S. BROVERMAN STUDY GUIDE FOR THE SOCIETY … 3MLC-BRO-12SSM-P... SOA Exam MLC Study Guide © S....
Transcript of S. BROVERMAN STUDY GUIDE FOR THE SOCIETY … 3MLC-BRO-12SSM-P... SOA Exam MLC Study Guide © S....
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S. BROVERMAN STUDY GUIDE
FOR THE
SOCIETY OF ACTUARIESEXAM MLC
2012 EDITION
EXCERPTS
Samuel Broverman, ASA, PHD
copyright © 2012, S. Broverman
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Excerpts:
Table of Contents
Introductory Note
Section 5 - Models for Survival and Mortality
Practice Exam 1 and Solutions
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S. BROVERMAN EXAM MLC STUDY GUIDE - VOLUME 1NOTES, EXAMPLES AND PROBLEM SETS
Introductory Note
TABLE OF CONTENTS
Section 1 - Review of Preliminary Topics 1 to 14
Problem Set 1 - Review of Preliminary Topics - Section 1 15 to 24
Section 2 - Discrete Time Markov Chains - One-Step Transitions 25 to 32
Section 3 - Discrete Time Markov Chains - Multi-Step Transitions 33 to 38
Section 4 - Continuous Time Markov Chains 39 to 42
Problem Set 2 - Markov Chains -Sections 2 to 4 43 to 46
Section 5 - Models for Survival and Mortality 47 to 54
Section 6 - The Force of Mortality 55 to 62
Section 7 - The Life Table 63 to 68
Section 8 - Mean and Variance of and 69 to 78X OB B
Section 9 - Parametric Survival Models 79 to 84
Section 10 - Fractional Age Assumptions 85 to 92
Section 11 - Select and Ultimate Mortality 93 to 100
Problem Set 3 - Survival and Mortality - Sections 5 to 11 101 to 124
Section 12 - One-Year Term Insurance Payable at the End of the Year of Death 125 to130
Section 13 - Term and Whole Life Insurance
Payable at the End of the Year of Death 131 to138
Section 14 - Pure Endowment and Other Life Insurances
Payable at the End of the Year of Death 139 to148
Section 15 - Insurance Payable at the Moment of Death 149 to 162
Section 16 - Additional Insurance Relationships 163 to 178
Problem Set 4 - Life Insurance - Sections 12 to 16 179 to 206
Section 17 - Discrete Whole Life Annuity-Due 203 to 208
Section 18 - Discrete Life Annuities 213 to 224
Section 19 - Continuous Life Annuities 225 to 234
Section 20 - Additional Annuity Relationships 235 to 242
Problem Set 5 - Life Annuities - Sections 17 to 20 243 to 262
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TABLE OF CONTENTS
Section 21 - The Loss at Issue Random Variable and
Principles of Premium Calculation 263 to 270
Section 22 - Equivalence Principle Premiums 271 to 284
Section 23 - Policy Expenses 285 to 290
Problem Set 6 - Annual Premiums - Sections 21 to 23 291 to 318
Section 24 - Introduction to Benefit Reserves 319 to 324
Section 25 - Prospective Form of Benefit Reserves 325 to 330
Section 26 - Additional Representations for Benefit Reserves 331 to 340
Section 27 - Reserves on Additional Policy Types 341 to 346
Section 28 - Expense Augmented Reserves 347 to 356
Section 29 - Recursion Relationships for Reserves
and Modified Reserves 357 to 370
Problem Set 7 - Reserves - Sections 24 to 29 371 to 394
Section 30 - The Joint Life Status 395 to 402
Section 31 - The Last-Survivor Status and the Common Shock Model 403 to 410
Section 32 - Multiple Life Insurances and Annuities 411 to 422
Section 33 - Contingent Probabilities and Insurances 423 to 432
Problem Set 8 - Multiple Life - Sections 30 to 33 433 to 452
Section 34 - Multiple Decrement Models 453 to 464
Section 35 - Associated Single Decrement Tables 463 to 478
Section 36 - Valuation of Multiple Decrement Benefits 479 to 480
Section 37 - Asset Shares, Non-Forfeiture Values and More General
Multi-State Models, Policy Profit Analysis 481 to 494
Problem Set 9 - Multiple Decrement Models - Sections 34 to 37 495 to 518
Section 38 - Variable Interest Rates 519 to 526
Problem Set 10 - Variable Interest Rates - Section 38 527 to 534
ILLUSTRATIVE LIFE TABLE
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SOA EXAM MLC STUDY GUIDE - VOLUME 2PRACTICE EXAMS
TABLE OF CONTENTS
PRACTICE EXAMS
Practice Exam 1 and Solutions PE-1 to PE-20
Practice Exam 2 and Solutions PE-21 to PE-40
Practice Exam 3 and Solutions PE-41 to PE-60
Practice Exam 4 and Solutions PE-61 to PE-80
Practice Exam 5 and Solutions PE-81 to PE-102
Practice Exam 6 and Solutions PE-103 to PE-122
Practice Exam 7 and Solutions PE-123 to PE-140
Practice Exam 8 and Solutions PE-141 to PE-164
Practice Exam 9 and Solutions PE-165 to PE-186
Practice Exam 10 and Solutions PE-187 to PE-206
Practice Exam 11 and Solutions PE-207 to PE-228
Practice Exam 12 and Solutions PE-229 to PE-250
ILLUSTRATIVE LIFE TABLE
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INTRODUCTORY NOTE
This study guide is designed to help in the preparation for Exam MLC of the Society ofActuaries (the life contingencies and probability exam).
The study guide is divided into two volumes. Volume 1 consists of review notes, examples andproblem sets. Volume 2 contains 12 practice exams of 30 questions each. There are over 170examples in the notes, over 360 problems in the problem sets and 360 questions in the 12practice exams. All of these (almost 900) questions have detailed solutions. The notes are brokenup into 38 sections Each section has a suggested approximate time frame.
Most of the examples in the notes and about half of the problems in the problem sets are fromolder SOA or CAS exams on the relevant topics. The practice exams have 30 questions each andare designed to be similar to actual 3-hour exams. Some of the questions on the practice examsare variations on actual exam questions. The SOA has posted on its website a sample questionfile for Exam MLC with solutions. Many of those questions are from old exams, and there maybe some overlap in the questions found in this study guide and those found in the SOA files, butI have attempted to limit that duplication. The SOA and CAS questions are copyrighted by theSOA and CAS, and I gratefully acknowledge that I have been permitted to include them in thisstudy guide.
Because of the time constraint on the exam, a crucial aspect of exam taking is the ability to workquickly. I believe that working through many problems and examples is a good way to build upthe speed at which you work. It can also be worthwhile to work through problems that havebeen done before, as this helps to reinforce familiarity, understanding and confidence. Workingmany problems will also help in being able to more quickly identify topic and question types. Ihave attempted, wherever possible, to emphasize shortcuts and efficient and systematic ways ofsetting up solutions. There are also occasional comments on interpretation of the language usedin some exam questions. While the focus of the study guide is on exam preparation, from time totime there will be comments on underlying theory in places that I feel those comments mayprovide useful insight into a topic.
It has been my intention to make this study guide self-contained and comprehensive for all ExamMLC topics, but there are may be occasional references to the books listed in the SOA examcatalog. While the ability to derive formulas used on the exam is usually not the focus of anexam question, it is useful in enhancing the understanding of the material and may be helpful inmemorizing 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 formuladerivations.
In order for the review notes in this study guide to be most effective, you should have somebackground at the junior or senior college level in probability and statistics. It will be assumedthat you are reasonably familiar with differential and integral calculus.
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Of the various calculators that are allowed for use on the exam, I think that theBA II PLUS is probably the best choice. It has several memories and has good financialfunctions. 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 consistof a standard normal distribution probability table and a life table. The tables are available fordownload from the Society of Actuaries website, but are included at the ends of both Volumes 1and 2 for convenience.
If you have any questions, comments, criticisms or compliments regarding this study guide, youmay contact me at the address below. I apologize in advance for any errors, typographical orotherwise, that you might find, and it would be greatly appreciated if you would bring them tomy attention. I will be maintaining a website for errata that can be accessed fromwww.sambroverman.com . It is my sincere hope that you find this study guide helpful and usefulin your preparation for the exam. I wish you the best of luck on the exam.
Samuel A. Broverman February, 2012Department of StatisticsUniversity of Toronto100 St. George StreetToronto, Ontario CANADA M5S 3G3 E-mail: [email protected] or [email protected]: www.sambroverman.com
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MLC SECTION 5 - MODELS FOR SURVIVAL AND MORTALITYMQR CHAPTER 5
The suggested time frame for covering this section is 2 hours.
The Survival Function
A standard way of describing survival and mortality is by using a random variable for the time
from birth until death. When an individual is born, the time until death is a random variable. The
Bowers book defines the following: time until death, or age at death of a newborn.\ œ X œ \!
is a continuous, positive random variable ( ). The of is\ ! \distribution function
J ÐBÑ œ J ÐBÑ œ T<Ò\ Ÿ BÓ œ T ÒX Ÿ BÓ\ ! !
probability that the newborn dies before or on attaining age .œ B
The or of issurvival function survival distribution function \
=ÐBÑ œ "J ÐBÑ œ T<Ò\ BÓ œ W ÐBÑ œ "J ÐBÑ\ ! !
œ Bprobability that the newborn dies after attaining age ,
the probability that the newborn survives to at least age .œ B
If then .B D T<ÒB \ Ÿ DÓ œ J ÐDÑ J ÐBÑ œ =ÐBÑ =ÐDÑ\ \
This is the probability that the newborn dies between ages and .B D
The following assumptions are made regarding the behavior of the cdf and survival function.
(i) and (the newborn is alive with probability 1) .J Ð!Ñ œ ! =Ð!Ñ œ "!
(ii) and (the probability of the newborn surviving forever is 0).lim limBp∞ BÄ∞
!J ÐBÑ œ " =ÐBÑ œ !
(iii) is a non-decreasing function and is a non-increasing function (as increases,J ÐBÑ =ÐBÑ B!
there is an increasing probability that the newborn dies by age ).B
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The upper age limit on a survival distribution is often denoted , which is usually a finite=
number (such as 100 or 110) for human survival, but for illustration purposes, some examples
have .= œ ∞
Time Until Death for ÐBÑ
The notation refers to an individual alive at age .ÐBÑ B
XB denotes the measuring the time until 's death from age continuous random variable ÐBÑ B
( described above is a special case of this with , a newborn). Note that given that a\ œ X B œ !!
newborn survives to age , the future lifetime measured from age is , so that timeB B X œ \ BB
until death from age is age at death minus , and can be regarded as the conditional randomB B XB
variable of additional number of years to be lived after age given that the individual hasB
survived to age . For instance, consider someone alive at age who dies at age 50. ForB B œ %!
this person, death occurs at age , but measured as time until death from age 40 we have\ œ &!
X œ X œ "! œ &! %! œ \ BB %! (of course, since this person was alive at age 40, it is also
true that this person was alive at age 20, so that for this person). When we areX œ $!#!
measuring , time until death from age , it is understood that the person is alive at age , andX B BB
hence can be represented as a conditional random variable in terms of :X \B
T ÒX Ÿ >Ó œ T Ò\ B Ÿ >l\ BÓ œ T Ò\ Ÿ B >l\ BÓB
.œ œ œTÒB\ŸB>Ó J ÐB>ÑJ ÐBÑ W ÐBÑW ÐB>Ñ
T Ò\BÓ "J ÐBÑ W ÐBÑ! ! ! !
! !
We will denote the distribution function of by (instead of XÐBÑ J Ð>Ñ œ T ÒX Ÿ >Ó J Ð>ÑÑßB B XB
which is the probability that ( ) will die within the next years (by age ) ; this probabilityB > B >
is denoted in actuarial notation by
> B B B; œ J Ð>Ñ œ T ÒX Ÿ >Ó
probability that ( ) will die within the next years (by age )œ B > B >
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The complement of is denoted , which is> B > B; :
> B > B B: œ " ; œ T ÒX >Ó
probability that survives at least to time (and dies some time after )œ ÐBÑ > >
By the conventions of actuarial notation, for mortality or survival for a 1 year period, we write
and " B B " B B; œ ; : œ :
It is assumed that (the probability of eventually dying is 1) and∞ B; œ "
∞ B ! B: œ ! ; œ ! ß (the probability of living forever is 0). It is also assumed that . In! B: œ "
the special case that we have and .B œ ! ß \ œ XÐ!Ñ : œ =Ð>Ñ> !
Factorization of Survival Probability and Deferred Mortality Probability
For the probability that survives at least -years can be factored in the following way:8 Ÿ > ß ÐBÑ >
> B 8 B >8 B8: œ : † : .
The verbal interpretation of this factorization is that in order for to survive years, mustÐBÑ > ÐBÑ
survive the first years (event , probability ), and then must survive more years8 E : > 88 B
from age (event , probability ). The probability of surviving years is theB 8 FlE : >>8 B8
event , and we use the probability rule . This rule appliesE ∩F T ÒE ∩ FÓ œ T ÒEÓ † T ÒFlEÓ
for any values of and , integer or fractional.8 >
As a special case, , and it follows that ; this allows us toB8 ! B ! 8 B: œ : † : 8 B: œW ÐB8ÑW ÐBÑ!
!
formulate survival/mortality probabilities from any age if we are given the probability of
survival from birth function .=ÐBÑ
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If is an integer, the survival probability over the -year period can be separated into consecutive> >
one-year periods to get the factorization form .> B B B" B>": œ : † : â:
An important probability in life contingencies is the .deferred mortality probability
The probability that survives years and then dies in the years following that isÐBÑ > ?
>l? B > B ? B> >? B > B > B >? B; œ : † ; œ ; ; œ : :
œ T Ò> X Ÿ > ?Ó œBW ÐB>ÑW ÐB>?Ñ
W ÐBÑ! !
!
This is a -year deferred, year mortality probability; it is the probability that survives > ? ÐBÑ >
years to age and then dies by age . Note that this is not simply equal to ,B > B > ? ;? B>
since must first survive to age and then from age must die in the following ÐBÑ B > B > ?
years. By notational convention, if the mortality period is one year, we write ; this >l" >lB B; œ ;
is the probability that survives years and then dies between ages and .ÐBÑ > B > B > "
B B > B > ?
> B ? B>: p ; p
Also note that , and . The interpretation of the last!l >l?B B >? B > B B; œ ; ; œ ; ;
expression is that in order for to die within years, either death occurs within the first ÐBÑ > ? >
years (probability ) or survives years and dies in the following years (prob. ).> B B>l?; ÐBÑ > ? ;
There is a great deal of actuarial notation that will be developed throughout the life
contingencies material. It is important to become familiar with and comfortable using the
notation. The notation is built up in a systematic way, with specific meanings for subscripts,
superscripts, etc. A good understanding of the underlying concepts will help in recognizing and
memorizing the actuarial notation system.
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Example 12: The following survival function is given: for .W ÐBÑ œ " ! Ÿ B Ÿ "! œ!B"!!
#
=
Find and .J ÐBÑ ß : ;! > % %#l#
Solution: J ÐBÑ œ " W ÐBÑ œ ß : œ œ œ " ß! ! > %B )>>"!! W Ð%Ñ )%
W Ð%>Ñ# #!
!
"
Þ)%
Ð%>Ñ#
"!!
J Ð>Ñ œ ; œ " : œ ! Ÿ > Ÿ ' œ "! B œ % ' œ B% > % > %)>>)%
#
for (since , if there are = =
years remaining in 's lifetime distribution). Therefore with this survival distribution, forÐ%Ñ
someone at age , " " is equal to , and for , % ∞ ' : œ !ß : œ ! 5 ' ; œ ß ; œ ß' % 5 % % # %* #!)% )%
$ % % % & % ' % 5 %; œ ß ; œ ß ; œ ß ; œ œ " ß ; œ " 5 ' ß$$ %) '& )%)% )% )% )% and for and
#l# % % % # %; œ ; ; œ œ œ#) Þ'%Þ$')% W Ð%Ñ Þ)%
W Ð'ÑW Ð)Ñ! !
!.
Example 13: For each of the following survival functions, find , and .J ÐBÑ : ;! > C C>l?
(a) , where .W ÐBÑ œ / + !!+B
(b) for .W ÐBÑ œ " ! Ÿ B Ÿ "!!!B"!!
(c) for .W ÐBÑ œ Ð" Ñ ! Ÿ B Ÿ "!!!#B
"!!
Solution: .J ÐBÑ œ " W ÐBÑ ß : œ ß ; œ : :! ! > C C > C >? C>l?W ÐC>ÑW ÐCÑ!
!
(a) .J ÐBÑ œ " / ß : œ œ / ß ; œ / /! > C C+B +> +> +Ð>?Ñ
>l?//
+ÐC>Ñ
+C
(b) J ÐBÑ œ ß : œ œ " ß! > CB >"!! Ð"!!CÑÎ"!! "!!C
Ð"!!C>ÑÎ"!!
>l? C; œ Ð" Ñ Ð" Ñ œ Þ> >? ?"!!C "!!C "!!C
(c) J ÐBÑ œ " Ð" Ñ ß : œ œ Ð" Ñ ß! > C# #B >
"!! Ð"!!CÑ Î"!! "!!CÐ"!!C>Ñ Î"!!# #
# #
>l? B# #; œ Ð" Ñ Ð" Ñ Þ> >?
"!!C "!!C
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Curtate Future Lifetime of ÐBÑ
X ÐBÑB denotes the exact time until 's death. In many situations, we only need to know the integer
part of , which is the whole number of years until death occurs. For someone alive at age ,X ÐBÑB
OB denotes the completed (integer) number of years until 's death (doesn't count theÐBÑ
fractional part of the year in which death occurs). is a non-negative discrete integer-valuedOB
random variable; (usually age is also an integer).O œ ! ß " ß # ß Þ Þ Þ BB
O œ ! ÐBÑ " B "B corresponds to not surviving complete year (dying before age ), which has
probability
T ÒO œ !Ó œ T Ò! X Ÿ "Ó œ ; œ ;B B B B!l
O œ " ÐBÑB denotes the event that survives 1 complete year and dies during the second year
(between ages and ) and has probabilityB " B #
T ÒO œ "Ó œ T Ò" X Ÿ #Ó œ ; œ ;B B B B"l" "l
For the non-negative integer , corresponds to surviving complete years and5 O œ 5 ÐBÑ 5B
dying during the -st year, which has probability5 "
T ÒO œ 5Ó œ T Ò5 X Ÿ 5 "ÓB B
œ ; œ ; œ : † ; œ : : œ ; ;5l" 5lB B 5 B B5 5 B 5" B 5" B 5 B
Note that .T ÒO Ÿ 5Ó œ T ÒX Ÿ 5 "ÓB B
The age of an individual is often clear from the context of a situation, so the notation mayOB
sometimes be simplified to if the age is clear and obvious.O B
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Note also, if are integers then8 >
(i) 8 B > B B B B B B B>l8> "l #l 8"l 5l5œ!
8"
; œ ; ; œ ; ; ; â ; œ ;and as a special case,
(ii) ." œ ; œ ; ; ; â œ ;∞ B B B B B"l #l 5l5œ!
∞The interpretation of (i) is that in order for to die within years, death must occur in eitherÐBÑ 8
the 1st year (prob. ), the 2nd year (prob. ), . . . , or the -th year (prob. ).; ; 8 ;B B B"l 8"l
The interpretation of (ii) is that must eventually die in some future year with probability 1.ÐBÑ
(ii) also is a verification of the probability rule for discrete random variables which states that the
sum of all individual point probabilities is 1; .5œ!
∞
T ÒO œ 5Ó œ "
A slight variation on is . is the , and it counts theO O œ O " OB B BB‡ time interval of failure
integer part of , and can be interpreted as counting the integer year, as measured from ageX OB B‡
B O O, in which death occurs. The MQR textbook tends to use , but has been the moreB‡
B
traditional curtate random variable and will be mostly used in this study guide. The difference
between the two is trivial and should not be a source of confusion or concern.
Example 12 (continued): Using for ,W ÐBÑ œ " ! Ÿ B Ÿ "! œ!B"!!
#
=
describe the probability distribution of .O%
Solution: T ÒO œ !Ó œ ; œ " : œ œ œ ß% % %W Ð%ÑW Ð&Ñ
W Ð%Ñ )%
*! !
!
)% (&"!! "!!
)%"!!
T ÒO œ "Ó œ ; œ ; ; œ : : œ œ ß% % # % % % # %"lW Ð&ÑW Ð'Ñ
W Ð%Ñ )%""! !
!
T ÒO œ #Ó œ ; œ ß T ÒO œ $Ó œ ; œ ß T ÒO œ %Ó œ ; œ ß% % % % % %#l $l %l"$ "& "()% )% )%
T ÒO œ &Ó œ ; œ T ÒO œ 'Ó œ ; œ !% % % %&l l"*)% . Note that since6
6l % ' % "! "! "!; œ : † ; œ ! † ; œ ! ;(in this survival model, is undefined since no one survives to
age ). For this probability distribution, . "! T ÒO 'Ó œ !%
Note that previous references used the notation for and for . I haveX ÐBÑ X OÐBÑ OB B
tried to update the notation in this study guide, but if you see or , they have theX ÐBÑ OÐBÑ
same meaning as and .X OB B
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MLC SECTION 5 - EXERCISES
1. You are given the survival function for .W ÐBÑ œ B !!"
Ð"BÑ#
Describe the event and calculate or formulate each of the following:
(i) , (ii) , (iii) ,"! "! "! > B"!l"!; ; :
(iv) , (v) , (vi) ,>l? B "! B; T ÒO œ "!Ó T ÒO œ 5Ó
(vii) , (viii) T ÒX Ÿ "!Ó T ÒO Ÿ "!Ó"! "!
2. Show that .T ÒX "!lX &Ó œ :#! #! & #&
MLC SECTION 5 - SOLUTION TO EXERCISES
1.(i) probability that dies by age 20 , .Ð"!Ñ ; œ œ " œ œ Þ(#&'"! "!W Ð"!ÑW Ð#!Ñ "Î#"
W Ð"!Ñ "Î"" %%"$#!! !
!
#
#
(ii) probability that survives to age 20 and dies by age 30,Ð"!Ñ
"!l"! "!; œ œ œ Þ"%)&W Ð#!ÑW Ð$!Ñ Ð"Î#" ÑÐ"Î$" Ñ
W Ð"!Ñ "Î""! !
!
# #
# .
(iii) probability that survives to at least age ,ÐBÑ B >
> B: œ œ œW ÐB>Ñ "ÎÐB>"Ñ ÐB"ÑW ÐBÑ "ÎÐB"Ñ ÐB>"Ñ!
!
# #
# # .
(iv) probability that survives to age and dies by age ,ÐBÑ B > B > ?
>l? B; œ œ œ W ÐB>ÑW ÐB>?Ñ "ÎÐB>"Ñ ÓÒ"ÎÐB>?"Ñ Ó ÐB"Ñ ÐB"Ñ
W ÐBÑ "ÎÐB"Ñ ÐB>"Ñ ÐB>?"Ñ! !
!
# # # #
# # # .
(v) probability that survives to age 20 and dies by age 21,Ð"!Ñ
T ÒO œ "!Ó œ ; œ œ œ Þ!#%%"! "!"!lW Ð#!ÑW Ð#"Ñ Ð"Î#" ÑÐ"Î## Ñ
W Ð"!Ñ "Î""! !
!
# #
# .
(vi) probability that survives to age and dies before age ,ÐBÑ B 5 B 5 "
T ÒO œ 5Ó œ ; œ œ œ B B5lW ÐB5ÑW ÐB5"Ñ "ÎÐB5"Ñ ÓÒ"ÎÐB5#Ñ Ó ÐB"Ñ ÐB"Ñ
W ÐBÑ "ÎÐB"Ñ ÐB5"Ñ ÐB5#Ñ! !
!
# # # #
# # #
(vii) probability that dies by age 20, same as (i) above.Ð"!Ñ
(viii) probability that survives at most 10 complete years, which is the same as theÐ"!Ñ
probability that dies before age 21 (before 11 complete years),Ð"!Ñ
"" "!; œ œ " œ Þ(&!!W Ð"!ÑW Ð#"Ñ "Î##
W Ð"!Ñ "Î""! !
!
#
# .
2. .T ÒX "!lX &Ó œ œ œ œ :#! #! & #&T ÒX "!ÓT ÒX &Ó : :
: : † :#!
#! & #! & #!
"! #! & #! & #&
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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 1
1. For a fully discrete 3-year endowment insurance of 1000 on , you are given:ÐBÑ
(i) is the prospective loss random variable at time .5P 5
(ii) (iii) 3 œ !Þ"! + œ #Þ(!")#ÞÞBÀ$l
(iv) Premiums are determined by the equivalence principle.
Calculate , given that survives to the end of the third year from issue.#P ÐBÑ
A) 540 B) 630 C) 655 D) 720 E) 910
2. For a double-decrement model:
(i) > %!wÐ"Ñ
: œ " ß ! Ÿ > Ÿ %!>"'!!
#
(ii) > %!wÐ#Ñ
: œ " ß ! Ÿ > Ÿ $!>*!!
#
Calculate ..%!#!Ð Ñ7
A) Less than .05 B) At least .05 but less than .10 C) At least .10 but less than .15
D) At least .15 but less than .20 E) At least .20
3. For independent lives (35) and (45):
(i) (ii) (iii) (iv) & $& & %& %! &!: œ !Þ*! : œ !Þ)! ; œ !Þ!$ ; œ !Þ!&
Calculate the probability that the first death of (35) and (45) occurs in the 6th year.
A) 0.0465 B) 0.0565 C) 0.0665 D) 0.0765 E) 0.0865
4. For a fully discrete whole life insurance of 100,000 on (35) you are given:
(i) Percent of premium expenses are 10% per year.
(ii) Per policy expenses are 25 per year.
(iii) Per thousand expenses are 2.50 per year.
(iv) All expenses are paid at the beginning of the year.
(v) The level annual expense loaded premium is 1234.
Calculate ."!!ß !!!T$&
A) Less than 800 B) At least 800 but less than 815 C) At least 815 but less than 830
D) At least 830 but less than 845 E) At least 845
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5. A pension plan provides a retirement annuity with annual rate of:a - 2% of the first 25,000 of final year salary for each of the first 25 completed years of serviceb - 2.5% of the amount of final year salary in excess of 25,000 for each year of the first 25completed years of servicec - 3% of the first 25,000 of final year salary for each completed year of service in excess of 25yearsd - 3.5% of the amount of final salary in excess of 25,000 for each completed year of service inexcess of 25 years.If has 10 years of service already and if , then can be expressedÐ%&Ñ GEW œ $&ß !!! TEF%& '&Þ&
in the form . Find .E † F E FWW'&
%&
A) 24,000 B) 24,250 C) 24,500 D) 24,750 E) 25,000
6. is the present-value random variable for a whole life insurance of payable at the moment^ ,
of death of . You are given:ÐBÑ
(i) (ii) $ .œ !Þ!% œ !Þ!# ß > !B>
(iii) The single benefit premium for this insurance is equal to .Z +<Ð^Ñ
Calculate .,
A) 2.75 B) 3.00 C) 3.25 D) 3.50 E) 3.75
7. For a special 3-year term insurance on (30), you are given:
(i) Premiums are payable semiannually.
(ii) Premiums are payable only in the first year.
(iii) Benefits, payable at the end of the year of death, are:
5 ,5"
! "!!!
" &!!
# #&!
(iv) Mortality follows the Illustrative Life Table.
(v) Deaths are uniformly distributed within each year of age.
(vi) 3 œ !Þ!)
Calculate the amount of each semiannual benefit premium for this insurance.
A) Less than 1.2 B) At least 1.2 but less than 1.3 C) At least 1.3 but less than 1.4
D) At least 1.4 but less than 1.5 E) At least 1.5
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8. For a Markov model with three states, Healthy (0), Disabled (1), and Dead (2):
(i) The annual transition matrix is given by
! " #
! !Þ(! !Þ#! !Þ"!
" !Þ"! !Þ'& !Þ#&
# ! ! "
(ii) There are 100 lives at the start, all Healthy. Their future states are independent.
Calculate the variance of the number of the original 100 lives who die in the second year.
A) Less than 8 B) At least 8 but less than 10 C) At least 10 but less than 12
D) At least 12 but less than 14 E) At least 14
9. An insurance company issues a special 3-year insurance to a high-risk individual. You are
given the following homogeneous Markov chain model:
(i) State 1: active
State 2: disabled
State 3: withdrawn
State 4: dead
Transition probability matrix:
" # $ %
" !Þ% !Þ# !Þ$ !Þ"
# !Þ# !Þ& ! !Þ$
$ ! ! " !
% ! ! ! "
(ii) Changes in state occur at the end of the year.
(iii) The death benefit is 1000, payable at the end of the year of death.
(iv) 3 œ !Þ!&
(v) The insured is active at the start of year 2.
Calculate the actuarial present value of the prospective death benefits at the beginning of the
second year.
A) Less than 200 B) At least 200 but less than 210 C) At least 210 but less than 220
D) At least 220 but less than 230 E) At least 230
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10. For a fully discrete whole life insurance of on , you are given:, ÐBÑ
(i) (ii) ; œ !Þ!#*!% 3 œ !Þ!$B*
(iii) The net amount at risk for policy year 10 is 872.
(iv) + œ "%Þ'&*('ÞÞB
(v) *Z œ #*'Þ"
Calculate the initial benefit reserve for policy year 10.
A) 295 B) 321 C) 343 D) 368 E) 388
11. For a fully discrete 2-year endowment insurance of 1000 on , you are given:ÐBÑ
(i) The level annual premium using the equivalence principle is 479.
(ii) . œ !Þ!'
An alternative premium payment scheme has benefit premium 668 in the first year.
Calculate the benefit premium in the second year in this alternative premium payment scheme.
A) 238 B) 258 C) 268 D) 288 E) 308
12. For an increasing 10-year term insurance, you are given:
(i) , , œ "!!ß !!!Ð" 5Ñ 5 œ !ß "ß ÞÞÞß *5"
(ii) Benefits are payable at the end of the year of death.
(iii) Mortality follows the Illustrative Life Table.
(iv) 3 œ !Þ!'
(v) The single benefit premium for this insurance on (40) is ."&ß &&"
Calculate the single benefit premium for this insurance on (41).
A) 15,220 B) 15,780 C) 16,220 D) 16,780 E) 17,220
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13. For a fully discrete whole life insurance of 1000 on :ÐBÑ
(i) Death is the only decrement.
(ii) The annual benefit premium is 80.
(iii) Expenses in year 1, payable at the start of the year, are 40% of contract premiums.
(iv) The starting asset share is 0 and the asset share at the end of the first year is 16.84.
(v) (vi) 3 œ !Þ"! "!!! Z œ %!" B
Calculate the level annual contract premium.
A) 100 B) 102 C) 104 D) 106 E) 108
14. You are given:
(i) is the future lifetime random variable for .X ÐBÑB
(ii) W œ " ß ! Ÿ B ŸBB= =
(iii) Z +<ÐX Ñ œ $&#Þ"$!
Calculate °/$!À%!lA) 26.7 B) 27.7 C) 28.7 D) 29.7 E) 30.7
15. A fully discrete 3-year term insurance of 10,000 on (40) is based on a double-decrement
model, death and withdrawal:
(i) Decrement 1 is death. (ii) ; œ !Þ!# ß 5 !B5Ð"Ñ
(iii) Decrement 2 is withdrawal, which occurs at the end of the year.
(iv) , (v) ; œ !Þ!% 5 œ !ß "ß # @ œ !Þ*&%!5wÐ#Ñ
Calculate the actuarial present value of the death benefits for this insurance.
A) 492 B) 502 C) 512 D) 522 E) 532
16. For a fully discrete 5-payment 10-year decreasing term insurance on (60), you are given:
(i) , œ "!!!Ð"! 5Ñ ß 5 œ !ß "ß #ß ÞÞÞß *5"
(ii) ; œ !Þ!# !Þ!!"5 ß 5 œ !ß "ß #ß ÞÞÞß *'!5
(iii) 3 œ !Þ!'
(iv) The benefit reserve at the end of year 2 is .#Z œ ((Þ''
Calculate the level annual benefit premium.
A) 188 B) 194 C) 202 D) 210 E) 218
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17. You are given:
(i) and are not independent.X XB C
(ii) , ; œ ; 5 œ !ß "ß #ß ÞÞÞB5 C5
(iii) , 5 BC 5 B 5 C: œ "Þ!# : : 5 œ "ß #ß $ß ÞÞÞ
(iv) ./ œ *Þ%%BÀC
Calculate .;B
A) .044 B) .046 C) .048 D) .050 E) .052
18. A bank offers the following choices for certificates of deposit: Term (in years) Nominal annual interest rate convertible quarterly 1 4.00% 3 5.00% 5 5.65%
The certificates mature at the end of the term. The bank does NOT permit early withdrawals.
During the next 6 years the bank will continue to offer certificates of deposit with the same terms
and interest rates. An investor initially deposits 10,000 in the bank and withdraws the principal
and interest at the end of 6 years. Calculate the maximum annual effective rate of interest the
investor can earn over the 6-year period.
A) 5.09% B) 5.22% C) 5.35% D) 5.48% E) 5.61%
19. A special annuity is set up based on a 2 decrement table. The payments start at age (x), and
a payment of 1 is made annually provided that either (i) (x) has not left the active group, (ii)or
(x) has left the active group due to cause 1 (with no regard to the future survival of (x) after
leaving the active group). Which of the following are correct formulations for the actuarial
present value of this annuity?
I. II. III. + "E.
BÐ#Ñ
+ + @ † ;ÞÞ ÞÞB B B
5œ!
∞5
5Ð#Ñ Ð Ñ Ð"Ñ7
A) All B) All but I C) All but II D) All but III E) None of A, B, C or D
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20. You are given:
(i) (ii) . $B> œ !Þ!$ ß > ! œ !Þ!&
(iii) is the future lifetime random variable. XB
(iv) is the standard deviation of .1 +X lB
Calculate .T< + 1 + + 1 B BX lB
A) 0.50 B) 0.54 C) 0.58 D) 0.62 E) 0.66
21. (50) is an employee of XYZ Corporation. Future employment with XYZ follows a double
decrement model:
(i) Decrement 1 is retirement
(ii) .&!>Ð"Ñ
œ!Þ!! ! Ÿ > &!Þ!# & Ÿ >
(iii) Decrement 2 is leaving employment with XYZ for all other causes
(iv) .&!>Ð#Ñ
œ!Þ!& ! Ÿ > &!Þ!$ & Ÿ >
(v) If (50) leaves employment with XYZ, he will never rejoin XYZ.
Calculate the probability that (50) will leave XYZ due to causes other than retirement before
age 60.
A) Less than .30 B) At least .30 but less than .32 C) At least .32 but less than .34
D) At least .34 but less than .36 E) At least .36
22. For a life table with a one-year select period, you are given:(i) °B j . j /ÒBÓ ÒBÓ ÒBÓB"
)! "!!! *! )" *#! *! )Þ#
(ii) Deaths are uniformly distributed over each year of age.
Calculate .°/Ò)!ÓA) 8.1 B) 8.2 C) 8.3 D) 8.4 E) 8.5
23. For a fully discrete 3-year endowment insurance of 1000 on :ÐBÑ
(i) (ii) 3 œ !Þ!& : œ : œ :B B"
(iii) the second year benefit reserve is .#Z œ &#'
Calculate .:
A) .62 B) .64 C) .66 D) .68 E) .70
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24. You are given: ; œ!Þ!& &! Ÿ B '!!Þ!% '! Ÿ B (!B
Calculate .%l"% &!;
A) 0.34 B) 0.36 C) 0.38 D) 0.40 E) 0.42
25. A fully discrete whole life insurance policy issued at age has face amount 1,000,000 .B
The policy expenses are:
1st Year Renewal Years
Percent of Premium 20% 5%
Face Amount 5 per 1000 per 1000"
Per Policy 1000 200
Settlement 1000 (at the end of the year of death)
You are given that and .3 œ Þ" + œ )Þ)")(ÞÞB
Find the expense-loaded premium using the equivalence principle (nearest 1000).
A) 22,000 B) 23,000 C) 24,000 D) 25,000 E) 26,000
26. For a special fully discrete 5-year deferred whole life insurance of 100,000 on (40), you are
given:
(i) The death benefit during the 5-year deferral period is return of benefit premiums paid without
interest.
(ii) The annual benefit premium is 3363 and is payable only during the deferral period..
(iii) Mortality follows the Illustrative Life Table.
(iv) 3 œ !Þ!'
Calculate ÐMEÑ%!À&l"
A) Less than .038 B) At least .038 but less than .040 C) At least .040 but less than .042
D) At least .042 but less than .044 E) At least .044
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27. You are pricing a special 3-year annuity-due on two independent lives, and . The B ÐCÑ
annuity pays 30,000 if both persons are alive 20,000 if dies first and 10,000 if dies first.ß ÐBÑ ÐCÑ
You are given:
(i) 5 : œ :5 B 5 C
" !Þ*"
# !Þ)#
$ !Þ(#
(ii) 3 œ !Þ!&
Calculate the actuarial present value of this annuity (nearest 100).
A) 78,300 B) 80,400 C) 82,500 D) 84,700 E) 86,800
28. Company ABC sets the contract premium for a continuous life annuity of 1 per year on ÐBÑ
equal to the single benefit premium calculated using:
(i) (ii) for for
$ .œ !Þ!$ œ!Þ!# > Ÿ "!!Þ!" > "!B>
However, a revised mortality assumption reflects future mortality improvement and is given by
.B> œ !Þ!# ß > !
Calculate the expected gain at issue for ABC (using the revised mortality assumption) as a
percentage of the contract premium.
A) 2% B) 8% C) 13% D) 18% E) 23%
29. A fully continuous whole life insurance of face amount 1 is based on the following
assumptions:
• constant force of interest of 6%
• constant force of mortality of .015 at all ages
• annual contract premium of .016
An insurer determines that with a portfolio of independent policies of this type, using the8
normal approximation, the probability of a positive loss on all policies combined is .05. How
many additional independent policies of the same type would be needed to reduce the probability
of a positive total loss to .025 (continuing to use the normal approximation)?
A) Less than 500 B) At least 500, but less than 600 C) At least 600, but less than 700
D) At least 700, but less than 800 E) At least 800
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30. For independent lives and , State 1 is that and are alive, and State 2 that isÐBÑ ÐCÑ ÐBÑ ÐCÑ ÐBÑ
alive but has died, State 3 is the is alive but has died, and State 4 that both andÐCÑ ÐCÑ ÐBÑ ÐBÑ
ÐCÑ have died.
You are given:
• W ÐBÑ œ Ð" Ñ ß ! Ÿ B Ÿ "!!B"Î#B
"!!
• W ÐCÑ œ " ß ! Ÿ C Ÿ "!!CC
"!!
• is the probability that and are in State at time given that they are in State: ÐBÑ ÐCÑ 4 8 "34Ð8Ñ
at time .3 8
• At time 0, is age 54 and is age 75.ÐBÑ ÐCÑ
Calculate .:"ß#Ð"!Ñ
A) Less than 0.026 B) At least 0.026, but less than 0.039
C) At least 0.039, but less than 0.052 D) At least 0.052, but less than 0.065
E) At least 0.065
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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 1 SOLUTIONS
1. The equivalence principle premium is
"!!!T œ "!!!Ð .Ñ œ "!!!Ð Ñ œ #(*Þ#"BÀ$l" " Þ"
+ #Þ(!")# "Þ"ÞÞBÀ$l
.
We are given that survives to the end of the third year from issue. Using the end of theÐBÑ
second year as a reference point, there will be the endowment benefit of 1000 paid one year later
(end of the third year) and there will be one premium received just at the start of the third year.
#P is the present value, value at the end of the second year, of the insurance payment minus the
present value of the future premiums. This will be . #P œ "!!!@ #(*Þ#" œ '#*Þ))
Answer: B
2. .%!> %! %! %!Ð Ñ Ð Ñ wÐ"Ñ wÐ#Ñ
> > >7 7
œ Þ : œ : † : œ Ð" ÑÐ" Ñ œ " :
:
> > > >"'!! *!! &(' "ß%%!ß!!!
..> > %!
Ð Ñ
> %!Ð Ñ
# # # %7
7
:
:
..> > %!
Ð Ñ
> %!Ð Ñ %!!
> %> #!#)) "ß%%!ß!!! #)) "ß%%!ß!!!
$
> ># %
&(' "ß%%!ß!!!
$#ß!!!
&(' "ß%%!ß!!!"'!ß!!!
7
7 œ p œ œ Þ""$
" %!#!Ð Ñ
" . Answer.
7 : C
3. The probability that the first death occurs in the 6th year is .&l $&À%&;
This can be formulated as
&l $&À%& & $&À%& ' $&À%& & $& & %& ' $& ' %& & $& & %& %! &!; œ : : œ : † : : † : œ : † : † Ð" : † : Ñ
From the given information we have and .: œ !Þ*( : œ !Þ*&%! &!
Then, . Answer: B&l $&À%&; œ ÐÞ*ÑÐÞ)ÑÒ" ÐÞ*(ÑÐÞ*&ÑÓ œ Þ!&'&#
4. We use the equivalence principle relationship
APV expense-loaded premium APV benefit plus expenses.œ
K+ œ "!!ß !!!E Þ"K+ #&+ #&!+ ÞÞÞ ÞÞ ÞÞ ÞÞ$& $& $& $& $&
Then ,"#$% œ "!!ß !!!T Þ"Ð"#$%Ñ #& #&!$&
so that . Answer: D"!!ß !!!T œ )$&Þ'!$&
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5. TEF œ'&Þ& (.02)(25,000)(25) + (.025) 35,000 25,000 (25) SS
65
45
+ (.03)(25,000)(5) + (.035) 35,000 25,000 (5) = 3750 + 28,000 SS
65
45 S
S65
45
. Answer: B.p E F œ #%ß #&!
6. The variance of the continuous whole life insurance with face amount is,
Z +<Ò^Ó œ , Ò E ÐE Ñ Ó œ Þ!% # # #
B B . Since the force of mortality is constant at .02 and ,$
E œ œ œ E œ œ œ
B B#. .
$ . $ . !%!# $ # Þ!)!# &Þ!# " Þ!# "
. and , so that
Z +<Ò^Ó œ , Ò Ð Ñ Ó œ ,E œ , †# #
B" " %, "& $ %& $
#
. The single benefit premium is . We are told that
, † œ , œ $Þ(&" %,$ %&
#
, from which we get . Answer: E
7. We assume that we are to find premiums based on the equivalence principle.
We will denote each of the two premiums as (assume to be paid the start of each half-yearU
during the first year). The APV of the premiums is .UÒ" @ † : ÓÞ&Þ& $!
The APV of the benefit is ."!!!@; &!!@ ; #&!@ ;$! $! $!# $"l #l
From the Illustrative Table, we have and .; œ Þ!!"&$ ß ; œ Þ!!"'" ; œ Þ!!"(!$! $" $#
Using UDD, the APV of premiums is .UÒ" @ Ð" Þ&ÐÞ!!"&$ÑÑÓ œ "Þ*'"&"%UÞ&
The APV of the benefit is
"!!!@ÐÞ!!"&$Ñ &!!@ ÐÞ**)%(ÑÐÞ!!"'"Ñ #&!@ ÐÞ**)%(ÑÐÞ**)$*ÑÐÞ!!"(!Ñ œ #Þ%%#!)* Þ# $
Then . Answer: EU œ œ "Þ#%&#Þ%%#!)*"Þ*'"&"%
8. Let denote the probability of dying during the second year. Then the number of deaths in; R
the second year has a binomial distribution based on trials and success (dying)7 œ "!!
probability . The variance of the binomial is .; Z +<ÒRÓ œ 7;Ð" ;Ñ œ "!!;Ð" ;Ñ
; ; œ T Ò Ó can be formulated as survive 1st year and die in 2nd year , which is equal to
œ U † U U † U œ ÐÞ(ÑÐÞ"Ñ ÐÞ#ÑÐÞ#&Ñ œ Þ"#Ð!ß!Ñ Ð!ß#Ñ Ð!ß"Ñ Ð"ß#Ñ
(this is the combination of staying healthy for the 1st year and dying in the 2nd year, or
becoming disabled in the 1st year and dying in the 2nd year).
Therefore . Answer: CZ +<ÒRÓ œ "!!ÐÞ"#ÑÐÞ))Ñ œ "!Þ&'
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9. At the beginning of year 2 the individual is active, there are still 2 years left on the 3-year
insurance policy. If the individual dies in the 2nd year, there will be a benefit of 1000 paid at that
time. The probability of this is (that is the probability of an active individual dyingU œ Þ"Ð"ß%Ñ
during the year). The APV at time 2 of the death benefit for death in the 2nd year is
"!!!@ÐÞ"Ñ œ *&Þ#% .
The individual can survive the 2nd year and die in the 3rd year, but the benefit will only be
payable if the individual is active or disabled at the start of the 3rd year. The probability of
remaining active to the start of the 3rd year and then dying in the 3rd year is
U † U œ ÐÞ%ÑÐÞ"Ñ œ Þ!%Ð"ß"Ñ Ð"ß%Ñ . The probability of becoming disabled as of the start of the 3rd
year and then dying in the 3rd year is The combined probabilityU † U œ ÐÞ#ÑÐÞ$Ñ œ Þ!'ÞÐ"ß#Ñ Ð#ß%Ñ
of surviving to the start of the 3rd year and not withdrawing, and then dying in the 3rd year, is
Þ!% Þ!' œ Þ"!. The APV at the beginning of the 2nd year of the death benefit for death in the
3rd year is . The total APV of the death benefit is"!!!@ ÐÞ"Ñ œ *!Þ(!#
*&Þ#% *!Þ(! œ ")&Þ*% . Answer: A
10. The initial benefit reserve for policy year 10 is (where is the*Z T œ #*'Þ" T T
benefit premium). T œ ,T œ ,Ð .Ñ œ Þ!$*!)), ÞB"+ÞÞB
The net amount at risk for policy year 10 is ., Z œ )(#"!
Using the net amount at risk form of the recursive relationship for benefit reserve, for year 10,
we have , which becomesÐ Z TÑÐ" 3Ñ Ð, Z Ñ; œ Z* "! B* "!
Ð#*'Þ" Þ!$*!)),ÑÐ"Þ!$Ñ Ð)(#ÑÐÞ!#*!%Ñ œ , )(# p , œ "#!! .
Then, T œ ,T œ "#!!ÐÞ!$*!))Ñ œ %'Þ*"ÞB
Then the initial benefit reserve for policy year 10 is . *Z T œ #*'Þ" %'Þ*" œ $%$
Answer: C
11. The level annual premium based on the equivalence principle is
%(* œ "!!!Ð .Ñ + œ " @: œ "Þ)&&$ÞÞ"
+ÞÞBÀ#l
, so that .BÀ#l B
Then, so that ." ÐÞ*%Ñ: œ "Þ)&&$ : œ Þ*!**B B
Then, , so that . Answer: B'') U@: œ %(*Ð" @: Ñ U œ #&)B B
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12. We are given . We wish to find ."!!ß !!!ÐMEÑ œ "&ß &&" "!!ß !!!ÐMEÑ%!À"!l %"À"!l" "
We use the relationship . This can be seen byÐMEÑ œ E @: ÐMEÑ 8@ ;BÀ8l BÀ8l" " "B B
B"À8l
8"8l
looking at the time line of possible death benefit payments; the first row is the sum of the second
and third rows.
B B " B # B $ Þ Þ Þ B 8 " B 8 B 8 "
ÐMEÑ o " # $ 8 " 8BÀ8l"
E o " " " " "BÀ8l"
@: ÐMEÑ o " # 8 # 8 " 8BB"À8l"
8@ ; 88"8l B
Then, ÐMEÑ œ E @: ÐMEÑ "!@ ;%!À"!l %!À"!l %"À"!l" " "%! %!
"""!l
E œ E @ : E œ ÐÞ"'"$#Ñ ÐÞ&$''(ÑÐÞ#%*!&Ñ œ Þ!#(''#%!À"!l" %! "! %! &!
"!
(we use values from the Illustrative Table, and note that ).@ : œ I"!"! %! "! %!
Also .@ ; œ @ † @ : † ; œ † ÐÞ&$''(ÑÐÞ!!&*#Ñ œ Þ!!#**("" "!"!l %! "! %! &!
""Þ!'
Then .Þ"&&&" œ Þ!#(''# † Ð" Þ!!#()Ñ ÐMEÑ "!ÐÞ!!#**(Ñ""Þ!' %"À"!l
"
Solving for results in ÐMEÑ ÐMEÑ œ Þ"'()%"À"!l %"À"!l" "
We then multiply by 100,000 to get . Answer: D"'ß ()!
13. The accumulation of asset share in the 1st year is
Ò! KÐÞ'ÑÓÐ"Þ"Ñ "!!!; œ : † Ð"'Þ)%ÑB B .
If we knew the value of , we could find .; KB
Using the recursive relationship for benefit reserve for the 1st year, we have
Ò! )!ÓÐ"Þ"Ñ "!!!; œ : Ð%!Ñ ; ; œ Þ!&B B B B , and solving for results in .
Then . Answer: AÞ'KÐ"Þ"Ñ "!!!ÐÞ!&Ñ œ ÐÞ*&Ñ † Ð"'Þ)%Ñ p K œ "!!
14. From the form of we see that survival from birth follows De Moivre's Law with upper=ÐBÑ
age . From we get .= =Z +<ÒX ÑÓ œ œ $&#Þ" œ *&$!Ð $!Ñ
"#= #
Under De Moivre's Law, so .> B > $!: œ " œ " : œ> > '&>B *&B '&=
/ œ : .> œ .> œ œ #(Þ'*#° . Answer: B$!À%!l ! !%! %!
> $!>œ!
%! '&>'& #Ð'&Ñ
Ð'&>Ñ#
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15. The APV of the death benefit is ."!ß !!!Ò@; @ ; @ ; Ó%! %! %!Ð"Ñ Ð"Ñ Ð"Ñ# $
"l #l
Since decrement 2 occurs at the end of the year, for each year, , and; œ ;B BÐ"Ñ wÐ"Ñ
; œ Ð" ; Ñ † ;B B BÐ#Ñ wÐ"Ñ wÐ#Ñ .
For decrement 1, we have for , since the force of decrement is; œ ; œ Þ!# B œ %!ß %"ß %#B BÐ"Ñ wÐ"Ñ
constant. Also for .: œ : † : œ ÐÞ*)Ñ † ÐÞ*'Ñ œ Þ*%!) B œ %!ß %"B B BÐ Ñ wÐ"Ñ wÐ#Ñ7
Then, and"l %! %! %"Ð"Ñ Ð Ñ Ð"Ñ
; œ : † ; œ ÐÞ*%!)ÑÐÞ!#Ñ œ Þ!"))"'7
#l %! %! %# %! %" %#Ð"Ñ Ð Ñ Ð"Ñ Ð Ñ Ð Ñ Ð"Ñ
##; œ : † ; œ : † : † ; œ ÐÞ*%!)Ñ ÐÞ!#Ñ œ Þ!"((!
7 7 7 .
The APV of the death benefit is
"!ß !!!Ò@ÐÞ!#Ñ @ ÐÞ!"))"'Ñ @ ÐÞ!"((!ÑÓ œ &"## $ . Answer: C
16. We use the recursive reserve formula, .Ð Z TÑÐ" 3Ñ , † ; œ : † Z5 5" B5 B5 5"
Also, for benefit reserves. For the first year, we have!Z œ !
Ð! TÑÐ"Þ!'Ñ "!ß !!!ÐÞ!#Ñ œ Þ*) † Z p Z œ "Þ!)"'T #!%Þ!)" " .
For the second year, we have
Ð"Þ!)"'T #!%Þ!) TÑÐ"Þ!'Ñ *ß !!!ÐÞ!#"Ñ œ Þ*(* † Z œ ÐÞ*(*ÑÐ((Þ''Ñ œ ('Þ!$#
. Answer: Ep T œ #")
17. is a valid relationship for all survival distributions of and ,/ œ / / / XÐBÑ X ÐCÑBC B C BC
whether or not they are independent. Since for all , it follows that for; œ Þ!& 5 : œ Þ*&B5 B5
all , and then . This follows from the fact that5 / œ : œ Ð: Ñ œB > B B>œ" >œ"
∞ ∞> :
":B
B
> B B B" B>"> # $: œ : † : â: œ ÐÞ*&Ñ " < < < â œ and ."
"<
Also, , since for all ./ œ ; œ ; 5C B5 C5:
":B
B
/ œ : œ "Þ!#Ð: Ñ œ "Þ!# † œ *Þ%%BC > BC B>œ" >œ"
∞ ∞#> Ð: Ñ
"Ð: ÑB
#
B# .
Then , and and . Answer: D: œ Þ*!#& : œ Þ*& ; œ Þ!&B#
B B
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18. The annual effective rate for the 1 year certificate is , for the 3-yearÐ"Þ!"Ñ " œ Þ!%!'%
certificate it is , and for the 5-year certificate it isÐ"Þ!"#&Ñ " œ Þ!&!*%
Ð"Þ!"%"#&Ñ " œ Þ!&((% . In order to withdraw the investment at the end of 6 years, the investor
must choose one of the following patterns of investment:
(i) 6 successive one-year certificates , annual effective rate is .0406 .
(ii) a 3 year certificate combined with 3 one-year certificates (in any order), annual effective rate
is (we have found the 6-year accumulation and then theÒÐ"Þ!"#&Ñ Ð"Þ!"Ñ Ó " œ Þ!%&)"# "# "Î'
equivalent annual effective rate that would compound to the same amount in 6 years).
(iii) Two 3-year certificates, annual effective rate .0509 .
(iv) A one-year certificate and a 5-year certificate (either order), annual effective rate is
ÒÐ"Þ!"%"#&Ñ Ð"Þ!"Ñ Ó " œ Þ!&%)#! % "Î' .
The maximum annual effective return is .0548 and is obtained with a 5-year and a 1-year
certificate, in either order. Answer: D
19. APV = v 1 q = v 1 q + q = v p + v q k=0 k=0 k=0 k=0
k k k kk k k k kx x x x x
( ) ( )∞ ∞ ∞ ∞
† Ð Ñ † Ð Ñ † †Ð#Ñ Ð"Ñ Ð"Ñ7 7
= a + v q , so that III is correct.¨( )x
k=0
kk x
7 ∞ †Ð"Ñ
Also, v 1 q = v v q = v q k=0 k=0 k=0 k=0 j=0
k k k kk kx x x
k–1
j
∞ ∞ ∞ ∞
±† Ð Ñ † Ð#Ñ Ð#Ñ Ð#Ñ 1
d
= v q = q = 1 v q = , so1 1 v 1d d d d
A † † † † j=0 k=j j=0 j=0
k jj j jx x x
1d
∞ ∞ ∞ ∞
± ± ±Ð#Ñ Ð#Ñ Ð#Ñ
jx Ð#Ñ
that I is correct. II has no meaning, and is incorrect. Answer: C
20. With constant force of mortality and force of interest , .. $œ Þ!$ œ Þ!& + œ œ "#Þ&B
"$ .
The variance of is+XÐBÑl
" " " Þ!$ Þ!$# ÐÞ!&Ñ Þ"Þ!$ Þ!&Þ!$$ $ $ . $ .. .
# # #Ò E ÐE Ñ Ó œ Ò Ð Ñ Ó œ Ò Ð Ñ Ó œ $'Þ!' # # # #
B B .
The standard deviation is .$'Þ!' œ 'Þ!!
We wish to find ] .T Ò"#Þ& 'Þ!! + "#Þ& 'Þ!!Ó œ T Ò 'Þ& + ")Þ& X l X lB B
We solve for , from the equation , so that , so that 8 + œ 'Þ& œ 'Þ& / œ Þ'(&8l
Þ!&8"/Þ!&
Þ!&8
(which is equivalent to years) and we solve for from the equation8 œ œ (Þ)' ß 768ÐÞ'(&Ñ
Þ!&
+ œ ")Þ& œ ")Þ& / œ Þ!(&7l
Þ!&7 , so that , so that "/Þ!&
Þ!&7
(which is equivalent to years)7 œ œ &"Þ)"68ÐÞ!(&Ñ
Þ!&
T Ò 'Þ& + ")Þ& œ T Ò(Þ)' XÐBÑ &"Þ)"Ó œ : :XÐBÑl (Þ)' B &"Þ)" B]
œ / / œ Þ(*! Þ#"" œ Þ&(*Þ!$Ð(Þ)'Ñ Þ!$Ð&"Þ)"Ñ . Answer: C
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21. The probability is , which can be written as ."! & & &&! &! &! &&Ð#Ñ Ð#Ñ Ð Ñ Ð#Ñ
; ; : † ;7
& >&! &!Ð#Ñ Ð Ñ Ð#Ñ
! !& &
B>Þ!&>; œ : .> œ / ÐÞ!&Ñ .> œ † ÐÞ!&Ñ œ Þ##"# 7
. "/Þ!&
Þ#&
;
alternatively, since for , we have .B>Ð"Ñ Ð#Ñ Ð Ñ
> >B BÞ!&>œ ! ! Ÿ > & ; œ ; œ " /
7
for , so that .! Ÿ > & ; œ " /& &!Ð#Ñ Þ#&
For , so that and! Ÿ > & ß œ œ Þ!& : œ /. . .&&> &&> &&> &&Ð Ñ Ð"Ñ Ð#Ñ Ð Ñ
>Þ!&>7 7
& >&& && &&>Ð#Ñ Ð Ñ Ð#Ñ
! !& & Þ!&>; œ : † .> œ / ÐÞ!$Ñ .> œ ÐÞ!$Ñ † œ Þ"$#(# 7
. "/Þ!&
Þ#&
.
Then, . Answer: C"! & & &&! &! &! &&Ð#Ñ Ð#Ñ Ð Ñ Ð#Ñ
; œ ; : † ; œ Þ##"# ÐÞ(())!ÑÐÞ"$#(#Ñ œ Þ$#%'7
22. With a one-year select period, , so that° °/ œ /Ò)"Ó" )#
/ œ / : † / œ : .> : † / œ Ð" ; Ñ .> : † /° ° ° ° °Ò)"Ó Ò)"Ó Ò)"Ó Ò)"Ó Ò)"Ó Ò)"ÓÒ)"ÓÀ"l )# > )# > )#! !" "
œ Ð" > † ; Ñ .> : † / œ Ð" >Ñ .> Ð Ñ † / œ Þ*&"" Þ*!## † / ! !" "
Ò)"Ó Ò)"Ó )# )# )#° ° °*! )$!*#! *#!
(using UDD and ). Therefore , so that°; œ œ Þ!*() )Þ# œ Þ*&"" Þ*!## † /Ò)"Ó )#*!*#!
/ œ )Þ!$& j œ j œ *"! j œ j œ )$!° . From the table we have and , so that)# )" )#Ò)!Ó" Ò)"Ó"
: œ ; œ)" )")$! )*"! *" and . Then, we use the relationship
/ œ / : † / : † / /° ° ° ° ° to solve for .Ò)!Ó Ò)!Ó Ò)!Ó Ò)!ÓÒ)!ÓÀ"l )"À"l # )#
From UDD we have and ,° °/ œ Ð" Þ!*>Ñ .> œ Þ*&& / œ Ð" >Ñ .> œ Þ*&'Ò)!ÓÀ"l )"À"l! !" " )
*"
: œ Þ*" ß : œ : † : œ ÐÞ*"ÑÐ Ñ œ Þ)$Ò)!Ó Ò)! Ò)!Ó# )")$*" . Then
/ œ Þ*&& ÐÞ*"ÑÐÞ*&'Ñ ÐÞ)$ÑÐ)Þ!$&Ñ / œ )Þ%*%° ° , so that . Answer: EÒ)!Ó Ò)!Ó
23. The 2nd year terminal reserve for a 3-year endowment insurance can be formulated as
# BÀ$lZ œ " ß+ÞÞ
+ÞÞB#À"l
BÀ$l
where and .+ œ " + œ " @: @ : œ " ÞÞ ÞÞB#À"l BÀ l B # B
#3
: :"Þ!& Ð"Þ!&Ñ
#
#
Then , so that .# BÀ$lZ œ " œ Þ&#' "Þ"!*( œ !"
"
: :Ð"Þ!&Ñ "Þ!&: :
"Þ!&
#
Ð"Þ!&Ñ#
#
#
Solving the quadratic equation, or substituting the possible answers results in .: œ Þ(
Answer: E
24. .%l"% &! % &! ") &!; œ : :
% &! ") &! "! &! ) '!% "! ): œ ÐÞ*&Ñ œ Þ)"%& ß : œ : † : œ ÐÞ*&Ñ ÐÞ*'Ñ œ Þ%$"* .
Then . Answer: C%l"% &!; œ Þ)"%& Þ%$"* œ Þ$)#'
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25. K+ œ "ß !!!ß !!!E Þ#K Þ!&K+ &ß !!! "!!! Ð"!!! #!!Ñ+ "!!!EÞÞB B B B B
K œ Þ"ß!!"ß!!!E 'ß!!!"#!!+
Þ*&+ Þ"&ÞÞB B
B
E œ " .+ œ " † Ð)Þ)")(Ñ œ Þ"*)$ ÞÞÞ
B BÞ""Þ"
K œ œ #&ß **&Þ!!"ß!!"ß!!!ÐÞ"*)$Ñ'ß!!!"#!!Ð(Þ)")(Ñ
Þ*&Ð)Þ)")(ÑÞ"& . Answer: E
26. If the annual benefit premium is , then .U U † + œ UÐMEÑ "!!ß !!! † EÞÞ%!À&l
%!À&l" &l %!
To find , we use the relationship . Using values from the+ + œ + I † +ÞÞ ÞÞ ÞÞ ÞÞ%!À&l %!À&l%! & %! %&
Illustrative Table, we have , so that ."%Þ)"'' œ + ÐÞ($&#*ÑÐ"%Þ""#"Ñ + œ %Þ%%!"ÞÞ ÞÞ%!À&l %!À&l
Also, .&l %! & %! %&E œ I † E œ ÐÞ($&#*ÑÐÞ#!"#!Ñ œ Þ"%(*%
Then , so that %Þ%%!"Ð$$'$Ñ œ Ð$$'$ÑÐMEÑ "!!ß !!!ÐÞ"%(*%Ñ ÐMEÑ œ Þ!%" Þ%!À&l %!À&l" "
Answer: C
27. The APV of the annuity is
$!ß !!!+ #!ß !!!Ð+ + Ñ "!ß !!!Ð+ + ÑÞÞ ÞÞ ÞÞ ÞÞ ÞÞBCÀ$l CÀ$l BCÀ$l BÀ$l BCÀ$l .
œ #!ß !!!+ "!ß !!!+ œ $!ß !!!+ÞÞ ÞÞ ÞÞCÀ$l BÀ$l BÀ$l .
+ œ " @: @ : œ #Þ'"!%$ÞÞBÀ$l B # B
#
so the APV is 78,312.9 . Answer: A
28. The loss at issue is the PVRV (present value random variable) of benefit to be paid minus the
PVRV of premium to be received. Since this is a single premium policy, the premium received is
a single payment equal to the contract premium. Based on the original mortality assumption, the
contract premium is
+ œ + @ : † + œ / / .> / / † B "! B B"!BÀ"!l
"! Þ!$> Þ!#> Þ!$Ð"!Ñ Þ!#Ð"!Ñ!"! "
Þ!$Þ!"
œ / † œ #$Þ!$"/ "Þ!& Þ!$Þ!"
Þ!&Ð"!ÑÞ!&Ð"!Ñ .
The loss at issue is , where is the PVRV of the continuous life annuity. TheP œ ] #$Þ!$ ]
expected loss at issue will be , where is calculated based on theIÒPÓ œ IÒ] Ó #$Þ!$ IÒ] Ó
revised mortality assumption. .IÒ] Ó œ + œ œ œ #!B
" " Þ!$Þ!#$ .
Then . Answer: CIÒPÓ œ #! #$Þ!$ œ $Þ!$
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29. The loss on one policy is ,P œ ^ U] œ ^ Þ!"']
where is the present value random variable for a continuous whole life insurance of 1 and is^ ]
the PVRV of a continuous life annuity of 1 per year.
IÒPÓ œ E Þ!"'+ œ ÐÞ!"'ÑÐ Ñ œ
B B"(&
Þ!"& "Þ!'Þ!"& Þ!'Þ!"& ,
Since , can be written as and then] œ œ P P œ Ð" Ñ^ "^ "^ Þ!"' Þ!"'Þ!' Þ!' Þ!'$
Z +<ÒPÓ œ Ð" Ñ Z +<Ò^Ó œ Ð" Ñ Ð E E Ñ Þ!"' Þ!"'
Þ!' Þ!'# # #
B B#
œ Ð" Ñ Ò Ð Ñ Ó œ Þ""%!*%Þ!"' Þ!"& Þ!"&Þ!' Þ"#Þ!"& Þ!'Þ!"&
# # .
If the total loss is denoted , then (sum of losses on each of the W W œ P P âP 8" # 8
policies), and , and .IÒWÓ œ 8IÒPÓ œ Z +<ÒWÓ œ 8Z +<ÒPÓ œ Þ""%!*%88(&
Using the normal approximation, .TÐW !Ñ œ TÐ ÑWIÒWÓ IÒWÓ
Z +<ÒWÓ Z +<ÒWÓ In order for this probability to be .05, we must have ,
IÒWÓ
Z +<ÒWÓ œ "Þ'%&
or equivalently, . Solving for results in . Ñ
Þ""%!*%8
( 8(& œ "Þ'%& 8 8 œ "($(
The number of independent policies needed to have a positive loss probability of .025, say ,7
must satisfy the equation , and solving for results in . Ñ
Þ""%!*%
(
m
m(& œ "Þ*' 7 7 œ #%''
The additional number of policies needed is . Answer: D#%'' "($( œ (#*
30. is the probability that for and , alive at age 64 and 85, will survive the year,: ÐBÑ ÐCÑ ÐBÑ"ß#Ð"!Ñ
but will die within the year. This probability that (64) survives the year is for , whichÐCÑ : ÐBÑ'%
is . The probability that (85) dies during the year is for , which is W Ð'&ÑW Ð'%ÑB
Bœ Þ*)'!"$ ; ÐCÑ)&
" œ Þ!'''( ÐCÑ œ "!! ÞW Ð)'ÑW Ð)&ÑC
C ( 's mortality follows De Moivre's Law with )=
Then, . Answer: E: œ : † ; œ ÐÞ*)'!"$ÑÐÞ!'''(Ñ œ Þ!''"ß# '% )&Ð"!Ñ B C