S. BROVERMAN STUDY GUIDE FOR THE SOCIETY OF ACTUARIES EXAM MLC 2012 EDITION EXCERPTS. Samuel Broverman, ASA, PHD

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1 S. ROVERMAN STUDY GUIDE FOR THE SOCIETY OF ACTUARIES EXAM MLC 2012 EDITION EXCERPTS Samuel roverman, ASA, PHD copyright 2012, S. roverman SOA Exam MLC Study Guide S. roverman 2012

2 Excerpts: Table of Contents Introductory Note Section 5 - Models for Survival and Mortality Practice Exam 1 and Solutions SOA Exam MLC Study Guide S. roverman 2012

3 S. ROVERMAN EXAM MLC STUDY GUIDE - VOLUME 1 NOTES, EXAMPLES AND PROLEM SETS Introductory Note TALE 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 X and O 69 to 78 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 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 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 to SOA Exam MLC Study Guide S. roverman 2012

4 TALE 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 to 318 Section 24 - Introduction to enefit Reserves 319 to 324 Section 25 - Prospective Form of enefit Reserves 325 to 330 Section 26 - Additional Representations for enefit 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 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 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 enefits 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 to 518 Section 38 - Variable Interest Rates 519 to 526 Problem Set 10 - Variable Interest Rates - Section to 534 ILLUSTRATIVE LIFE TALE SOA Exam MLC Study Guide S. roverman 2012

5 SOA EXAM MLC STUDY GUIDE - VOLUME 2 PRACTICE EXAMS TALE OF CONTENTS PRACTICE EXAMS Practice Exam 1 and Solutions Practice Exam 2 and Solutions Practice Exam 3 and Solutions Practice Exam 4 and Solutions Practice Exam 5 and Solutions Practice Exam 6 and Solutions Practice Exam 7 and Solutions Practice Exam 8 and Solutions Practice Exam 9 and Solutions Practice Exam 10 and Solutions Practice Exam 11 and Solutions Practice Exam 12 and Solutions PE-1 to PE-20 PE-21 to PE-40 PE-41 to PE-60 PE-61 to PE-80 PE-81 to PE-102 PE-103 to PE-122 PE-123 to PE-140 PE-141 to PE-164 PE-165 to PE-186 PE-187 to PE-206 PE-207 to PE-228 PE-229 to PE-250 ILLUSTRATIVE LIFE TALE SOA Exam MLC Study Guide S. roverman 2012

6 INTRODUCTORY NOTE This study guide is designed to help in the preparation for Exam MLC of the Society of Actuaries (the life contingencies and probability exam). The study guide is divided into two volumes. Volume 1 consists of review notes, examples and problem sets. Volume 2 contains 12 practice exams of 30 questions each. There are over 170 examples in the notes, over 360 problems in the problem sets and 360 questions in the 12 practice exams. All of these (almost 900) questions have detailed solutions. The notes are broken up 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 from older SOA or CAS exams on the relevant topics. The practice exams have 30 questions each and are designed to be similar to actual 3-hour exams. 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. ecause 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 books 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. roverman 2012

7 Of the various calculators that are allowed for use on the exam, I think that the A 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. 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 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. roverman February, 2012 Department of Statistics University of Toronto 100 St. George Street Toronto, Ontario CANADA M5S 3G3 sam@utstat.toronto.edu or 2brove@rogers.com Internet: SOA Exam MLC Study Guide S. roverman 2012

8 MLC SECTION 5 - MODELS FOR SURVIVAL AND MORTALITY MQR 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 owers book defines the following: \œx œ time until death, or age at death of a newborn. \ is a continuous, positive random variable (\ ). The distribution function of \ is J\ ÐÑœJ ÐÑœT<Ò\ŸÓœTÒX ŸÓ œ probability that the newborn dies before or on attaining age. The survival function or survival distribution function of \ is =ÐÑœ" J\ ÐÑœT<Ò\ ÓœW ÐÑœ" J ÐÑ œ probability that the newborn dies after attaining age, œ the probability that the newborn survives to at least age. If Dthen T<Ò \ŸDÓœJ ÐDÑ J ÐÑœ=ÐÑ =ÐDÑ. \ \ This is the probability that the newborn dies between ages and D. The following assumptions are made regarding the behavior of the cdf and survival function. (i) J ÐÑ œ and =ÐÑ œ " (the newborn is alive with probability 1). (ii) lim J ÐÑ œ " and lim =ÐÑ œ (the probability of the newborn surviving forever is 0). p Ä (iii) J ÐÑ is a non-decreasing function and =ÐÑ is a non-increasing function (as increases, there is an increasing probability that the newborn dies by age ). SOA Exam MLC Study Guide S. roverman 2012

9 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 ÐÑ The notation ÐÑ refers to an individual alive at age. X denotes the continuous random variable measuring the time until ÐÑ's death from age (\œx described above is a special case of this with œ, a newborn). Note that given that a newborn survives to age, the future lifetime measured from age is X œ \, so that time until death from age is age at death minus, and X can be regarded as the conditional random variable of additional number of years to be lived after age given that the individual has survived to age. For instance, consider someone alive at age œ % who dies at age 50. For this person, death occurs at age \œ&, but measured as time until death from age 40 we have X œ X% œ " œ & % œ \ (of course, since this person was alive at age 40, it is also true that this person was alive at age 20, so that X œ $ for this person). When we are measuring X, time until death from age, it is understood that the person is alive at age, and hence X can be represented as a conditional random variable in terms of \: TÒX Ÿ>ÓœTÒ\ Ÿ>l\ ÓœTÒ\Ÿ >l\ Ó T Ò \Ÿ >Ó J Ð >Ñ J ÐÑ W ÐÑ W Ð >Ñ T Ò\ Ó " J ÐÑ W œ œ œ. We will denote the distribution function of X ÐÑ by JÐ>Ñ œ T ÒX Ÿ >Ó (instead of JX Ð>ÑÑß which is the probability that ( ) will die within the next > years (by age > ) ; this probability is denoted in actuarial notation by > ; œjð>ñœtòx Ÿ>Ó œ probability that ( ) will die within the next > years (by age > ) SOA Exam MLC Study Guide S. roverman 2012

10 The complement of > ; is denoted > :, which is > : œ " > ; œ TÒX >Ó œ probability that ÐÑ survives at least to time > (and dies some time after > ) y the conventions of actuarial notation, for mortality or survival for a 1 year period, we write "; œ ; and ": œ : It is assumed that ; œ "(the probability of eventually dying is 1) and : œ (the probability of living forever is 0). It is also assumed that ; œ ß the special case that œß we have \œxðñ and > : œ=ð>ñ. : œ ". In Factorization of Survival Probability and Deferred Mortality Probability For 8Ÿ>ß the probability that ÐÑ survives at least > -years can be factored in the following way: > : œ 8: > 8: 8. The verbal interpretation of this factorization is that in order for ÐÑ to survive > years, ÐÑ must survive the first 8years (event E, probability 8: ), and then must survive > 8 more years from age 8(event FlE, probability > 8: 8). The probability of surviving > years is the event E F, and we use the probability rule T ÒE FÓ œ T ÒEÓ T ÒFlEÓ. This rule applies for any values of 8 and >, integer or fractional. As a special case, 8 : œ : 8:, and it follows that 8: œ WÐ 8Ñ WÐÑ ; this allows us to formulate survival/mortality probabilities from any age if we are given the probability of survival from birth function =ÐÑ. SOA Exam MLC Study Guide S. roverman 2012

11 If > is an integer, the survival probability over the > -year period can be separated into consecutive one-year periods to get the factorization form > : œ : : " â: > ". An important probability in life contingencies is the deferred mortality probability. The probability that ÐÑ survives > years and then dies in the? years following that is >l? ; œ > :?; > œ >? ; > ; œ > : >? : œtò> X Ÿ>?Óœ W Ð >Ñ W Ð >?Ñ WÐÑ This is a > -year deferred,? year mortality probability; it is the probability that ÐÑ survives > years to age > and then dies by age >?. Note that this is not simply equal to?; >, since ÐÑ must first survive to age > and then from age > must die in the following? years. y notational convention, if the mortality period is one year, we write >l" ; œ >l; ; this is the probability that ÐÑ survives > years and then dies between ages > and > ". > >? > : p?; > p Also note that l; œ ;, and >? ; œ > ; >l? ;. The interpretation of the last expression is that in order for ÐÑ to die within >? years, either death occurs within the first > years (probability ; ) or ÐÑ survives > years and dies in the following? years (prob. ; ). > >l? 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. SOA Exam MLC Study Guide S. roverman 2012

12 Example 12: The following survival function is given: WÐÑœ" " for ŸŸ"œ=. Find JÐÑß > :% and l;%. Ð% >Ñ WÐ% >Ñ " " )> > Solution: JÐÑœ" WÐÑœ " ß > :% œ W Ð%Ñ œ œ" )% ß Þ)% )> > JÐ>Ñœ ; œ" : œ for Ÿ>Ÿ' (since = œ", if œ% there are 'œ = % > % > % years remaining in )% Ð%Ñ's lifetime distribution). Therefore with this survival distribution, for * ' % 5 % % )% % )% $$ %) '& )% $ ;% œ ß %;% œ ß &;% œ ß ';% œ œ "ß 5 ;% œ " 5 ' ß someone at age %, " " is equal to ', : œß and : œ for 5 ', ; œ ß ; œ ß )% )% )% )% and for and ; œ ; ; œ œ œ l % % % % ) WÐ'Ñ WÐ)Ñ Þ'% Þ$' )% W Ð%Ñ Þ)%. Example 13: For each of the following survival functions, find JÐÑ, : and ;. (a) WÐÑœ/, where +. + (b) WÐÑœ" for ŸŸ". (c) WÐÑœÐ" Ñ for ŸŸ". " " WÐC >Ñ > C WÐCÑ >l? C > C >? C Solution: JÐÑœ" WÐÑß : œ ß ; œ : :. +ÐC >Ñ + / +> +> +Ð>?Ñ > C / +C >l? C (a) JÐÑœ" / ß : œ œ/ ß ; œ/ /. Ð" C >ÑÎ" > " > C Ð" CÑÎ" " C (b) JÐÑœ ß : œ œ" ß >l? C > >?? " C " C " C ; œ Ð" Ñ Ð" Ñ œ Þ Ð" C >Ñ Î" > " > C Ð" CÑ Î" " C (c) JÐÑœ" Ð" Ñ ß : œ œð" Ñ ß >l? > >? " C " C ; œ Ð" Ñ Ð" Ñ Þ > C >l? C SOA Exam MLC Study Guide S. roverman 2012

13 Curtate Future Lifetime of ÐÑ X denotes the exact time until ÐÑ's death. In many situations, we only need to know the integer part of X, which is the whole number of years until death occurs. For someone alive at age ÐÑ, O denotes the completed (integer) number of years until ÐÑ's death (doesn't count the fractional part of the year in which death occurs). is a non-negative discrete integer-valued random variable; O œß "ß ß ÞÞÞ (usually age is also an integer). O O œ corresponds to ÐÑ not surviving " complete year (dying before age " ), which has probability TÒO œóœtò X Ÿ"Óœ; œ ; l O œ " denotes the event that ÐÑsurvives 1 complete year and dies during the second year (between ages " and ) and has probability TÒO œ"óœtò" X ŸÓœ ; œ ; "l" "l For the non-negative integer 5, O œ 5 corresponds to ÐÑ surviving 5 complete years and dying during the 5 " -st year, which has probability TÒO œ5óœtò5 X Ÿ5 "Ó œ 5l" ; œ 5l; œ 5: ; 5 œ 5: 5 " : œ 5 " ; 5; Note that TÒO Ÿ5ÓœTÒX Ÿ5 "Ó. The age of an individual is often clear from the context of a situation, so the notation sometimes be simplified to O if the age is clear and obvious. O may SOA Exam MLC Study Guide S. roverman 2012

14 Note also, if 8 > are integers then 8 " (i) 8 ; œ > ; >l8 > ; œ ; "l ; l ; â 8 "l ; œ 5l ; 5œ and as a special case, (ii) " œ ; œ ; ; ; â œ ;. "l l 5l 5œ The interpretation of (i) is that in order for ÐÑ to die within 8 years, death must occur in either the 1st year (prob. ; ), the 2nd year (prob. "l; ),..., or the 8 -th year (prob. 8 "l; ). The interpretation of (ii) is that ÐÑ must eventually die in some future year with probability 1. (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; TÒO œ 5Ó œ ". 5œ A slight variation on O is O œ O ". O is the time interval of failure, and it counts the integer part of X, and O can be interpreted as counting the integer year, as measured from age, in which death occurs. The MQR textbook tends to use O, but O has been the more 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 WÐÑœ" " for ŸŸ"œ=, describe the probability distribution of O %. )% (& WÐ%Ñ WÐ&Ñ * % % % W Ð%Ñ )% )% " WÐ&Ñ WÐ'Ñ "" % "l % % % % % W Ð%Ñ )% "$ "& "( % l % )% % $l % )% % %l % )% "* % &l % )%. Note that % 6l % since " " Solution: TÒO œóœ; œ" : œ œ œ ß T ÒO œ "Ó œ ; œ ; ; œ : : œ œ ß TÒO œ Ó œ ; œ ß TÒO œ $Ó œ ; œ ß TÒO œ %Ó œ ; œ ß TÒO œ&óœ ; œ TÒO œ'óœ ; œ 6l ;% œ ':% ;" œ ;" œ (in this survival model, ;" age "). For this probability distribution, T ÒO % 'Ó œ. is undefined since no one survives to Note that previous references used the notation XÐÑfor X and OÐÑfor O. I have tried to update the notation in this study guide, but if you see X ÐÑ or OÐÑ, they have the same meaning as X and O. SOA Exam MLC Study Guide S. roverman 2012

15 MLC SECTION 5 - EXERCISES 1. You are given the survival function WÐÑœ for. " Ð" Ñ Describe the event and calculate or formulate each of the following: (i) ;, (ii) ;, (iii) :, " " "l" " > (iv) >l? ;, (v) T ÒO" œ "Ó, (vi) T ÒO œ 5Ó, (vii) T ÒX" Ÿ "Ó, (viii) T ÒO" Ÿ "Ó 2. Show that T ÒX "lx &Ó œ &:&. MLC SECTION 5 - SOLUTION TO EXERCISES 1.(i) probability that Ð"Ñ dies by age 20, ; œ œ " œ œ Þ(&'. " " (ii) probability that Ð"Ñ survives to age 20 and dies by age 30, "l" " W ÐÑ W Ð$Ñ W Ð"Ñ Ð"Î" Ñ Ð"Î$" Ñ "Î"" ; œ œ œ Þ"%)&. (iii) probability that ÐÑ survives to at least age >, : œ œ œ > W Ð >Ñ "ÎÐ > "Ñ Ð "Ñ W ÐÑ "ÎÐ "Ñ Ð > "Ñ. W Ð"Ñ W ÐÑ "Î" $ W Ð"Ñ "Î"" %%" (iv) probability that ÐÑ survives to age > and dies by age >?, ; œ œ œ >l? W Ð >Ñ W Ð >?Ñ "ÎÐ > "Ñ Ó Ò"ÎÐ >? "Ñ Ó Ð "Ñ Ð "Ñ W ÐÑ "ÎÐ "Ñ Ð > "Ñ Ð >? "Ñ (v) probability that Ð"Ñ survives to age 20 and dies by age 21,. T ÒO œ "Ó œ ; œ œ œ Þ%% " "l " W ÐÑ W Ð"Ñ W Ð"Ñ. Ð"Î" Ñ Ð"Î Ñ "Î"" (vi) probability that ÐÑ survives to age 5 and dies before age 5 ", TÒO œ 5Ó œ ; œ œ œ 5l W Ð 5Ñ W Ð 5 "Ñ "ÎÐ 5 "Ñ Ó Ò"ÎÐ 5 Ñ Ó Ð "Ñ Ð "Ñ W ÐÑ "ÎÐ "Ñ Ð 5 "Ñ Ð 5 Ñ (vii) probability that Ð"Ñ dies by age 20, same as (i) above. (viii) probability that Ð"Ñ survives at most 10 complete years, which is the same as the probability that Ð"Ñ dies before age 21 (before 11 complete years), ; œ œ " œ Þ(& "" " W Ð"Ñ W Ð"Ñ W Ð"Ñ "Î "Î"". T ÒX "Ó : : : TÒX &Ó : : & & & & " & & & 2. T ÒX "lx &Ó œ œ œ œ :. SOA Exam MLC Study Guide S. roverman 2012

16 S. ROVERMAN MLC STUDY GUIDE PRACTICE EXAM 1 1. For a fully discrete 3-year endowment insurance of 1000 on ÐÑ, you are given: (i) 5P is the prospective loss random variable at time 5. ÞÞ (ii) 3 œ Þ" (iii) + À$l œ Þ(") (iv) Premiums are determined by the equivalence principle. Calculate P, given that ÐÑsurvives to the end of the third year from issue. A) 540 ) 630 C) 655 D) 720 E) For a double-decrement model: wð"ñ > (i) > % "' wðñ > (ii) > % * Ð7Ñ. % : œ" ß Ÿ>Ÿ% : œ" ß Ÿ>Ÿ$ Calculate. A) Less than.05 ) 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 For independent lives (35) and (45): (i) &: $& œ Þ* (ii) &:%& œ Þ) (iii) ;% œ Þ$ (iv) ;& œ Þ& Calculate the probability that the first death of (35) and (45) occurs in the 6th year. A) ) C) D) E) 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 Calculate "ß T $&. A) Less than 800 ) 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 SOA Exam MLC Study Guide S. roverman 2012

17 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 service b - 2.5% of the amount of final year salary in excess of 25,000 for each year of the first 25 completed years of service c - 3% of the first 25,000 of final year salary for each completed year of service in excess of 25 years d - 3.5% of the amount of final salary in excess of 25,000 for each completed year of service in excess of 25 years. If Ð%&Ñ has 10 years of service already and if GEW%& œ $&ß, then T EF'&Þ& can be expressed W in the form E W '& F. Find E F. %& A) 24,000 ) 24,250 C) 24,500 D) 24,750 E) 25, ^ is the present-value random variable for a whole life insurance of, payable at the moment of death of ÐÑ. You are given: (i) $ œ Þ% (ii). > œ Þ ß > (iii) The single benefit premium for this insurance is equal to Z+<Ð^Ñ. Calculate,. A) 2.75 ) 3.00 C) 3.25 D) 3.50 E) 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) enefits, 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 ) 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 SOA Exam MLC Study Guide S. roverman 2012

18 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 ) 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 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 ) 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 SOA Exam MLC Study Guide S. roverman 2012

19 10. For a fully discrete whole life insurance of, on ÐÑ, you are given: (i) ; * œ Þ*% (ii) 3 œ Þ$ (iii) The net amount at risk for policy year 10 is 872. ÞÞ (iv) + œ "%Þ'&*(' (v) * Z œ *'Þ" Calculate the initial benefit reserve for policy year 10. A) 295 ) 321 C) 343 D) 368 E) For a fully discrete 2-year endowment insurance of 1000 on ÐÑ, you are given: (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 ) 258 C) 268 D) 288 E) For an increasing 10-year term insurance, you are given: (i), 5 " œ "ß Ð" 5Ñ, 5 œ ß "ß ÞÞÞß * (ii) enefits 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 ) 15,780 C) 16,220 D) 16,780 E) 17,220 SOA Exam MLC Study Guide S. roverman 2012

20 13. For a fully discrete whole life insurance of 1000 on ÐÑ: (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 (v) 3 œ Þ" (vi) " " Z œ % Calculate the level annual contract premium. A) 100 ) 102 C) 104 D) 106 E) You are given: (i) X is the future lifetime random variable for ÐÑ. = = (ii) W œ " ß Ÿ Ÿ (iii) Z +<ÐX Ñ œ $&Þ" $ Calculate / $À%l A) 26.7 ) 27.7 C) 28.7 D) 29.7 E) A fully discrete 3-year term insurance of 10,000 on (40) is based on a double-decrement model, death and withdrawal: Ð"Ñ (i) Decrement 1 is death. (ii) ; 5 œ Þ ß 5 (iii) Decrement 2 is withdrawal, which occurs at the end of the year. wðñ (iv) ; œ Þ%, 5 œ ß "ß œ Þ*& % 5 Calculate the actuarial present value of the death benefits for this insurance. A) 492 ) 502 C) 512 D) 522 E) For a fully discrete 5-payment 10-year decreasing term insurance on (60), you are given: (i), 5 " œ "Ð" 5Ñ ß 5 œ ß "ß ß ÞÞÞß * (ii) ;' 5 œ Þ Þ"5 ß 5 œ ß "ß ß ÞÞÞß * (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 ) 194 C) 202 D) 210 E) SOA Exam MLC Study Guide S. roverman 2012

21 17. You are given: (i) X and X are not independent. C (ii) ; œ;, 5 œß"ßßþþþ 5 C 5 (iii) 5: C œ "Þ 5: 5: C, 5 œ "ß ß $ß ÞÞÞ (iv) / ÀC œ *Þ%%. Calculate ;. A).044 ).046 C).048 D).050 E) A bank offers the following choices for certificates of deposit: Term (in years) Nominal annual interest rate convertible quarterly % % % 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% ) 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, or (ii) (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? ÐÑ " E ÞÞ ÞÞ 7 I. II. + III. + ;. ÐÑ Ð Ñ 5 Ð"Ñ 5 5œ A) All ) All but I C) All but II D) All but III E) None of A,, C or D SOA Exam MLC Study Guide S. roverman 2012

22 20. You are given: (i). > œ Þ$ ß > (ii) $ œ Þ& (iii) is the future lifetime random variable. X (iv) 1 is the standard deviation of + Xl. Calculate T< Xl. A) 0.50 ) 0.54 C) 0.58 D) 0.62 E) (50) is an employee of XYZ Corporation. Future employment with XYZ follows a double decrement model: (i) Decrement 1 is retirement Ð"Ñ (ii). & > œ Þ Ÿ > & Þ & Ÿ > (iii) Decrement 2 is leaving employment with XYZ for all other causes ÐÑ (iv). & > 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 ) 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 For a life table with a one-year select period, you are given: (i) j ÒÓ. ÒÓ j " / ÒÓ ) " * )" * * )Þ (ii) Deaths are uniformly distributed over each year of age. Calculate. / Ò)Ó A) 8.1 ) 8.2 C) 8.3 D) 8.4 E) For a fully discrete 3-year endowment insurance of 1000 on ÐÑ: (i) 3 œ Þ& (ii) : œ : œ : " (iii) the second year benefit reserve is Z œ &'. Calculate :. A).62 ).64 C).66 D).68 E).70 SOA Exam MLC Study Guide S. roverman 2012

23 Þ& & Ÿ ' 24. You are given: ; œ Þ% ' Ÿ ( Calculate %l"%;&. A) 0.34 ) 0.36 C) 0.38 D) 0.40 E) A fully discrete whole life insurance policy issued at age has face amount 1,000,000. The policy expenses are: 1st Year Renewal Years Percent of Premium 20% 5% Face Amount 5 per 1000 " per 1000 Per Policy Settlement 1000 (at the end of the year of death) ÞÞ You are given that 3 œ Þ" and + œ )Þ)")(. Find the expense-loaded premium using the equivalence principle (nearest 1000). A) 22,000 ) 23,000 C) 24,000 D) 25,000 E) 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 ) 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 SOA Exam MLC Study Guide S. roverman 2012

24 27. You are pricing a special 3-year annuity-due on two independent lives, and ÐCÑ. The annuity pays 30,000 if both persons are alive ß 20,000 if ÐÑ dies first and 10,000 if ÐCÑ dies first. You are given: (i) 5 5: œ 5: C " Þ*" Þ) $ Þ( (ii) 3 œ Þ& Calculate the actuarial present value of this annuity (nearest 100). A) 78,300 ) 80,400 C) 82,500 D) 84,700 E) 86, Company AC sets the contract premium for a continuous life annuity of 1 per year on ÐÑ equal to the single benefit premium calculated using: Þ for > Ÿ " (i) $ œþ$ (ii). > œ Þ" for > " However, a revised mortality assumption reflects future mortality improvement and is given by. > œþß> Calculate the expected gain at issue for AC (using the revised mortality assumption) as a percentage of the contract premium. A) 2% ) 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 8 independent policies of this type, using the 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 ) 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 SOA Exam MLC Study Guide S. roverman 2012

25 30. For independent lives ÐÑ and ÐCÑ, State 1 is that ÐÑ and ÐCÑ are alive, and State 2 that ÐÑ is alive but ÐCÑ has died, State 3 is the ÐCÑ is alive but ÐÑ has died, and State 4 that both ÐÑ and ÐCÑ have died. You are given: "Î W ÐÑ œ Ð" Ñ ß Ÿ Ÿ " " C " WCÐCÑ œ " ß Ÿ C Ÿ " Ð8Ñ : is the probability that ÐÑand ÐCÑare in State 4at time 8 " given that they are in State 34 3 at time 8. At time 0, ÐÑ is age 54 and ÐCÑ is age 75. Ð"Ñ Calculate : "ß. A) Less than ) At least 0.026, but less than C) At least 0.039, but less than D) At least 0.052, but less than E) At least SOA Exam MLC Study Guide S. roverman 2012

26 S. ROVERMAN MLC STUDY GUIDE PRACTICE EXAM 1 SOLUTIONS 1. The equivalence principle premium is "T œ "Ð ÞÞ.Ñ œ "Ð Ñ œ (*Þ". À$l " " Þ" + Þ(") "Þ" À$l We are given that ÐÑ survives to the end of the third year from issue. Using the end of the 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:. Ð7Ñ Ð7Ñ : Ð7Ñ wð"ñ wðñ.> 2.. % > œ > % Ð7Ñ Þ > : % œ > : % > : % œ Ð" ÑÐ" Ñ œ " : > %..> > Ð7Ñ > %> $ $ß % )) "ß%%ß Ð Ñ )) "ß%%ß Ð Ñ > > % % % "'ß > :% " &(' " "ß%%ß &(' "ß%%ß : % > > > > "' * &(' "ß%%ß 7 œ p. 7 œ œ Þ""$. Answer: C 3. The probability that the first death occurs in the 6th year is &l; $&À%&. This can be formulated as &l; $&À%& œ &: $&À%& ': $&À%& œ &: $& &:%& ': $& ':%& œ &: $& &:%& Ð" :% :& Ñ From the given information we have :% œ Þ*( and :& œ Þ*&. Then, &l; $&À%& œ ÐÞ*ÑÐÞ)ÑÒ" ÐÞ*(ÑÐÞ*&ÑÓ œ Þ&'&. Answer: 4. We use the equivalence principle relationship APV expense-loaded premium œ APV benefit plus expenses. ÞÞ ÞÞ ÞÞ ÞÞ K+ œ "ß E Þ"K+ &+ &+ Þ $& $& $& $& $& Then "$% œ "ß T$& Þ"Ð"$%Ñ & &, so that "ß T œ )$&Þ'. Answer: D $& SOA Exam MLC Study Guide S. roverman 2012

27 5. TEF'&Þ& œ (.02)(25,000)(25) + (.025) 35,000 25,000 (25) S + (.03)(25,000)(5) + (.035) 35,000 S 65 25,000 (5) = , p E F œ %ß &. Answer:. S S S S The variance of the continuous whole life insurance with face amount, is Z+<Ò^Óœ, ÒE ÐE ÑÓ. Since the force of mortality is constant at.02 and $ œþ%,. Þ ". Þ " E œ $. œ.% œ $ and E œ $. œ Þ) œ &, so that " " %, Z+<Ò^Óœ, Ò Ð ÑÓœ. The single benefit premium is " & $ %&,E œ,. We are told that $ " %,, œ, from which we get,œ$þ(&. 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 U (assume to be paid the start of each half-year Þ& during the first year). The APV of the premiums is : Ó. Þ& $ $ $ "l $ l $ The APV of the benefit is "@; &@ ; &@ ;. From the Illustrative Table, we have ; $ œ Þ"&$ ß ; $" œ Þ"'" and ; $ œ Þ"(. Þ& Using UDD, the APV of premiums is Ð" Þ&ÐÞ"&$ÑÑÓ œ "Þ*'"&"%U. The APV of the benefit is $ "@ÐÞ"&$Ñ &@ ÐÞ**)%(ÑÐÞ"'"Ñ &@ ÐÞ**)%(ÑÐÞ**)$*ÑÐÞ"(Ñ œ Þ%%)* Þ Þ%%)* "Þ*'"&"% Then U œ œ "Þ%&. Answer: E 8. Let ; denote the probability of dying during the second year. Then the number of deaths R in the second year has a binomial distribution based on 7 œ " trials and success (dying) probability ;. The variance of the binomial is Z +<ÒRÓ œ 7;Ð" ;Ñ œ ";Ð" ;Ñ. ; can be formulated as ; œ TÒsurvive 1st year and die in 2nd year Ó, which is equal to ÐßÑ ÐßÑ Ðß"Ñ Ð"ßÑ œu U U U œðþ(ñðþ"ñ ÐÞÑÐÞ&ÑœÞ" (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 Z +<ÒRÓ œ "ÐÞ"ÑÐÞ))Ñ œ "Þ&'. Answer: C SOA Exam MLC Study Guide S. roverman 2012

28 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 Ð"ß%Ñ time. The probability of this is U œ Þ" (that is the probability of an active individual dying 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 Ð"ß"Ñ Ð"ß%Ñ U U œ ÐÞ%ÑÐÞ"Ñ œ Þ%. The probability of becoming disabled as of the start of the 3rd Ð"ßÑ Ðß%Ñ year and then dying in the 3rd year is U U œ ÐÞÑÐÞ$Ñ œ Þ'Þ The combined probability of surviving to the start of the 3rd year and not withdrawing, and then dying in the 3rd year, is Þ% Þ' œ Þ". The APV at the beginning of the 2nd year of the death benefit for death in the 3rd year is "@ ÐÞ"Ñ œ *Þ(. The total APV of the death benefit is *&Þ% *Þ( œ ")&Þ*%. Answer: A 10. The initial benefit reserve for policy year 10 is *Z T œ *'Þ" T (where T is the " benefit premium). T œ,t œ,ð.ñ œ Þ$*)), Þ ÞÞ + 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 Ð* Z T ÑÐ" 3Ñ Ð, " Z Ñ; * œ " Z, which becomes Ð*'Þ" Þ$*)),ÑÐ"Þ$Ñ Ð)(ÑÐÞ*%Ñ œ, )( p, œ ". Then, T œ,t œ "ÐÞ$*))Ñ œ %'Þ*"Þ 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 " ÞÞ %(* œ "Ð ÞÞ.Ñ, so that + œ œ "Þ)&&$. + Àl Àl Then, " ÐÞ*%Ñ: œ "Þ)&&$ so that : œ Þ***. Then, '') U@: œ Ñ, so that U œ &). Answer: SOA Exam MLC Study Guide S. roverman 2012

29 12. We are given "ß ÐMEÑ œ "&ß &&". We wish to find "ß ÐMEÑ. We use the relationship " " %À"l %"À"l 8 " " " " À8l À8l 8l "À8l ÐMEÑ œ ÐMEÑ 8@ ;. This can be seen by looking at the time line of possible death benefit payments; the first row is the sum of the second and third rows. " $ ÞÞÞ 8 " 8 8 " ÐMEÑ" o " $ 8 " 8 À8l E" o " " " " " ÐMEÑ o " 8 8 " 8 " "À8l 8 " 8@ 8l ; 8 "" " " % " "l % %À"l %À"l %"À"l Then, ÐMEÑ œ ÐMEÑ "@ ; " E " œ ":% E& œ ÐÞ"'"$Ñ ÐÞ&$''(ÑÐÞ%*&Ñ œ Þ('' %À"l (we use values from the Illustrative Table, and note : œ I ). "" "l % " " % & " "Þ' " "Þ' " " % " % : ; œ ÐÞ&$''(ÑÐÞ&*Ñ œ Þ**(. Then Þ"&&&" œ Þ('' Ð" Þ()Ñ ÐMEÑ "ÐÞ**(Ñ. Solving for ÐMEÑ results in ÐMEÑ œ Þ"'() " " %"À"l %"À"l " %"À"l We then multiply by 100,000 to get "'ß (). Answer: D 13. The accumulation of asset share in the 1st year is Ò KÐÞ'ÑÓÐ"Þ"Ñ "; œ : Ð"'Þ)%Ñ. If we knew the value of ;, we could find K. Using the recursive relationship for benefit reserve for the 1st year, we have Ò )ÓÐ"Þ"Ñ "; œ : Ð%Ñ, and solving for ; results in ; œ Þ&. Then Þ'KÐ"Þ"Ñ "ÐÞ&Ñ œ ÐÞ*&Ñ Ð"'Þ)%Ñ p K œ ". Answer: A 14. From the form of =ÐÑ we see that survival from birth follows De Moivre's Law with upper Ð= $Ñ $ " > > '& > > = *& > $ '& % % % $À%l > $ '& > Ð'& >Ñ '& Ð'&Ñ >œ age =. From Z +<ÒX ÑÓ œ œ $&Þ" we get = œ *&. Under De Moivre's Law, : œ " œ " so : œ. / œ :.> œ.> œ œ (Þ'*. Answer: SOA Exam MLC Study Guide S. roverman 2012

30 Ð"Ñ Ð"Ñ $ Ð"Ñ % "l % l % Ð"Ñ wð"ñ ; œ ; 15. The APV of the death benefit is "ß ; Ó. Since decrement 2 occurs at the end of the year, for each year, ÐÑ wð"ñ wðñ. ; œ Ð" ; Ñ ; Ð"Ñ wð"ñ For decrement 1, we have ; œ ; œ Þ for œ %ß %"ß %, since the force of decrement is Ð7Ñ wð"ñ wðñ Ð"Ñ Ð7Ñ Ð"Ñ Then, "l % % %" Ð"Ñ Ð7Ñ Ð"Ñ Ð7Ñ Ð7Ñ Ð"Ñ l % % % % %" % constant. Also : œ : : œ ÐÞ*)Ñ ÐÞ*'Ñ œ Þ*%) for œ %ß %". ; œ : ; œ ÐÞ*%)ÑÐÞÑ œ Þ"))"' and ; œ : ; œ : : ; œ ÐÞ*%)Ñ ÐÞÑ œ Þ"((. The APV of the death benefit is, and $ "ß ÐÞ"((ÑÓ œ &". Answer: C 16. We use the recursive reserve formula, Ð5Z TÑÐ" 3Ñ, 5 " ; 5 œ: 5 5 " Z. Also, Z œ for benefit reserves. For the first year, we have Ð T ÑÐ"Þ'Ñ "ß ÐÞÑ œ Þ*) " Z p " Z œ "Þ)"'T %Þ). For the second year, we have Ð"Þ)"'T %Þ) T ÑÐ"Þ'Ñ *ß ÐÞ"Ñ œ Þ*(* Z œ ÐÞ*(*ÑÐ((Þ''Ñ œ ('Þ$ p T œ "). Answer: E 17. / C œ / / C / C is a valid relationship for all survival distributions of X ÐÑ and X ÐCÑ, whether or not they are independent. Since ; œ Þ& for all 5, it follows that : œ Þ*& for all 5, and then / œ : œ > : Ð: Ñ œ 5 > " : >œ" >œ" > $ > " > " : Also, / C œ " :, since ; 5 œ ; C 5 for all 5. / C œ > : C œ > Ð: Ñ "ÞÐ: Ñ œ "Þ " Ð: Ñ œ *Þ%%. >œ" >œ". This follows from the fact that " " < : œ : : â: œ ÐÞ*&Ñ and " < < < â œ. Then : œ Þ*&, and : œ Þ*& and ; œ Þ&. Answer: D 5 SOA Exam MLC Study Guide S. roverman 2012

31 % 18. The annual effective rate for the 1 year certificate is Ð"Þ"Ñ " œ Þ%', for the 3-year certificate it is % Ð"Þ"&Ñ " œ Þ&*, and for the 5-year certificate it is % 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 (ii) a 3 year certificate combined with 3 one-year certificates (in any order), annual effective rate is " " "Î' ÒÐ"Þ"&Ñ Ð"Þ"Ñ Ó " œ Þ%&) (we have found the 6-year accumulation and then the equivalent annual effective rate that would compound to the same amount in 6 years). (iii) Two 3-year certificates, annual effective rate (iv) A one-year certificate and a 5-year certificate (either order), annual effective rate is % "Î' ÒÐ"Þ"%"&Ñ Ð"Þ"Ñ Ó " œ Þ&%). The maximum annual effective return is.0548 and is obtained with a 5-year and a 1-year certificate, in either order. Answer: D ÐÑ Ð"Ñ Ð"Ñ k x k x k x k x k x k=0 k=0 k=0 k=0 19. APV = k v 1 q = k ( 7) v 1 q + q = k ( 7) v p + k Ð Ñ Ð Ñ v q Ð"Ñ x k x k=0 k 1 k ÐÑ k k ÐÑ 1 k k x k x ÐÑ Ð Ñ d j± x k=0 k=0 k=0 k=0 j=0 1 1 v 1 d k ÐÑ j d ÐÑ d d j ÐÑ j± x j± x j± x j=0 k=j j=0 j=0 7 = a ( ) + k v q, so that III is correct. Also, v 1 q = v v q = v q x = v q = q = 1 v q =, so that I is correct. II has no meaning, and is incorrect. Answer: C 1 A d ÐÑ " $. 20. With constant force of mortality. œ Þ$ and force of interest $ œ Þ&, + œ œ "Þ&. The variance of + XÐÑl is " ".. " Þ$ Þ$ ÒE ÐE ÑÓœ Ò Ð ÑÓœ Ò Ð ÑÓœ$'Þ' $ $ $. $. ÐÞ&Ñ Þ" Þ$ Þ& Þ$. The standard deviation is $'Þ' œ 'Þ. We wish to find T Ò"Þ& 'Þ + "Þ& 'ÞÓ œ T Ò 'Þ& + ")Þ&]. Xl 8l 68ÐÞ'(&Ñ Þ& " / Þ& We solve for 8, from the equation + œ 'Þ&, so that œ 'Þ&, so that / œ Þ'(& (which is equivalent to 8 œ œ (Þ)' years) ß and we solve for 7 from the equation 7l " / Þ& Þ&7 Þ&7 Þ&8 + œ ")Þ&, so that œ ")Þ&, so that / œ Þ(& 68ÐÞ(&Ñ (which is equivalent to 7 œ Þ& œ &"Þ)" years) T Ò 'Þ& + ")Þ&] œ T Ò(Þ)' X ÐÑ &"Þ)"Ó œ : : Þ$Ð(Þ)'Ñ Þ$Ð&"Þ)"Ñ œ / / œ Þ(* Þ"" œ Þ&(*. Answer: C XÐÑl (Þ)' &"Þ)" Xl Þ&8 SOA Exam MLC Study Guide S. roverman 2012

32 ÐÑ ÐÑ Ð7Ñ ÐÑ 21. The probability is ";&, which can be written as &;& &:& &;&&. ÐÑ & Ð Ñ ÐÑ & Þ&> ; œ : 7 Þ&.> œ " /. / ÐÞ&Ñ.> œ ÐÞ&Ñ œ Þ" ; & & > & > Þ& Ð"Ñ ÐÑ Ð7Ñ alternatively, since. > for, we have > > ÐÑ Þ& for Ÿ> &, so that &;& œ" /. Ð7Ñ Ð"Ñ ÐÑ Ð7Ñ Þ&> Ÿ > &ß.&& > œ.&& >. && > œ Þ& > :&& œ / ÐÑ & Ð7Ñ ÐÑ & Þ& & Þ&> && > " / &&.&& > Þ& ÐÑ ÐÑ Ð7Ñ ÐÑ " & & & & & & && œ Ÿ> & ; œ ; œ" / Þ&> For, so that and ; œ :.> œ / ÐÞ$Ñ.> œ ÐÞ$Ñ œ Þ"$(. Then, ; œ ; : ; œ Þ" ÐÞ(())ÑÐÞ"$(Ñ œ Þ$%'. Answer: C 22. With a one-year select period, / Ò)"Ó " œ / ), so that / œ / : / œ " :.> : / œ " Ð" ; Ñ.> : / Ò)"Ó Ò)"ÓÀ"l Ò)"Ó ) > Ò)"Ó Ò)"Ó ) > Ò)"Ó Ò)"Ó ) " " Ò)"Ó Ò)"Ó * )$ ) ) * * ) * ; Ò)"Ó œ œ Þ*() )Þ œ Þ*&"" Þ* / * ) œ Ð" > ; Ñ.> : / œ Ð" >Ñ.> Ð Ñ / œ Þ*&"" Þ* / (using UDD and ). Therefore, so that / œ )Þ$&. From the table we have j œ j œ *" and j œ j œ )$, so that ) Ò)Ó " )" Ò)"Ó " ) )$ ) )" *" and )" *". Then, we use the relationship : œ ; œ / Ò)Ó œ / : / : / to solve for / Ò)ÓÀ"l Ò)Ó )"À"l Ò)Ó ) Ò)Ó. " " From UDD we have / œ Ð" Þ*>Ñ.> œ Þ*&& and / œ ) Ð" >Ñ.> œ Þ*&', Ò)ÓÀ"l )"À"l )$ : Ò)Ó œ Þ*" ß : Ò) œ : Ò)Ó :)" œ ÐÞ*"ÑÐ *" Ñ œ Þ)$. Then / œ Þ*&& ÐÞ*"ÑÐÞ*&'Ñ ÐÞ)$ÑÐ)Þ$&Ñ, so that / œ )Þ%*%. Answer: E Ò)Ó Ò)Ó *" 23. The 2nd year terminal reserve for a 3-year endowment insurance can be formulated as ÞÞ + ZÀ$l œ " ÞÞ À"l + ß À$l ÞÞ ÞÞ : : where + œ" and : œ". À"l À3l "Þ& Ð"Þ&Ñ : : Ð"Þ&Ñ "Þ& Then Z œ " œ Þ&', so that "Þ"*( œ. À$l " " : : "Þ& Ð"Þ&Ñ Solving the quadratic equation, or substituting the possible answers results in :œþ(. Answer: E 24. %l"%;& œ %:& "):&. % " ) %:& œ ÐÞ*&Ñ œ Þ)"%& ß "):& œ ":& ):' œ ÐÞ*&Ñ ÐÞ*'Ñ œ Þ%$"*. Then %l"%;& œ Þ)"%& Þ%$"* œ Þ$)'. Answer: C SOA Exam MLC Study Guide S. roverman 2012

33 ÞÞ 25. K+ œ "ß ß E ÞK Þ&K+ &ß " Ð" Ñ+ "E Kœ E œ ".+ œ " Ð)Þ)")(Ñ œ Þ"*)$ Þ K œ "ß"ßE 'ß "+ ÞÞ Þ*&+ Þ"& Þ ÞÞ Þ" "Þ" "ß"ßÐÞ"*)$Ñ 'ß "Ð(Þ)")(Ñ Þ*&Ð)Þ)")(Ñ Þ"& œ &ß **&Þ. Answer: E ÞÞ 26. If the annual benefit premium is U, then U + œ UÐMEÑ "ß E. %À&l " &l % %À&l ÞÞ ÞÞ ÞÞ ÞÞ To find + %À&l, we use the relationship + % œ + %À&l & I % +%&. Using values from the ÞÞ ÞÞ Illustrative Table, we have "%Þ)"'' œ + ÐÞ($&*ÑÐ"%Þ"""Ñ, so that + œ %Þ%%". %À&l Also, &l E% œ & I % E%& œ ÐÞ($&*ÑÐÞ"Ñ œ Þ"%(*%. Then %Þ%%"Ð$$'$Ñ œ Ð$$'$ÑÐMEÑ "ß ÐÞ"%(*%Ñ, so that ÐMEÑ œ Þ%" Þ Answer: C " " %À&l %À&l %À&l 27. The APV of the annuity is ÞÞ ÞÞ ÞÞ ÞÞ ÞÞ $ß + ß Ð+ + Ñ "ß Ð+ + Ñ ÞÞ ÞÞ ÞÞ œ ß + CÀ$l "ß + À$l œ $ß + À$l. ÞÞ + : œ Þ'"%$ CÀ$l CÀ$l CÀ$l À$l CÀ$l. À$l so the APV is 78, 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. ased on the original mortality assumption, the contract premium is + œ " : + " œ Þ$> Þ> Þ$Ð"Ñ ÞÐ"Ñ " / /.> / / À"l " " Þ&Ð"Ñ " / Þ&Ð"Ñ " Þ& Þ$ Þ". Þ$ Þ" œ / œ $Þ$ The loss at issue is P œ ] $Þ$, where ] is the PVRV of the continuous life annuity. The expected loss at issue will be IÒPÓ œ IÒ] Ó $Þ$, where IÒ] Ó is calculated based on the " " $. Þ$ Þ revised mortality assumption. IÒ] Ó œ + œ œ œ. Then IÒPÓ œ $Þ$ œ $Þ$. Answer: C SOA Exam MLC Study Guide S. roverman 2012

34 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"'+ Þ"& " " œ ÐÞ"'ÑÐ Ñ œ Þ' Þ"& Þ' Þ"& (&, " ^ " ^ Þ"' Þ"' $ Þ' Þ' Þ' Þ"' Þ"' Þ' Þ' Þ"' Þ"& Þ"& Þ' Þ" Þ"& Þ' Þ"&. Since ]œ œ, Pcan be written as PœÐ" Ñ^ and then Z+<ÒPÓœÐ" ÑZ+<Ò^ÓœÐ" ÑÐE E Ñ œ Ð" Ñ Ò Ð Ñ Ó œ Þ""%*% If the total loss is denoted W, then W œp P â P (sum of losses on each of the 8 8 (& " 8 policies), and IÒWÓ œ 8IÒPÓ œ, and Z +<ÒWÓ œ 8Z +<ÒPÓ œ Þ""%*%8. W IÒWÓ Z+<ÒWÓ IÒWÓ Z+<ÒWÓ IÒWÓ Z+<ÒWÓ Using the normal approximation, T ÐW Ñ œ T Ð Ñ. In order for this probability to be.05, we must have œ "Þ'%&, ( 8 Ñ (& or equivalently, œ "Þ'%&. Solving for 8 results in 8 œ "($(. Þ""%*%8 The number of independent policies needed to have a positive loss probability of.025, say 7, ( m Ñ (& must satisfy the equation œ "Þ*', and solving for 7 results in 7 œ %''. Þ""%*% m The additional number of policies needed is %'' "($( œ (*. Answer: D Ð"Ñ "ß 30. : is the probability that for ÐÑ and ÐCÑ, alive at age 64 and 85, ÐÑ will survive the year, but ÐCÑ will die within the year. This probability that (64) survives the year is : for ÐÑ, which W Ð'&Ñ WÐ'%Ñ WCÐ)'Ñ WCÐ)&Ñ Ð"Ñ C "ß '% )& is œ Þ*)'"$. The probability that (85) dies during the year is ; for ÐCÑ, which is " œ Þ'''( ( ÐCÑ's mortality follows De Moivre's Law with = œ ") Þ Then, : œ : ; œ ÐÞ*)'"$ÑÐÞ'''(Ñ œ Þ''. Answer: E )& '% SOA Exam MLC Study Guide S. roverman 2012