TABLE OF CONTENTS
S. BROVERMAN EXAM MLC STUDY GUIDE - VOLUME 1
LIFE CONTINGENCIES
NOTES, EXAMPLES AND PROBLEM SETS
Introductory Note
TABLE OF CONTENTS - PART A
Section 1 - Review of Preliminary Topics
Problem Set 1 - Review of Preliminary Topics - Section 1
Section 2 - Models for Survival and Mortality
Section 3 - The Force of Mortality
Section 4 - The Life Table
Section 5 - Mean and Variance of XB and OB
1 to 14
15 to 24
25 to 32
33 to 40
41 to 46
47 to 56
Section 6 - Parametric Survival Models
Section 7 - Fractional Age Assumptions
Section 8 - Select and Ultimate Mortality
Problem Set 2 - Survival and Mortality - Sections 2 to 8
Section 9 - One-Year Term Insurance Payable at the End of the Year of Death
Section 10 - Term and Whole Life Insurance
Payable at the End of the Year of Death
Section 11 - Pure Endowment and Other Life Insurances
Payable at the End of the Year of Death
Section 12 - Insurance Payable at the Moment of Death
Section 13 - Additional Insurance Relationships
Problem Set 3 - Life Insurance - Sections 9 to 13
Section 14 - Discrete Whole Life Annuity-Due
Section 15 - Discrete Life Annuities
Section 16 - Continuous Life Annuities
Section 17 - Additional Annuity Relationships
Problem Set 4 - Life Annuities - Sections 14 to 17
Section 18 - The Loss at Issue Random Variable and
Principles of Premium Calculation
Section 19 - Equivalence Principle Premiums
Section 20 - Policy Expenses
Problem Set 5 - Annual Premiums - Sections 18 to 20
57 to 62
63 to 68
69 to 76
77 to 104
105 to 110
SOA Exam MLC Study Guide © S. Broverman, 2015
111 to 118
119 to 128
129 to 142
143 to 158
159 to 192
193 to 198
199 to 210
211 to 220
221 to 228
229 to 256
257 to 266
267 to 280
281 to 286
287 to 318
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TABLE OF CONTENTS
TABLE OF CONTENTS - PART B
Section 21 - Introduction to Net Reserves
Section 22 - Prospective Form of Net Reserves
Section 23 - Additional Reserve Representations
Section 24 - Reserves on Additional Policy Types
Section 25 - Expense-Augmented Reserves
Section 26 - Recursion Relationships for Reserves
Section 27 - Modified Reserves, Policy Profit
Problem Set 6 - Reserves - Sections 21 to 27
Section 28 - The Joint Life Status
Section 29 - The Last-Survivor Status and the Common Shock Model
Section 30 - Multiple Life Insurances and Annuities
Section 31 - Contingent Probabilities and Insurances
Problem Set 7 - Multiple Life - Sections 28 to 31
Section 32 - Multiple Decrement Models
Section 33 - Associated Single Decrement Tables
Section 34 - Valuation of Multiple Decrement Benefits , Asset Shares,
Non-Forfeiture Values and Policy Profit Analysis
Section 35 - Pension Mathematics
Problem Set 8 - Multiple Decrement Models - Sections 32 to 35
Section 36 - Discrete Time Markov Chains - One-Step Transitions
Section 37 - Discrete Time Markov Chains - Multi-Step Transitions
Section 38 - Continuous Time Markov Chains
Section 39 - Continuous Time Markov Chain Models
Problem Set 9 - Markov Chains -Sections 36 to 39
Section 40 - Term Structure, Yield Curves and Risk
Problem Set 10 - Variable Interest Rates - Section 40
Section 41 - Profit Testing an Insurance Policy
Section 42 - Participating Insurance and Universal Life Insurance
Problem Set 11 - Profit Testing, Participating
and Universal Life Insurance - Sections 41 and 41
319 to 324
325 to 330
331 to 340
341 to 346
347 to 356
357 to 368
369 to 380
381 to 3408
409 to 416
417 to 424
425 to 436
437 to 444
445 to 468
469 to 480
481 to 494
495 to 500
501 to 504
505 to 536
537 to 544
545 to 550
551 to 560
561 to 568
569 to 582
583 to 592
593 to 602
603 to 610
611 to 618
593 to 602
ILLUSTRATIVE LIFE TABLE
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SOA Exam MLC Study Guide © S. Broverman, 2015
INTRODUCTORY NOTE
INTRODUCTORY NOTE
This study guide is designed to help in the preparation for Exam MLC of the Society of
Actuaries.
The study guide is divided into two volumes. Volume 1 consists of review notes, examples and
problem sets. Volume 2 contains 4 practice exams of 20 multiple choice questions each and
written questions with a total point value of 56 points. Volume 2 also has 15 multiple choice
tests of 20 questions each. In Volume 1, there are close to 200 examples in the notes, over 375
multiple choice problems and almost 30 written answer problems in the problem sets. All of
these (about 1000) questions have detailed solutions. The notes are broken up into 42 sections
Each section has a suggested approximate time frame.
Many of the examples in the notes and about half of the problems in the problem sets are from
older SOA or CAS exams on the relevant topics. Some of the questions on the practice exams
are variations on actual exam questions. The SOA has posted on its website a sample question
file for Exam MLC with solutions. Many of those questions are from old exams, and there may
be some overlap in the questions found in this study guide and those found in the SOA files, but
I have attempted to limit that duplication. The SOA and CAS questions are copyrighted by the
SOA and CAS, and I gratefully acknowledge that I have been permitted to include them in this
study guide.
Because of the time constraint on the exam, a crucial aspect of exam taking is the ability to work
quickly. I believe that working through many problems and examples is a good way to build up
the speed at which you work. It can also be worthwhile to work through problems that have
been done before, as this helps to reinforce familiarity, understanding and confidence. Working
many problems will also help in being able to more quickly identify topic and question types. I
have attempted, wherever possible, to emphasize shortcuts and efficient and systematic ways of
setting up solutions. There are also occasional comments on interpretation of the language used
in some exam questions. While the focus of the study guide is on exam preparation, from time to
time there will be comments on underlying theory in places that I feel those comments may
provide useful insight into a topic.
It has been my intention to make this study guide self-contained and comprehensive for all Exam
MLC topics, but there are may be occasional references to the book listed in the SOA exam
catalog. While the ability to derive formulas used on the exam is usually not the focus of an
exam question, it is useful in enhancing the understanding of the material and may be helpful in
memorizing formulas. There may be an occasional reference in the review notes to a derivation,
but you are encouraged to review the official reference material for more detail on formula
derivations.
In order for the review notes in this study guide to be most effective, you should have some
background at the junior or senior college level in probability and statistics. It will be assumed
that you are reasonably familiar with differential and integral calculus.
SOA Exam MLC Study Guide © S. Broverman, 2015
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INTRODUCTORY NOTE
Of the various calculators that are allowed for use on the exam, I think that the
BA II PLUS is probably the best choice. It has several memories and has good financial
functions. I think that the TI-30X IIS would be the second best choice.
There is a set of tables that has been provided with the exam in past sittings. These tables consist
of a standard normal distribution probability table and a life table with insurance and annuity
values at a 6% rate of interest. The tables are available for download from the Society of
Actuaries website, but are included at the ends of both Volumes 1 and 2 for convenience.
If you have any questions, comments, criticisms or compliments regarding this study guide, you
may contact me at the address below. I apologize in advance for any errors, typographical or
otherwise, that you might find, and it would be greatly appreciated if you would bring them to
my attention. I will be maintaining a website for errata that can be accessed from
www.sambroverman.com . It is my sincere hope that you find this study guide helpful and useful
in your preparation for the exam. I wish you the best of luck on the exam.
Samuel A. Broverman
Department of Statistics
University of Toronto
100 St. George Street
Toronto, Ontario CANADA M5S 3G3
E-mail: [email protected] or [email protected]
Internet: www.sambroverman.com
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January, 2015
SOA Exam MLC Study Guide © S. Broverman, 2015
MLCS SECTION 14 - DISCRETE WHOLE-LIFE ANNUITY-DUE
MLC SECTION 14 - DISCRETE WHOLE LIFE ANNUITY-DUE
The suggested time frame for covering this section is 2 hours.
A life annuity is an ongoing series of payments made while someone remains alive. A life
annuity-due is an annuity with annual payments paid at the start of each year and payable over
some period related to an individual's lifetime. In the formulations that follow, we will make use
of the compound interest annuity-due,
7
ÞÞ
"@7
+7l œ "  @  @#  â  @7" œ "@
,
"@ œ .
where . is the annual effective rate of discount.
Whole Life Annuity-Due of 1 Per Year
We can describe the actuarial present value of a life annuity-due by considering the present value
of the "average" amount an insurer will have to pay. Suppose an insurer issues a life annuity-due
of 1 per year to jB individuals. Based on the expected numbers of survivors from year to year,
the amount that the insurer expects to pay can be described by the following time diagram.
Age/time
Insurer pays
1 to each of
BÎ!
l
B  "Î"
l
B  #Î#
l
jB
jB"
jB#
â
B  5Î5
l
â
â
jB5
â
The present value of the total expected payment by the insurer is
jB  @jB"  @# jB#  â  @5 jB5  â œ  @5 jB5 .
∞
5œ!
The present value per individual annuity is
 @5 5 :B .
Ð j" ÑÐjB  @jB"  @# jB#  â  @5 jB5  âÑ œ  @5 B5
jB œ
B
5œ!
5œ!
∞
j
∞
ÞÞ
This is the actuarial present value of the whole life annuity-due, and it is denoted +B .
∞
ÞÞ
The formulation +B œ  @5 5 :B may be referred to as the current payment form of the
5œ!
annuity. This life annuity is a sum of a series of pure endowments, one for each integer year.
The Exam MLC Illustrative Life Table provides a column of whole-life annuity-due values at
integer ages.
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193
MLC SECTION 14 - DISCRETE WHOLE-LIFE ANNUITY-DUE
ÞÞ
The annuity can also be considered from a stochastic point of view, with +B being the expected
value of a present value random variable. This annuity pays 1 at the start of each year as long as
ÐBÑ lives. If ÐBÑ survives O complete years and dies in the O  "-st year, then there would be
O  " payments made (payments at times !ß "ß #ß Þ Þ Þ ß O ).
O"
$ ÐO"Ñ
] œ "  @  @#  â  @O œ +¨ O"l œ "@.
œ "/ .
.
Note that ] is a discrete random variable that takes on the following possible values with
corresponding probabilities based on the listed events:
The PVRV is
Completed No. Years until Death, O
Oœ!
Oœ"
PV of annuity, ]
] œ +¨ "l
Probability
T ÒO œ !Ó œ ;B
] œ +¨ #l
T ÒO œ "Ó œ "l ;B
Oœ#
] œ +¨ $l
T ÒO œ #Ó œ #l ;B
ã
ã
ã
Oœ5
] œ +¨ 5"l
T ÒO œ 5Ó œ 5l ;B
ã
ã
ã
APV œ IÒ] Ó œ +¨ B œ +¨ "l † ;B  +¨ #l † "l ;B  +¨ $l † #l ;B  â  +¨ 5"l † 5l ;B  â
œ IÒ+¨ O"l Ó œ  +¨ 5"l 5 :B ;B5 (aggregate payment form)
∞
œ @
∞
5
œ  5 IB œ "  @:B  @# # :B  â (current payment form).
5œ!
∞
5 :B
5œ!
5œ!
In the following diagram, each row represents the payments made based on death occurring in a
particular year. The first row corresponds to death in the first year, O œ !, so only the payment
at age B is made, and the PV of the amount paid is +¨ "l (the first row in the table above, which
occurs with probability ;B ). The second row in the diagram corresponds to death in the second
year, O œ ", so only payments at ages B and B  " are made, and the PV of the amount paid is
+¨ #l (the second row of the table above, which occurs with probability "l ;B ), etc. Each row
represents one possible outcome related to the PVRV ] .
PV prob.
O
B
B" B# B$
B5"
â
l
|
|
|
|
+¨ "l ;B
!
"
+¨ #l
"l ;B
"
"
"
+¨ $l
#l ;B
#
"
"
"
ã
"
ã
"
"
"
â
"
"
"
"
"
â
"
ã
+¨ 5l
5"l ;B
ã
5"
+¨ 5"l
5l ;B
5
194
ã
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B5
l
â
"
SOA Exam MLC Study Guide © S. Broverman, 2015
MLCS SECTION 14 - DISCRETE WHOLE-LIFE ANNUITY-DUE
Current payment form is an expression for a life annuity APV as a sum of pure endowments, and
aggregate payment form is an expression using the basic formulation for the expected value
IÒ] Ó. The current payment form is usually more convenient to use for calculations.
The current payment form can be derived from the aggregate payment form by using the
relationship 5l ;B œ 5 :B ;B5 œ 5 :B  5" :B (this results in a "summation by parts").
In general, the APV of a discrete life annuity may be formulated as the sum of a collection of
pure endowments, with each pure endowment associated with a payment amount to be made at a
specific point in time contingent on survival to that point in time.
As 5 gets larger, so does +¨ 5"l , and as 5 p ∞ , the limit of +¨ 5"l is +¨ ∞l œ ." .
For instance, if 3 œ Þ!) , then . œ Þ!(%". The possible values of the PVRV ] are
+¨ "l œ " ß +¨ #l œ "  @ œ "Þ*#&* ß +¨ $l œ #Þ()$$ ß Þ Þ Þ . The insurer will have to make at least the
first payment of 1, but under no circumstances will the present value of the amount paid by the
insurer be any more than +̈∞l œ ." œ "$Þ&.
In the notes on life insurance, it was pointed out that in order to find probabilities for an event
related to a present value random variable, it is usually most efficient to translate the event into
one involving the time until death random variable. The same is true for life annuities. This is
illustrated in the following example.
Example 59: Survival is assumed to be based on the Illustrative Life Table. ] is the present
value random variable for a whole life annuity-due of 1 per year starting at age 50. The annual
effective rate of interest is .08.
Find (a) T Ò]  $Ó , (b) T Ò] "!Ó , and (c) T Ò] Ÿ "&Ó.
Solution: (a) From calculations above, we see that +¨ $l œ #Þ()$$ and +¨ %l œ $Þ&((".
] denotes the present value of the annuity payments, and therefore, T Ò]  $Ó is the probability
that Ð&!Ñ receives at most 3 payments. This occurs if Ð&!Ñ dies before receiving the 4th payment,
which is scheduled to be paid at age 53. Therefore,
T Ò]  $Ó œ $ ;&! œ "  jj&$ œ Þ!"*# (this is also equal to T ÒO  " Ÿ $Ó œ T ÒO Ÿ #Ó ; note that
&!
O œ # corresponds to 3 payments being made).
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195
MLC SECTION 14 - DISCRETE WHOLE-LIFE ANNUITY-DUE
ÞÞ
(b) We can proceed as we did in part (a) and calculate values of +8l , until we see that
ÞÞ
ÞÞ
+"(l œ *Þ)&  "!  "!Þ"# œ +")l . Therefore, T Ò] "!Ó is the probability that Ð&!Ñ receives at
least 18 payments. Since the 18th payment is made at time 17, Ð&!Ñ must survive at least 17
years in order to have ] "!. T Ò] "!Ó œ "( :&! œ jj'( œ Þ)!%' .
&!
ÞÞ
"@8
Note that we can solve for 8 in the equation +8l œ Þ!(%" œ "!. We get 8 œ 6868Þ#&*
@ œ "(Þ&.
Since 8 must be an integer, we must have that 8 is at least ") as before.
(c) The present value of the annuity payments can be no larger than +̈∞l œ ." œ "$Þ& ,
so ] must be Ÿ "& ; therefore, T Ò] Ÿ "&Ó œ ".
In (a) and (b), the event involving the annuity present value was translated into an event
involving the survival of Ð&!Ñ. Note that "small" values of ] correspond to "small" values of O .
Also, note that T ÒO  " Ÿ 5Ó œ 5 ;B and T ÒO  " 4Ó œ 4" :B .

IMPORTANT NOTE: There is an algebraic relationship linking the PVRV of the discrete
whole life annuity-due and the PVRV of the whole life insurance payable at the end of the year
of death. Recall that the PVRV for the whole life insurance of 1 payable at the end of the year of
death is ^ œ @O" .
O"
"EB
] œ +¨ O"l œ "@.
œ "^
p EB œ "  .+¨ B ß
¨ B œ IÒ] Ó œ IÒ "^
. ß p +
. Ӝ .
"
" #
#
and Z +<Ò] Ó œ Z +<Ò "^
. Ó œ . # Z +<Ò^ Ó œ . # Ò EB  (EB ) Ó.
Note that if the force of interest is doubled, the relationship EB œ "  .+¨ B becomes
#
EB œ "  # . † # +̈B , where # . œ #.  . # is the annual effective discount rate that results
when the force of interest is doubled. The variance of ] can be written in terms of annuity
functions:
Z +<Ò] Ó œ ."# Ò # EB  (EB )# Ó œ ."# Ò"  Ð#.  . # Ñ# +¨ B  Ð"  .+¨ B Ñ# Ó
œ .# Ò+¨ B  # +¨ B Ó  # +¨ B  +¨ #B œ IÒ] # Ó  ÐIÒ] ÓÑ# ,
where # +¨ B œ  @#5 5 :B œ  /#$ 5 5 :B .
∞
∞
5œ!
5œ!
Note also that IÒ] # Ó œ .# Ò+¨ B  # +¨ B Ó  # +¨ B .
196
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MLCS SECTION 14 - DISCRETE WHOLE-LIFE ANNUITY-DUE
The Exam MLC Illustrative Life Table insurance and annuity values are based on 3 œ Þ!'.
For example, for B œ &! , we have
"
"
#
#
#
Z +<Ò] Ó œ ."# Ò# EB  (EB )# Ó œ ÐÞ!&''Ñ
# Ò E&!  E&! Ó œ ÐÞ!&''Ñ# ÒÞ!*%('  ÐÞ#%*!&Ñ Ó œ "!Þ##.
Whole life annuity-due of 1 per year
O"
] œ "  @  @#  â  @O œ +¨ O"l œ "@.
œ "^
.
∞
ÞÞ
IÒ] Ó œ +B œ  @5 5 :B (current payment form)
œ  +¨ 5"l 5 :B ;B5 (aggregate payment form)
∞
5œ!
5œ!
Z +<Ò] Ó œ ."# Z +<Ò^ Ó œ ."# Ò# EB  (EB )# Ó
The current payment representation is most convenient when considering a life annuity
with payments that are varying. Suppose a life annuity-due issued to ÐBÑ has payment -5
at age B  5 , if alive. The actuarial present value in current payment form is
-!  -" @:B  -# @# # :B  â œ  -5 @5 5 :B .
∞
5œ!
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197
MLC SECTION 14 - DISCRETE WHOLE-LIFE ANNUITY-DUE
MLC SECTION 14 - EXERCISES
>
1. Suppose the survival model is W! Ð>Ñ œ "  "!!
for ! Ÿ > Ÿ "!! , and 3 œ Þ!).
ÞÞ
# ÞÞ
Find +*! , +*! and Z +<Ò] Ó , where ] is the PVRV for a whole life annuity-due of 1 issued to
Ð*!Ñ.
2. ] is the PVRV for a whole life annuity-due of 1 per year issued to Ð&!Ñ. Use the Exam MLC
ÞÞ
Illustrative Life Table at 6% to find the probability T Ò]  +&! Ó .
MLC SECTION 14 - SOLUTIONS TO EXERCISES
"!5
1. This is a DeMoivre model, so that 5 :*! œ "!!*!5
"!!*! œ "! .
∞
*
ÞÞ
+*! œ  @5 5 :*! œ  @5 Ð "!5 Ñ œ Ð " ÑÐ"!  *@  )@#  â  @* Ñ
5œ!
5œ!
"!
"!
"!+"!lÞ!)
ÞÞ
"
"
"
œ Ð "!
ÑÐH+Ñ"!lÞ!) œ Ð "!
ÑÐ
Ñ œ Ð "!
ÑÐ "!'Þ("!
.
Þ!(%" Ñ œ %Þ%%.
"
#
# #
# *
+*! œ  @#5 5 :*! œ  @#5 Ð "!5
"! Ñ œ Ð "! ÑÒ"!  *@  )Ð@ Ñ  â  Ð@ Ñ Ó
# ÞÞ
∞
*
5œ!
5œ!
#
"! +"!l
" # ÞÞ
"
"
œ Ð "!
ÑÐ H+Ñ"!lÞ!) œ Ð "!
ÑÐ #..# Ñ œ Ð "!
ÑÐ "!%Þ(#!
Þ"%#( Ñ œ $Þ(!.
Z +<Ò] Ó œ .# Ò+¨ *!  # +¨ *! Ó  # +¨ *!  +¨ #*! œ $Þ*( , or
#
+"!l
+"!l
"
#
Z +<Ò] Ó œ ."# Ò# E*!  E*!
Ó œ Ð Þ!(%"
Ñ# Ò "!  Ð "! Ñ# Ó œ $Þ*( .
ÞÞ
2. From the table we have +&! œ "$Þ#''). We solve the following equation for 5 :
5
68Ð""$Þ#'').Ñ
ÞÞ
+5l œ "@
œ #$Þ). Therefore, T Ò]  "$Þ#'')Ó
. œ "$Þ#'') ; 5 œ
68 @
is equal to the probability that Ð&!Ñ receives at least 24 payments. This is the probability that
Ð&!Ñ survives at least 23 years (the 24th payment is at age &!  #$ œ ($). Thus,
ÞÞ
T Ò]  +&! Ó œ T Ò]  "$Þ#'')Ó œ #$ :&! œ jj($ œ Þ''".
&!
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MLC - PROBLEM SET 4
MLC - PROBLEM SET 4
PROBLEM SET 4 -MULTIPLE CHOICE PROBLEMS
_
_
1. If +&! œ "#Þ(' when 3 œ Þ!' , and +&! œ (Þ(# when 3 œ Þ"##' , what is the variance
(nearest $1,000) of the present value random variable for a continuous life annuity of $100 per
year issued to Ð&!Ñ at 3 œ Þ!' ?
A) 100,000
B) 102,000
C) 104,000
D) 106,000
E) 108,000
ÞÞ
ÞÞ
2. If +&! œ "$Þ#'( , +(! œ )Þ&'* and +&!:#!| œ %&Þ'&& , then
A) .23
B) .24
C) .25
D) .26
#! I&!
œ
E) .27
3. A discrete 20 year temporary life annuity-due of $20,000 per year is issued to Ð'!Ñ . The
annual effective interest rate is 4% and mortality is on the basis of the MLC Illustrative Table.
What is the probability that the present value random variable (of the annuity-due) is greater than
$150,000 ?
A) .82
B) .83
C) .84
D) .85
E) .86
4. A special continuous life annuity issued to Ð&!Ñ is designed so that the payments continue
until 10 years after the death of Ð&!Ñ . If $ œ Þ!) and mortality follows the pattern .B œ Þ!%
for all B , what is the net single premium ?
A) 10.0
B) 10.2
C) 10.4
D) 10.6
E) 10.8
ÞÞ
ÞÞ
ÞÞ
5. For the three quantities V œ ÐM+ÑBÀ8l , W œ ÐH+ÑBÀ8l and X œ  8"
#  † +BÀ8l , which of the
following is the correct ranking regarding their sizes (assuming 3 !)?
A) R Ÿ S Ÿ T B) R Ÿ T Ÿ S C) S Ÿ R Ÿ T D) T Ÿ R Ÿ S
E) T Ÿ S Ÿ R
"!!
6. For a certain survival distribution, it is found that if $ œ Þ"!, then 
+ B œ B"
for B "! .
Find .&! .
A) .33
B) .36
C) .39
D) .42
E) .45
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MLC - PROBLEM SET 4
7. If +#!:%!| œ ##Þ!""# and =¨#!:%!| œ *$Þ#&$& , then the value of =¨#":$*| is nearest
A) 89.00
B) 89.15
C) 89.30
D) 89.45
E) 89.60
Ð"#Ñ
ÞÞ
ÞÞ
8. Given 3 œ Þ!), =_"|.!) œ "Þ!$'"', +B œ "!Þ'"#$ and # +B œ &Þ)*##, what is Z +<Ò^Ó ,
where ^ is the present value random variable for a whole life insurance benefit of 1 payable at
the end of the month of ÐBÑ's death? Assume UDD in each year of age.
A) .10
B) .11
C) .12
D) .13
E) .14
9. Which of the following expressions is equal to 1 ?
Ð7Ñ
ÞÞÐ7Ñ
ÞÞÐ7Ñ
ÞÞÐ7Ñ
Ð7Ñ
Ð7Ñ
Ð7Ñ
"
A) 3Ð7Ñ † +BÀ8l  EBÀ8l
B) 3Ð7Ñ † +BÀ8l  Ð"  37 Ñ † EBÀ8l
C) 3Ð7Ñ † +BÀ8l  7
  EBÀ8l
Ð7Ñ
ÞÞÐ7Ñ
Ð7Ñ
"
D) 3Ð7Ñ † +BÀ8l  7
E) None of A, B, C or D
 + Ð"  37 Ñ † EBÀ8l
10. You are given:

(i) EB and 
+ B are based on force of interest $ and force of mortality .B> .
w
(ii) EB and 
+ Bw are based on force of interest 5  $ and force of mortality .B> .
ww
(iii) EB and 
+ Bww are based on force of interest $ and force of mortality 5  .B> .
ww 
Determine EB  EB .
w 
w

B
w  $
A) 5+
B) EB  EB
C) EB  5EB D) Ð5  $ Ñ+
+B
E) $ (
+B 
+ Bw )
B
11. X is the random variable for future lifetime of ÐBÑ. Determine G9@Ò 
+ X l ß @X Ó .
# # 
#
#
E  E
EB  EB
#


#
A) B $ B
B) EB  # EB
C) 0
D) # EB  EB
E)
$
12. For ÐBÑ you are given:
(i) .B> œ !Þ!#%
68Ð!Þ%Ñ , for > !
(ii) $ œ !Þ!$
Calculate the probability that 
+ XB | will exceed 20.
A) 0.45
230
B) 0.55
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C) 0.67
D) 0.74
E) 0.82
SOA Exam MLC Study Guide © S. Broverman, 2015
MLC - PROBLEM SET 4
 Ó œ "!!
13. You are given: (i) Z +<Ò+
Xl
*
Calculate 5 .
A) .005
B) .010
C) .015
(ii) .B> œ 5 for all >
D) .020
(iii) $ œ %5
E) .025
14. You are given:
(i) .B> œ . for > !
(ii) force of interest is $
Which of the following is a correct expression for 
+ BÀ"l ?
Ð.$ Ñ
I. "/
Ð.$ Ñ#
A) None
II.
ÞÞ
+BÀ"l +BÀ"l
#
B) I only
ÞÞ
+BÀ"l +BÀ"l
III. 68 +
BÀ"l
C) II only
Ð.$ Ñ
IV. "/
68 +BÀ"l
D) III only
E) IV only
15. ] is the present value random variable for a continuous life annuity of 1 on ÐBÑ.
You are given:
+B œ &
+B œ %
(i) .B> œ .
(ii) 
(iii) #
Calculate Z +<Ð] Ñ.
A) 15
B) 25
C) 35
D) 45
E) 55
16. For a 10-year deferred life annuity-due of 1 per year on (60), you are given:
>
(i) W! Ð>Ñ œ "  "!!
for ! Ÿ > Ÿ "!!
(ii) 3 œ !
Calculate the probability that the sum of the payments made under the annuity will exceed the
actuarial present value, at issue, of the annuity.
A) 0.475
B) 0.500
C) 0.525
D) 0.550
E) 0.575
17. You are given:
(i) +̈B œ ) for all integral B
Calculate ) ;$! .
A) 0.263
B) 0.364
(ii) 3 œ !Þ!)
C) 0.537
SOA Exam MLC Study Guide © S. Broverman, 2015
D) 0.636
E) 0.737
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MLC - PROBLEM SET 4
18. You are given
ÞÞ
+5l
5
"
#
$
%
ÞÞ
Calculate +BÀ%l .
A) 1.6
B) 1.8
5"l ;B
"Þ!!
"Þ*$
#Þ)!
$Þ'#
C) 2.0
D) 2.2
!Þ$$
!Þ#%
!Þ"'
!Þ""
E) 2.4
19. Consider the following present value random variables, where O is the curtate future
ÞÞ
+
!ŸO8
ÞÞ
lifetime of ÐBÑ: ] œ +OB "l for O ! , ^ œ  ÞÞOB "l
.
+8l
O 8
You are given (i) 3 œ Þ!'
Calculate IÒ] Ó  IÒ^Ó.
A) 4
B) 5
C) 6
(ii) EB œ Þ#!(&&
D) 7
(iii) +BÀ8"l œ '
E) 8
20. ] is the present value random variable for a 30-year temporary life annuity of 1 payable at
the beginning of each year while ÐBÑ survives. You are given:
(i) 3 œ !Þ!&
(ii) $! :B œ !Þ(
(iii) # EBÀ$!l
œ !Þ!'*%
(iv) EBÀ$!l
œ !Þ"%%$
"
"
Calculate IÒ] # Ó.
A) 35.6
B) 47.1
21. You are given:
(i) "! I$! œ !Þ$&
C) 206.4
(ii) +$!À*l œ &Þ'
D) 218.0
E) 233.6
(iii) 3 œ !Þ"!
Calculate E
.
"
$!À"!l
A) 0.05
232
B) 0.10
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C) 0.15
D) 0.20
E) 0.25
SOA Exam MLC Study Guide © S. Broverman, 2015
MLC - PROBLEM SET 4
22. You are given:
(i) "! IB œ !Þ%!
(ii)
ÞÞ
Calculate +B .
A) 12.0
B) 12.5
"!l +B
ÞÞ
(iii) = BÀ"!l œ "&
œ(
C) 12.9
D) 13.0
E) 13.4
23. Which of the following expression is equivalent to 3+BÀ8l  Ð"  3ÑEBÀ8l  " ?
A) !
B) 3 † 8 IB
C) 3
D) 8 IB
E) "
24. You are given:
(i) "!!!EBÀ8l œ &'$
Calculate
A) 7.07
(ii) "!!!EB œ "#*
(iii) . œ !Þ!&(
(iv) "!!! 8 IB œ &%$
8l +B .
B) 7.34
C) 7.61
D) 7.78
E) 7.94
25. XB is the random variable for future lifetime of ÐBÑ. You are given:
(i) .B> œ !Þ!% ß > !
(ii) $ œ !Þ!'
`

(a) Find `8 8 IB (b) Find + BÀ8l
(c) Find the standard deviation of 
+ XB l
` _
(d) Find `B
+BÀ8l
26. Which of the following is true, regardless of the assumption about deaths within each year of
age?


$
I. EB œ $3 EB
II. 
+B œ
+ BÀ"l  $ :B
+ B"
III. 
+ B œ $3.# † +¨ B  3
IV. EB œ "  $
+B
$#
A) None
B) I only
C) II only

27. ÐM 8l
+ ÑB is equal to IÐ] Ñ where
You are given: (i) .B œ Þ!% for all B
.  
Calculate .8
ÐM 8l + ÑB .
A) 8/Þ"8
B) "!/Þ"8
D) III only

ÐM 
+ ÑX l
] œ   B
ÐM + Ñ8l  8Ð 8l
+ XB 8l Ñ
! Ÿ XB  8
.
XB 8
(ii) $ œ Þ!'.
C)  /Þ"8
SOA Exam MLC Study Guide © S. Broverman, 2015
E) IV only
D) /Þ"8
E) 10
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MLC - PROBLEM SET 4
28. Which of the following identities are correct?
I. 8l +¨ B œ +B  +BÀ8"l II. "  +BÀ8"l œ Ð"  3Ñ+BÀ8l
A) I
B) I, II
C) I, III
D) II, III
III. EBÀ8l œ @+¨ BÀ8l  +BÀ8"l
E) I, II, III
29. A company has just assumed responsibility for paying pensions to a group of 100 people
who have just turned 65. The pension will pay each of them $25,000 annually, payable
continuously while they are alive. Assume that all the lives are independent, and that the same
mortality table applies to all of them. In addition, you are given:

(i) $ œ !Þ!'
(ii) # E'& œ !Þ#
(iii) The actuarial present value of the portfolio of 100 pensions is $30,000,000.
Calculate the standard deviation of the total present value of the payments for those 100
pensions.
A) Less than $1,000,000
B) At least $1,000,000 but less than $1,250,000
C) At least $1,250,000 but less than $1,500,000
D) At least $1,500,000 but less than $1,750,000
E) At least $1,750,000
ÞÞ
30. M+ and M+ represent the standard increasing annuities. A person aged 20 buys a special fiveyear temporary life annuity-due, with payments of 1, 3, 5, 7, and 9.
Given:
ÞÞ
i) +#!À%l œ $Þ%"
ii) +#!À%l œ $Þ!%
ÞÞ
iii) ÐM+Ñ#!À%l œ )Þ!&
iv) ÐM+Ñ#!À%l œ 7.17
Calculate the net single premium.
A) Less than 18.0
B) At least 18.0, but less than 18.5
C) At least 18.5, but less than 19.0
D) At least 19.0, but less than 19.5
E) At least 19.5
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MLC - PROBLEM SET 4
31. A 10-year certain and whole life annuity of amount $10,000, payable continuously, is issued
to (65). The force of mortality is constant at .01, and the force of interest is constant at .1 . Let ]
be the random variable representing the present value of the benefits paid. Which of the
following graphs correctly shows the cumulative probability distribution function, J] ÐCÑ , for
the random variable?
A)
B)
C)
D)
E)
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MLC - PROBLEM SET 4
32. Given: @ œ !Þ*& ß
Find E "
#&À"!l
"! :#&
ÞÞ
ÞÞ
œ !Þ)( ß +#&À"&l œ *Þ)') ß +$&À&l œ %Þ$*#
Þ
A) Less than .16
B) At least .16, but less than .32
C) At least .32, but less than .48
D) At least .48, but less than .64
E) At least .64
33. You are given the following:
The probability that a newborn lives to be 25 is 70%.
The probability that a newborn lives to be 35 is 50%.
The following annuities-due each have actuarial present value equal to 60,000:
a life annuity-due of 7,500 on (25)
a life annuity-due of 12,300 on (35)
a life annuity-due of 9,400 on (25) that makes at most 10 payments
What is the interest rate?
A) 8.0%
B) 8.1%
C) 8.2%
D) 8.3%
E) 8.4%
34. A special 30 year annuity-due on a person age 30 pays 10 for the first 10 years, 20 for the
next 10 years, and 30 for the last 10 years. You are given:
..
..
..
#! I$! œ 7 , a $!À"!l œ ? , a $!À$!l œ @ , a &!À"!l œ A
Which of the following represents the actuarial present value of this annuity?
A) #!@  "!AÐ"  7Ñ
B) "!?  #!@  "!A
C) "!?  #!@  "!7A
D)  "!?  #!@  "!7A
E)  #!?  #!@  "!A
35. "Actuary Today", the magazine, offers 1-year subscriptions for $25, payable at the start of
the year. Given a renewal rate of 90% per year, what is the actuarial present value of a new
subscriber? (Assume a discount factor of 0.95).
A) Less than $150
B) At least $150, but less than $160
C) At least $160, but less than $170
D) At least $170, but less than $180
E) $180 or more
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MLC - PROBLEM SET 4
36. A 60-year-old lottery winner has the choice of receiving (i) a single lump sum payment of
T" in 10 years if she is still alive or (ii) a 20-year life annuity-due of T# per year beginning
today. Given:
• The payouts in (i) and (ii) are actuarially equivalent.
• . œ '%
• E " œ !Þ%"))
'!À"!l
#
• E
"
'!À"!l
œ !Þ#$$*
• E'!À#!l œ !Þ%%"(
• # E'!À#!l œ !Þ#$"#
Calculate the difference in the variances of the payouts in (i) and (ii).
A) !Þ!%T"#
B) !Þ"#T"# C) !Þ#!T"#
D) !Þ$$T"#
E) !Þ%"T"#
37. You are given:
•
The force of mortality, ., is constant.
•
The force of interest, $ , is constant.
Which of the following represents the probability that 
+ X l will exceed 
+ B for a life age B?
A) Ò.ÎÐ$  .ÑÓÐ$ Î.Ñ
D) Ò$ ÎÐ$  .ÑÓÐ.Î$ Ñ
B) Ò.ÎÐ$  .ÑÓÐ.Î$ Ñ
E) "ÎÐ$  .Ñ
C) Ò$ ÎÐ$  .ÑÓÐ$ Î.Ñ
38. The distribution of Jack’s future lifetime is a two-point mixture:
(i) With probability 0.60, Jack’s future lifetime follows the Illustrative Life Table, with deaths
uniformly distributed over each year of age.
(ii) With probability 0.40, Jack’s future lifetime follows a constant force of mortality . œ 0.02.
(iii) A fully continuous whole life insurance of 1000 is issued on Jack at age 62.
Calculate the APV for a continuous whole life annuity of 1 per year for Jack if 3 œ Þ!'.
39. The force of mortality is constant at . and the force interest is constant at $ and you are
_
ÞÞÐ"#Ñ
given that +B œ "#Þ5. Use the Woolhouse approximation to 3 terms to find +B .
A) Less than 12.20
B) At least 12.20 but less than 12.40
C) At least 12.40 but less than 12.60
D) At least 12.60 but less than 12.80
E) At least 12.80
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MLC - PROBLEM SET 4
40. For a group of individuals all of age B, of which 30% are smokers and 70% are nonsmokers, you are given:
smoker
non-smoker
(i) $ œ !Þ"!
(ii) EB
œ !Þ%%%
(iii) EB
œ !Þ#)'
(iv) XB is the future lifetime of ÐBÑ
smoker Ó œ )Þ)")
non-smoker Ó œ )Þ&!$
(v) Z +<Ò+
(vi) Z +<Ò+
X l
X l
B
B
 Ó for an individual chosen at random from this group.
Calculate Z +<Ò+
Xl
A) 8.5
B) 8.6
C) 8.8
D) 9.0
E) 9.1
41. You are given:

(i) .B> œ - ß > !
(ii) $ œ Þ!)
(iii) EB œ Þ$%%$
(iv) XB is the future lifetime random variable for ÐBÑ.
 Ñ.
Calculate Z +<Ð+
XB l
A) 12
B) 14
C) 16
D) 18
E) 20
42. For a three-year temporary life annuity due of 100 on (75), you are given:
B
(i) ! .> .> œ !Þ!"B"Þ# ß B  !
(ii) 3 œ !Þ""
Calculate the actuarial present value of this annuity.
A) 264
B) 266
C) 268
D) 270
E) 272
43. For a special fully discrete, 30-year deferred, annual life annuity-due of 200 on (30), you are
given:
(i) The net single premium is refunded without interest at the end of the year of death
if death occurs during the deferral period.
(ii) Mortality follows the Illustrative Life Table
(iii) 3 œ !Þ!'
Calculate the net single premium for this annuity.
A) 350
B) 360
C) 370
D) 380
E) 390
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MLC - PROBLEM SET 4
44. For a special 30-year deferred annual whole life annuity-due of 1 on (35):
(i) If death occurs during the deferral period, then G % net single premium is refunded without
interest at the end of the year of death.
(ii)
ä65 œ 9.90
(iii) E35:30| œ 0.21
(iv)
1
E35:30|
œ 0.07
(v)
The net single premium for this special deferred annuity is 1.4.
Calculate G .
A) Less than 10
B) At least 10 but less than 12
C) At least 12 but less than 14
D) At least 14 but less than 16
E) At least 16
45. You are given the following for every integer age B:
Assuming UDD over each year of age, calculate 
+
.
#!À"!l
A) 6.0
B) 6.1
C) 6.2
D) 6.3
E) 6.4
46. An insurance company has agreed to make payments to a worker age B who was injured at
work.
(i) The payments are 150,000 per year, paid annually, starting immediately and continuing for
the remainder of the worker’s life.
(ii) After the first 500,000 is paid by the insurance company, the remainder will be paid by a
reinsurance company.
(.7)> , 0 Ÿ > Ÿ 5.5
(iii) > :B œ 
0,
5.5  >
(iv)
3 œ 0.05
Calculate the actuarial present value of the payments to be made by the reinsurer.
A) Less than 50,000
B) At least 50,000, but less than 100,000
C) At least 100,000, but less than 150,000
D) At least 150,000, but less than 200,000
E) At least 200,000
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MLC - PROBLEM SET 4
47. For a whole life annuity-due of 1 on (B), payable annually:
(i) ;B œ 0.01
(ii) ;B1 œ 0.05
(iii) 3 œ 0.05
(iv) äB1 œ 6.951
Calculate the change in the actuarial present value of this annuity-due if :B1 is increased
by 0.03.
A) 0.16
B) 0.17
C) 0.18
D) 0.19
E) 0.20
PROBLEM SET 4 - WRITTEN ANSWER PROBLEMS
1. ^" is the present value random variable for a 10-year deferred continuous life-annuity of
$10,000 per year issued to Ð%!Ñ. ^# is the present value random variable for a 25-year
continuous certain and whole life annuity issued to Ð%!Ñ.
(a) (2 points) Show that the variance of ^" is the same as that of ^# . Formulate (but do not
calculate) the covariance between ^" and ^# .
(b) (2 points) Suppose that the force of interest is $ œ Þ!&, and the force of mortality is constant
at .01 for all ages. Calculate IÒ^" Ó and IÒ^# Ó.
(c) (4 points) With $ œ Þ!& and .%!> œ Þ!" for > !, formulate algebraically and sketch the
graph of the distribution function of ^" , ( J^" ÐCÑ œ T Ð^" Ÿ CÑ .
(d) (1 point) Describe in words the difference between the graph of J^" and J^# .
2. ^ is the present value random variable for a 20-year deferred life annuity-due of 1 per year
>
issued to Ð%!Ñ. Survival follows the model WÐ>Ñ œ "  "#!
, > ! (survival from birth).
Interest is at annual effective rate 4%.
(a) (3 points) Calculate IÒ^Ó, the net single premium for this annuity.
(b) (2 points) Formulate the random variable ^ in terms of O'! , the curtate future lifetime of the
annuitant,
(c) (3) Calculate the median of ^ .
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MLC - PROBLEM SET 4
3. For a group of individuals all of age B, of which 30% are smokers and 70% are non-smokers,
you are given:
non-smoker
(i) $ œ !Þ"!
(ii) .smoker
œ Þ!# for > !
(iii) .B>
œ Þ!" for > !
B>
(iv) XB is the future lifetime of ÐBÑ
An individual is chosen at random from this group. ] denotes the present value random variable
of the total payment made to the individual.
(a) (2 points) Calculate IÒ] Ó for this individual.
(b) (2 points) Calculate Z +<Ò] Ó for this individual.
(c) (2 points) Calculate T Ò] Ÿ &Þ!Ó
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241
MLC - PROBLEM SET 4
LIFE CONTINGENCIES - PROBLEM SET 4 SOLUTIONS
MULTIPLE CHOICE
X
#
_
_
#
X&!
Z +<Ò@ &! Ó
E&!  (E&! )
1. ] œ 
+ X&! l œ "@$ p Z +<Ò] Ó œ
œ
.
#
$
$#
_
_
#
#
But _E&! œ "  Ð#$ Ñ † +&! œ Þ"!!$"
_
and E&! œ "  $ † +&! œ Þ#&'%( p Z +<Ò] Ó œ "!Þ"( p Z +<Ò"!!] Ó œ "!"ß (!! . Answer: B.
2. +&! œ +&!:#!|  #!| +&! œ +&!:#!|  #! I&! † +(! p
p
#! I&!
œ
+&!
=&!:#!| +(!
œ
ÞÞ
+&! "
ÞÞ
=&!:#!| +(! "
œ
+&!
#! E&!
"#Þ#'(
%&Þ'&&(Þ&'*
œ =&!:#!|  +(!
œ Þ#$. Answer: A.
O'! "
ÞÞ
ÞÞ
3. T Ò#!ß !!! +O'! "l "&!ß !!!Ó œ T Ò+O'! "l (Þ&Ó œ T  "@.
(Þ&
œ T Ò@O'! " Ÿ Þ("#Ó œ T ÒÐO'!  "Ñ 68Ð@Ñ Ÿ 68 Þ("#Ó
œ T ÒO'! (Þ(Ó œ T ÒO )Ó œ ) :'! œ jj') œ (ß!")ß%$#
)ß"))ß!(% œ Þ)&( Þ Answer: E.
'!
_
_
_
_
_
_
4. SBP œ I  +X "!|  œ I  +X_ |  @X † +"!|  œ +&!  E&! † +"!| . But > :&! œ /–.04>
_
_
_
∞
"
p +&! œ  /–.08> † e–.04> .> œ Þ"#
œ )Þ$$$$ p E&! œ "  $ † +&! œ Þ$$$$ , and
!
–10$
–.8
_
+"!| œ "/$
œ "/
.08 œ 'Þ'))$% p SBP œ "!Þ'$. Answer D.
5. Since all three series have the same total paid (i.e., 8 ‚
8"
#
) , the sooner the payments are
made, the larger the present value of the series. The decreasing annuity has the largest payments
first and thus has the largest present value, and the increasing annuity has the smallest payments
first, and has the smallest present value. Thus, R Ÿ T Ÿ S . Answer: B.
_
. _
. _ _
"!!  "!!
6. .B
+x œ [.B  $ ] +x  " Ä .B œ ["  .B
+B ]/+B  $ œ ["  ÐB"Ñ
# ] [ B" ]  $
"
"
Ä .B œ B"
"!!  B"  $ Ä .&! œ Þ&"  &"  Þ"! œ Þ$*! .
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Answer: C.
SOA Exam MLC Study Guide © S. Broverman, 2015
MLC - PROBLEM SET 4
ÞÞ
ÞÞ
7. = #":$*| œ = #!:%!| 
p
%! I#!
ÞÞ
ÞÞ
. But +#!:%!| œ +#!:%!|  "  %! I#! œ %! I#! † = #!:%!|  "  %! I#!
+
"
ÞÞ
"
œ ÞÞ= #!:%!|" œ Þ#%%"% p = #":$*| œ *$Þ#&$&  Þ#%%"%
œ )*Þ"'. Answer: B.
1
%! I#!
#!:%!|
Ð"#Ñ
Ð"#Ñ
8. Z +<Ò^Ó œ # EB  ÐEB Ñ# .
Ð"#Ñ
Ð"#Ñ
ÞÞ
3
Under UDD EB œ 3Ð"#Ñ
† EB œ =_"| † Ð"  . † +x Ñ œ Þ##"'$.
#
3
# Ð"#Ñ
EB œ # 3Ð"#Ñ
† # EB , where # 3 œ Ð"  3Ñ#  " œ #"  3# œ Þ"''%,
# Ð"#Ñ
3
Ð"#Ñ
œ "# † Ò Ð"  3"# Ñ#  "Ó œ Þ"&*%# , since 3Ð"#Ñ œ
3
Ð"#Ñ
=_"|
œ Þ!((#!) ,
#
ÞÞ
and # EB œ "  # . † # +x œ Þ"&*%# , since # . œ "3# 3 œ Þ"%#''.
Ð"#Ñ
Thus, # EB œ Þ"("#$ and then Z +<Ò^Ó œ Þ"("#$  Ðß ##"'$Ñ# œ Þ"##. Answer: C.
Ð7Ñ
Ð7Ñ
Ð7Ñ
ÞÞÐ7Ñ
Ð7Ñ
ÞÞÐ7Ñ
9. EB:8_ | œ "  . Ð7Ñ † = B:8_ | p Ð"  37 Ñ † EB:8_ |  3Ð7Ñ † +B:8_ | œ "  37
Ð7Ñ
Ð7Ñ
ÞÞÐ7Ñ
"
p Ð"  37 Ñ † EB:8_ |  3Ð7Ñ † +B:8_ |  7
 œ ". Answer: D.
ww 
 
10. EB  EB œ "  $
+ Bww  Ð"  $
+ B Ñ œ $ Ð+
+ Bww Ñ Þ
B
∞

+ wwB œ ! /$ > > :Bww † .> , and
> ww
>
ww
> :B œ /B:Ò  ! .B= .=Ó œ /B:Ò  ! Ð5  .B= Ñ .=Ó
œ /B:Ò  5>  ! .B= .=Ó œ /5> † /B:Ò  ! .B= .=Ó œ /5> † > :B .
Then 
+ wwB becomes
∞
∞
∞

+ wwB œ ! /$ > † > :Bww .> œ ! /$ > /5> † > :B .> œ ! /Ð5$ Ñ> † > :B .> œ 
+ Bw .
ww 
 
 
Therefore, EB  EB œ $ Ð+
+ Bww Ñ œ $ Ð+
+ Bw Ñ.
Answer: E
B
B
>
>
X
"
"
X
X
X
X
11. G9@Ò 
+ X l ß @X Ó œ G9@Ò "@
$ ß @ Ó œ $ G9@Ò @ ß @ Ó œ $ Z +<Ò@ Ó
#
#
œ "
Answer: A
$ Ò E B  E B Ó.
12. T ÒX  >Ó œ > :B œ /.> œ /B:Ò  Ð !Þ!#%
68Ð!Þ%Ñ Ñ>Ó.
68ÐÞ%Ñ
"/$ X

T Ò+
 #!Ó œ T Ò/Þ!$X  Þ%Ó œ T ÒX  Þ!$ Ó
X ÐBÑl  #!Ó œ T Ò
$
68ÐÞ%Ñ
Þ)
œ /B:Ò  Ð !Þ!#%
œ Þ%&.
Answer: A
68Ð!Þ%Ñ ÑÐ Þ!$ ÑÓ œ /
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243
MLC - PROBLEM SET 4
_


13. Z +<Ò+X l Ó œ $"# Ò# EB  ÐEB Ñ# Ó. The force of mortality is constant, so we have


.
.
EB œ .$ and # EB œ .#$ Þ
_
.
.
"
5
5
#
Then Z +<Ò+X l Ó œ $"# † Ò .#$  Ð .$ Ñ# Ó œ Ð%5Ñ
# † Ò 5#Ð%5Ñ  Ð 5%5 Ñ Ó
"
"
" #
"
"'
"!!
œ "'5
# † Ò *  Ð & Ñ Ó œ "'5 # † ##& œ * .
Solving for 5 results in 5 œ Þ!#.
Answer: D
Ð$ .Ñ
"
"
"
14. 
+ BÀ"l œ ! /$ > † > :B .> œ ! /$ > † /.> .> œ ! /Ð$ .Ñ> .> œ "/$ . .
ÞÞ
+BÀ"l œ " ß +BÀ"l œ @:B œ /$ † /. œ /Ð$ .Ñ ß  68 +BÀ"l œ $  ..
III is correct.
Answer: D


15. Z +<Ò] Ó œ $"# Ò# EB  ÐEB Ñ# Ó. Since the force of mortality is a constant . for all ages,


.
.
we have EB œ .$ and # EB œ .#$ .
Also, 
+ B œ " œ & and # 
+ B œ " œ % , so that .  $ œ Þ# , .  #$ œ Þ#&.
. $
.#$
Using the two equations we can solve for . and $ ; . œ Þ"& ß $ œ Þ!&.


.
.
Þ"&
Þ"&
Then, EB œ .$ œ Þ"&Þ!&
œ Þ(& and # EB œ .#$ œ Þ"&Þ"
œ Þ' , so that


"
#
Z +<Ò] Ó œ $"# Ò# EB  ÐEB Ñ# Ó œ ÐÞ!&Ñ
# † ÒÞ'  ÐÞ(&Ñ Ó œ "&.
Note that it is also possible to formulate Z +<Ò] Ó directly in terms of annuities:
B  #
B Ñ# œ # Ð&  %Ñ  Ð&Ñ# œ #  #&. We still must find $ by solving
Z +<Ò] Ó œ #$ Ð+
+ B Ñ  Ð+
$
$
the two equations in . and $ that were given above, .  $ œ Þ# , .  #$ œ Þ#&.
#
Z +<Ò] Ó œ œ #$  #& œ Þ!&
 #& œ "&.
Answer: A
16. The APV is
¨ '!
"!l +
$*
='!"
"
œ  @5 5 :'! œ  "!!'!5
"!!'! œ %! Ð$!  #*  â  "Ñ
5œ"!
5œ"!
" Ð$!ÑÐ$"Ñ
œ %!
†
œ ""Þ'#&.
#
T Òsum of payments exceeds 11.625Ó œ T Òat least 12 paymentsÓ.
The 12th payment is made at age 81 (1st payment is at age 70), so that
T Òat least 12 paymentsÓ œ T Òsurvival to at least age 81Ó œ #" :'! œ "!!'!#"
"!!'! œ Þ%(&.
Answer: A
"
17. +¨ C œ "  @:C +¨ C" p ) œ "  "Þ!)
† :C † Ð)Ñ p :C œ Þ*%& (for any integral C).
Then, ) :$! œ :$! † :$" â:$( œ ÐÞ*%&Ñ) œ Þ'$' p ) ;$! œ Þ$'%.
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Answer: B
SOA Exam MLC Study Guide © S. Broverman, 2015
MLC - PROBLEM SET 4
18. The information is given in a form which appears to be more convenient for the application
of the aggregate payment formulation of the annuity:
$
ÞÞ
ÞÞ
ÞÞ
ÞÞ
ÞÞ
ÞÞ
ÞÞ
ÞÞ
+BÀ%l œ  +4"l † 4l ;B  +%l † % :B œ +"l † ;B  +#l † "l ;B  +$ l † #l ;B  +%l † $l ;B  +%l † % :B
4œ!
œ Ð"ÑÐÞ$$Ñ  Ð"Þ*$ÑÐÞ#%Ñ  Ð#Þ)!ÑÐÞ"'Ñ  Ð$Þ'#ÑÐÞ""Ñ  Ð$Þ'#Ñ † % :B .
We use the relationship  4l ;B œ 8 ;B , so that  4l ;B œ ;B  "l ;B  #l ;B  $l ;B œ Þ)% œ % ;B .
8"
$
4œ!
4œ!
Then % :B œ Þ"' and
ÞÞ
+BÀ%l œ Ð"ÑÐÞ$$Ñ  Ð"Þ*$ÑÐÞ#%Ñ  Ð#Þ)!ÑÐÞ"'Ñ  Ð$Þ'#ÑÐÞ""Ñ  Ð$Þ'#Ñ † ÐÞ"'Ñ œ #Þ##.
Answer: D
19. ] is the PVRV for a whole life annuity-due of 1 per year, so that
ÞÞ
"EB
B
IÒ] Ó œ +B œ "E
œ 3ÎÐ"3Ñ
œ "Þ#!(&&
.
Þ!'Î"Þ!' œ "% .
^ is the PVRV for an 8-year temporary life annuity-due of 1 per year, so that
ÞÞ
IÒ^Ó œ +BÀ8l œ "  +BÀ8"l œ "  ' œ (.
Then, IÒ] Ó  IÒ^Ó œ "%  ( œ (.
Answer: D
20. Z +<Ò] Ó œ IÒ] # Ó  ÐIÒ] ÓÑ# p IÒ] # Ó œ Z +<Ò] Ó  ÐIÒ] ÓÑ# .
For an 8-year temporary annuity due of 1 per year, Z +<Ò] Ó œ ."# Ò # EBÀ8l  ÐEBÀ8l Ñ# Ó
Therefore, Z +<Ò] Ó œ
.œ
#
3
"3
" #
. # Ò EBÀ$!l
 ÐEBÀ$!l Ñ# Ó.
œ Þ!%('"* ß EBÀ$!l œ E"
BÀ$!l
EBÀ$!l œ # E"
BÀ$!l
 #E
"
BÀ$!l
E
"
BÀ$!l
œ Þ"%%$  @$! $! :B œ Þ$!'$ Þ
œ Þ!'*%  @'! $! :B œ Þ"!'* ,
"
#
p Z +<Ò] Ó œ ÐÞ!%('"*Ñ
# ÒÞ"!'*  ÐÞ$!'$Ñ Ó œ &Þ(') Þ
IÒ] Ó œ +¨ BÀ$!l œ
"EBÀ$!l
.
œ "Þ$!'$
Þ!%('"* œ "%Þ&(.
IÒ] # Ó œ Z +<Ò] Ó  ÐIÒ] ÓÑ# œ &Þ((  Ð"%Þ&(Ñ# œ #").
21. E
"
$!À"!l
Answer: D
œ E$!À"!l  "! I$! œ "  .+¨ $!À"!l  "! I$! œ "  .Ð"  +$!À*l Ñ  "! I$! œ Þ!&.
Answer: A
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245
MLC - PROBLEM SET 4
œ "! IB † +B"!
ÞÞ
ÞÞ
p "!l +B œ "! IB † +B"! œ "! IB † Ð"  +B"! Ñ œ "! IB  "!l +B œ Þ%  ( œ (Þ%
ÞÞ
ÞÞ
ÞÞ
ÞÞ
ÞÞ
p +B œ +BÀ"!l  "!l +B œ "! IB † = BÀ"!l  "!l +B œ ÐÞ%ÑÐ"&Ñ  (Þ% œ "$Þ%. Answer: E
22.
"!l +B
ÞÞ
ÞÞ
23. 3+BÀ8l  Ð"  3ÑEBÀ8l  " œ 3 † Ò+BÀ8l  "  8 IB Ó  Ð"  3Ñ † Ð"  . † +BÀ8l Ñ  "
ÞÞ
ÞÞ
œ 3 † +BÀ8l  3  3 † 8 IB  "  3  3 † +BÀ8l  " œ 3 † 8 IB . Answer: B
24.
8l +B
œ 8 IB † +B8 œ 8 IB † Ð+¨ B8  "Ñ œ 8 IB † Ð
"EB8
.
 "Ñ.
From EBÀ8l œ EBÀ8l
 8 IB ß we get Þ&'$ œ EBÀ8l
 Þ&%$ p EBÀ8l
œ Þ!#.
"
"
"
Then EB œ EBÀ8l
 8 IB † EB8 p Þ"#* œ Þ!#  ÐÞ&%$ÑEB8 p EB8 œ Þ#!!( .
"
Finally,
8l +B
œ 8 IB † Ð
"EB8
.
 "Ñ œ ÐÞ&%$ÑÐ "Þ#!!(
Þ!&(  "Ñ œ (Þ!(.
Answer: A
`
` Þ"8
25.(a) 8 IB œ /$ 8 /.8 œ /Þ"8 p `8
œ  Þ"/Þ"8 Þ
8 IB œ `8 /
Þ"8
8
8
(b) 
+ BÀ8l œ ! @> > :B .> œ ! /$ > /.> .> œ "/Þ" .
#
.
.
"
Þ!%
Þ!% #
 Ó œ "# Ò # 
(c) Z +<Ò+
EBl  EB Ó œ $"# Ò #$ .  Ð $ . Ñ# Ó œ ÐÞ!'Ñ
# Ò Þ"'  Ð Þ" Ñ Ó œ #&
Xl
$
so standard deviation is 5Þ
` 
` "/Þ"8
(d) `B
+ BÀ8l œ `B
œ !.
Þ"
26. I and III require UDD. II is false (a corrected version of II is 
+B œ
+ BÀ"l  @:B
+ B" ).
IV is always true.
Answer: E
8


27. ÐM 8l
+ ÑB œ ÐM 
+ ÑBÀ8l  8 † 8l
+ B œ ! >/$ > † > :B .>  8/$ 8 † 8 :B † 
+ B8 Þ
8.
Since the force of mortality is constant at ., we have 8 :B œ /
, and 
+ B8 œ $ " . Þ
8

Then, ÐM 8l
+ ÑB œ ! >/Ð$ .Ñ> .>  8/8Ð$ .Ñ † $ " .
8

p . ÐM 8l
+ ÑB œ . Ò >/Ð$ .Ñ> .>  8/8Ð$ .Ñ † " Ó
.8
œ 8/
.8
Ð$ .Ñ8
/
$ .
!
8Ð$ .Ñ
†
"
$ .
 8/
8Ð$ .Ñ
œ/
8Ð$ .Ñ
. B
We have used the calculus rule .B
+ 0 Ð>Ñ .> œ 0 ÐBÑ.
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† $ " . œ "!/Þ"8 .
Answer: B
SOA Exam MLC Study Guide © S. Broverman, 2015
MLC - PROBLEM SET 4
28. I.
¨B
8l +
œ 8"l +B œ +B  +BÀ8"l . True.
II. "  +BÀ8"l œ +¨ BÀ8l Á Ð"  3Ñ+BÀ8l . False
III. EBÀ8l œ "  .+¨ BÀ8l œ "  Ð"  @Ñ+¨ BÀ8l œ @+¨ BÀ8l  Ð+¨ BÀ8l  "Ñ œ @+¨ BÀ8l  +BÀ8"l . True.
Answer: C

#
29. The variance of an annuity with annual payment of 1 is $"# Ò # E'&  E'& Ó.
'& Ñ œ $!ß !!!ß !!! , so that 
We are given Ð"!!ÑÐ#&ß !!!+
+ '& œ "#, and

E'& œ "  $
+ '& œ Þ#). The variance of an annuity with annual payment of 1 is then
"
ÐÞ!'Ñ# Ò Þ#
 ÐÞ#)Ñ# Ó œ $$Þ(). The standard deviation of the present value of 100 annuities each
with annual payment of 25,000 is "!!Ð#&ß !!!Ñ# Ð$$Þ()Ñ œ "ß %&$ß !!!.
30. The schedule of payments is
Age #!
#"
Pmt. "
$
##
&
#$
(
#%
*
#$
'
"
#%
)
"
Answer: C
The schedule of payments can be expressed as
Age
Pmt.

#!
!
"
#"
#
"
##
%
"
The net single premium is the combination of the two actuarial present values in the second
ÞÞ
representation. The APV is #ÐM+Ñ#!À%l  +#!À&l .
ÞÞ
ÞÞ
We use the relationship +#!À5l œ "  +#!À5"l , so that +#!À&l œ "  +#!À%l œ %Þ!%.
The net single premium is #Ð(Þ"(Ñ  %Þ!% œ ")Þ$).
Answer: B
31. The present value random variable for the annuity is ] œ 
"!ß!!! 
+ "!l for !X Ÿ"!
"!ß!!! 
+
X l for X "!
.
The 10-year certain period guarantees that the pv is at least "!ß !!! 
+ "!l œ '$ß #"# , and the
minimum payment is made if death occurs within 10 years; this has probability
œ "  /"!ÐÞ!"Ñ œ Þ!*&. Therefore, J] ÐCÑ œ ! for C  '$ß #"# , and
J] Ð'$ß #"#Ñ œ Þ!*&. There is a probability mass of .095 at C œ '$ß #"#. This corresponds to a
"jump" in the graph of J] ÐCÑ from 0 to .095 at that point. This occurs only in graph D.
Answer: D
"! ;'&
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247
MLC - PROBLEM SET 4
32. We use the relationship
E"
25À10l
EBÀ8l
œ EBÀ8l  @8 8 :B to get
"
œ E25À10l  @10 10 :25 œ E25À10l  ÐÞ*&Ñ"! ÐÞ)(Ñ œ E25À10l  Þ&#!* Þ
ÞÞ
We then use the relationships E25À10l œ "  . +25À10l and
ÞÞ
ÞÞ
ÞÞ
+25À1&l œ +25À10l  @10 10 :25 +$5À&l Þ
ÞÞ
We get +25À10l œ *Þ)')  ÐÞ*&Ñ"! ÐÞ)(ÑÐ%Þ$*#Ñ œ (Þ&)!#,
and then E25À10l œ "  Ð"  @ÑÐ(Þ&)!#Ñ œ "  ÐÞ!&ÑÐ(Þ&)!#Ñ œ Þ'#"! Þ
Finally, E "
25À10l
œ E25À10l  Þ&#!* œ Þ'#"!  Þ&#!* œ Þ"! Þ Answer: A
33. We are given: #& :! œ Þ(! ß $& :! œ Þ&! ß
ÞÞ
ÞÞ
ÞÞ
(ß &!!+#& œ '!ß !!! ß "#ß $!!+$& œ '!ß !!! ß *ß %!!+#&À"!l œ '!ß !!!
ÞÞ
ÞÞ
ÞÞ
We use the relationships +#& œ +#&À"!l  @"! "! :#& +$& and $& :! œ #& :! † "! :#& Þ
We first get
"! :#&
"!
Solving for @
œ
Þ&
Þ(
œ
&
(
, and then )Þ! œ 'Þ$)$!  @"! ‚
results in @"! œ Þ%'%" . Then. @ œ
"
"3
&
(
‚ %Þ)()! Þ
œ ÐÞ%'%"ÑÞ" œ Þ*#', so that 3 œ Þ!)! Þ
Answer: A
34. The series of payments is
Age 30
...
39 40 ...
Pmt 10
...
10 20 ...
49 50 ...
20 30 ...
59 60
30
There are various ways in which we can regroup the payments.
From the information given, we have the actuarial pv of a 10 year annuity-due starting at age 30.
ÞÞ
ÞÞ
ÞÞ
ÞÞ
We can also use the relationship +$!À$!l œ +$!À#!l  #! I$! † +&!À"!l to solve for +$!À#!l .
ÞÞ
ÞÞ
ÞÞ
+$!À#!l œ +$!À$!l  #! I$! † +&!À"!l œ @  7A.
The series of payments can be formed in the following combination of payments:
Age
Series 1
Series 2
Series 3
Net Total
248
30
20
-10
10
...
...
...
...
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39 40
20 20
-10
10
20
...
...
49 50 ...
20
59 60
...
30 ...
20 30 ...
30
30
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MLC - PROBLEM SET 4
34. continued
ÞÞ
The APV of series 1 is #!+$!À#!l œ #!Ð@  7AÑ.
ÞÞ
The APV of series 2 is  "!+$!À"!l œ  "!?.
ÞÞ
The APV of series 3 is $! † #! I$! † +&!À"!l œ $!7A.
The total APV of the three series of payments is
#!Ð@  7AÑ  "!?  $!7A œ  "!?  #!@  "!7A.
Answer: D
35. The actuarial present value of a subscriber is the actuarial present value of all future
subscription payments. This forms a life annuity, with constant probability :B œ : œ Þ*!.
The probability of continual renewal to 5 years from now is 5 :C œ :5 .
#&
#&
The APV is #&Ò"  @:  @# :#  âÓ œ "@:
œ "ÐÞ*&ÑÐÞ*Ñ
œ "(#Þ%". Answer: D
36. The actuarial present value of payout (i) is T" † E
"
'!À"!l
The variance of the payout in (i) is T"# † Ò # E
"
'!À"!l
œ Þ%"))T" .
#
" Ñ Ó
'!À"!l
 ÐE
Since (i) and (ii) are actuarially equivalent, we have T" † E
"
'!À"!l
œ Þ!&)&T"# .
ÞÞ
œ T# † +'!À#!l .
"E'!À#!l
ÞÞ
ÞÞ
We use the relationship +'!À#!l œ
to get +'!À#!l œ *Þ#))$.
.
Then, Þ%"))T" œ *Þ#))$T# p T# œ Þ!%&"T" .
The variance of an 8-year life annuity-due of 1 per year is ."# † Ò# E'!À#!l  ÐE'!À#!l Ñ# Ó ,
so the variance of payout (ii) is the variance of a 20 year life annuity-due of T# per year, which is
T##
.#
"
† Ò# E'!À#!l  ÐE'!À#!l Ñ# Ó œ ÐÞ!%&"T" Ñ# † Þ!!$'
† ÒÞ#$"#  ÐÞ%%#(Ñ# Ó œ Þ!"**T"# .
The difference in the two variances is Þ!&)&T"#  Þ!"**T"# œ Þ!$*T"# .
Answer: A
37. With constant force of mortality ., we have 
+ B œ $ " . .
 
  " Ó œ T Ò "/$ X  " Ó œ T Ò/$ X  . Ó.
T Ò+
+
Ó
œ
T
Ò+
B
Xl
Xl
$ .
$
$ .
$ .
.8
With constant force of mortality ., we also have 8 :B œ /
œ T ÒX  8Ó Þ
.
.
$ X
Then, T Ò/
 $ . Ó œ T Ò  $ X  68Ð $ . ÑÓ
.
.
.
œ T ÒX   "$ † 68Ð $ . ÑÓ œ /B:Ò  . † Ò  $" † 68Ð $ . ÑÓ Ó œ Ð $ . Ñ.Î$ . Answer: B
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MLC - PROBLEM SET 4
38. The present value random variable for Jack's insurance is "!!!/$ X , and the
actuarial present value is
IÒ"!!!/$ X Ó œ IÒ"!!!/$ X lIllustrative TableÓ † T ÒIllustrative TableÓ
 IÒ"!!!/$ X lconstant forceÓ † T Òconstant forceÓ
.
Þ!#
Þ!'
MX
œ "!!!Ò $3 E'#
† ÐÞ'Ñ  .
$ † ÐÞ%ÑÓ œ "!!!Ò 68 "Þ!' † ÐÞ$*'(ÑÐÞ'Ñ  Þ!#68 "Þ!' † ÐÞ%ÑÓ
œ $%(Þ$.

'#

+ '# œ "E
œ "Þ$%($
$
68 "Þ!' œ ""Þ#.
39. With constant force of interest $ and constant force of mortality ., we have
_
ÞÞ
"
+B œ $ " . œ "#Þ5. It follows that $  . œ Þ!). Then, +B œ "@:
œ "/"$ †/. œ "$Þ!!''(.
Using Woolhouse's approximation with 7 œ "# we have
ÞÞÐ"#Ñ ÞÞ
"## "
+B œ +B  "#"
Answer: C
#‚"#  "#‚"## Ð.  $ Ñ œ "#Þ&%#.
40. We use the relationship for the variance of a continuous whole life annuity of 1 per year:

 Ó œ "# † Ò# 
Z +<Ò+
EB  ÐEB Ñ# Ó.
Xl
$
The survival distribution is a mixture of the survival distributions for smokers and non-smokers.

smoker
non-smoker
EB œ Þ$EB
 Þ(EB
œ ÐÞ$ÑÐÞ%%%Ñ  ÐÞ(ÑÐÞ#)'Ñ œ Þ$$$%.
#
# smoker
# non-smoker
EB œ Þ$ EB
 Þ( EB
.
smoker
non-smoker
We find # EB
and # EB
from the given annuity variances.

"
"
smoker
# smoker
# smoker

)Þ)") œ Z +<Ò+ X l Ó œ $ # † Ò EB
 ÐEB Ñ# Ó œ ÐÞ"Ñ
 ÐÞ%%%Ñ# Ó ,
# † Ò EB
smoker
from which we get # EB
œ Þ#)&$.
non-smoker

"
# smoker
non-smoker Ó œ "# † Ò# 
Similarly, )Þ&!$ œ Z +<Ò+
EB
 ÐEB Ñ# Ó œ ÐÞ"Ñ
 ÐÞ#)'Ñ# Ó ,
# † Ò EB
$
Xl
non-smoker
from which we get # EB
œ Þ"'').
#
# smoker
# non-smoker
Then, EB œ Þ$ EB
 Þ( EB
œ Þ#!#% , and finally,


 Ó œ "# † Ò# EB  ÐEB Ñ# Ó œ " # † ÒÞ#!#%  ÐÞ$$$%Ñ# Ó œ *Þ" . Answer: E 
Z +<Ò+
Xl
$
ÐÞ"Ñ
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MLC - PROBLEM SET 4
#
 Ñ œ "# † Ð# 
41. Z +<Ð+
EB  EB Ñ
XB l
$

Since the force of mortality is constant at -, we have EB œ $ and

Therefore, from EB œ Þ$%%$ œ Þ!), we get - œ Þ!%# ,

Þ!%#
and then # EB œ #ÐÞ!)ÑÞ!%#
œ Þ#!(*.
"
#

Z +<Ð+
X ÐBÑl Ñ œ ÐÞ!)Ñ# † ÒÞ#!(*  ÐÞ$%%$Ñ Ó œ "$Þ*'.
#
EB œ #$-- .
Answer: B
ÞÞ
42. +(&À$l œ "  @:(&  @# # :(& .
From (i) we get 8 :(& œ /(& .> Ñ .> œ /Þ!"ÒÐ(&8Ñ (& Ó .
"Þ#
"Þ#
Therefore, :(& œ /Þ!"Ò(' (& Ó œ Þ*("* and
Þ!"Ò(("Þ# (&"Þ# Ó
œ Þ*%%&.
# :(& œ /
Þ*%%&
The APV of the annuity is "  Þ*("*
"Þ""  Ð"Þ""Ñ# œ #Þ'%.
(&8
"Þ#
"Þ#
Answer: A
ÞÞ
43. The equivalence principle equation is U œ #!! $!l +$!  UE " .
$!À$!l
ÞÞ
ÞÞ
$!l + $! œ #! I$! † "! I&! † +'! œ ÐÞ#*$(%ÑÐÞ&"!)"ÑÐ""Þ"%&%Ñ œ "Þ'(#.
E$! œ E "
$!À$!l
E"
$!À$!l
 #! I$! † "! I&! † E'! , so that
œ Þ"!#%)  ÐÞ#*$(%ÑÐÞ&"!)"ÑÐÞ$'*"$Ñ œ Þ!%(!*.
#!!Ð"Þ'(#Ñ
Solving for U results in U œ "Þ!%(!* œ $&".
Answer: A
ÞÞ
ÞÞ
$!l +$&
44. U œ Þ!"G † UE "  $!l +$& p U œ "Þ!"G†E
"
$&À$!l
.
$&À$!l
We are given E " œ !Þ!(.
$&À$!l
ÞÞ
ÞÞ
ÞÞ
$!
$!l + $& œ @ $! :$& † + '& . We are given +'& œ *Þ*! .
We also have @$! $! :$& œ E$&À$!l  E
"
$&À$!l
ÐÞ"%ÑÐ*Þ*!Ñ
œ Þ#"  Þ!( œ Þ"%.
Then, "Þ% œ "ÐÞ!"GÑÐÞ!(Ñ . Then G œ "%Þ$
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Answer: D
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251
MLC - PROBLEM SET 4
45. We use the relationship 
+B œ
+ BÀ8l  @8 8 :B 
+ B8 :

+ #! œ 
+ #!À"!l  @"! "! :#! 
+ $! .

""
From UDD we have EB œ $3 EB œ 68Þ!&
"Þ!& † #" œ Þ&$') for all B ,

and then 
+ B œ "EB œ *Þ%*% for all B.
$
Therefore *Þ%*% œ 
+ #!À"!l  @"! "! :B † Ð*Þ%*%Ñ œ 
+ #!À"!l 
and then 
+
œ 'Þ#.
Þ&')!
Ð"Þ!&Ñ"!
† Ð*Þ%*%Ñ
#!À"!l
Note that since @ œ
#!
#"
and :B œ Þ*%& , we have @:B œ Þ* , so that
@"! "! :B œ ÐÞ*Ñ"! œ Þ$%)'(). Answer: C
46. The reinsurer's first potential payment is 100,000 and will be made 3 years from now (the
primary insurer would pay 150,000 now and for the next 2 years, and 50,000 in three years to
reach the limit of 500,000). Starting 4 years from now, the reinsurer would 150,000 per year, as
long as the worker survives. The actuarial present value of the reinsurer's payments is
"!!ß !!!@$ $ :B  "&!ß !!!Ò@% % :B  @& & :B  @' ' :B  âÓ
Since > :B œ ! for >  &Þ& , and > :B œ ÐÞ(Ñ> for > Ÿ &Þ& , this becomes
ÐÞ(Ñ$
Þ( %
Þ( &
"!!ß !!! Ð"Þ!&Ñ$  "&!ß !!!ÒÐ "Þ!&
Ñ  Ð "Þ!&
Ñ Ó œ (*ß !"# Þ
Answer: B
ÞÞ
ÞÞ
47. +B œ "  @:B +B" œ (Þ&&$) ß
ÞÞ
ÞÞ
ÞÞ
+B" œ "  @:B" +B# p +B# œ 'Þ*&""
@:B" œ 'Þ&((%#
The revised values (superscript R ) are found as follows:
ÞÞ
ÞÞR
R ÞÞR
R ÞÞ
+R
B œ "  @:B + B" ß +B" œ "  @:B" +B#
ÞÞ
ÞÞ
R
:BR œ :B œ Þ** , :B"
œ Þ*) , +R
B# œ +B# (only change is at age B  ")
ÞÞ
Þ*)
R ÞÞ
p +R
B" œ "  @:B" + B# œ "  "Þ!& Ð'Þ&((%#Ñ œ (Þ"$)*
ÞÞ
Þ**
R ÞÞR
p +R
B œ "  @:B + B" œ "  "Þ!& Ð(Þ"$)*Ñ œ (Þ($".
Change is (Þ($"  (Þ&&% œ Þ"((.
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Answer: C
SOA Exam MLC Study Guide © S. Broverman, 2015
MLC - PROBLEM SET 4
WRITTEN ANSWER
1.(a) ^# œ ^"  "!ß !!! 
+ "!l . Since ^" and ^# differ by a constant, they must have the same
variance.
G9@Ò^" ß ^# Ó œ G9@Ò^" ß ^"  
+ #&l Ó œ G9@Ò^" ß ^" Ó œ Z +<Ò^" Ó .
(b) IÒ^" Ó œ "!ß !!! "!l
+ %! œ "!ß !!!/"!$ "! :%! ‚ 
+ &! .
∞
∞
"
"!‚Þ!"
, and 
+ &! œ ! /$ > > :&! .> œ ! /Þ!&> ‚ /Þ!"> .> œ Þ!'
.
"! :%! œ /
IÒ^" Ó œ *"%Þ'* .

+ "!l œ
"/"!$
$
(c) ^" œ 
œ
œ (Þ)'* , so that IÒ^# Ó œ *"%Þ'*  "!ß !!! ‚ (Þ)'* œ (*ß '!* .
!
"!ß !!! /
Þ& 
+ X%! "!l
!
#!!ß !!! Ð/
Þ&
/
Þ!&X%!
Ñ
!
X%! Ÿ "!
Þ!&ÐX%! "!Ñ
œ
X%!  "!
"!ß !!! /Þ& "/
Þ!&
X%! Ÿ "!
X%!  "!
X%! Ÿ "!
X%!  "!
T Ð^" œ !Ñ œ T ÐX%! Ÿ "!Ñ œ "! ;%! œ "  "! :%! œ "  /"!‚Þ!" œ Þ!*&# .
For A  ! , T Ð^" Ÿ CÑ œ T Ð^" œ !Ñ  T Ð!  ^" Ÿ CÑ
œ Þ!*&#  T Ò!  #!!ß !!! Ð/Þ&  /Þ!&X%! Ñ Ÿ CÓ
We can rewrite the inequality !  #!!ß !!! Ð/Þ&  /Þ!&X%! Ñ Ÿ C in the form
C
"!  X%! Ÿ  #! 68Ð /Þ&  #!!ß!!! Ñ .
From T Ð?  X%! Ÿ @Ñ œ ? :%!  @ :%! , we get
C
T Ò!  #!!ß !!! Ð/Þ&  /Þ!&X%! Ñ Ÿ CÓ œ T Ò"!  X%! Ÿ  #! 68Ð /Þ&  #!!ß!!! ÑÓ ,
C
so ? œ "! , and @ œ  #! 68Ð /Þ&  #!!ß!!! Ñ Þ
Also, with contstant force of mortality .01, we have > :B œ /Þ!"> .
C
Then T Ò!  #!!ß !!! Ð/Þ&  /Þ!&X%! Ñ Ÿ CÓ œ /Þ"  Ð /Þ&  #!!ß!!! ÑÞ# for A  !.
The maximum value of A is "!ß !!! "!l
+ ∞l œ #!!ß !!!/Þ& œ "#"ß $!' .
Finally, for A  ! , J^" ÐCÑ œ T Ð^" œ !Ñ  T Ð!  ^" Ÿ CÑ
C
œ "  Ð /Þ&  #!!ß!!! ÑÞ# .
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MLC - PROBLEM SET 4
The graph of the distribution function is
(d) Since ^# œ ^"  "!ß !!! 
+ "!l , the graph of J^# is the same shape as that of J^" , but it is
shifted to the right by an amount of "!ß !!! ‚ 
+ œ ()ß '*! .
"!l
2.(a) The APV is
¨ '!
#!l +
$*
&*
5
œ  @5 5 :'! œ  "#!'!5
"#!'! @
5œ#!
œ
"
'!
"
'!
#"
&)
%!+%!lÞ!%
Þ!%
œ $Þ**'$ .
!
(b) ^ œ  #! ÞÞ
@ +O'! "#!l
O'!  #!
O'! #!
œ
5œ#!
&*
"
‚ Ò%!@  $*@  â  #@  @ Ó œ '!
‚ @"* ‚ ÐH+Ñ%!lÞ!%
#!
"*
‚ @Þ!%
‚
(c) The median of ^ is the smallest - for which T Ò^ Ÿ -Ó Þ& ,
WÐB5Ñ
5
From the nature of the distribution, we see that 5 ;B œ "  WÐBÑ œ "#!B
,
5
so that T ÒO'! Ÿ 5  "Ó œ 5 ;'! œ '!
Þ& is equivalent to 5 $! or 5  " #*,
ÞÞ
Since ^ œ @#! +O'! "#!l it follows that
ÞÞ
T Ò^ Ÿ -Ó œ T Ò@#! +O'! "#!l Ÿ -Ó œ T ÒO'! Ÿ 5  "Ó , where
ÞÞ
@#! +5"#!l œ - . But since 5 œ $! is the smallest 5 for which T ÒO'! Ÿ 5  "Ó Þ&
ÞÞ
it follows that - œ @#! +#*"#!l œ $Þ)%*( is the smallest value of - for which
T Ò^ Ÿ -Ó Þ& . The median value of ^ is $Þ)%*( ,
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MLC - PROBLEM SET 4
3. ] is a mixture of ] = and ] 8= , with mixing weights .3 and .7, respectively.
(a) IÒ] Ó œ Þ$ ‚ IÒ] = Ó  Þ( ‚ IÒ] 8= Ó
For a mortality model with constant force of mortality . with constant force of interest $ , the net
single premium for a continuous whole life annuity is ." $ , so
"
"
IÒ] = Ó œ Þ!#Þ"
œ )Þ$$$$ and IÒ] 8= Ó œ Þ!"Þ"
œ *Þ!*!*
and IÒ] Ó œ Þ$ ‚ )Þ$$$  Þ( ‚ *Þ!*!* œ )Þ)'% .
(b) Since ] œ "^
$ , where ^ is the PVRV for a continuous whole life insurance of 1, it
follows that Z +<Ò] Ó œ $"# ‚ Z +<Ò^Ó œ "!! ‚ Z +<Ò^Ó ., and Z +<Ò^Ó œ IÒ^ # Ó  IÒ^Ó# .

.
From the mixture distribution and the fact that EB œ .$ for the constant force model, we get
IÒ^Ó œ Þ$ ‚ IÒ^ = Ó  Þ( ‚ IÒ^ 8= Ó œ Þ$ ‚
(note that this is consistent with IÒ] Ó œ
Þ!#
Þ!#Þ"
 Þ( ‚
"IÒ^Ó
$
#
À )Þ)'% œ
From the mixture distribution and the fact that EB œ
IÒ^ # Ó œ Þ$ ‚ IÒÐ^ = Ñ# Ó  Þ( ‚ IÒÐ^ 8= Ñ# Ó œ Þ$ ‚
Þ!"
Þ!"Þ"
.
.#$
Þ!#
Þ!#Þ#
#
œ Þ""$'
"Þ""$'
Þ"
).
for the constant force model, we get
 Þ( ‚
Þ!"
Þ!"Þ#
œ Þ!'!' .
Then Z +<Ò^Ó œ IÒ^ # Ó  IÒ^Ó# œ Þ!'!'  ÐÞ""$'Ñ œ Þ!%(( .
Finally, Z +<Ò] Ó œ $"# ‚ Z +<Ò^Ó œ "!! ‚ Þ!%(( œ %Þ(( .
Note that we could formulate Z +<Ò] Ó as IÒ] # Ó  IÒ] Ó# , and it continues to be true for the
mixture distribution that IÒ] # Ó œ Þ$ ‚ IÒÐ] = Ñ# Ó  Þ( ‚ IÒÐ] 8= Ñ# Ó , but it is easier to find
IÒÐ^ = Ñ# Ó directly than to find IÒÐ] = Ñ# Ó directly.
(c) For the mixture distribution we can use conditioning to find probability.
T Ò] Ÿ &Ó œ Þ$ ‚ T Ò] = Ÿ &Ó  Þ( ‚ T Ò] 8= Ÿ &Ó .
$ XB=
T Ò] = Ÿ &Ó œ T Ò "/$
Ÿ &Ó œ T ÒXB= Ÿ  "! 68 Þ&Ó œ T ÒXB= Ÿ 'Þ*$Ó
=
œ 'Þ*$ ;B= œ "  'Þ*$ :B œ "  /'Þ*$.B œ "  /'Þ*$‚Þ!# œ Þ"#* .
In a similar way, T Ò] 8= Ÿ &Ó œ T ÒXB8= Ÿ 'Þ*$Ó œ "  /'Þ*$‚Þ!" œ Þ!'( .
Then T Ò] Ÿ &Ó œ Þ$ ‚ Þ"#*  Þ( ‚ Þ!'( œ Þ!)&' .
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MLC - PROBLEM SET 4
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MLC - MULTIPLE CHOICE TEST 6
S. BROVERMAN MLC STUDY GUIDE
MULTIPLE CHOICE TEST 6
1. You are given:
B,
(i) The survival function for males is W! ÐBÑ œ "  =
!  B  =.
>
(ii) Female mortality follows the survival model W! Ð>Ñ œ "  )&
(De Moivre).
(iii) At age 60, the female force of mortality is 60% of the male force of mortality.
For two independent lives, a male age 65 and a female age 60, calculate the expected time until
the second death.
A) 4.33
B) 5.63
C) 7.23
D) 11.88
E) 13.17
2. Mr. Ucci has only 3 hairs left on his head and he won’t be growing any more.
(i) The future mortality of each hair follows 5l ;B œ !Þ"Ð5 "Ñß 5 œ !ß "ß #ß $
and B is Mr. Ucci’s age
(ii) Hair loss follows the hyperbolic assumption at fractional ages.
(iii) The future lifetimes of the 3 hairs are independent.
Calculate the probability that Mr. Ucci is bald (has no hair left) at age B  #Þ&Þ
A) 0.098
B) 0.103
C) 0.108
D) 0.113
E) 0.118
3. A government creates a fund to pay this year’s lottery winners.
You are given:
(i) There are 100 winners each age 40.
(ii) Each winner receives payments of 10 per year for life, payable annually, beginning
immediately.
(iii) Mortality follows the Illustrative Life Table.
(iv) The lifetimes are independent.
(v) 3 œ !Þ!'Þ
(vi) The amount of the fund is 15,092.
Using the normal approximation, find the probability that the fund is sufficient to make all
payments.
A) 82%
B) 85%
C) 88%
D) 91%
E) 95%
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MCT-67
MLC - MULTIPLE CHOICE TEST 6
4. You are given:
(i) TB œ !Þ!*! (ii) TBÀ8l
œ Þ!)&
"
Calculate 8 ZB Þ
A) 0.50
B) 0.52
(iii) TBÀ8l" œ !Þ!!)'%
C) 0.54
D) 0.56
E) 0.58
5. Don, age 50, is an actuarial science professor. His career is subject to two decrements:
(i) Decrement 1 is mortality. The associated single decrement table follows the model
>
W! Ð>Ñ œ "  "!!
(DeMoivre).
Ð#Ñ
(ii) Decrement 2 is leaving academic employment, with .&!> œ !Þ!&ß > !
Calculate the probability that Don dies between ages 55 and 60 while still employed as a
professor.
A) 0.05
B) 0.06
C) 0.07
D) 0.08
E) 0.09
6. For a 20-year term life insurance on ÐBÑ, you are given:
• 3œ!
Ð"Ñ
>
• .B> œ #!
, the force of mortality due to accident
>
• .B> œ "!
, the force of mortality due to other causes
• The benefit is paid at the moment of death.
• The benefit of 2 is paid if death occurs by accident, and a benefit of 1 is paid if death occurs by
other causes.
Calculate the actuarial present value of this insurance.
A) #$ Ð"  /#! Ñ
B) "  /#!
C) "  /$!
D) #$ Ð"  /$! Ñ
E) %$ Ð"  /$! Ñ
Ð#Ñ
7. Lee, age 63, considers the purchase of a single premium whole life insurance of 10,000 with
death benefit payable at the end of the year of death.
The company calculates net premiums using:
(i) Mortality based on the Illustrative Life Table,
(ii) 3 œ !Þ!&
The company calculates contract premiums as 112% of net premiums. The single contract
premium at age 63 is U. Lee decides to delay the purchase for two years and invests the U. The
annual rate of return that the investment earns is 3.0% and two years later it provides exactly the
amount equal to the single contract premium at age 65. Calculate U.
A) Less than 5000
B) At least 5000 but less than 5100
C) At least 5100 but less than 5200
D) At least 5200 but less than 5300
E) At least 5300
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MLC - MULTIPLE CHOICE TEST 6
8. For a special fully discrete 20-year endowment insurance on (55):
(i) Death benefits in year 5 are given by ,5 œ Ð#"5Ñß 5 œ "ß #ß á ß #!Þ
(ii) The maturity benefit is 1.
(iii) Annual net premiums are level.
(iv) 5 Z denotes the net reserve at the end of year 5ß 5 œ "ß #ß á ß #!Þ
(v) "! Z œ &Þ!
(vi) "* Z œ !Þ'
(vii) "" Z œ &Þ$
(viii) 3 œ !Þ!)
Calculate ;&& .
A) Less than .06
B) At least .06 but less than .07
C) At least .07 but less than .08
D) At least .08 but less than .09
E) At least .09
9. Mortality follows Gompertz's law with F œ !Þ!!!" and -  !.
You are given that for a pair of independent lives both aged B, the joint life mortality random
variable XBB has a distribution that is identical to that of XB% , the single life mortality random
variable an individual aged B  %Þ Determine the value of :&! .
A) Less than .50
B) At least .50 but less than .60
C) At least .60 but less than .70
C) At least .70 but less than .80
C) At least .80
10. For a special fully discrete 35-payment whole life insurance on (30):
(i) The death benefit is G for the first 20 years and is 5 thereafter.
(ii) The initial net premium paid during the each of the first 20 years is one fifth of the net
premium paid during each of the 15 subsequent years.
(iii) Mortality follows the Illustrative Life Table.
(iv) 3 œ !Þ!'Þ
(v) E$!À#!l œ 0.32307
(vi) +̈$!À$&l œ 14.835
(vii) The initial annual net premium is .016.
Calculate G .
A) 1.5
B) 1.7
C) 1.5
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E) 1.9
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MCT-69
MLC - MULTIPLE CHOICE TEST 6
11. The Boiler Room Sales company ranks each of its sales people at the end of each month
according to the sales made by that sales person. There are three rankings:
Excellent , Good , Poor.
A salesperson who had poor sales for the month is immediately fired and never again rehired.
Some sales people die during the month (perhaps because of the high pressure under which
they have to work). The BRS company has created a homogeneous Markov Chain model to
describe transitions in a salesperson's ranking from one month to the next. The model has three
states:
I - excellent sales for the month just ended,
K - good sales for the month just ended, and
T - poor sales or died in the month just ended.
The one-step transition matrix is
 :IßI
 :KßI
 :T ßI
:IßK
:KßK
:T ßK
:IßT 
 Þ#

:KßT œ Þ"
:T ßT   !
Þ'
Þ&
!
Þ# 
Þ% .
"
A salesperson has just received a ranking of K.
BRS company pays a salesperson a bonus of $1000 whenever that person has two consecutive
months with a ranking of I . At a monthly interest rate of 1%, find the combined actuarial present
values of the bonus payments to be made at the end of the next three months to a salesperson whose
current rating is K, and a salesperson whose current rating is I .
A) Less than 300
B) At least 300 but less than 320
C) At least 320 but less than 340
D) At least 340 but less than 360
E) At least 360
12. Under the assumption of UDD in the multiple decrement table, what is the formulation for
the APV of a disability income benefit of $C per month payable to age 75, issued at age 40,
with a 6-month waiting period, and with coverage expiring at age 60. Assume that disability
occurs at mid-year.
A) "#G  @5 # 5 :%! ;%!5 +¨ %"5À$%5l
"*
"
Ð7 Ñ Ð3Ñ
Ð"#Ñ3
C) "#G  @5" 5 :%! ;%!5 +¨ %"5À$%5l
5œ!
"*
Ð7 Ñ Ð3Ñ
E) "#G  @5" 5 :%! ;%!5 +¨
5œ!
"*
5œ!
MCT-70
Ð7 Ñ Ð3Ñ
Ð"#Ñ3
B) G  @5 # 5 :%! ;%!5 +¨ %"5À$%5l
"*
"
Ð7 Ñ Ð3Ñ
Ð"#Ñ3
D) G  @5" 5 :%! ;%!5 +¨ %"5À$%5l
5œ!
"*
Ð7 Ñ Ð3Ñ
Ð"#Ñ3
5œ!
Ð"#Ñ3
%"5À$% "# 5l
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MLC - MULTIPLE CHOICE TEST 6
13. You have calculated the actuarial present value of a last-survivor whole life insurance of 1
on ÐBÑ and ÐCÑ. You assumed:
(i) The death benefit is payable at the moment of death.
(ii) You assume that the future lifetimes of ÐBÑ and ÐCÑ follow a common shock model in which
each life has a constant force of mortality with . œ !Þ!' and with a common shock component
with constant force 0.02.
(iii) $ œ !Þ!&.
Your supervisor points out that these are independent future lifetimes, and each mortality
assumption for ÐBÑ and for ÐCÑ is correct.
Calculate the change in the actuarial present value over what you originally calculated.
A) .039 larger
B) .039 smaller
C) .093 larger
D) .093 smaller
E) no change
14. A decreasing term life insurance on (80) pays (20  5 ) at the end of the year of death if (80)
dies in year 51, for 5 œ 0, 1, 2, á , 19.
You are given:
(i) 3 œ 0.06
(ii) For a certain mortality table with ;80 œ 0.2, the single net premium for this insurance is 13.
(iii) For this same mortality table, except that ;80 œ G , the single net premium for this insurance
is "#Þ$!.
Calculate G .
A) Less than .10
B) At least .10 but less than .11
C) At least .11 but less than .12
D) At least .12 but less than .13
E) At least .13
15. Mortality for Audra, 25 years old, follows a constant force of mortality of . œ Þ!" at all
ages. If she takes up hot air ballooning for the coming year, her assumed mortality will be
adjusted so that for the coming year only, she will have a mortality probability of 0.1 based on
uniform distribution of death with that year. Calculate the decrease in the 11-year temporary
complete life expectancy for Audra if she takes up hot air ballooning.
A) 0.60
B) 0.70
C) 0.80
D) 0.90
E) 1.00
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MCT-71
MLC - MULTIPLE CHOICE TEST 6
16. For a special 20-year term life insurance on Ð%!Ñ, you are given:
(i) The death and endowment benefits are 10,000.
(ii) The death benefit is payable at the moment of death.
(iii) During the 5th year the gross premium is 750 paid continuously at a constant rate
(iv) The force of mortality follows Gompertz's law with F œ !Þ!!!!% and - œ "Þ"
(v) The force of interest is 4%.
(vi) Expenses are 5% of premium payable continuously and 100 payable at the moment of death.
(vii) At the end of the 5th year the expected value of the present value of future losses
random variable is 3000.
Euler's method with steps of 2 œ !Þ#& years is used to calculate a numerical solution to Thiele's
differential equation. Calculate the expected value of the present value of future losses random
variable at the end of 4.5 years.
A) Less than 2450
B) At least 2450 but less than 2550
C) At least 2550 but less than 2650
D) At least 2650 but less than 2750
E) At least 2750
"7. ÐBÑ and ÐCÑ are two lives with identical expected mortality. You are given:
TB œ TC œ !Þ" ,
TBC œ !Þ"), where TBC is the annual net premium for a fully discrete insurance of 1 on ÐBCÑ.
. œ !Þ!'
Calculate the premium TBC , the annual net premium for a fully discrete insurance of 1 on ÐBCÑ.
A) 0.04
B) 0.06
C) 0.08
D) 0.10
E) 0.12
18. For a multiple decrement model on (60):
Ð"Ñ
(i) .'!> , > !, follows the Illustrative Life Table.
(ii)
Ð7 Ñ
Ð"Ñ
.'!> œ - † .'!> ,
> !
Ð7 Ñ
"! ;'!
(iii)
œ Þ&($(#
Calculate - .
A) 2.0
B) 2.5
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C) 3.0
D) 3.5
E) 4.0
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MLC - MULTIPLE CHOICE TEST 6
19. For a special fully discrete 3-year term insurance on ÐBÑ:
(i) Level net premiums are paid at the beginning of each year.
(ii)
5
,5"
;B5
0 200,000 0.03
1 150,000 0.06
2 100,000 0.09
(iii) 3 œ !Þ!'
Calculate the initial net reserve for year 3 (nearest 100).
A) 6,500
B) 7,500
C) 8,500
D) 9,500
E) 10,500
20. For a special fully discrete whole life insurance on ÐBÑ:
(i) The level premium is determined using the equivalence principle.
(ii) Death benefits are given by ,5" œ Ð"3Ñ5" where 3 is the interest rate.
(iii) P is the loss random variable at > œ ! for the insurance.
(iv) O is the future curtate lifetime random variable of ÐBÑÞ
Which of the following expressions is equal to P?
Ð@O" EB Ñ
Ð@O" EB Ñ
A)
B) Ð@O"  EBÑÐ" EBÑ
C)
Ð" EB Ñ
Ð" EB Ñ
D) Ð@O"  EBÑÐ" EBÑ
E)
Ð@O"EB Ñ
Ð"EB Ñ
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MCT-73
MLC - MULTIPLE CHOICE TEST 6
S. BROVERMAN MLC STUDY GUIDE
MULTIPLE CHOICE TEST 6 SOLUTIONS
1. The male survival distribution is De Moivre's Law with = . The force of mortality for a male
"
at age 60 is ='!
. Female mortality follows De Moivre's Law with upper age limit 85.
"
"
"
Female force of mortality at age 60 is )&'!
œ #&
œ Þ'Ð ='!
Ñ p = œ (& .
"!
(&'&
)&'!
/° '&Q À'!J œ /° '&Q  /° '!J  /° '&Q À'!J œ

 ! Ð> :'&Q ÑÐ> :'!J Ñ .> Þ
#
#
)&'!>
#&>
œ
ß > :'!J œ )&'! œ #&
> :'&Q œ
"!
"!
#&>
"
#
 "!
p ! Ð> :'&Q ÑÐ> :'!J Ñ .> œ ! Ð "!>
"! ÑÐ #& Ñ .> œ Ð"!ÑÐ#&Ñ ! Ð#&!  $&>  > Ñ .>
"
"!!!
°
œ Ð"!ÑÐ#&Ñ
Ò#&!!  $&Ð "!!
# Ñ  $ Ó œ %Þ$$$ p /'&Q À'!J œ &  "#Þ&  %Þ$$$ œ "$Þ"'
(&'&>
(&'&
"!>
"!
.
Answer: E
2. For a particular hair, the probability that this hair has been lost by time 2.5 is #Þ& ;B . The
probability of being bald by time 2.5 is T Ò(lose hair 1) ∩ (lose hair 2) ∩ (lose hair 3)Ó .
Since the hairs are independent of one another, this probability is
T Ò(lose hair 1)Ó † T Ò(lose hair 2)Ó † T Ò(lose hair 3)Ó œ Ð#Þ& ;B Ñ$
(we use the rule for independent events E and F that says T ÒE ∩ FÓ œ T ÒEÓ † T ÒFÓ ).
From the given information we have ;B œ !l ;B œ Þ" ß
# ;B
œ ;B  "l ;B œ Þ"  Þ# œ Þ$ ß $ ;B œ ;B  "l ;B  #l ;B œ Þ' .
#l ;B
Then # :B œ Þ( so that ;B# œ : œ $( .
# B
#Þ& ;B œ "  #Þ& :B œ "  # :B † Þ& :B# œ "  ÐÞ(ÑÐ"  Þ& ;B# Ñ .
Þ&†;B#
From the hyperbolic assumption, we have Þ& ;B# œ "Ð"Þ&ц;
B#
Then,
#Þ& ;B
œ Þ#(#( .
œ œ "  ÐÞ(ÑÐ"  Þ#(#(Ñ œ Þ%*" , and Ð#Þ& ;B Ñ$ œ ÐÞ%*Ñ$ œ Þ"") .
Answer: E
3. This problem involves the variance of a life annuity present value random variable and the
normal approximation. The expected present value of payment to the winners is
ÞÞ
IÒWÓ œ Ð"!!ÑÐ"!ÑÐ+%! Ñ œ "%ß )"'Þ' (for 100 annuitants). Since the prize winners are
independent, the combined variance of the present value random variables is the sum of the
variances of the separate present value random variables. For one prize winner, the variance of
the present value random variable is
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MLC - MULTIPLE CHOICE TEST 6
Z +<Ò"!] Ó œ "!!Z +<Ò] Ó œ "!!Ð ."# ÑÒ# E%!  ÐE%! Ñ# Ó œ (!&Þ&& .
The variance of the total present value paid to the winners is Z +<ÒWÓ œ "!!Ð(!&Þ&&Ñ œ (!ß &&&Þ
We assume that W is approximately normal to find the probability that T ÒW Ÿ "&ß !*#Ó .
We standardize the probability to get
T ÒW Ÿ "&ß !*#Ó œ T Ò 
WIÒWÓ
Z +<ÒWÓ
Ÿ
"&ß!*#IÒWÓ
Z +<ÒWÓ Ó
œ T Ò^ Ÿ
"&ß!*#"%ß)"(
(!ß&&& Ó
œ FÐ"Þ!$&Ñ œ Þ)& (from the normal table). Answer: B
4. 8 ZB œ
TB TBÀ8l
"
T
"
BÀ8l
5. We wish to find
p 8 ZB œ Þ!*Þ!)&
Þ!!)'% œ Þ&()(. Answer: E
Ð"Ñ
&l& ;&!
œ & :&! ‚ & :&& œ & :&! ‚ & :&! ‚ ! > :&& † .&&> .> Þ
Ð 7Ñ
Ð"Ñ
Under De Moivre's Law with = œ "!! we have
wÐ"Ñ
wÐ"Ñ
& :&!
wÐ#Ñ
&
wÐ#Ñ
Ð7 Ñ
Ð"Ñ
œ "!!&!&
"!!&! œ Þ* and
and under constant force of decrement & :&! œ /&ÐÞ!&Ñ œ /Þ#& .
wÐ"Ñ
Ð"Ñ
"
Under De Moivre's Law we also have > :&& œ "!!&&>
"!!&& and .&&> œ "!!&&> , so
Ð7 Ñ
Ð"Ñ
wÐ"Ñ
wÐ#Ñ
Ð"Ñ
"!!&&>
"
Þ!&>
‚ "!!&&>
> :&& † .&&> œ > :&& ‚ > :&& † .&&> œ "!!&& ‚ /
Þ!&>
œ / %& .
Þ#&
& Ð7 Ñ
& Þ!&>
Ð"Ñ
Ð"Ñ
Then & :&& œ ! > :&& † .&&> .> œ ! / %& .> œ "/
Þ!&‚%& .
Þ#&
Ð"Ñ
Ð 7Ñ
Ð"Ñ
Finally, &l& ;&! œ & :&! ‚ & :&& œ Þ* ‚ /Þ#& ‚ "/
Þ!&‚%& œ Þ!')* . Answer: C
6. 8 :B œ /B:Ò  ! .B> .>Ó œ /B:Ò  ! Þ"&> .>Ó œ /Þ!(&8 .
#
#!
#!
Ð7 Ñ
Ð"Ñ
APV of accidental death benefit is #! > :B † .B> .> œ #! /Þ!(&> †
Ð7 Ñ
8 Ð7 Ñ
8
#
>
#!
.>
Using the substitution, = œ Þ!(&># , .= œ Þ"&> .> , and the integral becomes
 /=
"
#!ÐÞ"&Ñ
>œ#!
#
.= , so the APV is # ‚ "$ Ð  /Þ!(&> 
Ñ œ #$ Ð"  /$! Ñ .
>œ!
The APV of the benefit for death due to other causes is
#
#!
Ð#Ñ
!#! > :BÐ7 Ñ † .B>
.> œ ! /Þ!(&> † > .> . We see that this integral is twice as large as
!#! /Þ!(&># † > .> , so !#! /Þ!(&># † > .> œ # Ð"  /$! Ñ .
#!
"!
$
%
$!
The total APV is $ Ð"  / Ñ .
Answer: E
"!
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MCT-75
MLC - MULTIPLE CHOICE TEST 6
7. This problem involves single net premium for life insurance with death benefit payable at the
end of the year of death, and it also involves the recursive relationship for insurance valuation.
The accumulated value of the U invested for 2 years at 3.0% is Ð"Þ!$Ñ# U. The single net
premium for the insurance at age 65 (2 years after age 63) is "!ß !!!E'& , and the contract
premium is Ð"Þ"#ÑÐ"!ß !!!ÑE'& . We cannot use the EB values in the Illustrative Life Table,
since they are based on an interest rate of 6%. The single net premium at age 63 is
U œ "!ß !!!E'$ . We use the relationship EB œ @;B  @# † "l ;B  @# † # :B † EB# . From the
Illustrative Life Table, we have ;'$ œ Þ!"())ß ;'% œ Þ!"*&# . The single net premium at age 63
U
is "Þ"# Þ Therefore,
U
"!ß!!!
"!ß!!!
"!ß!!!
"Þ"# œ "Þ!& † ÐÞ!"())Ñ  Ð"Þ!&Ñ# † ÐÞ*)#"#ÑÐÞ!"*&#Ñ  Ð"Þ!&Ñ# † ÐÞ*)#"#ÑÐÞ*)!%)Ñ † E'&
p "!ß !!!E'& œ "Þ!###&!U  $*%Þ!&
The contract premium for the insurance at age 65 will be
Ð"Þ"#ÑÐ"!ß !!!ÑE'& œ Ð"Þ"#ÑÐ"Þ!###&!U  $*%Þ!&Ñ œ Ð"Þ!$Ñ# U Þ
Solving for U results in U œ &#&# .
Answer: D
8. Since we are given "! Z we use the relationship
Ð"! Z  T ÑÐ"  3Ñ  ,"" † ;B"! œ :B"! † "" Z ,
where T is the annual net premium and ,"" is the death benefit payable at the end of the 11th
year for death in the 19-th year. The insurance is a 20-year decreasing insurance, so that
,"" œ #"  "" œ "! . Also, the issue age is B œ &&, so that the equation becomes
Ð&Þ!  T ÑÐ"Þ!)Ñ  "!;'& œ :'& † "" Z , which is Ð&Þ!  T ÑÐ"Þ!)Ñ  "!;'& œ :'& Ð&Þ$Ñ .
In order to find ;'& we need to know T . We can find T by using the recursive relationship for
the final (20th) year: Ð"* Z  T ÑÐ"  3Ñ  ,#! † ;B"* œ :B"* † #! Z .
When a policy matures, the maturity amount is the reserve at the end of the policy term.
Therefore, #! Z œ " , since we are told that this is a 20-year policy with maturity value 1.
The death benefit is also ,#! œ " in the 20th year. The recursive relationship becomes
Ð!Þ'  T ÑÐ"Þ!)Ñ  ;'% œ :'% † " , which becomes ÐÞ'  T ÑÐ"Þ!)Ñ œ " , since
;B  :B œ " for any B. We then get T œ Þ$#&* , and then ;'& œ Þ"! . Answer: E
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MLC - MULTIPLE CHOICE TEST 6
9. .B œ F- B p .BÀB œ .B  .B œ F- B  F- B œ .B% œ F- B% .
It follows that # œ - % , from which we get - œ "Þ")*# .
Under Gompertz law,
"
"
:&! œ /B:Ò  ! .&!> .>Ó œ /B:Ò  ! Þ!!!" † "Þ")*#&!> .>Ó
&"
"Þ")*#&!
œ /B:Ò  Þ!!!" † "Þ")*#
Ó œ Þ&$ .
68 "Þ")*#
Answer: B
"0. This problem involves the equivalence principle for finding a net premium.
For the first 20 years, the net premium is T , and for the next 15 years it is &T . The death benefit
is G for the first 20 years and then 5 after that. The equivalence principle equation is
ÞÞ
ÞÞ
ÐÞ!"'ÑÒ+$!À#!l  5@#! † #! :$! † +&!À"&l Ó œ GE "
 &@#! † #! :$! † E&! .
$!À#!l
ÞÞ
ÞÞ
ÞÞ
We use the relationships @ † #! :$! † +&!À"&l œ +$!À$&l  +$!À#!l , and
"E$!À#!l
ÞÞ
+$!À#!l œ
, and E "
œ E$!À#!l  @#! † #! :$! .
.
#!
$!À#!l
From the Illustrated Life Table, we get @#! † #! :$! œ Þ#*$(% .
"E$!À#!l
ÞÞ
Then, +$!À#!l œ
œ ""Þ*&* ß
.
ÞÞ
ÞÞ
ÞÞ
@#! † #! :$! † +&!À"&l œ +$!À$&l  +$!À#!l œ #Þ)(' ß and
E"
$!À#!l
œ E$!À#!l  @#! † #! :$! œ Þ!#*$ . Then, G œ "Þ* .
Answer: E
11. The following sequences of rankings for the next three months result in two consecutive
I -I ratings (including the current ranking of K):
Sequence that pay bonus at time 2: K  I  I , prob. ÐÞ"ÑÐÞ#Ñ œ Þ!2
Sequences that pay bonus at time 3:
K  K  I  I , prob. ÐÞ&ÑÐÞ"ÑÐÞ#Ñ œ Þ!" , K  I  I  I , prob. ÐÞ"ÑÐÞ#ÑÐÞ#Ñ œ Þ!!% ,
total prob. of bonus payment at time 3 is .014.
Actuarial present value is "!!!ÒÞ!#@#  Þ!"%@$ Ó œ $$Þ"* .
The following sequences of rankings for the next three months result in two consecutive
I -I ratings (including the current ranking of I ):
Sequence that pays bonus at time 1: I  I , prob. .2
Sequence that pay bonus at time 2: I  I  I , prob. ÐÞ#ÑÐÞ#Ñ œ Þ!%
Sequences that pay bonus at time 3:
I  K  I  I , prob. ÐÞ'ÑÐÞ"ÑÐÞ#Ñ œ Þ!"# , I  I  I  I , prob. ÐÞ#ÑÐÞ#ÑÐÞ#Ñ œ Þ!!) ,
total prob. of bonus payment at time 3 is .0#.
Actuarial present value is "!!!ÒÞ#@  Þ!%@#  Þ!#@$ Ó œ #&'Þ'% .
Total APV is 289.82 .
Answer: A
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MCT-77
MLC - MULTIPLE CHOICE TEST 6
12. The annual rate of payment is "#G , so answers B and D are incorrect. Since disability
occurs at mid-year of age, and it begins "# -year later, at the next year of age and continues to one
month before age 75, and the valuation is on age 40. Answer: C
13. This problem involves the common shock model for dependence of two lives.


Þ!'
'
The insurance single net premium for each of ÐBÑ and ÐCÑ is EB œ EC œ Þ!'Þ!&
œ ""
.
Since .B œ Þ!' œ .‡B  - and - œ Þ!# , it follows that ÐBÑ's force of mortality is .‡B œ Þ!%
before common shock, and for ÐCÑ it is similarly true that .‡C œ Þ!%. The force of failure for the
joint life status is .‡B  .C‡  - œ Þ!%  Þ!%  Þ!# œ Þ"! , so the single net premium for
Þ"!
a joint life insurance is EBC œ Þ"!Þ!&
œ "!
"& . Then, the single net premium for the last survivor
insurance, under the original common shock model is




'
'
EBC œ EC  EC  EBC œ ""
 ""
 "!
"& œ Þ%#% .
If it assumed that the lives are independent with .B œ .C œ Þ!' , then the force of failure for the
joint life status is .B  .C œ Þ"# , and the joint life insurance net premium is
Þ"#
EBC œ Þ"#Þ!&
œ "#
"( . Then the single net premium for the last survivor insurance, under the




'
'
independence model is EBC œ EC  EC  EBC œ ""
 ""
 "#
"( œ Þ$)& .
The new value is Þ%#%  Þ$)& œ Þ!$* smaller than the original value.
14. The APV at age 80 is ÐHEÑ
"
)!À#!l
œ #!@;)!  @:)! † ÐHEÑ
"
)"À"*l
We are given that 3 œ Þ!' , ;)! œ Þ# and ÐHEÑ
"
)!À#!l
Therefore, ÐHEÑ
"
)"À"*l
œ
"$#!@ÐÞ#Ñ
@ÐÞ)Ñ
If ;)! is changed to G , then ÐHEÑ
the premium ÐHEÑ8/A
"
Answer: B
.
œ "$ .
œ "#Þ##& .
"
)"À"*l
œ 12.225 remains unchanged, and the revised value of
is #!@G  @Ð"  GÑ † Ð"#Þ##&Ñ œ "#Þ30 .
)!À#!l
Solving for G results in G œ Þ"!%' .
MCT-78
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Answer: B
SOA Exam MLC Study Guide © S. Broverman, 2015
MLC - MULTIPLE CHOICE TEST 6
15. If she does not take up hot-air ballooning, Audra's 11-year temporary complete life
""
""
expectancy will be /° #&À""l œ ! > :#& .> œ ! /Þ!"> .> œ "!Þ%"'' . The expectation can also be
written as
"
"!
/° #&À""l œ /° #&À"l  :#& † /° #'À"!l œ ! > :#& .>  :#& † ! /Þ!"> .>
"
œ ! > :#& .>  :#& † Ð*Þ&"'$Ñ.
If Audra does take up hot-air ballooning for one year (year of age 25), her survival probability
for year of age 25 will be > :B‡ œ "  Þ"> for ! Ÿ >  " (UDD for the year).
Her revised 11-year temporary complete life expectancy will be
‡
‡
‡
" ‡
‡
‡
/° #&À""l œ /° #&À"l  :#&
† /° #'À"!l œ ! > :#&
.>  :#&
† Ð*Þ&"'$Ñ
œ Þ*&  ÐÞ*ÑÐ*Þ&"'$Ñ œ *Þ&"%'( .
Her expectation would decrease "!Þ%"''  *Þ&"%'( œ Þ*! .
Answer: D
16. From Gompertz's law we have .%& œ F- %& œ Þ!!!!% ‚ "Þ"%& œ Þ!!#*"' , and
.%%Þ(& œ !!#)%(. Then

%Þ(& Z œ $!!!  Þ#&Ð(&!ÐÞ*&Ñ  Þ!%Ð$!!!Ñ  Þ!!#*"'Ð#!ß "!!  $!!!ÑÑ œ #)!%Þ$% ,
and

%Þ& Z œ #)!%Þ$%  Þ#&Ð(&!ÐÞ*&Ñ  Þ!%Ð#)!%Þ$%Ñ  Þ!!#)%(Ð#!ß "!!  #)!%Þ$%Ñ œ #'"!Þ%)
Answer: C .
E
"7. We wish to find TBC œ ÞÞ+ BC œ
BC
TBC . From TB œ
"
ÞÞ
+B
ÞÞ
".+BC
ÞÞ
+BC
ÞÞ
œ ÞÞ+"  . , so if we can find +BC , we can find
BC
ÞÞ
"
 . , we get +B œ T ". œ Þ"'
, and also
B
ÞÞ
E
".+
ÞÞ
"
+C œ T ". œ Þ"'
. Also, TBC œ ÞÞ+ BC œ ÞÞ+ BC œ ÞÞ+"  . , so that
C
BC
BC
BC
ÞÞ
"
"
+BC œ T . œ Þ#% Þ
BC
ÞÞ
ÞÞ
ÞÞ
ÞÞ
ÞÞ
"
"
"
However, since +BC œ +B  +C  +BC , we get +BC œ Þ"'
 Þ"'
 Þ#%
œ )Þ$$$
"
Then, TBC œ ÞÞ+"  . œ )Þ$$$
 Þ!' œ Þ!' . Answer: B
BC
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MCT-79
MLC - MULTIPLE CHOICE TEST 6
18. This problem relies on the following relationship 8 :B œ /! .B> .> .
Ð"Ñ
Since we are told that .'!> follows the Illustrative Life Table, it follows that
Ð7 Ñ
8 Ð7 Ñ
/! .'!> .> œ 8 :'! (from the Illustrative Life Table).
8 Ð"Ñ
Ð7 Ñ
Ð"Ñ
Then, since .'!> œ - † .'!> , we get
Ð7 Ñ
8 :B
œ /! .B> .> œ /! - .'!> .> œ /- ! .'!> .>
8 Ð7 Ñ
8
8 Ð"Ñ
Ð"Ñ
œ Ò/! .'!> .> Ó- œ Ð8 :'! Ñ- (from the Illustrative Life Table).
8 Ð"Ñ
From the Illustrative Table we have
"! :'!
'ß'"'ß"&&
œ jj(! œ )ß"))ß!(% œ Þ)!)!#$ .
'!
It follows that "  Þ&($(# œ Þ%#'#) œ ÐÞ)!)!#$Ñ- , and solving for - results in
68ÐÞ%#'#)Ñ
- œ 68ÐÞ)!)!#$Ñ œ %Þ! .
Answer: E
19. This problem can be solved with the accumulation relationship for net reserves.
The initial net reserve for a particular year is the terminal reserve for the previous year just ended
plus the annual net premium:
year >  " initial net reserve œ > Z  net premium .
Ð# Z  T ÑÐ"  3Ñ  ,$ † ;B# œ :B# † $ Z , and the initial net reserve for year 3 is
# Z  T . It is always the case that $ Z œ ! for a 3-year term insurance, and for this problem,
3 œ Þ!'ß ;B# œ Þ!$ß and ,$ œ "!!ß !!! .
Then Ð# Z  T ÑÐ"  Þ!'Ñ  "!!ß !!!ÐÞ!*Ñ œ ! , so that # Z  T œ )ß %*" . Answer: C
20. This problem involves the loss-at-issue random variable. The loss-at-issue is
P œ present value random variable of benefit
 present value random value random variable of premiums .
The PVRV of benefit is ,O" † @O" œ Ð"  3ÑO" @O" œ " and the PVRV of premium is
ÞÞ
U+O"l , where U is the equivalence principle premium. U is found from the equivalence
ÞÞ
principle equation APV benefit œ APV premium . This equation is IÒ"Ó œ U † +B , so that
U œ ÞÞ+" . The loss random variable becomes
B
ÞÞ
O"
+O"l
Ð"@O" ÑÎ.
ÞÞ
EB
P œ "  U+X l œ "  ÞÞ+ œ "  Ð"E ÑÎ. œ @ "E
. Answer: A
B
MCT-80
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B
B
SOA Exam MLC Study Guide © S. Broverman, 2015