
l x 
2000

d x 
20
d x 
60
1920
1840
30
40
50
because 2000  20  60  1920 ;
1
2
1920  30  50  1840
Let premium = P
1840 2 
 2000 1920
EPV premiums = 

v
v  P  2.749 P
2000 
 2000 2000
30 2
40 3 
 20
EPV benefits = 1000 
v
v 
v   40.41
2000
2000 
 2000
40.41
P
 14.7
2.749
MLC-09-13
112
Question #203
Answer: A

10
a30   e0.08t e0.05dt  10 Ex  e0.08t e0.08t dt
0
10 0.13t
 e
0
dt  e
0

1.3
0.16
0 e
e0.13t 10
e0.16t
 e1.3
0.13 0
0.16
1.3
1.3
e
1
e



0.13 0.13 0.16
= 7.2992


dt

0

A30   e0.08t e0.05t  0.05  dt  e1.3  e0.16t  0.08  dt
10
0
0
 1
e 
e
 0.05 

   0.08 
0.16
 0.13 0.13 
 0.41606
1.3
1.3
= P  A30  
A30 0.41606

 0.057
a30 7.29923
1
1
a40 

0.08  0.08 0.16
A40  1   a40
 1   0.08/ 0.16  0.5
10V
 A30   A40  P  A30  a40
 0.5 
 0.057   0.14375
0.16
*Question #204
Answer: C
Let T be the future lifetime of Pat, and K be the curtate future lifetime.
L  100,000 vT  1600 aK 1
0  T 10
 100,000vT  1600 a10
10  t  20
1600a10
Minimum is 1600a10
 12,973
MLC-09-13
20<t
when evaluated at i = 0.05
113
Question #205 - Removed
Question #206
Answer: A
Pax:3  EPV (stunt deaths)
 2500  2486 /1.08  2466 / 1.08 2 
 4 /1.08  5/ 1.082  6 / 1.08 3 

P
  500000 


2500
2500




P  2.77   2550.68
 p  921
*Question #207
Answer: D
x2 
1

dx
S  x  dx 30  10, 000 

30 0


2
S0  30 
 30 
1 

 100 

x3  80
x

30, 000  30


0.91
33.833

0.91
 37.18
80
e30:50
80 
Question #208
Answer: B
A60  v   p60  A61  q60 
 1/1.06    0.98  0.440  0.02 
 0.42566
a60  1  A60  / d
 1  0.42566  /  0.06 /1.06 
 10.147
1000 10V  1000 A60  1000 P50  a60
 425.66  10.147  25
 172
MLC-09-13
114
Question #209
Answer: E
Let Y65  present value random variable for an annuity due of one on a single life
age 65.
Thus E Y65   a65
Let Y75  present value random variable for an annuity due of one on a single life
age 75.
Thus E Y75   a75
E (X)  50  2  a65  30 1 a75
 100  9.8969   30  7.217   1206.20
Var (X)  50  22Var Y65   30 1 Var Y75   200 13.2996   30 11.5339   3005.94
2
where Var Y65  
and Var Y75  
1
d2

1
d2

2
2

2
A65  A65


2
A75  A75

1
 0.05660
1
 0.05660
2
2
0.23603   0.43982   13.2996


0.38681   0.59149 2   11.5339


95th percentile  E  X   1.645 Var  X
 1206.20  1.645 54.826
 1296.39
Question #210
Answer: C

a   e t  e t dt 
0
EPV  50, 000 
1
0.5
1
 
1
 0.5    d   100, 000  ln   1  ln   0.5
1
 0.045  1 
 100,000  ln 

 0.045  0.5 
 65,099
Question #211 - Removed
Question #212 - Removed
MLC-09-13
115
Question #213 - Removed
Question #214
Answer: A
Let  be the net premium at issue.
  10,000
A45:20
a45:20
0.20120  0.25634  0.43980   0.25634
 10,000 
14.1121  0.25634  9.8969 
 297.88
The expected prospective loss at age 60 is
10,00015V45:20  10,000 A60:5  297.88 a60:5
 10,000  0.7543  297.88  4.3407
 6250
1
 0.36913  0.68756  0.4398  0.06674
where A60:5
A60:51  0.68756
A60:5  0.06674  0.68756  0.7543
a60:5  11.1454  0.68756  9.8969  4.3407
1
 (Reduced Amount) A60:51
After the change, expected prospective loss = 10,000 A60:5
Since the expected prospective loss is the same
6250  10,000 0.06674   Reduced Amount  0.68756
Reduced Amount = 8119
MLC-09-13
116
Question #215
Answer: D
Ax  Ax1:5  5 Ex Ax15 :7  12 Ex Ax12
where
5 0.040.02 
 0.7408
5 Ex  e
0.04
Ax1:5 
 1  0.7408  0.1728
0.04  0.02
7  0.050.02 
 0.6126
7 Ex  5  e
0.05 

Ax 1 5 :7  
 1  0.6126   0.2767
 0.05  0.02 
12 Ex  5 Ex  7 Ex 5  0.7408  0.6126  0.4538
0.05
Ax12 
 0.625
0.05  0.03
Ax  0.1728   0.7408 0.2767    0.4538 0.625
= 0.6614
Question #216
Answer: A
EPV of Accidental death benefit and related settlement expense =
0.004
 89.36
 2000 1.05 
0.004  0.04  0.05
0.04
EPV of other DB and related settlement expense = 1000 1.05 
 446.81
0.094
EPV of Initial expense = 50
3
EPV of Maintenance expense =
 31.91
0.094
100
EPV of future premiums 
 1063.83
0.094
EPV of 0 L  89.36  446.81  50  31.91 1063.83
 445.75
MLC-09-13
117
Question #217
Answer: C
One approach is to enumerate all the paths from state 1 to state 3 and evaluate
their probabilities. The paths are:
113 – (0.8)(0.05) = 0.04
123 – (0.15)(0.05) = 0.0075
133 – (0.05)(1) = 0.05
The sum is 0.0975.
Alternatively, the required probability is in row 1 column 3 of the square of the
transition probability matrix. Multiplying the first row of the matrix by each column
produces the first row of the square, [0.6475 0.255 0.0975]. While only the third
calculation is required, doing all three provides a check as the numbers must sum
to one.
The number of the 50 members who will be in state 3 has a binomial distribution,
so the variance is 50(0.0975)(0.9025) = 4.40.
MLC-09-13
118
Question #218
Answer: C
 0.6 0.3 0.1


Q0   0
0
1 
 0
0
1 

 0.36 0.18 0.46 


Q0  Q1   0
0
1 
 0
0
1 

 0 0.108 0.892 


Q0  Q1  Q2   0
0
1 
0
0
1 

Note that after four transitions, everyone must be in state 2. For premiums, the
probability is the sum of the first two entries in row 1 (with probability 1 that a
premium is paid at time 0). Then,
APV  Premiums   1  0.9v  0.54v2  0.108v3  2.35 .
For benefits, the probabilities are the second entry in row 1. Then,
APV  Benefits   4 0.3v  0.18v2  0.108v3  2.01


Difference = 2.35 – 2.01 = 0.34
Note that only the first row of each of the products needs to be calculated.
It is also possible to enumerate the transitions that produce the required events,
though for this problem the list is relatively long:
Premium at time 1:
00 (0.6)
01 (0.3)
Total = 0.9
Premium at time 2:
000 (0.6)(0.6) = 0.36
001 (0.6)(0.3) = 0.18
Total = 0.54
Premium at time 3:
0001 (0.6)(0.6)(0.3) = 0.108
Benefit at time 1:
01 (0.3)
Benefit at time 2:
001 (0.6)(0.3) = 0.18
Benefit at time 3:
0001 (0.6)(0.6)(0.3) = 0.108
MLC-09-13
119
Question #219
Answer: E
0.25 1.5 qx

0.25
px  1.75 px
Let  be the force of mortality in year 1, so 3  is the force of mortality in year 2.
Probability of surviving 2 years is 10%
 0.10  px px 1  e  e3  e4 


ln  0.1
 0.5756
 
4

0.25
1
px  e 4 0.5756  0.8660

1.75
px  px  0.75 px1  e e
1.5 qx  0.25

0.251.5 qx
0.25

px
0.25

3
 3 
4
e

13
 0.5756
4
px  1.75 px 0.866  0.154

 0.82
0.866
0.25 px
Question #220
Answer: C
500
1

500(110  x) 110  x
2
 1  xS   xS 
2
110  x
 xNS 
 lxS  l0S 110  x  [see note below]
2
Thus
S
t p20 
NS
t p25 
MLC-09-13
 0.1540
S
l20
 90  t 
t

S
l20
902
2
NS
l25
85  t 
t

NS
85
l25
120
e20:25  

85
0 t
p20:25dt
85
pS
0 t 20 t
NS
p25
dt
2
90  t   85  t 


dt
0
 90 2 85
85
85
1
2
90  t   90  t  5  dt


688,500 0
1
 85  90  t 3 dt  5 85  90  t 2 dt

0
688,500   0

85
   90  t 4 5  90  t 3 
1




688,500 
4
3
 0

1

 156.25  208.33  16, 402,500  1, 215,000
688,500
= 22.1
[There are other ways to evaluate the integral, leading to the same result].
The S0 ( x) form is derived as S0 ( x)  e
x 2 
 
dt
0  110t 
The l x form is equivalent.
Question #221
Answer: B
a30:20  a30:10  10 E30  a40:10
15.0364  8.7201  10 E30  8.6602
 15.0364  8.7201 /8.6602  0.72935
Expected present value (EPV) of benefits =
1
1
 1000  A40:10
 2000  10 E30  A50:10
10 E30
 16.66  2  0.72935  32.61  64.23
EPV of premiums =   a30:10  2  0.72935  a40:10
   8.7201  2    0.72935  8.6602
 21.3527
  64.23/ 21.3527  3.01
MLC-09-13
121
e
2ln 110t 
x
0
 110  x 


 110 
2
*Question #222
Answer: A
15V
 P25 s25:15 
1
P25:15

1
A25:15
15 E25
P25:15  P25:151 
(this is the retrospective reserve calculation)
0.05332  0.05107
 0.00225

0.05107  P25:151 
1
A25:15
15 E25

a25:15
15 E25
a25:15
1
A25:15
/ a25:15
15 E25
1
A25:15
/ a25:15


1
s25:15
0.00225
 0.04406
0.05107
0.01128
 0.22087
0.05107
25,00015V  25,000  0.22087  0.04406  25,000  0.17681  4420
P25 s25:15 
There are other ways of getting to the answer, for example writing
A: the retrospective reserve formula for 15V .
B: the retrospective reserve formula for the 15th net premium reserve for a
15-year term insurance issued to (25), which = 0
Subtract B from A to get
1
P25  P25:15
s25:15  15V


Question #223
Answer: C
ILT:
We have p70  6,396,609/ 6,616,155  0.96682
2 p70  6,164,663/ 6,616,155  0.93176
e70:2  0.96682  0.93176  1.89858
CF:
0.93176  2 p70  e2     0.03534
Hence e70:2  p70  2 p71  e   e2   1.89704
DM: Since l70 and 2 p70 for the DM model equal the ILT, therefore l72 for the DM
model
MLC-09-13
122
also equals the ILT. For DM we have l70  l71  l71  l72  l71
DM 
 6,390,409
Hence e70:2  6,390,409 / 6,616,155  6,164,663/ 6,616,155  1.89763
So the correct order is CF < DM < ILT
You could also work with p’s instead of l’s. For example, with the ILT,
p70  1  0.03318  0.96682
2
p70   0.96682 1  0.03626   0.93176
Note also, since e70:2  p70  2 p70 , and 2 p70 is the same for all three, you could just
order p70 .
Question #224
Answer: D
 
l60
 1000

 
l61
 1000  0.99  0.97  0.90   864.27

 
d60
 1000  864.27  135.73

 
d60
 135.73 
3
 ln  0.9 
 ln  0.99  0.97  0.9  
 135.73 
0.1054
 98.05
0.1459
 
l62
 864.27  0.987  0.95 0.80   648.31

 
d61
 864.27  648.31  215.96

 
d61
 215.96 
3
 ln  0.80 
 ln  0.987  0.95 0.80  
 215.96 
 
 
 d61
 98.05  167.58  265.63
So d60
3
MLC-09-13
3
123
0.2231
 167.58
0.2875
Question #225
Answer: B
t
t
p40  e0.05t
p50   60  t  / 60
50t  1/  60  t 
10
0
t
p40:50 50t dt  
10 e
10
0.05t
0
1 e0.05t
dt  
60
60  0.05  0



20
1  e0.5  0.13115
60
Question #226
Answer: A
Actual payment (in millions) 
q3  1 
1 q3

3
5
 2  6.860
1.1 1.1
0.30
 0.5
0.60
0.30  0.10
 0.333
0.60
 0.5 0.333 
Expected payment = 10 

 7.298
2 
 1.1 1.1 
Ratio
6.860
 94%
7.298
Question #227
Answer: E
At duration 1
Kx
1
>1
1L
v  Px1:2
Prob
qx 1
0  Px1:2
1  qx 1
So Var  1 L   v2qx1 1  qx1   0.1296
MLC-09-13
124
That really short formula takes advantage of
Var  a X  b   a 2Var ( X ) , if a and b are constants.
Here a = v; b  Px1:2 ; X is binomial with p  X  1  qx1 .
Alternatively, evaluate Px1:2  0.1303
 0.9  0.1303  0.7697 if K x  1
1 L  0  0.1303  0.1303 if K x  1
1L
E  1 L    0.2 0.7697    0.8 0.1303  0.0497
 
E 1 L2   0.2  0.7697    0.8 0.1303  0.1320
2
2
Var  1 L   0.1320   0.0497   0.1295
2
Question #228
Answer: C
P  Ax  
Ax
Ax
 Ax  0.1  13 



 0.05
ax  1  Ax  1  Ax
1  13 
  


 P  Ax  

Var  L   1 





1  0.05 
 1 

5  0.10 

2
2

2
2

2
Ax  Ax2
Ax  Ax2



Ax  Ax2  0.08888
 
Var  L  1  
 
2

2
Ax  Ax2

16 
 
 1 
  0.08888
45  0.1 
2
 

1 
 4
 0.1 
  0.1
2
MLC-09-13
125
Question #229
Answer: E

Seek g such that Pr aT
 40 

 g  0.25
aT is a strictly increasing function of T.
Pr T  40   60  0.25 since

 Pr aT
 40
60
p40 
100  40
 0.25
120  40

 a60  0.25
g  a60  19.00
Question 230
Answer: B
 0.05 
A51:9  1  da51:9  1  
  7.1  0.6619
 1.05 
11V
  2000  0.6619   100  7.1  613.80
 10V  P  1.05  11V  q50  2000  11V 
 10V  100 1.05  613.80   0.011 2000  613.80
10V
 499.09
where q50   0.00110    0.001  0.011
Alternatively, you could have used recursion to calculate A50:10 from A51:9 , then
a50:10
from A50:10 , and used the prospective reserve formula for 10V .
MLC-09-13
126
Question #231
Answer: C
1000 A81  1000 A80  1  i   q80 1000  A81 
689.52   679.80 1.06   q80 1000  689.52 
q80 
720.59  689.52
 0.10
310.48
q80  0.5q80  0.05
1000 A80  1000vq80  vp80 1000 A81
 1000 
0.05
0.95
 689.52 
 665.14
1.06
1.06
Question #232
Answer: D
l x 
776
752
d x 
8
8

42
43
d x 
16
16
1
2


1
2
 
 
l42
and l43
came from lx 1  lx   d x   d x 



2000 8v  8v 2  1000 16v  16v 2
EPV Benefits =
776
  76.40
 776  752v 
EPV Premiums  34 
   34 1.92  = 65.28
776


2V
 76.40  65.28  11.12
MLC-09-13
127
*Question #233
Answer: B
pxx  1  qxx  0.96
px  0.96  0.9798
px1:x1  1  qx1:x1  0.99
px1  0.99  0.995
ax:3  1  vpx  v 2  2 px  1 
0.9798  0.9798 0.995

1.05
1.052
 2.8174
0.96  0.96  0.99 

 2.7763
1.05
1.052
 6000 axx:3
axx:3  1  vpxx  v 2  2 pxx  1 
EPV = 2000 ax:3  2000 ax:3
  4000  2.8174    6000  2.7763
 27,927
MLC-09-13
128
Question #234
Answer: B
t
1
1
1
px   x   t   qx   0.20
t
2
2
px   1  tqx   1  0.08t
t
3
3
px   1  tqx   1  0.125t
qx   
1
1
0t
1

1
2
3
1
1
px   x   t  dt   t px  t px  t px   xt dt
0
  1  0.08t 1  0.125t  0.20  dt
1
0
1


 0.2 1  0.205t  0.01t 2 dt
0
1
 0.205t 2 0.01t 3 
 0.2 t 


2
3 0

0.01 

  0.2  1  0.1025 
 0.1802
3 

MLC-09-13
129
Question #235
Answer: B
1
d 
 w
G  0.1G  1.50 1  1.06   1000q40
 2.93  q40

AS 
 
 
1  q40
 q40
d

w
 0.9G  1.50 1.06   1000  0.00278   2.93 0.2 
1  0.00278  0.2

0.954G  1.59  2.78  0.59
0.79722
 1.197G  6.22
2
d 
 w

1 AS  G  0.1G  1.50 1  1.06   1000q41  2 CV  q41
AS 
 
 
1  q41
 q41
d


w
1.197G  6.22  G  0.1G  1.501.06  1000 0.00298  2 CV  0
1  0.00298  0
 2.097G  7.721.06  2.98
0.99702
 2.229G  11.20
2.229G  11.20  24
G  15.8
MLC-09-13
130
Question #236
Answer: A
5
1
 2

4 AS  G 1  c4   e4  1  i   1000qx  4  5 CV  q x  4
AS 
1  qx4  qx 4
1


2
396.63  281.77 1  0.05  7  1  i   90  572.12  0.26
1  0.09  0.26
 657.311  i   90  148.75
0.65
 694.50
 657.311  i   90  148.75   0.65 694.50 
690.18
 1.05
657.31
i  0.05
1 i 
Question #237 - Removed
Question #238 – Removed
MLC-09-13
131
Question #239
Answer: B
Let P denote the annual premium
The EPV of benefits is 25,000 Ax:20  25,000(0.4058)  10,145 .
The EPV of premiums is Pax:20  12.522P
The EPV of expenses is
25, 000
 (15  3)  3ax:20
1, 000
 0.20 P  0.6261P  194.025  12  37.566  0.8261P  243.591
 0.25  0.05 P  0.05 P ax:20
  2.00  0.50   0.50 ax:20 
Equivalence principle:
12.522 P  10,145  0.8261 P  243.591
10,389.591
12.522  0.8261
 888.31
P
Question #240
Answer: D
Let G denote the premium.
Expected present value (EPV) of benefits = 1000A40:20
EPV of premiums = G a40:10
EPV of expenses   0.04  0.25 G  10   0.04  0.05 G a40:9  5a40:19
 0.29G  10  0.09G a40:9  5a40:19
 0.2G  10  0.09G a40:10  5a40:19
(The above step is getting an a40:10 term since all the answer choices have one. It
could equally well have been done later on).
MLC-09-13
132
Equivalence principle:
G a40:10  1000 A40:20  0.2G  10  0.09G a40:10  5 a40:19


G a40:10  0.2  0.09 a40:10  1000 A40:20  10  5 a40:19
G
1000 A40:20  10  5 a40:19
0.91a40:10  0.2
Question #241 - Removed
Question #242
Answer: C
11 AS 

 10 AS  G  c10 G  e10  1  i   10,000qxd10  11CV qxw10
1  qx10  qx10
d
w
1600  200   0.04 200  70 1.05  10,000 0.02  1700 0.18

1  0.02  0.18
1302.1
0.8
 1627.63
MLC-09-13
133
*Question #243
Answer: E
The net premium reserve at the end of year 9 is the certain payment of the benefit
one year from now, less the premium paid at time 9. Thus, it is 10,000v – 76.87.
The gross premium reserve adds expenses paid at the beginning of the tenth year
and subtracts the gross premium rather than the net premium. Thus it is 10,000v +
5 + 0.1G – G where G is the gross premium.
Then,
10, 000v  76.87  (10, 000v  5  0.9G)  1.67
0.9G  81.87  1.67
0.9G  83.54
G  92.82
Question #244
Answer: C
4 AS 
 3 AS  G  c4G  e4  1  i   1000qxd3  4 CVqxw3
1  qx 3  qx3
d
w
Plugging in the given values:
4 AS 

 25.22  30   0.02 30  5 1.05  1000  0.013  75 0.05
1  0.013  0.05
35.351
0.937
= 37.73
MLC-09-13
134
With higher expenses and withdrawals:
4 AS
revised

25.22  30  1.2    0.02  30   5  1.05   1000  0.013   75 1.2  0.05 
1  0.013  1.2  0.05 


 48.51.05  13  4.5
0.927
33.425
0.927
= 36.06
4 AS
 4 AS revised  37.73  36.06
= 1.67
Question #245
Answer: E
Let G denote the gross premium.
EPV (expected present value) of benefits  100010 20 A30 .
EPV of premiums  Ga30:5 .
EPV of expenses   0.05  0.25 G  20 first year
  0.05  0.10 G  10 a 30:4 years 2-5
10 5 a 35:4 years 6-10 (there is no premium)
 0.30G  0.15G a30:4  20  10 a30:4  10 5 a 30:5
 0.15G  0.15Ga30:5  20  10 a30:9
(The step above is motivated by the form of the answer. You could equally well put it that
form later).
Equivalence principle:
Ga30:5  100010 20 A30  0.15G  0.15G a30:5  20  10 a30:9
G

MLC-09-13
1000
10 20 A30
 20  10 a 30:9

 20  10 a30:9

1  0.15 a30:5  0.15
1000
10 20 A30
0.85 a30:5  0.15
135
Question #246
Answer: E
Let G denote the gross premium
EPV (expected present value) of benefits
  0.1 3000  v   0.9  0.2  2000  v 2   0.9  0.8 1000v 2

300 360
720


 1286.98
2
1.04 1.04 1.042
EPV of premium = G
EPV of expenses  0.02G  0.03G  15   0.9  2  v
 0.05G  16.73
Equivalence principle: G  1286.98  0.05G  16.73
1303.71
G
 1372.33
1  0.05
Question #247
Answer: C
EPV (expected present value) of benefits = 3499 (given)
EPV of premiums  G   0.9  G  v
G
0.9G
 1.8571G
1.05
EPV of expenses, except settlement expenses,
 25   4.510   0.2 G    0.9  10  1.5 10   0.1G  v   0.9 0.85  10  1.5 10  v2
0.9  25  0.1G  0.765  25

1.05
1.052
=108.78+0.2857G
 70  0.2G 
MLC-09-13
136
Settlement expenses are 20  110   30 , payable at the same time the death benefit is
paid.
 30 
So EPV of settlement expenses  
 EPV of benefits
 10,000 
  0.003 3499 
 10.50
Equivalence principle:
1.8571G  3499  108.78  0.2857G  10.50
3618.28
G
 2302.59
1.8571  0.2857
Question #248
Answer: D
a50:20  a50  20 E50 a70
 13.2668   0.23047  8.5693
 11.2918
 0.06 
A50:20  1  d a50:20  1  
 11.2918 
 1.06 
 0.36084
Expected present value (EPV) of benefits  10,000A50:20
 3608.40
EPV of premiums  495a50:20
 5589.44
EPV of expenses   0.35 495  20  1510   0.05 495  5  1.50 10  a50:19
 343.25   44.75 11.2918  1
 803.81
E(L) = EPV benefits + EPV expenses – EPV premiums
= 3608.40 + 803.81 – 5589.44
= –1177.23
MLC-09-13
137
Question #249
Answer: B
1
1
0
0
q1xy   t pxy  x t dt   t px t p y  x t dt
1
  qx e0.25t dt (under UDD, t px  x t  qx )
0
1
0.125  qx (4e 0.25t )  qx (4)(1  e 0.25 )  0.8848qx
0
qx  0.1413
Question #250
Answer: C
2
p[11x ]1  p[11x ]1 p[11x ] 2  p[12x ]1 p[21x ]2
0.1 
0.1  
0.1 
0.2 

  0.7 
 0.7 
   0.3 
 0.4 

2 
3  
2 
3 

 0.75(0.7333)  0.25(0.3333)  0.6333
Note that Anne might have changed states many times during each year, but the
annual transition probabilities incorporate those possibilities.
Questions #251-260 – Removed
Question #261
Answer: A
The insurance is payable on the death of (y) provided (x) is already dead.

E ( Z )  Axy2   e  t t qx t p y  y t dt
0

 e
0.06 t
0
(1  e 0.07 t )e 0.09 t 0.09dt

 0.09 e 0.15t  e 0.22t dt
0
1 
 1
 0.09 

  0.191
 0.15 0.22 
MLC-09-13
138
Question #262
Answer: C
t
px 
95  x  t
1
,  x t 
,
95  x
95  x  t
t
p y  e  t
Pr(x dies within n years and before y)  
n
0 t

n
0
px t p y  x t dt
n
95  x  t  t
1
1
1  e n
 t
e
dt 
e
dt

95  x
95  x  t
95  x 0
 (95  x)
Question #263
Answer: A
0.25
2
q30.5:40.5

0.25
0
t

0.25
0.4
0.6t
dt
1  0.5(0.4) 1  0.5(0.6)
0
p30.5 30.5 t t q40.5 dt
0.4(0.6) t 2

0.8(0.7) 2
0.25
 0.0134
0
Question #264 – Removed
Question #265
Answer: D
t
t
t
2
px  exp    5rdr   e 2.5t
 0

t
2
p y  exp    rdr   e 0.5t
 0

1
1
0
0
1
qx1: y   t p y t px  x t dt   e 0.5t e 2.5t 5tdt  5 e 3t tdt
2
2
0
1
5 2
5
 e 3t  (1  e 3 )  0.7918
6
6
0
MLC-09-13
139
2
Question #266
Answer: B
5
G
5
0
1 t
t2
1
dt 

30 25
30(25)(2) 0 60
H  5 p80:85  10 p80:85 
GH 
25 20 20 15 200 4



30 25 30 25 750 15
1 16 17


 0.2833
60 60 60
Question #267
Answer: D
t
t
S0 (t )  exp    (80  x) 0.5 dx   exp  2(80  x) 0.5   exp 2((80  t ) 0.5  800.5 ) 
0
 0


F  S0 (10.5)  exp  2(69.50.5  800.5 )   0.29665
S0 (10)  0.31495
S0 (11)  0.27935
G  S0 (10.5)exp  [0.31495(0.27935)]0.5  0.29662
F  G  0.00003
Question #268
Answer: A
4
4
E ( Z )  500 0.2(1  0.25t )dt  1000  0.25(0.2t )dt
0
0
 500(0.2)  t  0.125t 2   1000(0.25) (0.1t 2 )
4
4
0
0
 100(4  2)  250(1.6)  600
Question #269
Answer: A
10
q30:50  10 q30 10 q50  (1  e0.5 )(1  e0.5 )  0.1548
MLC-09-13
140
Question #270
Answer: C
e30:50  e30  e50  e30:50

e30  e50   e 0.05t dt  20
0

e30:50   e 0.10t dt  10
0
e30:50  20  20  10  30
Question #271
Answer: B


0
0
1
A30:50
  e t t p30 t p50 30t dt   e0.03t e0.05t e0.05t 0.05dt 
0.05
 0.3846
0.13
Question #272
Answer: B
T30:50 has the exponential distribution with parameter 0.05 + 0.05 = 0.10 and so its
mean is 10 and its variance is 100.
Question #273
Answer: D



Cov[T30:50 , T30:50 ]   e30  e30:50  e50  e30:50   (20  10)(20  10)  100



See solution to Question #270 for the individual values.
Question #274
Answer: E
V  ( 2V   3 )(1  i3 )  qx  2 (b3  3V )
3
96  (84  18)(1.07)  qx  2 (240  96)
qx  2  13.14 /144  0.09125
MLC-09-13
141
Question #275
Answer: A
( 3V   4 )(1  i4 )  qx 3b4 (96  24)(1.06)  0.101(360)

 101.05
px  3
0.899
V
4
Question #276
Answer: D
Under UDD:
0.5qx 3
0.5(0.101)

 0.0532
0.5 qx 3.5 
1  0.5qx 3 1  0.5(0.101)
Question #277
Answer: E
V  v 0.5 0.5 p3.5 4V  v 0.5 0.5 q3.5b4
3.5
 1.060.5 (0.9468)(101.05)  1.060.5 (0.0532)(360)
 111.53
Question #278
Answer: D
1  10 p30:40  1  10 p30 10 p40  1 
60 50 2

70 60 7
Question #279
Answer: A
10
2
q30:40

10
0
t
p30 30t (1  t p40 )dt  
10
0
1 t
50
dt 
 0.0119
70 60
70(60)
Question #280
Answer: A

20
10 t

p30 30t t q40 dt  
20
10
20
10 t
p40 40t t q30 dt
20 1 t
1 t
1 400  100 1 400  100
dt  
dt 

 0.0714
10
70 60
60 70
70 2(60)
60 2(70)
MLC-09-13
142
Question #281
Answer: C
140, 000
30
t
0
p30 30t t p40 dt  180, 000
30
0
t
p40 40 t t p30 dt
30 1 70  t
1 60  t
dt  180, 000
dt
0 70 60
0 60 70
1 602  302
1 702  402
 140, 000
 180, 000
 115, 714
70 2(60)
60 2(70)
 140, 000
30
Question #282
Answer: B
P
20
0

t
p30 t p40 dt  P 
20
0
70  t 60  t
P 20
dt 
4200  130t  t 2 dt

0
70 60
4200
P
[4200(20)  130(200)  8000 / 3]  14.444P
4200
Question #283
Answer: A
Note that this is the same as Question 33, but using multi-state notation rather
than multiple-decrement notation.
The only way to be in State 2 one year from now is to stay in State 0 and then
make a single transition to State 2 during the year.
1
1
1
p   t p  dt   e
02
x
0
00
x
MLC-09-13
02
x t
0
 (0.3 0.5 0.7) t
e1.5t
1
0.5dt  0.5
 (1  e1.5 )  0.259
1.5 0 3
143
Question #284
Answer: C
m  1 m2  1

(  80 ) . Then,
2m 12m2
2  1 22  1
4  1 42  1
(2)
(4)
a80
 a80
 a80 

(



)

a


(  80 )
80
80
2(2) 12(2) 2
2(4) 12(4) 2
8.29340  8.16715  (1/ 4)  (3 / 8)  [(3 / 48)  (15 /192)](  80 )
(m)
Woolhouse’s formula to three terms is a80
 a80 
0.00125  0.015625(  80 )
  80  0.08
(2)
a80  a80
 1/ 4  (3 / 48)(0.08)  8.5484.
The answer is:
(12)
a80
 a80 
12  1 122  1
11 143

(  80 )  8.54840  
(0.08)  8.08345.
2
2(12) 12(12)
24 1728
Question #285
Answer: B
1
| a70:2  v 2 2 p70  v 3 3 p70
 B  x t

c ( c 1)
 At  ln c 
t
p70  e
2
p70  e0.0002(2) (0.9754483)1.211  0.9943956
3
p70  e 0.0002(3) (0.9754483)1.3311  0.9912108
1
e
e
 0.000003  70
t

1.1 (1.1 1)
0.0002 t  ln(1.1) 
e
 e 0.0002t (0.9754483)1.1 1
t
| a70:2  0.9943956 /1.052  0.9912108 /1.053  1.7582.
Question #286
Answer: E
1
A50:2
 vq50  v 2 p50 q51
px  e
 B  x

 c ( c 1)
 ln c 
e
 0.000005  x

1.2 (0.2)
 ln(1.2) 
 e 0.0000054848(1.2)
50
p50  0.951311
p51  0.941861
1
A50:2
 0.048689 /1.03  0.951311(0.058139) /1.032  0.09940.
MLC-09-13
144
*Question #287
Answer: E
The reserve at the end of year 2 is 10,000vq52   3 where  3 is the net premium in year 3.
From the equivalence principle at time zero:
10, 000(vq50  v 2 p50 q51  v 3 p50 p51q52 )  0.5 3  0.5 3vp50   3v 2 p50 p51
0.95 0.95(0.94) 
 0.05 0.95(0.06) 0.95(0.94)(0.07)  
10, 000 



   0.5  0.5
3
2
3
1.04
1.04
1.04
1.04 2 
 1.04
 
1563.4779  1.78236 3
 3  877.1953
V  10, 000(0.7) /1.04  877.1953  204.12
2
*Question #288
Answer: C
The net premium in year one is vq50 . By the equivalence principle:
10, 000(vq50  v 2 p50 q51  v3 p50 p51q52 )  10, 000vq50  (vp50  v 2 p50 p51 )
 0.95(0.06) 0.95(0.94)(0.07)   0.95 0.95(0.94) 
10, 000 




2
1.043
1.042 
 1.04
  1.04
1082.708665  1.739090
  622.57
V  10, 000(0.07) /1.04  622.57  50.51.
2
Note that this is a full preliminary term reserve without that term being used.
Question #289
Answer: E
t-th
year
0*
1
2
3
V
P
E
I
EDB
E tV
Pr
0
0
700
700
0
14500
14500
14500
1000
100
100
100
0
864
906
906
0
14000
15000
16000
0
690.2
689.5
0
-1000
573.8
316.5
6
t 1
t 1
px
1.000
1.000
0.986
0.971

-1000.0
573.8
312.1
5.8
NPV  1000  573.8v  312.1v2  5.8v3  216.08 using a 10% discount rate.
*The 1000 at time 0 is neither accumulated nor discounted. The value is treated as
occurring at the end of time 0 and not as occurring at the beginning of year 1.
*Question #290
MLC-09-13
145
Answer: C
Expected present value of future profits
 140v  0.95(135) p67 v2  (0.95)2 (130) 2 p67v3  314.09 at a 10% discount rate.
*Question #291
Answer: B
Solution
Profit  (6, 000  700  10)(1.06)1/12  0.002(200, 000)  0.0015(50, 000  15, 000)
 0.9965(6, 200)  46.7.
*Question #292
Answer: C
245  p40 274 and 300  2 p40 395
Present value of expected premiums:
1000[1  (245 / 274)(1/1.12)  (300 / 395)(1/1.122 )]  2403.821.
Present value of expected profits:
400  150 /1.12  245 /1.122  300 /1.123  142.775.
PV Profit / PV premium = 5.94%
Question #293
Answer: B
P[1  0.97(0.98624)  0.92(0.98624)(0.98499)]
 100, 000[0.97(0.01376)  0.92(0.98264)(0.01501)  0.87(0.98264)(0.98499)(0.01638)]
P  1431.74.
MLC-09-13
146
Question #294
Answer: A
Let IF equal 1 if the index drops below its current level in the next year and equals
0 otherwise.
E[ X N | I F  1]  N (1000)vqx  48.54 N
E[ X N | I F  0]  0
E[ X N ]  0.1(48.54 N )  0  4.854 N
Var[ X N | I F  1]  (1000v) 2 Nqx (1  qx )  44, 773.31N
Var[ X N | I F  0]  0
E[Var ( X N | I F )]  0.1(44, 773.31N )  0  4, 477.33 N
Var[ E ( X N | I F )]  0.1(48.54 N ) 2  0  (4.854 N ) 2  212.05 N 2
Var[ X N ]  4, 477.33 N  212.05 N 2
Var ( X 10 )
10
lim
 25.69
Var ( X N )
N 
 lim
N 
N
25.69  14.56  11.13.
4, 477.33 N  212.05 N 2
 212.05  14.56
N
*Question #295
Answer: E
The actuarial present value of the death benefit is
( s62 )(4)d 62(2) v 0.5  ( s63 )(4)d 63(2) v1.5  ( s64 )(4)d 64(2) v 2.5
100, 000
s62l62
 100, 000
3.589(4)(213)v 0.5  3.643(4)(214)v1.5  3.698(4)(215)v 2.5
3.589(52,860)
 4,585.
MLC-09-13
147
Question #296
Answer: A
Policy 1:
AV371  [ AV36  G  E  (100, 000  AV371 ) 1/12 q63 /1.004](1.004)
 ( AV36  G  E )1.004  100, 000 1/12 q63  AV371 1/12 q63
 [( AV36  G  E )1.004  100, 000 1/12 q63 ] / (1  1/12 q63 )
Policy 2
AV372  [ AV36  G  E  100, 000 1/12 q63 /1.004](1.004)
 ( AV36  G  E )1.004  100, 000 1/12 q63
Because the starting account value, G and E are identical for both policies:
AV361 / AV362  1/ (1  1/12 q63 )  1/ [1  (1/12)(0.01788)]  1.0015.
*Question #297
Answer: E
The recursive formula for the account values is:
AVk 1  ( AVk  0.95G  50)1.06  (100,000  AVk 1 )q50k .
This is identical to the recursive formula for net premium reserves for a 20-year
term insurance where the net premium is 0.95G – 50. Because the net premium
reserve is zero after 20 years, using this premium will ensure that the account
value is zero after 20 years. Therefore,
0.95G  50  100, 000
1
A50:20
a50:20

A50  20 E50 A70
0.24905  0.23047(0.51495)
 100, 000
 1154.55
a50  20 E50 a70
13.2668  0.23047(8.5693)
G = (1154.55+50)/0.95 = 1,267.95.
MLC-09-13
148
Question #298
Answer: A
2
 5, 000 
E ( Z )  (5, 000 /1.030) (0.005)  
 (1  0.005)(0.006)
1.030(1.032) 
2
2
2


5, 000

 (1  0.005)(1  0.006)(0.007)
1.030(1.032)(1.035) 
 392,917
Question #299
Answer: E
Note: This solution used the derivatives at 4.50 and 4.75. The question did not
specify whether this should be done as a forward or a backward recursion. Either
would have been fine. Using the derivatives at 4.75 and 5.00 produces an answer
of 922.3. See sample questions 306 and 307, where the derivatives to use are
specified.
44.75  0.00004(1.144.75 )  0.002847
44.5  0.00004(1.144.5 )  0.002780
1000  E ( 4.75 L)  0.25{0.04 E ( 4.75 L)  150  150(0.05)  0.002847[10, 000  100  E ( 4.75 L)]}
E ( 4.75 L)  {1000  0.25[150  150(0.05)  0.002847(10, 000  100)]} / [1  0.25(0.04  0.002847)]
 961.27
961.27  E ( 4.5 L)  0.25{0.04 E ( 4.5 L)  150  150(0.05)  0.002780[10, 000  100  E ( 4.5 L)]}
E ( 4.5 L)  {961.27  0.25[150  150(0.05)  0.002780(10, 000  100)]} / [1  0.25(0.04  0.002780)]
 922.795
Question #300
Answer: B
Assumed profit =
800[(83.30  87(1  0.03))(1.05)  (0.0035)(10,000)  (1  0.0035)(141.57)]  0 .
This is expected because it should be zero with gross premium reserves and the
gross premium reserve assumptions
Actual profit =
800[(83.30  87(1  0.025))(1.04)  (0.0025)(10,000)  (1  0.0025)(141.57)]  6907.14.
Gain from all sources = 6,907.14 – 0 = 6,907.14.
MLC-09-13
149
Question #301
Answer: E
The death benefit is max(100,000,1.3  85,000  110,500)  110,500 .
The withdrawal benefit is 85,000  1,000  84,000 .
(death)
(withdrawal)
( 9 AS  P  E )(1  i )  110,500q79
 84, 000q79
AS

10
(death)
(withdrawal)
1  q79
 q79
(75, 000  9, 000  900)(1.08)  110,500(0.01)  84, 000(0.03)
1  0.01  0.03
 89, 711.
Note – companies find asset share calculations less useful for universal life
policies than for traditional products. This is because policyholders who choose
different premium payment patterns have different asset shares. However, for any
pattern of premiums, an asset share can be calculated.

Question #302
Answer: B
10V 

( 9V  G  E )(1  i )  1000qx(death)
 10 CV (1  qx(death)
)qx(withdrawal)
9
9
9
(death)
(withdrawal)


1  qx(death)

(1

q
)
q
9
x 9
x 9
(115  16  3)(1.06)  1000(0.01)  110(1  0.01)(0.10)
 128.83.
1  0.01  (1  0.01)(0.10)
Expected deaths = 1000(0.01) = 10; actual deaths = 15.
Expected withdrawals = 1000(1 – 0.01)(0.1) = 99; actual withdrawals = 100.
Gain from mortality and withdrawals is equal to
(Expected deaths – Actual deaths)(Death benefit – End of Year Reserve)
+ (Expected withdrawals – Actual withdrawals)(Withdrawal benefit – End of Year
Reserve)
= (10 – 15)(1000 – 128.83) + (99 – 100)(110 – 128.83) = -4337.
Question #303
Answer: C
Because there are no cash flows at the beginning of the year, the only item
earning interest is the reserve from the end of the previous year. The gain is
1000(10,994.49)(0.05 – 0.06) = –109,944.90
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Question #304
Answer: B
Expenses are incurred for all who do not die. Because gain from mortality has not
yet been calculated, the anticipated experience should be used. Thus, expenses
are assumed to be incurred for 1000(1 – 0.01) = 990 survivors. The gain is 990(50
– 60) = -9,900.
Question #305
Answer: E
The tenth reserve is
(d )
9V (1  i )  (1  q74 )(1000  50)
V

10
(d )
1  q74
10,994.49(1.06)  (1  0.01)(1050)
1  0.01
V

10,
721.88.
10
V
10
Note that reserves are prospective calculations using anticipated experience.
There were 2 more deaths than expected. For each extra death, an annuity benefit
is not paid, an expense is not paid and a reserve does not have to be maintained.
Thus, the saving is 1000 + 60 + 10,721.88 = 11,781.88. The total gain is
2(11,781.88) = 23,563.76. Because the gain from expenses has already been
calculated, the actual value is used.
As an aside, the total gain from questions 303-305 is -96,281. The actual profit can
be determined by first calculating the assets at the end of the year. Begin with
1000(10,994.49) = 10,994,490. They earn 5% interest, to accumulate to
11,544,215. At the end of the year, expenses are 60 for each of 988 who did not
die, for 59,280. Annuity benefits of 1000 are paid to the same 988 people, for
988,000. Assets at the end of the year are 11,544,215 – 59,280 – 988,000 =
10,496,935. Reserves must be held for the 988 continuing policyholders. That is,
988(10,721.88) = 10,593,217. The difference, 10,496,935 – 10,593,217 = -96,282.
Note also that gain is actual profit minus expected profit. If reserves are gross
premium reserves, as in this problem, expected profit = 0 and thus total gain =
actual profit. Problem 300 illustrates a case where total gain is different from actual
profit.
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Question #306
Answer: E
d
tV  G   ( tV )  t (bt  tV )
dt
At t = 4.5,

4.5V  25  0.05( 4.5V )  0.02(4.5)(100  4.5V )  16  0.14( 4.5V )
Euler’s formula in this case is 5.0V  4.5V  (5.0  4.5) 4.5V  .
V'
t
Because the endowment benefit is 100, 5.0V  100 and thus,
100  4.5V  0.5( 4.5V )  4.5V  0.5[16  0.14( 4.5V )]  8  1.07( 4.5V )
V  85.981.
Similarly,

4.5V  4V  (4.5  4.0) 4V and

4.0V  25  0.05( 4.0V )  0.02(4.0)(100  4.0V )  17  0.13( 4.0V )
85.981  4.0V  0.5( 4.0V )  4.0V  0.5[17  0.13( 4.0V )]  8.5  1.065( 4.0V )
4.5
V  72.752.
4.0
Note that if smaller step sizes were used (which would be inappropriate for an
exam question, where the step size must be specified), the estimate of the time 4
reserve converges to its true value of 71.96.
Question #307
Answer: A
d
tV  G   ( tV )  t (bt  tV )
dt
At t = 5.0,

5.0V  25  0.05( 5.0V )  0.02(5.0)(100  5.0V )  15  0.15( 5.0V )
V'
t
Because the endowment benefit is 100, 5.0V  100 and thus,

5.0V  15  0.15(100)  30 .
Euler’s formula in this case is 4.5V  5.0V  (4.5  5.0) 5.0V   100  0.5(30)  85.
Similarly,

4.0V  4.5V  (4.0  4.5) 4.5V and

4.5V  25  0.05( 4.5V )  0.02(4.5)(100  4.5V )  16  0.14( 4.5V )  16  0.14(85)  27.9 .
4.0V  85  (4.0  4.5)(27.9)  71.05.
Note that if smaller step sizes were used (which would be inappropriate for an
exam question, where the step size must be specified), the estimate of the time 4
reserve converges to its true value of 71.96.
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Question #308
Answer: A
t
t
0 s ds  e 0 (0.05 0.02 s ) ds  e  (0.05t 0.01t 2 )
t p0  e

1
p0  e0.06  0.94176
4
APV   80(1  0.25t )e  t t p0 t dt
0
f (1)  80(1  0.25)e0.05 (0.94176)(0.05  0.02)  3.7625.
Question #309
Answer: D
Let P be the net premium. There are three ways to approach this problem. The
first two are intuitive:
The actuarial present value of the death benefit of 1000 is 1000 10| A40 . The return of
net premium benefit can be thought of as two benefits. First, provide a ten-year
temporary annuity-due to everyone, with actuarial present value Pa40:10 . However,
those who live ten years must then return the accumulated value of the premiums.
This forms a pure endowment with actuarial present value Ps10 10 E40 . The total
actuarial present value of all benefits is 100010| A40  Pa40:10  Ps10 10 E40 . Setting this
equal to the actuarial present value of net premiums ( Pa40:10 ) and solving gives
1000 10| A40  Pa40:10  Ps10 10 E40  Pa40:10
1000 10| A40  Ps10
P
1000 10| A40
s10 10 E40

1000 10 E40 A50 1000 A50

.
s10 10 E40
s10
The second intuitive approach examines the reserve at time 10. Retrospectively,
premiums were returned to those who died, so per survivor, the accumulated
premiums are only the ones paid by the survivors, that is Ps10 . There are no other
past benefits so this is the reserve (it is easy to show this using a recursive
formula) Prospectively, the reserve is the actuarial present value of benefits (there
are no future premiums), or 1000A50 . Setting the two reserves equal to each other
produces the premium:
1000A50
.
P
s10
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The final approach is to work from basic principles:
APV (net premiums) = Pa40:10 .
9 9 k
APV benefits = 1000 10| A40  P v k  j 1 k p40 (1  i) j 1 j| q40 k
k 0 j 0
In that double sum, k is the time the premium is paid, with probability k p40 that it is
paid. j is the curtate time, from when the premium was paid, until death. The
amount of premium refunded, with interest is P(1  i) j 1 . The probability, given that
it was paid, that it will be refunded at time j+1 is j | q40 k . The interest discount
factor, from the date of refund to age 40 is v j  k 1 vk+j+1. Then,
9 9k
9 9k
P  v k  j 1 k p40 (1  i ) j 1 j| q40 k  P  v k k p40 j| q40 k
k 0 j 0
k 0 j 0
 P  v k  k p40  10 p40   P  a40:10  10 p40 a10 
9
k 0
Setting APV net premiums = APV benefits:
Pa40:10  1000 10| A40  P(a40:10  10 p40 a10 )
P 10 p40 a10  1000 10| A40
P
10
E40 (1000 A50 ) v10 10 p40 (1000 A50 ) 1000 A50


.
p
a
p
a
s
10 40 10
10 40 10
10
The numerical answer is
249.05
P
 17.83.
13.97164
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Question #310
Answer: B
Final average salary
50, 000
(1.04) 26  (1.04) 25  (1.04) 24   133,360.2
3 
Annual retirement benefit
 0.017(27)(final average salary)(0.85)  52,030
Note that the factor of 0.85 is based on an interpretation of the 5% reduction as
producing a factor of 1 – 3(0.05) = 0.85. This is the approach used in AMLCR. An
alternative approach is to use 0.953. Upon rounding, the same answer choice
results.
Question #311
Answer: E
Retirement Benefit
20%
Age at
Retirement
63
64
65
Salary
253,094
263,218
273,747
Interest
discount
e0.05(3)
e0.05(4)
e0.05(5)
Prob
0.06
0.05
0.43
Years
of Service
0.6
0.8
1.0
Reduction
0.40
0.20
0.0
Sum
Total
APV
4,705
6,896
91,674
103,275
APV above
= Salary  Interest discount  prob  (20%  Years of service)  (1  reduction)
Death Benefit
Age at
Death
60
61
62
63
64
Salary
225,000
234,000
243,360
253,094
263,218
Interest
discount
e0.05
e 0.05(2)
e 0.05(3)
e 0.05(4)
e0.05(5)
prob
0.01
0.01
0.01
0.01
0.01
sum
total
2,140
2,117
2,095
2,072
2,050
10,474
Total APV of both benefits = 103,275 + 10,474 = 113,749.
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Question #312
Answer: B
a 81   a 80  1  1  i  / p 80
  5.89  1  1.06  / 1  0.077   5.616
A81  1  d a 81  1   0.06/1.06   5.616  0.6821
Modified renewal net premium
 1000  0.6821/ 5.616  121.5
A90  1  d a 90  1   0.06/1.06   3.65  0.7934
V FPT  1000  0.7934  3.65 121.5  350
10
Note: Other reserve formulas could also be used, including 1000  1  a 90 / a 81  .
MLC-09-13
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Question #313
Answer: E
Gross premium = G
  2000 
   2000 

Ga 45  2000 A45  1
  20    0.5 
  10  a 45  0.20G  0.05Ga 45
  1000 
   1000 

 0.95a
G
22
45
 0.20 G  2000 A45  22  11a 45
2000 A45  22  11a 45
0.95a 45  0.20

11
2000(0.20120)  22  11(14.1121)
 43.89
0.95(14.1121)  0.20
There are two ways to proceed. The first is to calculate the gross premium reserve
and the benefit reserve and take the difference.
2000 A45 2000(0.2012)
The net premium is

 28.52.
a45
14.1121
The net premium reserve is
2000 A55  28.52a55  2000(0.30514)  28.52(12.2758)  260.17.
The gross premium reserve is
2000 A55  [0.05(43.89)  0.5(2000 /1000)  10]a55  43.89a55
 2000(0.30514)  (13.19  43.89)(12.2758)  233.41.
Expense reserve is 233.41 – 260.17 = –27
The second is to calculate the expense reserve directly based on the pattern of
expenses. The first step is to determine the expense premium.
The present value of expenses is
[0.05G  0.5(2000 /1000)  10]a45  0.20G  1.0(2000 /1000)  20
 13.1945(14.1121)  30.778  216.98.
The expense premium is 216.98/14.1121=15.38
The expense reserve is the expected present value of future expenses less future
expense premiums, that is,
[0.05G  0.5(2000 /1000)  10]a55  15.38a55  2.1855(12.2758)  27
There is a shortcut with the second approach based on recognizing that expenses
that are level throughout create no expense reserve (the level expense premium
equals the actual expenses). Therefore, the expense reserve in this case is
created entirely from the extra first year expenses. They occur only at issue so the
expected present value is 0.20(43.89)+1.0(2000/1000)+20 = 30.778. The expense
premium for those expenses is then 30.778/14.1121 = 2.181 and the expense
reserve is the present value of future non-level expenses (0) less the present
value of those future expense premiums, which is 2.181(12.2758) = 27 for a
reserve of –27.
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Question #314
Answer: B
1000 A40 161.32

 10.89.
a40
14.8166
2.78
The valuation premium in year 1 is 1000q40v 
 2.62.
1.06
Let P 2 denote the valuation premium in years 2-20. By the equivalence principle:
1000 A40  2.62  P2 (a40:20  1)  10.89 20| a40
The valuation premium in years 21 and later is
where
20|
a40 
20
E40 a60  0.27414(11.1454)  3.0554 and
a40:20  a40  20| a40  14.8166  3.0554  11.7612.
Then,
161.32  2.62  (11.7612  1) P 2  10.89(3.0554)
P 2  11.66
V mod  1000 A55  11.66a55:5  10.89 5| a55
15
where
5| a55  0.70810(11.1454)  7.8921
a55:5  12.2758  7.8921  4.3837
V mod  305.14  11.66(4.3837)  10.89(7.8921)  168.08
15
Question #315
Answer: D
px  0.9, 2 px  0.81
Let v(k ) denote the present value at time 0 of a payment of 1 at time k. Thus,
1
1
1
v(1) 
and v(2) 

.
1  year 1 rate
1  year 1 rate 1  year 2 rate
This notation is used in Chapter 11 of ALMCR. There all the rates are known
based on the initial yield curve. Here the v(k ) are conditional upon the future
interest rates. For the up/up path, the year 1 rate is 0.07 and the year 2 rate is
0.09. Then,
v(1)  0.93458, v(2)  0.85741, ax:3  1  1 px v(1)  2 px v(2)
1
 1  0.9(0.93458)  0.81(0.85741)  2.53562.
For the other paths:
v(1)
Path
Year 1 rate Year 2 rate
Up/down
Down/up
Down/down
MLC-09-13
0.07
0.03
0.03
0.05
0.05
0.01
0.93458
0.97087
0.97087
158
v(2)
ax:3
0.89008
0.92464
0.96126
2.56209
2.62274
2.65240
The unconditional expected present value is the weighted average of the
conditional ones:
Y  100, 000[0.75(0.75)(2.53562)  0.75(0.25)(2.56209)
0.25(0.75)(2.62274)  0.25(0.25)(2.65240)]
 256, 422.
Question #316
Answer: C
E40  (1  q40 ) /1.04  (1  0.00278) /1.04  0.95887
If i becomes 0.06 starting at age 41,
APV benefits  1000q40v  1 E40 (1000) A41
1
 1000(0.00278) /1.04  0.95877(168.69)  164.41
APV premiums of P  P(1  1 E40 a41 )  P[1  0.95887(14.6864)]  15.0823P
If i remains at 0.04,
1  A40
1  0.27345
a40 

 18.8911
(0.04 /1.04)
0.03846
Unconditional APV of benefits  0.75(273.45)  0.25(164.41)  246.19
Unconditional APV of premiums  0.75(18.8911P)  0.25(15.0823P)  17.9389 P
By the equivalence principle, P  246.19 /17.9389  13.72
Question #317
Answer: D
Profit  ( 20V  G  E )(1  i )  10, 000q  21V (1  q)
 (2, 750  200  20)(1.07)  10, 000(0.01)  2,920(0.99)
 144.30
Dividend  0.8(144.30)  115.44
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Question #318
Answer: B
Profit  ( 20V  G  E )(1  i )  (Total face)q  21V (1  q)
 (3, 600  200  20)(1.07)  (10, 000  2, 000)(0.01)  3,800(0.99)
 162.60
Dividend  0.8(162.60)  130.08
130.08
Reversionary bonus amount 
 296.31
0.439
296.31
Compound reversionary bonus rate 
 0.025
10, 000  2, 000
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