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S. BROVERMAN STUDY GUIDE FOR THE SOCIETY OF ACTUARIES EXAM MLC 2008 EDITION VOLUME 2 POISSON PROCESSES, MARKOV CHAINS, AND PRACTICE EXAMS Samuel Broverman, ASA, PHD [email protected] www.sambroverman.com

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Page 1: Broverman MLC 2008 Volume 2

S. BROVERMAN STUDY GUIDE

FOR THE

SOCIETY OF ACTUARIESEXAM MLC

2008 EDITION

VOLUME 2POISSON PROCESSES,

MARKOV CHAINS,AND PRACTICE EXAMS

Samuel Broverman, ASA, PHD

[email protected]

Page 2: Broverman MLC 2008 Volume 2

(copyright © S. Broverman, 2008)

[email protected]

Page 3: Broverman MLC 2008 Volume 2

SOA Exam MLC Study Guide © S. Broverman, 2008 www.sambroverman.com

SOA EXAM MLC STUDY GUIDE - VOLUME 2POISSON PROCESSES,

MARKOV CHAINS, AND PRACTICE EXAMS

TABLE OF CONTENTS

POISSON PROCESSESNotes and Examples PP-1 to PP-14Problem Set PP-15 to PP-30

MULTI-STATE TRANSITION MODELS (MARKOV CHAINS) (MC)Notes and Examples MC-1 to MC-16Problem Set MC-17 to MC-34

PRACTICE EXAMSPractice Exam 1 and Solutions PE-1 to PE-20Practice Exam 2 and Solutions PE-21 to PE-40Practice Exam 3 and Solutions PE-41 to PE-60Practice Exam 4 and Solutions PE-61 to PE-78Practice Exam 5 and Solutions PE-79 to PE-102Practice Exam 6 and Solutions PE-103 to PE-124Practice Exam 7 and Solutions PE-125 to PE-142Practice Exam 8 and Solutions PE-143 to PE-166Practice Exam 9 and Solutions PE-167 to PE-188Practice Exam 10 and Solutions PE-189 to PE-210Practice Exam 11 and Solutions PE-211 to PE-232Practice Exam 12 and Solutions PE-233 to PE-254

MAY 2007 SOA EXAM MLC AND SOLUTIONS MLC07-1 to MLC07-22REFERENCE BY TOPIC FOR MAY 2007 EXAM MLC

Page 4: Broverman MLC 2008 Volume 2

S. BROVERMAN A.S.A. , Ph.D.PREPARATION SEMINARS FOR ACTUARIAL EXAMINATIONS

[email protected] www.sambroverman.com

Professor Samuel Broverman, Professor of Actuarial Science at the University of Toronto, will beconducting exam preparation seminars for the Fall 2008 examination period. Seminars will take place inOctober 2008 for SOA Exams FM, MLC and C. Seminars will be held in New York and Chicago.

Schedule of Fall 2008 Seminars

New York (Midtown area) Chicago (O'Hare area)Exam FM (2 days) October 18-19 Exam MLC (3 days) October 11-13 September 26-28Exam C (4 days) October 2-5 October 14-17

Fall 2008 seminar fees: Exam FM - $295 , Exam MLC - $575 , Exam C - $675

To register for a seminar or for more information please contact Prof. Broverman either by telephone at416-966-9111, or by e-mail at [email protected] . A registration form and updated information aboutthe seminars will also be available at Prof. Broverman's website: www.sambroverman.com .

Dates for seminars in Spring 2009 have not yet been finalized. Please call or e-mail Prof. Broverman forthat information.

Prof. Broverman has over 30 years experience in actuarial education and is the author of "Mathematicsand Investment and Credit", which is one of the catalog references for Exam FM (CAS 2) and the EA1exam. He is also the author of the ACTEX study guides for Exam P and Exam C .

The recommended study material for the Exam C seminar is the ACTEX manual written by Prof.Broverman. For the Exam FM and MLC seminars, Prof. Broverman will provide electronic versions ofstudy guides to accompany the seminars. The emphasis in the class will be on developing and refiningproblem solving skills and identifying and classifying recurring problem types, with time also devoted toreviewing and reinforcing fundamental principles and understanding. Prof. Broverman will be availableboth before and after the seminar to answer questions via e-mail.

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SOA Exam MLC Study Guide © S. Broverman, 2008 www.sambroverman.com

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 material for Exam MLC is divided into one large topic and two smaller topics. The largetopic is life contingencies, and the smaller topics are Poisson processes and multi-state transition(Markov Chain) models. I think that the proper order in which to study the topics is the order inwhich they are listed in the previous sentence.

The study guide is divided into two volumes. Volume 1 consists of review notes, examples andproblem sets for life contingencies. Volume 2 covers the other topics with review notes,examples and problem sets. Volume 2 also contains 12 practice exams of 30 questions eachalong with the May 2007 MLC exam and solutions. There are over 160 examples, over 300problems in the problem sets and 390 questions in the 12 practice exams and May 2007 SOAexam. All of these (about 850) questions have detailed solutions. The notes are broken up intosections (32 sections for life contingencies, and one section each for Poisson Processes andMarkov Chains). Each section has a suggested time frame.

Most of the examples in the notes and almost half of the problems in the problem sets are fromolder SOA or CAS exams (pre-2007) on the relevant topics. The 12 practice exams in Volume 2include many questions from SOA exams released from 2000 to 2006. The practice exams have30 questions each and are designed to be similar to actual 3-hour exams. The SOA and CASquestions are copyrighted by the SOA and CAS, and I gratefully acknowledge that I have beenpermitted to include them in this study guide.

Because of the time constraint on the exam, a crucial aspect of exam taking is the ability to 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 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 examquestion, 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.

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www.sambroverman.com SOA Exam MLC Study Guide © S. Broverman, 2008

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.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 should beavailable for download from the Society of Actuaries website. It is recommended that you havethem available while studying.

Based on the weight applied to topics on recent actual exams, I have created the practice examsto include about 24 questions on life contingencies and 3 each on Poisson processes and multi-state transition models.

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 May, 2008Department 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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PRACTICE EXAMS

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PRACTICE EXAM 1

SOA Exam MLC Study Guide © S. Broverman, 2008 www.sambroverman.com PE-1

S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 1

1. For a fully discrete 3-year endowment insurance of 1000 on , you are given: (i) is the prospective loss random variable at time . (ii) (iii)

(iv) Premiums are determined by the equivalence principle.Calculate , given that dies in the second year from issue.A) 540 B) 630 C) 655 D) 720 E) 910

2. For a double-decrement model: (i) (ii)

Calculate .A) 0.025 B) 0.038 C) 0.050 D) 0.063 E) 0.075

3. For independent lives (35) and (45): (i) (ii) (iii) (iv) Calculate the probability that the last death of (35) and (45) occurs in the 6th year.A) 0.0095 B) 0.0105 C) 0.0115 D) 0.0125 E) 0.0135

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) .Calculate the level annual expense-loaded premium using the equivalence principle.A) 930 B) 1041 C) 1142 D) 1234 E) 1352

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PE-2 www.sambroverman.com SOA Exam MLC Study Guide © S. Broverman, 2008

5. Kings of Fredonia drink glasses of wine at a Poisson rate of 2 glasses per day. Assassinsattempt to poison the king's wine glasses. There is a 0.01 probability that any given glass ispoisoned. Drinking poisoned wine is always fatal instantly and is the only cause of death. Theoccurrences of poison in the glasses and number of glasses drunk are independent events.Calculate the probability that the current king survives at least 30 days.A) 0.40 B) 0.45 C) 0.50 D) 0.55 E) 0.60

6. is the present-value random variable for a whole life insurance of payable at the momentof death of . You are given: (i) (ii) (iii) The single benefit premium for this insurance is equal to .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: (iv) Mortality follows the Illustrative Life Table.(v) Deaths are uniformly distributed within each year of age.(vi)Calculate the amount of each semiannual benefit premium for this insurance.A) 1.3 B) 1.4 C) 1.5 D) 1.6 E) 1.7

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SOA Exam MLC Study Guide © S. Broverman, 2008 www.sambroverman.com PE-3

8. For a Markov model with three states, Healthy (0), Disabled (1), and Dead (2):(i) The annual transition matrix is given by (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 within the first two years.A) 11 B) 14 C) 17 D) 20 E) 23

9. An insurance company issues a special 3-year insurance to a high-risk individual. You aregiven the following homogeneous Markov chain model:(i) State 1: active State 2: disabled State 3: withdrawn State 4: deadTransition probability matrix: (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)(v) The insured is disabled at the end of year 1.Calculate the actuarial present value of the prospective death benefits at the beginning of year 2.A) 440 B) 528 C) 634 D) 712 E) 803

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PE-4 www.sambroverman.com SOA Exam MLC Study Guide © S. Broverman, 2008

10. For a fully discrete whole life insurance of on , you are given:(i) (ii) (iii) The initial benefit reserve for policy year 10 is 343.(iv) The net amount at risk for policy year 10 is 872.(v)Calculate the terminal benefit reserve for policy year 9.A) 280 B) 288 C) 296 D) 304 E) 312

11. For a special fully discrete 2-year endowment insurance of 1000 on , you are given:(i) The first year benefit premium is 668.(ii) The second year benefit premium is 258.(iii)Calculate the level annual premium using the equivalence principle.A) 469 B) 479 C) 489 D) 499 E) 509

12. For an increasing 10-year term insurance, you are given:(i) , (ii) Benefits are payable at the end of the year of death.(iii) Mortality follows the Illustrative Life Table.(iv)(v) The single benefit premium for this insurance on (41) is 16,736.Calculate the single benefit premium for this insurance on (40).A) 12,700 B) 13,600 C) 14,500 D) 15,500 E) 16,300

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SOA Exam MLC Study Guide © S. Broverman, 2008 www.sambroverman.com PE-5

13. For a fully discrete whole life insurance of 1000 on :(i) Death is the only decrement.(ii) The annual benefit premium is 80.(iii) The annual contract premium is 100.(iv) Expenses in year 1, payable at the start of the year, are 40% of contract premiums.(v) (vi) Calculate the asset share at the end of the first year.A) 17 B) 18 C) 19 D) 20 E) 21

14. A fully discrete 3-year term insurance of 10,000 on (40) is based on a double-decrementmodel, death and withdrawal:(i) Decrement 1 is death. (ii) (iii) Decrement 2 is withdrawal, which occurs at the end of the year.(iv) , (v) Calculate the actuarial present value of the death benefits for this insurance.A) 487 B) 497 C) 507 D) 517 E) 527

15. You are given:(i) (ii) °

(iii) is the future lifetime random variable for .Calculate .A) 332 B) 352 C) 372 D) 392 E) 412

16. For a fully discrete 5-payment 10-year decreasing term insurance on (60), you are given:(i)(ii) Level benefit premiums are payable for five years and equal 218.15 each.(iii)(iv)Calculate , the benefit reserve at the end of year 2.A) 70 B) 72 C) 74 D) 76 E) 78

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PE-6 www.sambroverman.com SOA Exam MLC Study Guide © S. Broverman, 2008

17. You are given:(i) and are not independent.(ii) , (iii) , Into which of the following ranges does , the curtate expectation of life of the last survivorstatus, fall?A) B) C) D) E) B)

18. Subway trains arrive at your station at a Poisson rate of 20 per hour. 25% of the trains areexpress and 75% are local. The types and number of trains arriving are independent. An expressgets you to work in 16 minutes and a local gets you there in 28 minutes. You always take thefirst train to arrive. Your co-worker always takes the first express. You are both waiting at thesame station. Calculate the conditional probability that you arrive at work before your co-worker, given that a local arrives first.A) 37% B) 40% C) 43% D) 46% E) 49%

19. Beginning with the first full moon in October deer are hit by cars at a Poisson rate of 20 perday. The time between when a deer is hit and when it is discovered by highway maintenance hasan exponential distribution with a mean of 7 days. The number hit and the times until they arediscovered are independent. Calculate the expected number of deer that will be discovered in thefirst 10 days following the first full moon in October.A) 78 B) 82 C) 86 D) 90 E) 94

20. You are given:(i) (ii) (iii) is the future lifetime random variable. (iv) is the standard deviation of .

Calculate .

A) 0.53 B) 0.56 C) 0.63 D) 0.68 E) 0.79

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SOA Exam MLC Study Guide © S. Broverman, 2008 www.sambroverman.com PE-7

21. (50) is an employee of XYZ Corporation. Future employment with XYZ follows a doubledecrement model:(i) Decrement 1 is retirement

(ii)

(iii) Decrement 2 is leaving employment with XYZ for all other causes

(iv)

(v) If (50) leaves employment with XYZ, he will never rejoin XYZ.Calculate the probability that (50) will retire from XYZ before age 60.A) 0.069 B) 0.074 C) 0.079 D) 0.084 E) 0.089

22. For a life table with a one-year select period, you are given:(i) °

(ii) Deaths are uniformly distributed over each year of age.Calculate .°

A) 8.0 B) 8.1 C) 8.2 D) 8.3 E) 8.4

23. For a fully discrete 3-year endowment insurance of 1000 on :(i) (ii) Calculate the second year terminal benefit reserve.A) 526 B) 632 C) 739 D) 845 E) 952

24. You are given:

Calculate .A) 0.38 B) 0.39 C) 0.41 D) 0.43 E) 0.44

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PE-8 www.sambroverman.com SOA Exam MLC Study Guide © S. Broverman, 2008

25. For a fully discrete whole life insurance of 1000 on (50), you are given:(i) The annual per policy expense is 1.(ii) There is an additional first year expense of 15.(iii) The claim settlement expense of 50 is payable when the claim is paid.(iv) All expenses, except the claim settlement expense, are paid at the beginning of the year.(v) Mortality follows DeMoivre's law with .(vi)Calculate the level expense-loaded premium using the equivalence principle.A) 27 B) 28 C) 29 D) 30 E) 31

26. For a special fully discrete 5-year deferred whole life insurance of 100,000 on (40), you aregiven:(i) The death benefit during the 5-year deferral period is return of benefit premiums paid withoutinterest.(ii) Annual benefit premiums are payable only during the deferral period.(iii) Mortality follows the Illustrative Life Table.(iv) (v)

Calculate the annual benefit premium.A) 3300 B) 3320 C) 3340 D) 3360 E) 3380

27. You are pricing a special 3-year annuity-due on two independent lives, both age 80. Theannuity pays 30,000 if both persons are alive and 20,000 if only one person is alive.You are given:(i) (ii)Calculate the actuarial present value of this annuity.A) 78,300 B) 80,400 C) 82,500 D) 84,700 E) 86,800

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SOA Exam MLC Study Guide © S. Broverman, 2008 www.sambroverman.com PE-9

28. Company ABC sets the contract premium for a continuous life annuity of 1 per year on equal to the single benefit premium calculated using:(i) (ii) However, a revised mortality assumption reflects future mortality improvement and is given by

forfor

Calculate the expected loss at issue for ABC (using the revised mortality assumption) as apercentage of the contract premium.A) 2% B) 8% C) 15% D) 20% E) 23%

29. A group of 1000 lives each age 30 sets up a fund to pay 1000 at the end of the first year foreach member who dies in the first year, and 500 at the end of the second year for each memberwho dies in the second year. Each member pays into the fund an amount equal to the singlebenefit premium for a special 2-year term insurance, with:(i) Benefits: (ii) Mortality follows the Illustrative Life Table.(iii)The actual experience of the fund is as follows: Interest Rate Earned Number of Deaths Calculate the difference, at the end of the second year, between the expected size of the fund asprojected at time 0 and the actual fund.A) 840 B) 870 C) 900 D) 930 E) 960

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PE-10 www.sambroverman.com SOA Exam MLC Study Guide © S. Broverman, 2008

30. For independent lives and , State 1 is that and are alive, and State 2 that isalive but has died, State 3 is the is alive but has died, and State 4 that both and

have died.You are given:••• is the probability that and are in State at time given that they are in State

at time .• At time 0, is age 54 and is age 75.Calculate .A) Less than 0.026 B) At least 0.026, but less than 0.039C) At least 0.039, but less than 0.052 D) At least 0.052, but less than 0.065E) At least 0.065

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SOA Exam MLC Study Guide © S. Broverman, 2008 www.sambroverman.com PE-11

S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 1 SOLUTIONS

1. The equivalence principle premium is .

We are given that dies in the second year. Using the end of the first year as a reference point,there will be the death benefit of 1000 paid one year later (end of the second year) and there willbe one premium received just at the start of the second year. is the present value, value at theend of the first year, of the insurance payment minus the present value of the future premiums.This will be . Answer: B

2.

. Answer: E

3. The probability that the last death occurs in the 6-th year is .This can be formulated as (for independent lives, ).From the given information .Also, and .Then . Answer: B

4. We use the equivalence principle relationshipAPV expense-loaded premium APV benefit plus expenses.

Solving for results in .

Answer: D

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PE-12 www.sambroverman.com SOA Exam MLC Study Guide © S. Broverman, 2008

5. denotes the Poisson process of the number of glasses of wine drunk in days; hasparameter per day. Each glass of wine drunk has a .01 chance of being poisoned.

denotes the number of glasses of wine that are poisoned in days; is a Poissonprocess with parameter per day. The number of glasses of wine that arepoisoned in 30 days has a Poisson distribution with a mean of . The current kingwill survive at least 30 days if no glasses of wine are poisoned in the next 30 days. Theprobability of this is . Answer: D

6. The variance of the continuous whole life insurance with face amount is . Since the force of mortality is constant at .02 and ,

. .and , so that

. The single benefit premium is . We are told that

, 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-yearduring the first year). The APV of the premiums is .The APV of the benefit is .From the Illustrative Table, we have and .Using UDD, the APV of premiums is .The APV of the benefit is

Then . Answer: A

8. Let denote the probability of dying within the first two years. Then the number of deaths in the first two years has a binomial distribution based on trials and success (dying)probability . The variance of the binomial is . can be formulated as die in the 1st year survive 1st year and die in 2nd year .

die in the 1st year ,survive 1st year and die in 2nd year

(this is the combination of staying healthy for the 1st year and dying in the 2nd year, orbecoming disabled in the 1st year and dying in the 2nd year).Therefore, and the . Answer: C

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SOA Exam MLC Study Guide © S. Broverman, 2008 www.sambroverman.com PE-13

9. At the beginning of year 2 the individual is disabled, there are still 2 years left on the 3-yearinsurance policy. If the individual dies in the 2nd year, there will be a benefit of 1000 paid at thattime. The probability of this is (that is the probability of a disabled individual dyingduring the year). The APV at time 2 of the death benefit for death in the 2nd year is

.The individual can survive the 2nd year and die in the 3rd year, but the benefit will only bepayable if the individual is active or disabled at the start of the 3rd year. The probability ofremaining disabled to the start of the 3rd year and then dying in the 3rd year is

. The probability of returning to active as of the start of the 3rdyear and then dying in the 3rd year is The combined probabilityof surviving to the start of the 3rd year and not withdrawing, and then dying in the 3rd year, is

. The APV at the beginning of the 2nd year of the death benefit for death in the3rd year is . The total APV of the death benefit is

. Answer: A

10. The initial benefit reserve for policy year 10 is (where is the benefitpremium). The net amount at risk for policy year 10 is .Using the net amount at risk form of the recursive relationship for benefit reserve, for year 10,we have , which becomes

.Then, . Then , so that

. Answer: C

11. The level annual premium based on the equivalence principle is , where .

If we find we can find the premium.A key point in solving the problem is .

This follows from

We are given .

Using , we solve for .Then , and . Answer: B

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PE-14 www.sambroverman.com SOA Exam MLC Study Guide © S. Broverman, 2008

12. We are given . We wish to find .

We use the relationship . This can be seen by

looking at the time line of possible death benefit payments; the first row is the sum of the secondand third rows.

Then,

(we use values from the Illustrative Table, and notice that ).Also .Then .

We then multiply by 100,000 to get . Answer: D

13. Assuming a starting asset share of 0, the accumulation of asset share in the 1st year is .

If we knew the value of , we could find .Using the recursive relationship for benefit reserve for the 1st year, we have

, and solving for results in .Then . Answer: A

14. The APV of the death benefit is .Since decrement 2 occurs at the end of the year, for each year, , and

.For decrement 1, we have for , since theforce of decrement is constant. Also for .Then, and

.The APV of the death benefit is

. Answer: C

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SOA Exam MLC Study Guide © S. Broverman, 2008 www.sambroverman.com PE-15

15. From the form of we see that survival from birth follows DeMoivre's Law with upper

age . Once is found, . Under DeMoivre's Law, ° .

Solving for results in . Then . Answer: B

16. We use the recursive reserve formula, .Also, for benefit reserves. For the first year, we have

.For the second year, we have

. Answer: E

17. is a valid relationship for all survival distributions of and ,whether or not they are independent. Since for al , it follows that for

all , and then . This follows from the fact that

and .Also, , since for all .

.

Then . Answer: D

18. Given that a local train arrives first, you will get to work 28 minutes after that local trainarrives, since you will take it. Your co-worker will wait for first express train. You will get towork before your co-worker if the next express train (after the local) arrives more than 12minutes after the local. We expect 5 express trains per hour, so the time between express trains isexponentially distributed with a mean of of an hour, or 12 minutes. Because of the lack ofmemory property of the exponential distribution, since we are given that the next train is local,the time until the next express train after that is exponential with a mean of 12 minutes.Therefore, the probability that after the local, the next express arrives in more than 12 minutes is

, where has an exponential distribution with a mean of 12. This probability is (37%). Answer: A

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19. We expect a deer to be hit by a car every days. For the deer that is expected to behit at days, the chance of being discovered within the first 10 days is the probability ofbeing discovered within days after being hit. Since time of discovery after being hithas an exponential distribution with mean 7 days, this probability is (the prob.

, where is exponential with mean 7). The expected number of deer

discovered within 10 days following the first full moon in October is ,

since each term in the sum is the expected number of deer discovered for the one deer hit at time. The sum goes to 199 since the 200-th deer is expected to be hit just at time 10, and cannot be

discovered before time 10.

(round up to the next integer value 94) . Answer: E

20. With constant force of mortality and force of interest , .

The variance of is

.

The standard deviation is .We wish to find ] .

We solve for , from the equation , so that , so that

(which is equivalent to years).If dies exactly at that time , then , so it follows that is equal to

(the present value of the annuity is if lives atleast years). Since , it follows that .Therefore, ] . Answer: E

21. The probability is , which can be written as .Since the force of decrement for retirement is 0 to age 55, .Also, for , so that .For , so that and

.Then, . Answer: A

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22. With a one-year select period, , so that° °° ° ° ° °

° °

(using UDD and ).From the table we have and , so that and .Then, we use the relationship to solve for .° ° ° ° °

From UDD we have and ,° °

. Then° ° , so that .

Finally . Answer: C° ° °

23. The 2nd year terminal reserve for a 3-year endowment insurance can be formulated as

where and .3

Then . For face amount 1000, the reserve is 526. Answer: A

24. . .

.We must split the probability because of the change in force of mortality at age 60.

. Answer: A

25. APV expense-loaded premium APV benefit plus expense .

From DeMoivre's law, , so that

.

Then .

The premium equation becomes so that . Answer: E

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26. If the annual benefit premium is , then .

To find , we use the relationship . Using values from the

Illustrative Table, we have , so that .

Also, .Then , so that . Answer: D

27. The APV of the annuity can be formulated as an annuity of 20,000 per year while at leastone is alive, combined with an additional 10,000 per year while both are alive (this makes for atotal payment of 30,000 per year while both are alive). The APV is

.

Both and are 80. and

. Since the two lives are independent, we have

, and , so that .

The APV of the annuity is . Answer: B

28. The loss at issue is the PVRV (present value random variable) of benefit to be paid minus thePVRV of premium to be received. Since this is a single premium policy, the premium received isa single payment equal to the contract premium. Based on the original mortality assumption, thecontract premium is (the is the continuous annuity value for a

constant force of mortality). The loss at issue is , where is the PVRV of thecontinuous life annuity. The expected loss at issue will be , where iscalculated based on the revised mortality assumption.

.

We have split the whole life annuity into age intervals over which the force of mortality isconstant. Then , which is 15% of the contract premium. Answer: C

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29. The single benefit premium per person is .Using the Illustrative Table, we have and .The single benefit premium is . Since this is a single benefit premium, the expected sizeof the fund at the end of the 2-year term is 0. The actual fund progresses in the following way.

is collected at the start of the first year.At the end of the first year, with interest minus death benefit, the fund value is

.With interest minus death benefits to the end of the second year, the fund value is

. Answer: C

30. is the probability that for and , alive at age 64 and 85, will survive theyear, but will die within the year. This probability that (64) survives the year is for ,which is . The probability that (85) dies during the year is for , which

is ( 's mortality follows DeMoivre's Law with )

Then, . Answer: E

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 2

1. For a special whole life insurance on , you are given:(i) is the present value random variable for this insurance.(ii) Death benefits are paid at the moment of death.(iii) , (iv) (v) , CalculateA) 0.075 B) 0.080 C) 0.085 D) 0.090 E) 0.0995

2. For a whole life insurance of 1 on , you are given:(i) Benefits are payable at the moment of death.(ii) Level premiums are payable at the beginning of each year.(iii)(iv)(v)Calculate the 10th year terminal benefit reserve for this insurance.A) 0.18 B) 0.25 C) 0.26 D) 0.27 E) 0.30

3. A special whole life insurance of 100,000 payable at the moment of death of includes adouble indemnity provision. This provision pays during the first ten years an additional benefitof 100,000 at the moment of death for death by accidental means. You are given:(i) , (ii) , , where is the force of decrement due to death by accidentalmeans.(iii)Calculate the single benefit premium for this insurance.A) 1640 B) 1710 C) 1790 D) 1870 E) 1970

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4. Kevin and Kira are modeling the future lifetime of (60).(i) Kevin uses a double decrement model 60 1000 120 80 61 800 160 80 62 560 - -(ii) Kira uses a non-homogeneous Markov model: (a) The states are 0 (alive), 1 (death due to cause 1), 2 (death due to cause 2). (b) is the transition matrix from age 60 to 61; is the transition matrix from age 61 to 62.(iii) The two models produce equal probabilities of decrement.Calculate .

A) B) C)

D) E)

5. A certain species of flower has three states: sustainable, endangered and extinct. Transitionsbetween states are modeled as a non-homogeneous Markov chain with transition matrices asfollows: Sustainable Endangered Extinct

SustainableEndangeredExtinct

Calculate the probability that a species endangered at the start of year 1 will ever become extinct.A) 0.45 B) 0.47 C) 0.49 D) 0.51 E) 0.53

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6. For a special 3-year term insurance:(i) Insureds may be in one of three states at the beginning of each year: active, disabled, or dead.All insureds are initially active. The annual transition probabilities are as follows: Active Disabled DeadActive 0.8 0.1 0.1Disabled 0.1 0.7 0.2Dead 0.0 0.0 1.0

(ii) A 100,000 benefit is payable at the end of the year of death whether the insured was active ordisabled.(iii) Premiums are paid at the beginning of each year when active. Insureds do not pay anyannual premiums when they are disabled.(iv)Calculate the level annual benefit premium for this insurance.A) 9,000 B) 10,700 C) 11,800 D) 13,200 E) 20,800

7. Customers arrive at a bank according to a Poisson process at the rate of 100 per hour. 20% ofthem make only a deposit, 30% make only a withdrawal and the remaining 50% are there only tocomplain. Deposit amounts are distributed with mean 8000 and standard deviation 1000.Withdrawal amounts have mean 5000 and standard deviation 2000. The number of customersand their activities are mutually independent. Using the normal approximation, calculate theprobability that for an 8-hour day the total withdrawals of the bank will exceed the total deposits.A) 0.27 B) 0.30 C) 0.33 D) 0.36 E) 0.39

8. A Mars probe has two batteries. Once a battery is activated, its future lifetime is exponentialwith mean 1 year. The first battery is activated when the probe lands on Mars. The secondbattery is activated when the first fails. Battery lifetimes after activation are independent. Theprobe transmits data until both batteries have failed. Calculate the probability the probe istransmitting data three years after landing.A) 0.05 B) 0.10 C) 0.15 D) 0.20 E) 0.25

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9. For a special fully discrete 30-payment whole life insurance on 45, you are given:(i) The death benefit of 1000 is payable at the end of the year of death.(ii) The benefit premium for this insurance is equal to for the first 15 years followed byan increased level annual premium for the remaining 15 years.(iii) Mortality follows the Illustrative Table.(iv)Calculate .A) 16.8 B) 17.3 C) 17.8 D) 18.3 E) 18.8

10. For a special fully discrete 2-year endowment insurance on :(i) The pure endowment is 2000.(ii) The death benefit for year is plus the benefit reserve at the end of year , .(iii) is the level annual benefit premium.(iv)(v) , CalculateA) 1027 B) 1047 C) 1067 D) 1087 E) 1107

11. For a group of 250 individuals age , you are given:(i) The future lifetimes are independent.(ii) Each individual is paid 500 at the beginning of each year, if living.(iii)(iv)(v)Using the normal approximation, calculate the size of the fund needed at inception in order to be90% certain of having enough money to pay the life annuities.A) 1.43 million B) 1.53 million C) 1.63 million D) 1.73 million E) 1.83 million

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12. For a double decrement table, you are given Age 40 1000 60 55 41 70 42 750 Each decrement is uniformly distributed over each year of age in the double decrement table.Calculate .A) 0.077 B) 0.078 C) 0.079 D) 0.080 E) 0.081

13. The actuarial department for the SharpPoint Corporation models the lifetime of pencilsharpeners from purchase using a generalized DeMoivre model with ,for and .A senior actuary examining mortality tables for pencil sharpeners has determined that theoriginal value of must change. You are given:(i) The new complete expectation of life at purchase is half what it was previously.(ii) The new force of mortality for pencil sharpeners as 2.25 times the previous force of mortalityfor all durations.(iii) remains the same.Calculate the original value of .A) 1 B) 2 C) 3 D) 4 E) 5

14. You are given:(i) is the future lifetime random variable.(ii) , (iii) .Calculate .°

A) 2.6 B) 5.4 C) 6.3 D) 9.5 E) 10.0

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15. For a fully discrete 15-payment whole life insurance of 100,000 on ( ), you are given:(i) The expense-loaded level annual premium using the equivalence principle is 4669.95 .(ii)(iii)

(iv)(v) Expenses are incurred at the beginning of the year.(vi) Percent of premium expenses are 10% in the first year and 2% thereafter.(vii) Per policy expenses are in the first year and 5 in each year thereafter until death.Calculate .A) 10.0 B) 16.5 C) 23.0 D) 29.5 E) 36.5

16. For the future lifetimes of and :(i) With probability 0.4, (i.e. deaths occur simultaneously)(ii) With probability 0.6, the joint density function is , , .Calculate .A) 0.30 B) 0.32 C) 0.34 D) 0.36 E) 0.38

17. For an insurance on and :(i) Upon the first death, the survivor receives the single benefit premium for a whole lifeinsurance of 10,000 payable at the moment of death.(ii) while both are alive(iii)(iv) After the first death, for the survivor.(v)Calculate the actuarial present value of this insurance on and .A) 4500 B) 5400 C) 6000 D) 7100 E) 7500

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18. For a population whose mortality follows DeMoivre's law you are given:(i) (ii) ° ° ° °Calculate .A) 3.0 B) 3.5 C) 4.0 D) 4.5 E) 5.0

19. Kevin and Kira are in a history competition:(i) In each round, every child still in the contest faces one question. A child is out as soon as heor she misses one question. The contest will last at least 5 rounds.(ii) For each question, Kevin's probability and Kira's probability of answering that questioncorrectly are each 0.8; their answers are independent.Calculate the conditional probability that both Kevin and Kira are out by the start of round 5given that at least one of them participates in round 3.A) 0.13 B) 0.16 C) 0.29 D) 0.22 E) 0.25

20. For a special increasing whole life annuity-due on , you are given:(i) is the present value random variable.(ii) Payments are made once every 30 years, beginning immediately.(iii) The payment in year 1 is 10, and payments increase by 10 every 30 years.(iv) Mortality follows DeMoivre's Law with .(v)Calculate .A) 10.5 B) 11.0 C) 11.5 D) 12.0 E) 12.5

21. For a fully discrete whole life insurance of 1000 on , you are given: Calculate .A) 279 B) 282 C) 284 D) 286 E) 288

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22. For a fully discrete whole life insurance of 1000 on , you are given:(i) The expenses, payable at the beginning of the year, are: Expense Type First Year Renewal Years % of Premium 20% 6% Per Policy 8 2(ii) The level expense-loaded premium is 41.20.(iii)Calculate the value of the expense augmented loss variable , , if the insured dies in the 3rdpolicy year.A) 770 B) 790 C) 810 D) 830 E) 850

23. For a special 3-year term insurance on , you are given:(i) is the present-value random variable for this insurance.(ii) , (iii) The following benefits are payable at the end of the year of death: (iv)Calculate .A) 9,600 B) 10,000 C) 10,400 D) 10,800 E) 11,200

24. The graph of a piecewise linear survival function, , consists of 3 line segments withendpoints .Calculate .

A) 0.69 B) 0.71 C) 0.73 D) 0.75 E) 0.77

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25. For a 3-year fully discrete term insurance of 1000 on , subject to a double decrementmodel:(i) (ii) Decrement 1 is death. Decrement 2 is withdrawal.(iii) There are no withdrawal benefits.(iv)Calculate the level annual benefit premium for this insurance.A) 14.3 B) 14.7 C) 15.1 D) 15.5 E) 15.7

26. For a fully continuous whole life insurance of 1 on , you are given:(i) The force of mortality is 0.05 in the first 10 years and 0.08 thereafter.(ii)Calculate the benefit reserve at time 10 for this insurance.A) 0.144 B) 0.155 C) 0.166 D) 0.177 E) 0.188

27. For a 10-payment, 20-year term insurance of 100,000 on Pat:(i) Death benefits are payable at the moment of death.(ii) Contract premiums of 1600 are payable annually at the beginning of each year for 10 years.(iii)(iv) is the loss random variable at the time of issue.Calculate the minimum value of as a function of the time of death of Pat.A) B) C) D) E)

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28. Which of the following is/are true?1. A counting process is said to possess independent increments if the number of events thatoccur between times and is independent of the number of events that occur between time and for all .2. All Poisson processes have stationary and independent increments.3. The assumption of stationary and independent increments is essentially equivalent toasserting that at any point in time the process probabilistically restarts itself.A) 1 only B) 2 only C) 3 only D) 1 and 2 only E) 2 and 3 only

29. For a fully discrete 20-year term life insurance of 1 on a life aged 90, you are given:• 10% for the first two years and 6% thereafter.• Mortality follows the Illustrative Life Table.Calculate .

A) Less than 0.71 B) At least 0.71, but less than 0.73 C) At least 0.73, but less than 0.75D) At least 0.75, but less than 0.77 E) At least 0.77

30. You are given the following 2-year Select-and-Ultimate table: Calculate .A) Less than 0.066 B) At least 0.066, but less than 0.067C) At least 0.067, but less than 0.068 D) At least 0.068, but less than 0.069E) At least 0.069

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 2 SOLUTIONS

1. With constant force of mortality of .02, survival probability is .

Then . Answer: C

2. This is a semicontinuous whole life insurance. The 10th year reserve isa . Under UDD, ...

From the given information, 2 .

Then, . Answer: C

3. The basic death benefit has APV ; this is the APV of a

continuous insurance with constant force of mortality .001 and force of interest .06.The APV of the additional accidental death benefit is

.The total single benefit premium is . Answer: C

4. According to Kevin's two-decrement model, the survival probability at age 61 is

. This is the transition probability from state 0 (alive) at age

61 to state 0 at age 62. The only matrix at age 61 with is D. Answer: D

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5. Note that the notation used in this question is not consistent with the notation in the Danielstudy note. In the Daniel study note denotes the transition matrix from time to but inthis question it denotes the transition matrix from to . We see that a species that is eithersustainable or endangered at the start of the 4th year will never become extinct, because thetransition probability to extinct is 0. Therefore, a flower that is endangered at the start of the 1styear can only become extinct in the 1st, 2nd or 3rd year. We denote the states Sustainable,Endangered and Extinct as states 1, 2 and 3, respectively. Then is the probability that aflower endangered at the start of the first year is extinct at the start of the 4th year, and so this isthe probability of ever becoming extinct. is the (2,3)-entry of the 3-step transition matrixfor the first 3 years. The 3-step matrix is . Since we only want the (2,3)-entry,we "multiply" the 2nd row of by the 3rd column of . The second row of is

Then, the (2,3) entry of is . Answer: C

6. Suppose the level benefit premium is . The APV of premiums iswhere is the probability of being active at the start of the 2nd year, and is the probability ofbeing active at the start of the 3rd year. . The probability of being active at the start ofthe 3rd year is 8 1 (this is a combination of two cases: (a) active atthe start of the second year and active at the start of the 3rd year, and (b) disabled at the start ofthe 2nd year and active at the start of the 3rd year). The APV of premiums is

(note that ). The probability of dying inthe 1st year is .1, and the probability of dying in the 2nd year is (beactive at the start of the 2nd year and then die in 2nd year, or be disabled at the start of the 2ndyear and then die in the second year). The probability of dying in the 3rd year is

(these are the probabilities of thecombinations Time: 0 1 2 3 Prob. State Active Active Active Dead State Active Active Disabled Dead State Active Disabled Active Dead State Active Disabled Disabled Dead The APV of the insurance is .To find the annual benefit premium we set APV of premium equal to APV of benefit, so that

. This results in . Answer: B

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7. Because of independence, we can separate the deposit process and the withdrawal process astwo independent processes. The rate per hour at which depositors arrive is , andthe rate per hour of withdrawers arriving is 30. The number of depositors arriving in an 8-hourday has a Poisson distribution with a mean of , and the number of withdrawersarriving in an 8-hour day has a Poisson distribution with a mean of .The total amount deposited in a day has a compound Poisson distribution with Poisson parameter160, and individual deposit amount (severity distribution) with mean 8000 and standarddeviation 1000. The mean and variance of the total deposit in an 8-hour day is

and

.In a similar way, we get the mean and variance of the total withdrawals in an 8-hour day:

and

.We wish to find using the normal approximation. ,and since and are independent,

.Then, using the normal approximation,

. Answer: A

8. There are a few ways to approach this problem. One approach is the convolution approach tofinding the distribution function of the sum of random variables and . If and arecontinuous independent non-negative random variables, the

. In this case, and are both exponential with mean 1,and , this becomes

. This is the probability that the total time until failure of bothbatteries is . The probability that total time until failure is is .An alternative solution is based on the observation that if and are independent exponentialrandom variables both with mean 1, then has a gamma distribution with and

, and the pdf of is .Then .

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8. continuedYet another approach is to note that with exponential inter-event time with mean 1 year, thenumber of failures forms a Poisson process with a rate of 1 per year. The probe will betransmitting in 3 years if there is at most one battery failure in the 3 year period. The number offailures in a 3-year period, , is Poisson with mean 3, so the probability is

. Answer: D

9.

Also, .

Therefore, and .

From the Illustrative Table, , , and

.Then . Answer: B

10. We use the recursive reserve relationship .At time 0, , and at the time of endowment , endowment amount .For the first year ( ) , since , we have

, so that .Then, for the second year , , so that

.Solving for results in .An alternative approach uses the following general relationship for reserves:

.

With for this policy, we have and , so that

so that . Answer: A

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11. Let denote the aggregate present value random variable of the 250 annuities.

Then , where each , and is the PVRV for a life annuity-due of 1 per

year. Then , and

.Then, , and

, and from the independence of the lives, .

We want the initial fund amount so that . Applying the normalapproximation, we get .

From the normal table, we get , so that . Answer: A

12. Under the UDD in multiple table assumption, .In this problem we have , and

2 , so that .Then, , , and , so that and . Answer: A

13. Under the generalized DeMoivre model, the force of mortality at age is , and the

complete expectation at age is .°

Suppose that the new parameter for is denoted . Then the new force of mortality at age is , which we are told is . Therefore, .

The new complete expectation at purchase (age 0) is , which we are told is of .

Therefore , so that .

Solving for results in . Answer: D

14. Constant force of mortality is equivalent to having an exponential distribution withmean and . , so that .

° . Answer: C

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15. APV expense-loaded premium APV benefit APV expenses . and

This equation becomes .

Solving for results in . Answer: A

16. The event is the same as the event .Therefore,

.Since the joint density is the conditional joint density given that

, and it is constant on the rectangle , .Therefore, the probability is equal to

(area of the region of probability) . In the following graph, the shaded regionrepresents the region of probability .

The area of the shaded region is (area of non-shaded region) .The probability becomes . Answer: D

17. After the moment of the first death, the force of mortality for the survivor is .1.The value of an insurance of 10,000 based on force of mortality .1 is .

This is the amount paid at the moment of the first death. Since the force of failure for the jointlife status is constant at .12, the value of an insurance of 7,143 paid at the moment of the firstdeath (failure of the joint life status) is . Answer: B

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18. For DeMoivre's Law, . The question doesn't explicitly state it, but we must°

assume that the lives are independent.° .Note that if two independent lives have the same age , the joint life status has force offailure This is the same as a generalized DeMoivre'sLaw with and the same . Therefore In the same way, .°

Then, ° ° ° °

and .° ° ° °

Therefore, , from which we get .

Then, , so that . °

If , another representation for is .°

This tends to be awkward unless , in which it is , and underDeMoivre's Law this is Answer: E

19. Let be the event that both are out by round 5, and let be the event that at least one ofthem participates in round 3. We are to find .

is the probability that Kevin correctly answers questions in a row, and the same istrue for Kira. is the probability that at least one is participating in round 3, which is

. is the event that at least one participates in round 3. This is the disjoint union of

both participate in round 3 , only Kevin participates in round 3, andonly Kira participates in round 3.

Then .

.

.Then, .Finally, . Answer: E

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20. There will definitely be a payment of 10 now, and there is a possible payment of 20 in 30years, and a possible payment of 30 in 60 years. Those are the only payments.Only the first payment is if death is within 30 years, and the first two payments are made if deathis between 30 and 60 years. All three payments are made if death is after 60 years.

Therefore prob. .prob.

prob.Then and and .Answer: E

21. We use the equivalent recursive relationships for reserves, and .For we have .We can find if we know and .For we have , so that and for we have , so that .Then, , and we get , and then .

Then, . Answer: D

22. present value of benefit expenses present value of premiums .If death occurs in the third year, then the benefit is payable at the end of the third year, and thepresent value is . There will be 3 years of expenses,

at the start of the first year, and at the startof the second and third years. The present value of the expenses is

.The present value of the 3 premiums is .Then . Answer: A

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23. .

. .

. Answer: C

24. so that .

. From piecewise linearity on the interval from to ,

we see that since and , it follows that .since 15 is .6 of the way from 0 to 25.Similarly, , since 35 is .2 of the way from 25 to 75 ,and , .Therefore, . Answer: A

25. The annual benefit premium satisfies the equivalence principle relationship , which becomes

Solving for (notice that the 2000 can be cleared in the denominator first) results in .Answer: B

26. We will assume that there is a level continuous annual benefit premium .Then ,

where and .

Then, , and the benefit reserve at time 10 is

Note also that and the prospective benefit reserve at time 10 is

. Answer: A

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27. PVRV benefit PVRV premiumThe later death occurs, the smaller the PV of benefit and the larger the PV of premium received.Therefore, the minimum occurs at the latest date of death. If Pat survives beyond the 20 yearterm then no benefit is paid but 10 premiums were received, so that the loss is

. This is minimum possible loss. Answer: C

28. 1. The time intervals must be disjoint for independence. False.2. There can be non-homogeneous Poisson processes. False.3. True. Answer: C

29.

is calculate using 10% interest rate, so

is found using 6% interest, so we can use insurance values from the Illustrative Table and

the following identity

(since the Illustrative Table last age is 110, the 18-year term insurance at

age 92 is almost exactly equal to the whole life insurance at age 92).Then . Answer: C

30. . Answer: C

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 3

1. Michael, age 45, is a professional motorcycle jumping stuntman who plans to retire in threeyears. He purchases a three-year term insurance policy. The policy pays 500,000 for death froma stunt accident and nothing for death from other causes. The benefit is paid at the end of theyear of death. You are given:(i)(ii) where represents deaths from stunt accidents and represents deaths from other causes.(iii) Level annual benefit premiums are payable at the beginning of each year.(iv) Premiums are determined using the equivalence principle.Calculate the annual benefit premium.A) 920 B) 1030 C) 1130 D) 1240 E) 1350

2. You are given the survival function , Calculate , the 50-year temporary complete expectation of life of .°

A) 27 B) 30 C) 34 D) 37 E) 41

3. For a fully discrete whole life insurance of 1000 on , you are given:(i)(ii)(iii)(iv)CalculateA) 170 B) 172 C) 174 D) 176

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4. For a pension plan portfolio, you are given:(i) 80 individuals with mutually independent future lifetimes are each to receive a whole lifeannuity-due.(ii)(iii) Age Number of Annual annuity of annuitants payment 65 50 2 9.8969 0.43980 0.23603 75 30 1 7.2170 0.59149 0.38681Using the normal approximation, calculate the 95th percentile of the distribution of the presentvalue random variable of this portfolio.A) 1220 B) 1239 C) 1258 D) 1277 E) 1296

5. The time elapsed between claims processed is modeled such that represent the timeelapsed between processing the and the claim. ( time until the first claim isprocessed).You are given:(i) are mutually independent.(ii) The pdf of each is , , where is measured in minutes.Calculate the probability of at least two claims being processed in a ten minute period.A) 0.2 B) 0.3 C) 0.4 D) 0.5 E) 0.6

6. A casino has a game that makes payouts at a Poisson rate of 5 per hour and the payoutamounts are 1,2,3,... without limit. The probability that any given payout is equal to is .Payouts are independent. Calculate the probability that there are no payouts of 1, 2, or 3 in agiven 20 minute period.A) 0.08 B) 0.13 C) 0.18 D) 0.23 E) 0.28

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7. You arrive at a subway station at 6:15. Until 7:00, trains arrive at a Poisson rate of 1 train per30 minutes. Starting at 7:00, they arrive at a Poisson rate of 2 trains per 30 minutes. Calculateyour expected waiting time until a train arrives.A) 24 minutes B) 25 minutes C) 26 minutes D) 27 minutes E) 28 minutes

8. For a fully discrete 20-year endowment insurance of 10,000 on that has been in force of15 years, you are given:(i) Mortality follows the Illustrative Life Table.(ii)(iii) At issue, the benefit premium was calculated using the equivalence principle.(iv) When the insured decides to stop paying premiums after 15 years, the death benefit remainsat 10,000 but the pure endowment value is reduced such that the expected prospective loss at age60 is unchanged.Calculate the reduced pure endowment value.A) 8120 B) 8500 C) 8880 D) 9260 E) 9640

9. For a whole life insurance of 1 on with benefits payable at the moment of death, you aregiven:

(i)

(ii)

Calculate the present value of this insurance.A) 0.59 B) 0.61 C) 0.64 D) 0.66 E) 0.68

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10. For a fully continuous whole life insurance on , you are given:(i) The benefit is 2000 for death by accidental means (decrement 1).(ii) The benefit is 1000 for death by other means (decrement 2).(iii) The initial expense at issue is 50.(iv) Settlement expenses are 5% of the benefit, payable at the moment of death.(v) Maintenance expenses are 3 per year, payable continuously.(vi) The gross or contract premium is 100 per year, payable continuously.(vii)(viii)(ix)Calculate the actuarial present value of the insurer's expense augmented loss random variable forthis insurance.A) 446 B) C) 0 D) 223 E) 446

11. A homogeneous Markov model has three states representing the status of the members of apopulation.State 1 healthy, no benefitsState 2 disabled, receiving Home Health Care benefitsState 3 disabled, receiving Nursing Home benefits

The annual transition matrix is given by

Transitions occur at the end of each year. At the start of year 1, there are 50 members, all in state1, healthy. Calculate the variance of the number of those 50 members who will be receivingNursing Home benefits during year 3.A) 2.3 B) 2.7 C) 4.4 D) 4.5 E) 4.6

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12. A non-homogeneous Markov model has:(i) Three states: 0, 1, and 2(ii) Annual transition matrix as follows:

for , and for

An individual starts out in state 0 and transitions occur mid-year.An insurance is provided whereby:(i) A premium of 1 is paid at the beginning of each year that an individual is in state 0 or 1.(ii) A benefit of 4 is paid at the end of any year that the individual is in state 1 at the end of theyear.(iii)Calculate the actuarial present value of premiums minus the actuarial present value of benefits atthe start of this insurance.A) 0.17 B) 0.00 C) 0.34 D) 0.50 E) 0.66

13. You are given the following information on participants entering a special 2-year programfor treatment of a disease:(i) Only 10% survive to the end of the second year.(ii) The force of mortality is constant within each year.(iii) The force of mortality for year 2 is three times the force of mortality for year 1.Calculate the probability that a participant who survives to the end of month 3 dies by the end ofmonth 21.A) 0.61 B) 0.66 C) 0.71 D) 0.75 E) 0.82

14. In a population, non-smokers have a force of mortality equal to one half that of smokers. Fornon-smokers, , .Calculate for a smoker and a non-smoker with independent future lifetimes.°A) 18.3 B) 20.4 C) 22.1 D) 24.5 E) 26.8

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15. For a special fully discrete 20-year term insurance on :(i) The death benefit is 1000 during the first ten years and 2000 during the next ten years.(ii) The benefit premium, determined by the equivalence principle, is for each of the first tenyears and for each of the next ten years.(iii)

(iv)

Calculate .A) 2.9 B) 3.0 C) 3.1 D) 3.2 E) 3.3

16. For a fully discrete whole life insurance of 25,000 on , you are given:(i)(ii)

(iii)

Calculate .A) 4420 B) 4460 C) 4500 D) 4540 E) 4580

17. You are given 3 mortality assumptions:(i) Illustrative Life Table (ILT),(ii) Constant force model (CF), where (iii) DeMoivre model (DM), where .For the constant force and DeMoivre models, is the same as for the Illustrative Life Table.Rank for these 3 models.

A) ILT CF DM B) ILT DM CF C) CF DM ILTD) DM CF ILT E) DM ILT CF

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18. A population of 1000 lives age 60 is subject to 3 decrements, death (1), disability (2), andretirement (3). You are given:(i) The following absolute rates of decrement: (ii) Decrements are uniformly distributed over each year of age in the multiple decrement table.Calculate the expected number of people who will retire before age 62.A) 248 B) 254 C) 260 D) 266 E) 272

19. You are given:(i) The future lifetimes of and are independent.(ii) The survival function for is based on a constant force of mortality, .(iii) The survival function for follows DeMoivre's Law with .Calculate the probability that dies within 10 years and dies before .A) 10% B) 13% C) 16% D) 19% E) 25%

20. Oil wells produce until they are dry. the survival function for a well is given by (years)

An oil company owns 10 wells age 3. It insures them for 1 million each against failure for twoyears where the loss is payable at the end of the year of failure. You are given:(i) is the present-value random variable for the insurer's aggregate losses on the 10 wells.(ii) The insurer actually experiences 3 failures in the first year and 5 in the second year.(iii)Calculate the ratio of the actual value of to the expected value of .A) 0.94 B) 0.96 C) 0.98 D) 1.00 E) 1.02

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21. For a fully discrete 2-year term insurance on :(i)(ii)(iii) is the prospective loss random variable at time 1 using the premium determined by theequivalence principle.Calculate .A) 0.05 B) 0.07 C) 0.09 D) 0.11 E) 0.13

22. For a fully continuous whole life insurance of 1 on :(i)(ii)(iii) is the loss at issue random variable using the premium based on the equivalenceprinciple.(iv)(v) is the loss at issue random variable using the premium .(vi) .Calculate .A) 0.05 B) 0.08 C) 0.10 D) 0.12 E) 0.15

23. You are given:(i) is the present value random variable for a continuous whole life annuity of 1 per year on

.(ii) Mortality follows DeMoivre's Law with .(iii)Calculate the 75th percentile of the distribution of .A) 12.6 B) 14.0 C) 15.3 D) 17.7 E) 19.0

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24. For a special fully discrete 20-year endowment insurance on :(i) The death benefit is 1000 for the first 10 years and 2000 thereafter. The pure endowmentbenefit is 2000.(ii) The annual benefit premium, determined using the equivalence principle, is 40 for each ofthe first 10 years and 100 for each year thereafter.(iii)(iv)(v)

Calculate the 10th year terminal reserve using the benefit premium.A) 490 B) 500 C) 530 D) 550 E) 560

25. For a whole life insurance of 1000 on , with death benefits payable at the end of the yearof death, you are given:(i) Mortality follows a select and ultimate mortality table with a one-year select period.(ii)(iii)(iv)(v)Calculate .A) 655 B) 660 C) 665 D) 670 E) 675

26. For a fully-discrete 4-year term insurance on , who is subject to a double decrementmodel:(i) The benefit is 2000 for decrement 1 and 1000 for decrement 2.(ii) The following is an extract from the double-decrement table for the last 3 years of thisinsurance: (iii)(iv) The benefit premium, based on the equivalence principle, is 34.Calculate , the benefit reserve at the end of year 2.A) 8 B) 9 C) 10 D) 11 E) 12

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27. You are pricing a special 3-year life annuity-due on two lives each age , with independentfuture lifetimes. The annuity pays 10,000 if both persons are alive and 2000 if exactly one personis alive.You are given:(i)(ii)(iii)Calculate the actuarial present value of this annuity.A) 27,800 B) 27,900 C) 28,000 D) 28,100 E) 28,200

28. For a triple decrement table, you are given:(i) Each decrement is uniformly distributed over each year of age in its associated singledecrementtable.(ii)(iii)(iv)Calculate .A) 0.177 B) 0.180 C) 0.183 D) 0.186 E) 0.189

29. The city of Stevens Crossing wishes to insure a 50-year-old bridge against collapse for thenext 25 years. The present mayor, concerned with the tax burden of the premium on the 2000residents of Stevens Crossing, proposes the following scheme:• A benefit of $10 million is payable at the end of the year that the bridge collapses.• Each resident of Stevens Crossing would pay an annual premium of payable at thebeginning of the year for the next 10 years and for the remaining 15 years of the term.• In the event of bridge collapse, premium payments cease.The city's actuary has determined the following:• The number of residents of Stevens Crossing will remain steady over the next 25 years.• The bridge mortality follows the Illustrative Table with 6%.Calculate , the initial annual premium per resident.A) Less than 45 B) At least 45, but less than 50 C) At least 50, but less than 55D) At least 55, but less than 60 E) At least 60

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30. For a non-homogeneous Markov Chain with States X, Y, and Z, the following matricesshow the probability of movement between states at times 1, 2, and 3.

Calculate the probability that a subject in State X at time 0 will be in State Z at time 3.A) Less than 0.16 B) At least 0.16, but less than 0.18 C) At least 0.18, but less than 0.20D) At least 0.20, but less than 0.22 E) At least 0.22

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 3 SOLUTIONS

1. The actuarial present value of the benefit is

.The APV of three level premiums of each is

.Then from , we get . Answer: A

2. °

. Answer: D

3. We can use the formula .

.

.

.Then, . Answer: B

4. The APV of one annuity to a 65-year old is , andthe APV of one annuity to a 75-year old is 7.2170. The total APV of all annuities is

.The variance of the present value random variable for one annuity to a 65-year old is

.

The variance of the present value random variable for one annuity to a 75-year old is .

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4 continuedThe total variance of the PVRV's for all 80 annuities is

.The 95th percentile of the present value random variable is

. Answer: E

5. Any 10-minute period can be considered. In order for there to be at least two claims in a 10minute period, there must be a first claim, say at time , and then a second claim within

minutes after the first claim. We will call the event that there are two claims within the10 minute period .

1st claim is at time 2nd claim is within minutes after 1st claim) 2nd claim within minutes after 1st claim 1st claim at

We use the impendence of successive claim processing times, to get that2nd claim within minutes after 1st claim 1st claim at

2nd claim within minutes after 1st claim .The integral is .Alternatively, because the time between successive claims processed is exponential with mean

, the number of claims processed follows a Poisson process with rate per minute.The number of claims processed in a 10 minute period has a Poisson distribution with a mean of2. The probability of at least 2 claims being processed in a 10 minute period is the complementof 0 or 1 claims processed, which is

. Answer: E

6. When a payout occurs, it is 1, 2 or 3 with probability .The number of payouts that are 1, 2 or 3 follows a Poisson process with an hourly rate of

. The expected number of payouts that are 1, 2 or 3 in 20 minutes, say , has a

Poisson distribution with mean . The probability that there are no payouts of 1, 2, or3 in a given 20 minute period is the probability that , which is .Answer: D

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7. Let be the time until the next train arrives. Then , where , where is the number of trains arriving by time .

Measuring time in hours, for , has a Poisson distribution with mean , and for, has a Poisson distribution with mean

For , and for , .Then, hours.This is minutes. Note that the rate at which train arrivals occur follows a non-homogeneous Poisson process with

for , and for . Answer: D

8. The reduced pure endowment value must satisfy the equation .

From the Illustrative Table at 6%, we have , and

, and

Then, .

Now, .Answer: A

9. 1st 5 years of benefit next 7 years of benefit remaining benefit

. Answer: D

10. The APV at issue of the expense-augmented loss random variable for this insurance isAPV benefit APV expenses APV premiums.APV benefit .APV expenses APV benefit

.APV premiums .APV of expense augmented loss at issue is . Answer: A

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11. The number of people receiving Nursing Home benefits in year 3 of the original 50 has abinomial distribution with and . The probability can be found by matrixmultiplication. will be the entry in the matrix product . That entry is foundby "multiplying" row 1 of Q by column 3 of , which results in

.The variance of the number in State 3 in during year 3 is . Answer: C

12. The APV of premiums can be formulated as the APV of the first 4 years of premium plusthe APV of premiums paid from time 4 and later. A premium is paid if not in state 2. The APVof the first 3 premiums is .At time 3, the state must be 1 or 2, since there is no transition into state 1 from time 2 to 3. Then,at time 4 and on the state must be 2, so the last possible premium is at time 3.

,

so .

so .APV of premium is .

The APV of benefits is (as noted above, from time 4 and on,the state must be 2). APV of benefits is .

APV premium APV benefits . Answer: C

13. Force of mortality for 1st year is , force of mortality for 2nd year is . 2-year survivalprobability is . We wish to find (.25 years is 3months, and 21 months is 1.5 years after 3 months).

. . Answer: E

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14. Non-smokers have DeMoivre mortality with , and have force of mortality at age of . Smokers have force of mortality that is double, so for a smoker at age , the force ofmortality is ; smokers have generalized DeMoivre mortality with and .Survival probability for a smoker is .Survival probability for a nonsmoker is .° (the integral goes to 85 since the 25 year old has a maximum of85 years until death). The integral becomes .

With the substitution ,, , ,

so . Answer: C° , ,

15.

From , we get

, so that .Then, ,from which we get . Answer: B

16. We use the relationship . Retrospective reserve is

.

. Answer: A

17. From the ILT, .

For the CF model, For the DM model, .

Rankings will be based on ranking of . ILT CF DM

CF DM ILT . Answer: C

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18. From UDD in the multiple decrement table, we have .

The expected number of people who will retire before age 62 is .

, and and .

, .Then .The expected number of people who retire before age 62 is . Answer: D

19. .

Since 's mortality is based on DeMoivre's Law with , we have for . For , .

The integral becomes . Answer: B

20. The actual value of (in millions) is .The actuarial present value for the insurance cost of one insured well is

. The actuarial present value of the insurance cost for 10 wells is .The ratio of actual to expected cost is . Answer: A

21. is the same as . For the 2-year term insurance, the conditional

distribution of given is prob.prob.

The equivalence principle premium is .

Then , and

.Answer: E

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22. Since the policy is fully continuous whole life of amount 1,

and . Then , so .

, and then . Answer: C

23. The 75th percentile of is , where . .

From DeMovire's Law with , we know that .Therefore, if , then . There is a 75% chance of dying within 60 years,;stated another way, .Therefore, there is a 75% chance that is .

This is the 75-th percentile of . We have translated probability involving into

probability involving ; . Answer: E.

24. APV benefit APV premium (as of time 10).APV benefit .

.

APV of benefit is .APV of premium .

. Answer: B

25. With a one-year select period, .From , we get ,and solving for results in . Then, , and

. . Answer: C

26. APV benefit APV premium (as of time 2).At the end of the 2nd year, there are still 2 premiums remaining;APV of premium ) .APV of benefit

. . Answer: D

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27. APV is .

.

APV .Answer: B

28. With UDD in the associated single decrement tables, we have . Answer: B

29. .

We find the annuity and insurance values from , and

, and .6

From the Illustrative Table, we get ,

, and

6 .

Then, . Answer: E

30. is the entry in the matrix .This will be row 1 of "multiplied by" column 3 of .Row 1 of is .Then, . Answer: D

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 4

1. For a special whole life insurance on , payable at the moment of death:

(i) ,

(ii)

(iii) The death benefit at time is ,

(iv) is the present value random variable for this insurance at issue.CalculateA) 0.038 B) 0.041 C) 0.043 D) 0.045 E) 0.048

2. For a fully continuous whole life insurance of 1 on , you are given:

(i) The forces of mortality and interest are constant.

(ii)

(iii)

(iv) is the loss-at-issue random variable based on the benefit premium.Calculate .A) 0.20 B) 0.21 C) 0.22 D) 0.23 E) 0.24

3. A fully discrete whole life policy issued to (x) has net annual premium P, and gross annualpremium G . The expenses are E in the first year, and the renewal expenses are level at E . Thecovariance Cov[ L , L ] can be expressed as Var[ v ] multiplied bye

K+1

A) (1 + ) B) (1 + ) C) (1 + )(1 + ) D) (1 + )(1 + ) E) (1 + )(1 + )P P+E G P G–E P G–E P+Ed d d d d d d d

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4. An automobile insurer has a driver rating scheme based on a homogeneous Markov Chain inwhich drivers are classified as low risk, medium risk and high risk. The rating is updated at theend of each year. The one-step transition probability matrix for rating changes is L M H L .7 .2 .1 M .5 .4 .1 H .1 .3 .6Find the probability that a driver currently rated as low risk will not be rated as high risk for thenext three years.A) .512 B) .729 C) .810 D) .900 E) .910

5. For a special fully continuous last survivor insurance of 1 on and , you are given:

(i) and are independent.

(ii) ,

(iii) ,

(iv)

(v) is the annual benefit premium payable until the first of and dies.Calculate .A) 0.055 B) 0.080 C) 0.105 D) 0.120 E) 0.150

6. For a special fully discrete whole life insurance of 1000 on :(i) The contract premium for the first 4 years is equal to the level benefit premium for a fullydiscrete whole life insurance of 1000 on .(ii) The contract premium after the fourth year is equal to the level benefit premium for a fullydiscrete whole life insurance of 1000 on .(iii) Mortality follows the Illustrative Life Table (found at the end of the study guide).(iv)(v) is the prospective loss random variable at time 3, based on the contract premium.(vi) is the curtate future lifetime of .Calculate E[ .A) 27 B) 31 C) 44 D) 48 E) 52

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7. Your company is competing to sell a life annuity-due with an actuarial present value of500,000 to a 50-year old individual. Based on your company’s experience, typical 50-year oldannuitants have a complete life expectancy of 25 years. However, this individual is not ashealthy as your company’s typical annuitant, and your medical experts estimate that his completelife expectancy is only 15 years. You decide to price the benefit using the issue age thatproduces a complete life expectancy of 15 years. You also assume:(i) For typical annuitants of all ages, mortality follows De Moivre’s Law with the same limitingage .(ii)Calculate the annual benefit that your company can offer to this individual.A) 38,000 B) 41,000 C) 46,000 D) 49,000 E) 52,000

8. For a double decrement table, you are given:(i) (ii)(iii) Each decrement is uniformly distributed over each year of age in the double decrementtable. Calculate .A) 0.020 B) 0.031 C) 0.042 D) 0.053 E) 0.064

9 For a Markov model for an insured population:(i) Annual transition probabilities between health states of individuals are as follows: Healthy Sick TerminatedHealthy 0.7 0.1 0.2Sick 0.3 0.6 0.1Terminated 0.0 0.0 1.0

(ii) The mean annual healthcare cost each year for each health state is: Healthy Sick Terminated 500 3000 0

(iii) Transitions occur at the end of the year.(iv)Calculate the expected future healthcare costs for this year and the next two years for an insuredindividual whose current state is healthy.A) 1,700 B) 1,725 C) 1,750 D) 1,775 E) 1,800

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10. Which of the following statements is false?Assume that all forces of mortality and interest are .A) B) C) D) E)

11. A $100,000 endowment policy issued to (x) has cash values equal to benefit-plus-expensereserves. Expenses are $100 plus 25% of gross premium in the first year, and $25 plus 10% ofgross premium in the second year. Given:(i) i = .06 , (ii) q = .001223 , (iii) q = .01 , (iv) q = .0012735 , (v) q = .015 ,x x

(d) (w) (d) (w)x+1 x+1

(vi) V = 518.35 , (vii) V = 1,292.48 , (viii) AS = 0 and (ix) AS = 1,464.64 , what is G?e e

A) 1030 B) 1032 C) 1034 D) 1036 E) 1038

12. The mortality of and follows a common shock model with components , and .(i) , and are independent and have exponential distributions with respective forces

and .(ii) The probability that survives 1 year is 0.96.(iii) The probability that survives 1 year is 0.97.(iv)Calculate the probability that both and survive 5 years.A) 0.65 B) 0.67 C) 0.70 D) 0.72 E) 0.74

13. For a fully discrete whole life insurance of 100,000 on each of 10,000 lives age 60, you aregiven:(i) The future lifetimes are independent.(ii) Mortality follows the Illustrative Life Table.(iii) .(iv) is the premium for each insurance of 100,000.Using the normal approximation, calculate , such that the probability of a positive totalloss is 1%.A) 3340 B) 3360 C) 3380 D) 3390 E) 3400

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14. For a special fully discrete 3-year endowment insurance on (75), you are given:(i) The maturity value is 1000.(ii) The death benefit is 1000 plus the benefit reserve at the end of the year of death.(iii) Mortality follows the Illustrative Life Table.(iv) .Calculate the level benefit premium for this insurance.A) 321 B) 339 C) 356 D) 364 E) 373

15. For a special fully discrete 3-year term insurance on (55), whose mortality follows a doubledecrement model:(i) Decrement 1 is accidental death; decrement 2 is all other causes of death.(ii) (iii)(iv) The death benefit is 2000 for accidental deaths and 1000 for deaths from all other causes.(v) The level annual contract premium is 50.(vi) is the prospective loss random variable at time 1, based on the contract premium.(vii) is the curtate future lifetime of (55).Calculate .A) 5 B) 9 C) 13 D) 17 E) 20

16. A 25-year fully discrete endowment insurance policy issued to has the death benefitequal to the terminal reserve and endowment amount 100,000.You are given . Find .A) 0 B) 200 C) 29,530 D) 50,000 E) 100,000

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17. Customers arrive at a store at a Poisson rate that increases linearly from 6 per hour at 1:00p.m. to 9 per hour at 2:00 p.m.Calculate the probability that exactly 2 customers arrive between 1:00 p.m. and 2:00 p.m.A) 0.016 B) 0.018 C) 0.020 D) 0.022 E) 0.024

18. For (80) and (84), whose future lifetimes are independent: 80 0.50 81 0.40 82 0.60 83 0.25 84 0.20 85 0.15 86 0.10Calculate the change in value of if is decreased from 0.60 to 0.30.A) 0.03 B) 0.06 C) 0.10 D) 0.16 E) 0.19

19. At interest rate :(i)(ii) The actuarial present value of a 2-year certain and life annuity-due of 1 on is

(iii) (iv) Calculate .A) 0.077 B) 0.079 C) 0.081 D) 0.083 E) 0.084

20. You are given:(i) The future lifetimes of (50) and (50) are independent.(ii) Mortality follows the Illustrative Life Table.(iii) Deaths are uniformly distributed over each year of age.Calculate the force of failure at duration 10.5 for the last survivor status of (50) and (50).A) 0.001 B) 0.002 C) 0.003 D) 0.004 E) 0.005

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21. For a deferred whole life annuity-due on (25) with annual payment of 1 commencing at age60, you are given:(i) Level benefit premiums of each are payable at the beginning of each year during thedeferral period.(ii) During the deferral period, a death benefit equal to the benefit premium already paid,without interest is payable at the end of the year of death.

Which of the following is a correct expression for the benefit reserve at the end of the 20th year?

A) B)

C) D)

E)

22. For a special whole life insurance:(i) The benefit for accidental death is 50,000 in all years.(ii) The benefit for non-accidental death during the first 2 years is return of the single benefitpremium without interest.(iii) The benefit for non-accidental death after the first 2 years is 50,000.(iv) Benefits are payable at the moment of death.(v) Force of mortality for accidental death: (vi) Force of mortality for non-accidental death: (vii)Calculate the single benefit premium for this insurance.A) 1,000 B) 4,000 C) 7,000 D) 11,000 E) 15,000

23. A special whole life insurance on pays 10 times salary if the cause of death is anaccident and 500,000 for all other causes of death. You are given:(i)(ii) accident

(iii) Benefits are payable at the moment of death.(iv)(v) Salary of at time is Calculate the actuarial present value of the benefits at issue.A) 78,000 B) 83,000 C) 92,000 D) 100,000 E) 108,000

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24. Z is the present value random variable for a 15-year pure endowment of 1 on :(i) The force of mortality is constant over the 15-year period.(ii)(iii)Calculate .A) 0.020 B) 0.025 C) 0.030 D) 0.035 E) 0.040

25. Using the Illustrative Life Table, calculate the average number of complete years livedbetween ages 60 and 65.A) Less than 4.70 B) At least 4.70, but less than 4.75 C) At least 4.75, but less than 4.80D) At least 4.80, but less than 4.85 E) At least 4.85

26. Using the Illustrative Life Table and the uniform distribution of death assumption, which ofthe following are true? 1.

A) 1 only B) 3 only C) 1 and 2 only D) 2 and 3 only E) 1, 2, and 3

27. For a 5-year deferred whole life insurance of 1 on , you are given•• 4• The benefit is paid at the moment of death.• is the present value random variable of the insurance benefit.Calculate .A) Less than 0.05 B) At least 0.05, but less than 0.06 C) At least 0.06, but less than 0.07D) At least 0.07, but less than 0.08 E) At least 0.09

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28. A customer service operator accepts calls continuously throughout the work day. The lengthof each call is exponentially distributed with an average of 3 minutes. Calculate the probabilitythat at least one call will be completed in the next 2 minutes.A) Less than 0.50 B) At least 0.50, but less than 0.55 C) At least 0.55, but less than 0.60D) At least 0.60, but less than 0.65 E) At least 0.65

29. Customers arrive to buy lemonade according to a Poisson distribution with , where istime in hours, as follows:

At 9:00 a.m., is 0. Calculate the number of customers expected to arrive between 10:00a.m.and 2:00 p.m.A) Less than 63 B) At least 63, but less than 65 C) At least 65, but less than 67D) At least 67, but less than 69 E) At least 69

30. Robin is initially classified as a Standard driver. At the beginning of each subsequent year,Robin will transition between the classes (1) Preferred, (2) Standard, and (3) Non-Standardaccording to the following transition matrix:

Standard premium is $500. A discount of $50 is applied to Preferred premium and a surcharge of$75 is applied to Non-Standard premium. Premiums are paid at the beginning of each year. Theannual interest rate is 3%. Calculate the actuarial present value of the premium paid by Robin inthe first three years.A) Less than 1430 B) At least 1430, but less than 1460 C) At least 1460, but less than1490D) At least 1490, but less than 1520 E) At least 1520

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 4 SOLUTIONS

1. With constant force of mortality of .05, survival probability is .

Then . Answer: D

2. Suppose that the force of interest is and the force of mortality is . Then .

. Then , and

.

We have used the relationship

.Answer: A

3. G = (A + E E + E a )/a = P + E + . L = ( 1 + ) v ¨ ¨0 0K+1E E

a d dP P0

0 e 0 0K+1 K+1

K+1| K+1|L = v + E E + E a G a = ( 1 + ) v + E E + ¨ ¨ G E G Ed d

Cov[ L , L ] = Cov[ (1+ ) v , (1 + ) v ] = (1 + ) (1 + ) Var[v ] .0 0 eK+1 K+1 K+1P G E P G E

d d d dAnswer: D.

4. The 3-step transitions that do not involve a high risk rating for the next three years, and theirrelated probabilities are Ratings sequence for the next three years Probability Total probability is

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4 continuedA simpler solution follows from the observation that when the current rating is either L or M,there is a .9 chance that it will still be L or M. Therefore, if the current rating is L or M, theprobability that the next three transitions avoid the H rating is . Answer: B

5. .Since and are independent, .Then for constant force of mortality , and8

, and .

Also for constant force of mortality, .

Then . Answer: A

6. At age 45, there is one more premium of amount , and the premiums after that willbe . From the Illustrative Life Table at 6%, we have ,

and . Then

APV benefit APV contract premium as of age 45.APV premium .APV benefit .The expected value is . Answer: B

7. We wish to find the annual payment so that , where is the APV ofthe life annuity to the individual with higher than typical mortality. For typical mortality for a50-year old, the life expectancy under DeMoivre's law with upper age limit , is . We aretold that life expectancy for a typical 50-year old is 25 years. Therefore, for typicalmortality. An individual at age 70 would have life expectancy 15 years, since

. We price the annuity to the higher mortality 50-year old as if his age was 70.Therefore, , and ...

Under DeMoivre's Law, ,

so that .

Then . Finally, . Answer: E

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8. .Since we are assuming UDD in the double decrement table, it follows that

and , and .We have .From the UDD in double decrement table assumption, we have

.

Then , and then , so that . Answer: D

9. The sequences of transitions, their probabilities and their costs are as follows. Sequence Probability Cost The expected cost is . Answer: E

10. . (C) Is false. Answer: C

11. ( AS + .9G 25)(1.06) 127.35 (1292.48)(.015)1 .0162735

1 = 1,464.64But, = AS AS = .80402G 236.13 G = 1032 .(.75G 100)(1.06) 122.30 5.1835

1 .011224 1 1

Answer: B.

12. Under the common shock model , where is the constant hazard rateof the common shock. We are has constant force , so that

.Therefore, . In a similar way, .We wish to find .Since is exponential, , and similarly,

. Then, . Answer: E

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13. The (issue date) loss on one policy will be ,since , with expected value andvariance .The total loss on 10,000 independent policies will be

, and the mean and variance of are] and

.We wish to find so that using the normal approximation to .After standardizing , the probability becomes

.

Therefore, , so that . This equation becomes] ] .

This can be written as .

Solving for results in . Answer: C

14. The equation for this insurance is , where

is the endowment amount at the end of 3 years (the maturity date of the policy).Using the Illustrative Table and an interest rate of 5%, we have

.

Solving for results in . Answer: C

15. The expected value is APV benefit APV premium as of age 56.APV benefit

.

(we have used the relationships and ).APV premium .The expected prospective loss is . Answer: D

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16. We use the recursive relationship for variance of the loss. .

For since is the constant endowmentamount. Then

since in any year. This will be true at ., so that for all . Answer: A

17. The intensity function for arrivals is for . The expectednumber of arrivals between 1PM and 2PM is . Therefore the numberof arrivals between 1PM and 2PM has a Poisson distribution with a mean of 7.5 . The

probability of exactly 2 arrivals between 1PM and 2PM is . Answer: A

18. (by independence we have ). The original values are

, and , so that the original value of is

.Since only appears as a factor of , the updated value of is

, and the updated value of is . The change in value is .

Answer: B

19. From the relationship we get , and then we get . The 2-year certain and life annuity-due has APV

.

From the given values and , we get , andsolving for results in , and . Note that if we rounded to .9505,

we would get . Answer: B

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20. .

.This is true by UDD, for for , so that

. The force of failure is .

From the Illustrative table, , .

and then . Answer: B

21. Retrospectively, the accumulated premium is (20 years since the policy was

issued at age 25). Since the annuity has not yet begun, the accumulated cost of insurance is thereturn of premium for the first 20 years, which forms an increasing benefit. The accumulated

cost of insurance is , so the retrospective form of the reserve is .

The prospective representation would be . Answer: D

22. The single benefit premium must be equal to the APV of benefits.The combined force of decrement is , so .The accidental death benefit has APV

.The non-accidental death benefit is

.Then the equivalence principle sets premium equal to APV of benefits on the issue date, so

, from which we get . Answer: D

23. APV of accidental death benefit is . This is .

APV of non-accidental death benefit is.

Total APV is . Answer: D

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24. For the 15-year pure endowment, , and

.

Since the force of mortality is constant, for all , so .We are given , and .Therefore, , so that , and , so that .Answer: B

25. The average number of complete years lived between ages 60 and 65 is .

. Answer: C

26. Under UDD, we have for and

for .1.

. False.2. (compare denominators). True.

3. Constant density within each year of age under UDD. True. Answer: D

27. .

.

We have used the relationship for constant force of mortality.

Similarly, .

. Answer: B

28. Since calls are accepted continuously, we assume that the operator starts a call as soon as theprevious one is completed. From the lack of memory property of the exponential distribution, itis irrelevant how long the current call has taken, the remaining time until the call is completed isexponential with a mean of 3 minutes. . Answer: A

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29. The number of cars arriving between 10 a.m. and 2 p.m. has a Poisson distribution withmean . Answer: C

30. The first year premium is $500, since Robin is initially classified as Standard.The expected second year premium is

,so the APV of the second year premium is .The expected third year premium is

.We get these probabilities by "multiplying" the second row of by each column of :

.The expected third year premium is ,and the APV of the third year premium is .

The APV of the first three years of premiums is .Answer: B

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 5

1. For two independent lives now age 30 and 34, you are given: 6 Assuming UDD in each year of age, calculate the probability that is the 2nd of these twolives to die, and 's death occurs during the 3rd year.A) 0.054 B) 0.104 C) 0.124 D) 0.144 E) 0.164

2. For a whole life insurance of 1000 on payable at the moment of death:

(i)

(ii)

Calculate the single benefit premium for this insurance.A) 379 B) 411 C) 444 D) 519 E) 594

3. For a special fully discrete insurance of 1000 on (40):(i) The level benefit premium for each of the first 20 years is .

(ii) The benefit premium payable thereafter is , (iii) Mortality follows that Illustrative Life Table.(iv)Calculate .A) 4.79 B) 5.11 C) 5.34 D) 5.75 E) 6.07

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Questions 4 and 5 relate to the following information.A 5-year endowment insurance policy with face amount 100,000 and discrete annual premiumshas issue age 50. The policy has the following schedule of expenses: 40% of premium in thefirst year, 10% of premium in renewal years and a per policy expense of 250 in the first year and50 in each renewal year. Mortality follows the Illustrative Life Table in the Exam M Tablesfound at the end of this study guide, with an annual interest rate of 6%.

4. What is the equivalence principle expense-loaded premium for the policy?A) 20,157 B) 20,257 C) 20,357 D) 20,457 E) 20,557

5. What is the 4th year terminal expense reserve?A) 2443 B) 1443 C) 0 D) 1443 E) 2443

6. A whole life policy provides that on accidental death as a passenger on an airplane a benefitof 1,000,000 will be paid. If death occurs from other accidental causes, a benefit of 500,000 willbe paid. If death occurs from a cause other than accident, a benefit of 250,000 will be paid.You are given:(i) Death benefits are payable at the moment of death.

(ii) where (1) indicates accidental death as a passenger on an airplane

(iii) where (2) indicates death from other accidental causes

(iv) where (3) indicates non-accidental death

(iv)Calculate the single benefit premium for this insurance.A) 450 B) 460 C) 470 D) 480 E) 490

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7. For an annuity payable semiannually, your are given:(i) Deaths are uniformly distributed over each year of age.

(ii)

(iii)

(iv)

Calculate .A) 8.35 B) 8.47 C) 8.59 D) 8.72 E) 8.85

8. For a sequence, is defined by the following recursion formula:

for

(i)

(ii)

(iii)Which of the following is equal to ?

A) B) C) D) E) 1

9. Subway trains arrive at a station at a Poisson rate of 20 per hour. 25% of the trains are expressand 75% are local. The type of each train is independent of the types of precedingtrains. An express gets you to the stop for work in 16 minutes and a local gets you there in 28minutes. You always take the first train to arrive. Your co-worker always takes the first express.You both are waiting at the same station. Calculate the probability that the train you take willarrive at the stop for work before the train your co-worker takes.A) 0.28 B) 0.37 C) 0.50 D) 0.56 E) 0.75

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10. A new disease has the following characteristics:(i) Changes in state occur only at the end of the year.(ii) 30% of those who are acutely ill in a given year are in remission in the following year and10% are cured or dead.(iii) 20% of those who are in remission in a given year become acutely ill in the following yearand 30% are cured or dead.(iv) Those who are cured do not become acutely ill or in remission again.Calculate the probability that someone currently in remission is cured or dead at the end of 3years.A) .505 B) .525 C) .545 D) .565 E) .585

11. For a fully discrete whole life insurance of 1000 on (40), the contract premium is the levelannual benefit premium based on the mortality assumption at issue. At time 10, the actuarydecides to increase the mortality rates for ages 50 and higher. You are given:(i)(ii) Mortality assumptions: At issue Revised prospectively at time 10 45

(iii) is the prospective loss random variable using the contract premium.

Calculate using the revised mortality assumption.A) Less than 225 B) At least 225, but less than 250C) At least 250, but less than 275 D) At least 275, but less than 300E) At least 300

12. For a space probe to Mars:(i) The probe has three radios, whose future lifetimes are independent, each with mortalityfollowing where time 0 is the moment the probe lands on Mars(ii) The failure time of each radio follows UDD assumption within each year.(iii) The probe will transmit until all three radios have failed.Calculate the probability that the probe is no longer transmitting 2.25 years after landing onMars.A) 0.069 B) 0.067 C) 0.063 D) 0.059 E) 0.053

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13. The RIP Life Insurance Company specializes in selling a fully discrete whole life insuranceof 10,000 to 65 year olds by telephone. For each policy:(i) The annual contract premium is 500.(ii) Mortality follows the Illustrative Life Table.(iii) i = 0.06The number of telephone inquiries RIP receives follows a Poisson process with mean 50 per day.20% of the inquiries result in the sale of a policy. The number of inquiries and the futurelifetimes of all the insureds who purchase policies on a particular day are independent.Using the normal approximation, calculate the probability that S, the total prospective loss atissue for all the policies sold on a particular day, will be less than zero.A) 0.33 B) 0.50 C) 0.67 D) 0.84 E) 0.99

14. For a special fully discrete whole life insurance on (40):(i) The death benefit is 1000 for the first 20 years; 5000 for the next 5 years; 1000 thereafter.(ii) The annual benefit premium is for the first 20 years; for the next 5 years;

thereafter.(iii) Mortality follows the Illustrative Life Table.(iv)Calculate , the benefit reserve at the end of 21 years for this insurance.A) 255 B) 259 C) 263 D) 267 E) 271

15. For a whole life insurance of 1 on (41) with death benefit payable at the end of year of death,you are given:

(i)

(ii)

(iii)

(iv)(v) is the present-value random variable for this insurance.Calculate .A) .023 B) .024 C) .025 D) .026 E) .027

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16. For a perpetuity-immediate with annual payments of 1:(i) the sequence of annual discount factors follows a Markov chain with the following threestates: State Number 0 1 2 Annual Discount Factor 0.95 0.94 0.93

(ii) The transition matrix for the annual discount factors is: .

is the present value of the perpetuity payments when the initial state is 1.Calculate .A) 15.67 B) 15.71 C) 15.75 D) 16.82 E) 16.86

17. John is an investment advisor and convinces a client to purchase today an option to buystock A and another option to buy stock B, even though the market price today of each stock isbelow the option strike price. Each option gives the client the right to buy the stock at a specificprice, the strike price, at the end of 3 days. On day ( ), for each stock

market price strike price depends only on the relationships on day as follows: Relationships of stocks on day market price strike price on day

Market price strike price for neither stock Market price strike price for exactly one stock Market price strike price for both stocks

We assume that at time , the relationship of the market price of stock A to its strike price isindependent of the relationship of the market price of stock B to its strike price given the state attime . At the end of 3 days John will receive a bonus of 100 from his investment firm ifexactly one of the stocks has a market price in excess of its option strike price, and he willreceive a bonus of 500 if both stocks have market prices in excess of their strike prices.Calculate John's expected bonus at the end of 3 days.A) Less than 125 B) At least 125 but less than 150C) At least 150 but less than 175 D) At least 175 but less than 200E) At least 200

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18. A member of a high school math team is practicing for a contest. Her advisor has given herthree practice problems: #1, #2, and #3. She randomly chooses one of the problems, and workson it until she solves it. Then she randomly chooses one of the remaining unsolved problems,and works on it until solved. Then she works on the last unsolved problem. She solves problemsat a Poisson rate of 1 problem per 5 minutes. Calculate the probability that she has solvedproblem #3 within 10 minutes of starting the problems.A) 0.18 B) 0.34 C) 0.45 D) 0.51 E) 0.59

19. For a double decrement table you are given:

(i) ,

(ii) ,

(iii)

Calculate .A) 0.45 B) .53 C) .58 D) 0.64 E) 0.73

20. For :(i) is the curtate future lifetime random variable.

(ii)Calculate .A) 1.1 B) 1.2 C) 1.3 D) 1.4 E) 1.5

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21. Michel, age 45, is expected to experience higher than standard mortality only at age 64.For a special fully discrete whole life insurance of 1 on Michel, you are given:(i) The benefit premiums are not level.(ii) The benefit premium for year 20, , exceeds for a standard risk by 0.010.(iii) Benefit reserves on his insurance are the same as benefit reserves for a fully discrete whole life insurance of 1 on (45) with standard mortality and level benefit premiums.(iv)

(v)

Calculate the excess of for Michel over the standard .A) 0.012 B) 0.014 C) 0.016 D) 0.018 E) 0.020

22. For a block of fully discrete whole life insurances of 1 on independent lives age , you aregiven:

(i)

(ii)

(iii)(iv) , where is the contract premium for each policy.(v) Losses are based on the contract premium.Using the normal approximation, calculate the minimum number of policies the insurer mustissue so that the probability of a positive total loss on the policies issued is less than or equal to0.05.A) 25 B) 27 C) 29 D) 31 E) 33

23. Your company currently offers a whole life annuity product that pays the annuitant 12,000 atthe beginning of each year. A member of your product development team suggests enhancing theproduct by adding a death benefit that will be paid at the end of the year of death. Using adiscount rate, , of 8%, calculate the death benefit that minimizes the variance of the presentvalue random variable of the new product.A) 0 B) 50,000 C) 100,000 D) 150,000 E) 200,000

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24. For , the future lifetime random variable for (0):(i)(ii)(iii)(iv) .°

Calculate the complete expectation of life at 40.A) 30 B) 35 C) 40 D) 45 E) 50

. For a fully continuous whole life insurance of 1 issued to , the expense-augmented lossvariable is given as , where , and .

You are given(i)(ii)(iii)(iv)(v)Benefit premiums and expense-loaded premiums are calculated according to the equivalenceprinciple. Calculate .A) 0.252 B) 0.263 C) 0.278 D) 0.293 E) 0.300

26. Two actuaries use the same mortality table to price a fully discrete 2-year endowmentinsurance of 1000 on (x).(i) Kevin calculates non- level benefit premiums of 608 for the first year and 350 for the secondyear.(ii) Kira calculates level annual benefit premiums of .(iii) .Calculate .A) 482 B) 489 C) 497 D) 508 E) 517

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27. For a fully discrete 10-payment whole life insurance of 100,000 on , you are given:(i)(ii)(iii)(iv)(v) The level annual benefit premium is 2078.(vi) The benefit reserve at the end of year 9 is 32,535.Calculate .A) 34,100 B) 34,300 C) 35,500 D) 36,500 E) 36,700

28. You are given:(i) Mortality follows DeMoivre's Law with .(ii) (45) and (65) have independent future lifetimes.Calculate .°(A) 33 (B) 34 (C) 35 (D) 36 (E) 37

29. For a fully discrete whole life insurance of 1 on , you are given:(i)(ii) Mortality follows the Exam M Table.(iii)CalculateA) 0.0019 B) 0.0020 C) 0.0021 D) 0.0022 E) 0.0023

30. For two independent lives (55) and (60), you are given:••••Calculate the probability that at least one of the lives survives 20 years.A) Less than 0.75 B) At least 0.75, but less than 0.80 C) At least 0.80, but less than 0.85D) At least 0.85, but less than 0.90 E) At least 0.90

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 5 SOLUTIONS

1. This event occurs in the following two cases.(i) dies within 2 years, and survives 2 years and dies in the 3rd year. This probabilityis .(ii) and both survive 2 years, and is the 2nd to die in the 3rd year. This probabilityis .

.

The probability for case (ii) is .The total probability is . Answer: A

2. The single benefit premium is .This can be formulated as

.In this expression, is the force of interest during the first 10 years and is theforce of interest after 10 years. For the constant force of mortality of .06 during the first 10years, we have , and for the constant force of mortality of .07 after 10 years, wehave . The single benefit premium becomes

. Answer: E

3. Since premiums are benefit premiums, they are based on the equivalence principle, so on theissue date we have the equivalence principle relationship:actuarial present value of benefit actuarial present value of premium .

APV of benefit is .

APV of premiums is

The second factor in the APV of premiums is , which is the same as the

second factor in the expression for the APV of benefit.

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3 continuedWhen we set APV of benefit equal to APV of premium we get

, which simplifies to

. Therefore, .

To find we use the relationship , so that

40 From the Illustrative Table we get

.

Solving for results in

We use a similar relationship for annuities to find : .40

Again, from the Illustrative Table we getso that .

Then The problem has been constructed so that the premiums starting at age 60 exactly match thebenefit cost from age 60 on. Therefore, for the equivalence principle, we must match the first 20years of premium with the first 20 years of benefit. Answer: B

4. G a = 100,000 A + .3G + .1G a + 200 + 50 a¨ ¨ ¨50 : 5| 50 : 5| 50 : 5| 50 : 5|

G = , where a = 4.41137 ,¨100,000 A + 200 + 50 a

.9 a .350 : 5| 50 : 5|

50 : 5| 50 : 5|

A = 1 d a = .75030 G = 20,557.44 Answer: E.¨50 : 5| 50 : 5|

5. The benefit premium for the policy is 100,000 P = 17,008.33 . The level annual50 : 5|

equivalence principle expense loading is e = 20,557.44 17,008.33 = 3,549.11 . The 4th yearterminal expense reserve is as of t = 4 APV future (equiv. princ.) expenses APV future (equiv. princ.) expense loading = .1 (20,557.44) + 50 3,549.11 = 1443.37 . Answer: B.

6.Since the force of decrement for all causes is constant, it follows that .The single benefit premium is

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6 continued

. Answer: B

7. Under the UDD assumption, .Therefore, .Since the interest rate is 6%, from the Illustrative Tables we have

and .The relationships we will use to find are , (both of whichare always true) and (which is true under the UDD assumption).We are given . Since , we have and

Then , and ,

and , and finally, . Answer: C

8. From the recursion relationship we have (70)

1 .

Going back one more year, we have ( )

.

In general, this recursion will be of the form (endowment insurance)

with

.

8 continuedTherefore, .

(Note that if we had been given , then would be .

Continuing with the recursion would result in , and then .) Answer: C

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9. If the next train is an express train the you and your co-worker arrive at the same time. Inorder for you to arrive at work earlier than your co-worker, the next train must be a local train,and the next express train must arrive more than 12 minutes after the next local train. If and

denote the times (in minutes) until the next local and express trains respectively, then theprobability that you arrive before your co-worker is . The number of localtrains arriving per minute has a Poisson distribution with a mean of , and thenumber of express trains arriving per hour has a Poisson distribution with a mean of

. We are told that the type of a train is independent of types of preceding trains,so that the numbers of local and express trains arriving per hour are independent of each other.

The Poisson distribution is related to the exponential distribution. If is the number of eventsof a certain type occurring per hour and if has a Poisson distribution with mean , we define

to be the time between events. Then has an exponential distribution with mean .Therefore, the time until the next local train, has an exponential distribution with a mean of 4minutes and the time until the next express train, has an exponential distribution with a meanof 12 minutes. Since the numbers of local and express trains are independent, the arrival timesof the next local and express trains and are also independent. The density functions of and are and , and the joint density function of and is

(since and are independent, the joint density is the product ofthe two individual densities). Then,

Another way of looking at this situation is that since 25% of the trains are express and 75% arelocal, and train types are independent of one another, there is a chance that the next train islocal (and chance that it is express). If the next train is local, then it will get you to workbefore the next express train if the next express train arrives at least 12 minutes after the local.Because of the lack of memory property. once the next local arrives, the time until the nextexpress after that is still exponential with mean 12 minutes, so .Therefore, the probability that you arrive at work before your coworker is

next train is local first express train arrives more than 12 minutes after next local . Answer: A

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10. The three states and transition probabilities can be modeled as a Markov chain. The one-

step transition probability matrix is , where the first row and column

correspond to the state "acutely ill" (state 1), the second row and column correspond to the state"in remission" (state 2), and the third row and column correspond to the state "cured or dead"(state 3). The 3-step transition matrix is , and we want the (2,3)-entry in

, which is , the probability of transferring from state 2 (in remission) to state 3 (curedor dead) from time 0 to time 3. To do the calculation for the (2,3)-entry in , weonly need the middle row of to multiply by the third column of . Therefore, we do not needto calculate the full matrix . We do the following calculation

, and then

. . Answer: E

11. The original mortality assumption is DeMoivre's law with , since the deferredmortality probabilities are constant for 50 more years from age 40. The contract premium is thewhole life benefit premium .

Under DeMoivre's law , so that

( ). Then .

, where the insurance and annuity values arecalculated under the revised mortality assumption. Under the revised mortality assumption atage 50, survival from age 50 follows DeMoivre's law with (constant deferred mortalityfor 25 years from age 50). Therefore, ,

and . Finally, . Answer: E

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12. For a particular radio, the probability that the radio has stopped transmitting by time 2.25 is . The probability that the probe is no longer transmitting by time 2.25 is

(radio 1 has failed) (radio 2 has failed) (radio 3 has failed) .Since the radios are independent of one another, this probability is

(radio 1 has failed) (radio 2 has failed) (radio 3 has failed)(we use the rule for independent events and that says ).

From the given information we have so that .

From the given information we have , but we also know ,from which it follows that .Then from UDD, we have , and .We now get , so that .The probability in question is . Answer: E

13. We will denote by the loss-at-issue random variable for one insurance policy sold. Then , where is the present value random variable for a discrete whole life

insurance of 1 at age 65, and is the present value random variable for a discrete whole lifeannuity-due of 1 at age 65. The number of telephone enquiries in one day resulting in a sale,say , has a Poisson distribution with a mean of . Then

, and has a compound Poisson distribution. has mean and variance . To find the mean and the variance of we

note that , so that

(note that ). Then , and

.The mean of the compound distribution variable is , andthe variance is .Applying the normal approximation we have

(this found from the normal table by

interpolating between and ). Answer: C

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14. We use the retrospective form of the reserve, as it does not require finding .We first find and then apply the recursive reserve relationship to find . Since the interestrate is .06, the insurance and annuity values can be found from the Illustrative Table.

Since the benefit for the first 20 years is 1000 and the premium for the first 20 years is ,it follows that (the fact that the premiums and benefits change after time 20does not change the fact that the reserve at the end of 20 years is the same as ). Wecan use any method available to find , and the easiest is usually the annuity form(which is valid for whole life and endowment policies). =1000( ) .

We now use the recursive reserve relationship for .

The premium at time 20 (age 60) is , and ,

and the death benefit is 5000 at age 61, so that .

Solving for results in

Note again that this solution did not require us to find the value of . Answer: A

15. We use the relationship to get. Since we are given , we have

.

Similar relationships hold for : .Then and results in

.

Finally, . Answer: C

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16. From the transition matrix we see that after one transition the chain will move to either state0 or state 2, and then must move back to state 1. Every second transition will be back to state 1.The present value factor for the first year is .94 . The present value factor for the second year iseither .95 (prob. .9) or .93 (prob. .1). At the end of two years, the state is again 1, and the presentvalue at the end of two years of the future perpetuity immediate is again since the processhas "started over".Year 1 2 3 4 ...PV factor .94 ? .94 ? ...

. where is the present value factor forthe 2nd year, and is the present value at the end of the 2nd year (just after the 2nd payment) ofthe rest of the perpetuity. Since after the 2nd year the state must be 1, it is as if the process hasstarted over, and . Therefore.

. . Then, .

Alternatively, we see that the present value factor for the 1st, 3rd, 5th, ... years is always .94(because the process must return to state 1 every second transition). The expected present valuefactor in the 2nd, 4th, 6th, ... years is , so that

. Answer: D

17. The day to day behavior of stock prices can be modeled as a Markov chain with 3 states:state 1 - market price strike price for neither stock ,state 2 - market price strike price for exactly one stock ,state 3 - market price strike price for both stocks .The state of the chain when the stocks are purchase (time 0) is state 1.John will receive a bonus of 100 if the chain is in state 2 at time 3, and he will receive a bonus of500 if the state is 3 at time 3. John's expected bonus at time 3 is .

This is a homogeneous Markov Chain. If the process is in state 1 at time , then there is aprobability of that stock A's market price will be less than its strike price on day , and thesame is true for stock B. The probability that at time neither market price is greater than itsstrike price is ; thus, . In a similar way, given state 1 at time , the

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17 continuedprobability that both stocks have market price that are greater than their strike price (state 3) attime is . Since , it follows that

(alternatively, if the state is 1 at time , then there is a probability thatstock A market price is greater than its strike price and stock B market price is not greater thanits strike price, and there is probability that stock B market price is greater than its strikeprice and stock A market price is not greater than its strike price, for a combined probability of

that exactly one of the two market prices is greater than the strike price).Similar reasoning results in the following one-step transition probability matrix:

We wish to find and .

We have not calculate the full matrix multiplication, we have just performed the matrixmultiplications needed to get and .The expected bonus is . Answer: C

18. The number of problems solved in 5 minutes has a Poisson distribution with a mean of 1.Therefore, the number of problems solved within 10 minutes, say , has a Poisson distributionwith a mean of 2 (the number of problems solved by time forms a Poisson process).

We are looking for #3 is solved within 10 minutes . This can be formulated as#3 is solved within 10 min. #3 is first problem chosen

#3 is solved within 10 min. #3 is second problem chosen#3 is solved within 10 min. #3 is third problem chosen

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18 continued#3 is solved within 10 min. #3 is first problem chosen #3 is first problem chosen#3 is solved within 10 min. #3 is second problem chosen #3 is second problem chosen#3 is solved within 10 min. #3 is third problem chosen #3 is third problem chosen .

Since the problems are chosen randomly, there is a 1/3 chance that problem # 3 is chosen first,and the same probability of being chosen second and also third.

#3 is solved within 10 min. #3 is first problem chosen .

#3 is solved within 10 min. #3 is second problem chosen .

#3 is solved within 10 min. #3 is third problem chosen

.Then, #3 is solved within 10 minutes . Answer: E

19. We use the exponential formulation for .,

so that .

In the two decrement table, since it follows that , since . We use the integral of density form to represent multiple decrement

probabilities: and .It then follows from that

.

We find from ,and we note that , so that .Finally, .

We could have solved the problem another (more tedious) way. From we cansolve for , , so that , and therefore

2 . Then, .Then, .This integration requires a change of variable to simplify. Answer: D

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20. The probability function for is , and also we have for the curtate future lifetime.

if if if if which is also equal to .

. Note that .

( )

.

( ) . Answer: A

21. We are given that , and we are asked to find , where is themortality probability at age 64 for Smith and is the standard mortality probability at age 64.We are told that for every , for this policy is the same as (whole life policy reservebased on standard mortality). The recursive relationship for reserves for this policy in the 20thyear is and the recursive relationship for reservesfor the policy based on standard mortality in the 20th year is .

We are told that the policy reserves on the high mortality policy are the same as those on thestandard mortality policy, so that and . Then, subtracting thestandard mortality recursive equation from the high mortality recursive equation we have

. Then sinceand , we get . Answer: D

22. The loss for one policy is , where is the present value random variablefor a whole life insurance of 1 for , and is the present value random variable for a wholelife annuity-due of 1 for . Then and .The mean of is .The variance of is .

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22 continuedSuppose that there are independent policies. The total loss is

, and , .

The probability of a positive total loss is , and under the normal approximation thisbecomes .

In order for this to be less than or equal to 0.05, we must have ,or equivalently, . Since , follows that we must have

, or equivalently, . Answer: B

23. If the death benefit is , then the present value random variable of the combined annuity andinsurance is , Since is a whole life annuity-due (of 1 per year) presentvalue random variable and is a discrete whole life insurance of 1 present value randomvariable, we have . Then ,with variance .This variance will be 0 if . Answer: D

24. We then use the relationship .° ° °

Then .° ° ° ° °

From (iv) we have .°

Then solving for results in . Answer: E° °

25.

Answer: D

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26. With discount rate , the one-year present value factor is .We note that the 2-year discrete endowment insurance has actuarial present value

] .

Based on Kevin's premiums, we have , so that . It follows that .

Then, for Kira's premium, we have , so that , and then . Answer: B

27. Using the recursive relationship for reserves from time 9 to time 10, we have , so that

.Solving for results in .

Since this is a 10-payment policy, it is paid up as of time 10 so that the prospective form of thereserve is . But we have just found that this is 35,636 .Therefore, .

We now use the recursive relationship for discrete insurance, .Therefore, , which becomes

, and then . Answer: E

28. .° ° ° °Under DeMoivre's Law, , so that and .° ° °°

.Then, . Answer: B°

29. , .

. Answer: C

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30. .

Then, . Answer: E

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 6

1. Given the survival function , where

Calculate .A) .45 B) .55 C) .80 D) 1.00 E) 1.20

2. For a triple decrement table for which decrements are uniformly distributed in the multipledecrement table in each year of age, you are given:

(i)

(ii)

(iii) Calculate .A) Less than .2 B) At least .2 but less than .3 C) At least .3 but less than .4D) At least .4 but less than .5 E) At least .5

3. You are given: (i) the following select-and-ultimate table with a 3-year select period: (ii) Calculate , the actuarial present value of a 2-year deferred 2-year term insurance on .A) .156 B) .160 C) .186 D) .190 E) .195

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4. You are given:

(i)

(ii)

(iii) Calculate .A) 12.5 B) 13.0 C) 13.4 D) 13.9 E) 14.3

5. Lucky Tom finds coins on his way to work at a Poisson rate of 0.5 coins/minute.The denominations are randomly distributed: (i) 60% of the coins are worth 1 each (ii) 20% of the coins are worth 5 each (iii) 20% of the coins are worth 10 each.Calculate the probability that in the first ten minutes of his walk he finds at least 2 coinsworth 10 each, and in the first twenty minutes finds at least 3 coins worth 10 each.A) 0.08 B) 0.12 C) 0.16 D) 0.20 E) 0.24

6. For a fully discrete whole life insurance of 1000 on (60), the annual benefit premium wascalculated using the following: (i) (ii) (iii) (iv) A particular insured is expected to experience a first- year mortality rate ten times the rate usedto calculate the annual benefit premium. The expected mortality rates for all other years are theones originally used. Calculate the expected loss at issue for this insured, based on the originalbenefit premium.A) 72 B) 86 C) 100 D) 114 E) 128

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7. For a fully discrete whole life insurance of 1000 on (40), you are given: (i) (ii) Mortality follows the Illustrative Life Table (iii)

(iv)

(v)

At the end of the tenth year, the insured elects an option to retain the coverage of 1000 forlife, but pay premiums for the next ten years only. Calculate the revised annual benefit premiumfor the next 10 years.A) 11 B) 15 C) 17 D) 19 E) 21

8. For a double-decrement table where cause 1 is death and cause 2 is withdrawal, you are given:(i) Deaths are uniformly distributed over each year of age in the single-decrement table.(ii) Withdrawals occur only at the end of each year of age.(iii)(iv)(v)Calculate .A) .51 B) .53 C) .55 D) .57 E) .59

9. You intend to hire 200 employees for a new management-training program. To predict thenumber who will complete the program, you build a multiple decrement table. You decide thatthe following associated single decrement assumptions are appropriate:(i) Of 40 hires, the number who fail to make adequate progress in each of the first three years is10, 6, and 8, respectively.(ii) Of 30 hires, the number who resign from the company in each of the first three years is 6, 8,and 2, respectively.(iii) Of 20 hires, the number who leave the program for other reasons in each of the first threeyears is 2, 2, and 4, respectively.(iv) You use the uniform distribution of decrements assumption in each year in the multipledecrement table.Calculate the expected number who fail to make adequate progress in the third year.A) 4 B) 8 C) 12 D) 14 E) 17

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10. Bob is an overworked underwriter. Applications arrive at his desk at a Poisson rate of 60 perday. Each application has a 1/3 chance of being a “bad” risk and a 2/3 chance of being a “good”risk. Since Bob is overworked, each time he gets an application he flips a fair coin. If it comesup heads, he accepts the application without looking at it. If the coin comes up tails, he acceptsthe application if and only if it is a “good” risk. The expected profit on a “good” risk is 300 withvariance 10,000. The expected profit on a “bad” risk is –100 with variance 90,000. Calculate thevariance of the profit on the applications he accepts today.A) 4,000,000 B) 4,500,000 C) 5,000,000 D) 5,500,000 E) 6,000,000

11. For a temporary life annuity- immediate on independent lives (30) and (40): (i) Mortality follows the Illustrative Life Table. (ii) .Calculate .

A) 6.64 B) 7.17 C) 7.88 D) 8.74 E) 9.86

12. For a special whole life insurance on (35), you are given:(i) The annual benefit premium is payable at the beginning of each year.(ii) The death benefit is equal to 1000 plus the return of all benefit premiums paid inthe past without interest.(iii) The death benefit is paid at the end of the year of death.

(iv)

(v)(vi)Calculate the annual benefit premium for this insurance.A) 73.66 B) 75.28 C) 77.42 D) 78.95 E) 81.66

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13. For a fully discrete whole life insurance of 100,000 of , you are given:(i) Expenses, paid at the beginning of the year, are as follows:Year Percent of Premium Per 1000 Expenses Per Policy Expenses Expenses 1 50% 2.0 150 2+ 4% 0.5 25(ii)(iii)(iv) Death is the only decrementCalculate the expense-loaded premium using the equivalence principle.A) 5800 B) 5930 C) 6010 D) 6120 E) 6270

14. Subway trains arrive at a station at a Poisson rate of 20 per hour. 25% of the trains areexpress and 75% are local. The types of each train are independent. An express gets you to workin 16 minutes and a local gets you there in 28 minutes. You always take the firsttrain to arrive. Your co-worker always takes the first express. You both are waiting at thesame station. Which of the following is true?A) Your expected arrival time is 6 minutes earlier than your co-worker’s.B) Your expected arrival time is 4.5 minutes earlier than your co-worker’s.C) Your expected arrival times are the same.D) Your expected arrival time is 4.5 minutes later than your co-worker’s.E) Your expected arrival time is 6 minutes later than your co-worker’s.

15. For a special fully continuous whole life insurance of 1 on the last-survivor of and ,you are given:(i) and are independent.(ii) , (iii) .(iv) Premiums are payable until the first death.Calculate the level annual benefit premium for this insurance.A) 0.04 B) 0.07 C) 0.08 D) 0.10 E) 0.14

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16. For a fully discrete whole life insurance of 1000 on , you are given: (i) (ii) (iii) (iv) (v) Calculate .A) 0.024 B) 0.025 C) 0.026 D) 0.027 E) 0.028

17. For a fully discrete whole life insurance of 1000 on , you are given:(i) .(ii) Mortality follows the Illustrative Life Table, except that there are extra mortality risks at age60 such that .Calculate the annual benefit premium for this insurance.A) 31.5 B) 32.0 C) 32.1 D) 33.1 E) 33.2

18. Homerecker Insurance Company classifies its insureds based on each insured’s creditrating, as one of Preferred, Standard or Poor. Individual transition between classes each year ismodeled as a discrete Markov process with a transition matrix as follows: Preferred Standard Poor Preferred 0.95 0.04 0.01 Standard 0.15 0.80 0.05 Poor 0.00 0.25 0.75This year, the company has 1000 policies, of which 600 are standard, 300 are preferred and 100are poor. Assuming no new policies are issued for the next two years, calculate the percentage ofinsureds in the Preferred class in two years.A) 33% B) 43% C) 50% D) 63% E) 67%

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19. Nancy reviews the interest rates each year for a 30-year fixed mortgage issued onJuly 1. She models interest rate behavior by a homogeneous Markov model assuming:(i) Interest rates always change between years.(ii) The change in any given year is dependent on the change in prior years as follows:

from year to from year to Probability that year willyear year increase from year Increase Increase 0.10Decrease Decrease 0.20Increase Decrease 0.40Decrease Increase 0.25

She notes that interest rates decreased from year 2000 to 2001 and from year 2001to 2002. Calculate the probability that interest rates will decrease from year 2003 to 2004.A) 0.76 B) 0.79 C) 0.82 D) 0.84 E) 0.87

20. For a 20-year deferred whole life annuity-due of 1 per year on , you are given:(i) Mortality follows De Moivre’s law with .(ii)Calculate the probability that the sum of the annuity payments actually made will exceed theactuarial present value at issue of the annuity.A) 0.425 B) 0.450 C) 0.475 D) 0.500 E) 0.525

21. and are independent lives with . satisfies UDD and satisfies thehyperbolic assumption in each year of age. Find .

A) .0948 B) .0950 C) .0952 D) .0954 E) .0956

22. XYZ Co. has just purchased two new tools with independent future lifetimes. Each tool hasits own distinct De Moivre survival pattern. One tool has a 10-year maximum lifetime and theother a 7-year maximum lifetime. Calculate the expected time until both tools have failed.A) 5.0 B) 5.2 C) 5.4 D) 5.6 E) 5.8

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23. XYZ Paper Mill purchases a 5-year special insurance paying a benefit in the event itsmachine breaks down. If the cause is “minor” (1), only a repair is needed. If the cause is“major” (2), the machine must be replaced. Given:(i) The benefit for cause (1) is 2000 payable at the moment of breakdown.(ii) The benefit for cause (2) is 500,000 payable at the moment of breakdown.(iii) Once a benefit is paid, the insurance contract is terminated.(iv) and , for (v)Calculate the actuarial present value of this insurance.A) 7840 B) 7880 C) 7920 D) 7960 E) 8000

24. You are given:

(i)

(ii) (iii) is a constant such that Determine an expression for .

A) B)

C) D)

E)

25. A large machine in the ABC Paper Mill is 25 years old when ABC purchases a 5- year terminsurance paying a benefit in the event the machine breaks down. Given:(i) Annual benefit premiums of 6643 are payable at the beginning of the year.(ii) A benefit of 500,000 is payable at the moment of breakdown.(iii) Once a benefit is paid, the insurance contract is terminated.(iv) Machine breakdowns follow De Moivre’s law with .(v)Calculate the benefit reserve for this insurance at the end of the third year.A) 91 B) 0 C) 163 D) 287 E) 422

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26. For a whole life insurance of 1 on , you are given:(i) The force of mortality is .(ii) The benefits are payable at the moment of death.(iii)(iv)Calculate the revised actuarial present value of this insurance assuming is increased by

for all and is decreased by .A) 0.5 B) 0.6 C) 0.7 D) 0.8 E) 0.9

27. A maintenance contract on a hotel promises to replace burned out light bulbs at the end ofeach year for three years. The hotel has 10,000 light bulbs. The light bulbs are all new. If areplacement bulb burns out, it too will be replaced with a new bulb.You are given:(i) For new light bulbs, , (ii) Each light bulb costs 1.(iii) .Calculate the actuarial present value of this contract.A) 6700 B) 7000 C) 7300 D) 7600 E) 8000

28. A hospital is considering the purchase of a power supply system using two new independentgenerators. System X provides power as long as both generators are working. System Y providespower as long as at least one generator is working. For all generators and for ,

.Calculate the difference in expected lifetimes of system X and system Y.A) Less than 1.5 years B) At least 1.5 years, but less than 3.0 yearsC) At least 3.0 years, but less than 4.5 years D) At least 4.5 years, but less than 6.0 yearsE) At least 6.0 years

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29. An appliance store sells microwave ovens with a 3-year warranty against failure. At the timeof purchase, the customer may buy a 2-year extended warranty that would pay half of theoriginal purchase price at the moment of failure. You are given:• The extended warranty begins exactly 3 years after the time of purchase, but only if the ovenhas not failed by then.• Any failure is considered permanent.• 4%• Failure of ovens follows the mortality table below, with uniform distribution of failure withineach year.Age 0 1 2 3 4 5

.008 .015 .026 .042 .063 .089Calculate the actuarial present value of the extended warranty as a percent of the purchase price.A) Less than 3.8% B) At least 3.8%, but less than 4.1% C) At least 4.1%, but less than4.4%D) At least 4.4%, but less than 4.7% E) At least 4.7%

30. A retirement residence has the facilities to accommodate individuals who can liveindependently, require assisted living or have a temporary disability. The facility models theresidents using a multi-state transition model approach with four states:

- living independently, -requiring assisted living accommodation,-living in accommodation for those with a disability, and -out of the facility.

A resident's state is updated on Jan. 1 of each year.The one-step transition probability matrices for Jan. 1, 2006 to Jan. 1, 2007 and Jan. 1, 2007 to

Jan. 1, 2008 are and

(the states for the matrices are in the order from top to bottom and left to right).

Annual health care costs for a resident is 300 for someone living independently, 700 for someonein assisted living, and 2000 for someone who is disabled. On January 1, 2006 there are 200residents in the facility, of whom 150 are living independently, 40 are in assisted living and 10are disabled. Find the expected health care costs for these residents for years 2006, 2007 and2008 combined (nearest 10,000).A) 250,000 B) 270,000 C) 290,000 D) 310,000 E) 330,000

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 6 SOLUTIONS

1. . Answer: E

2.

. Answer: B

3.

. Answer: D

4. Since the mortality basis changes after 5 years, we write the actuarial present value of theannuity as . .

Since the force of mortality is constant at .01 for the first 5 years, we have (so that ), and

.Since the force of mortality is constant at .02 from age on, we have

.

Then, . Answer: B

5. The number of coins with denomination 10 being found forms a Poisson processwith mean rate per minute. We measure time units in minutes.

We are asked to find .The event can be formulated as a disjoint union

.

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5 continuedThe probability is then .

has a Poisson distribution with mean , so that

.Also, since forms a Poisson process, we have

.

Then , and has a Poisson distribution with

mean 1, so that , and .

Finally, . Answer: D

6. The original benefit premium is based on the original insurance value. .

Based on the modified mortality probability in the first year, the actuarial present value of theinsurance is .The actuarial present value of the premium is

.The expected loss at issue isAPV benefit APV premium . Answer: D

7. The original policy reserve at the end of the 10th year plus the actuarial present value of 10more premiums must be equal to the remaining benefit of a life insurance of amount 1000 at age50. The original policy reserve at the end of 10 years is

(the annuity values are taken from

the Illustrative Life Table). The revised premium must satisfy.

From the Illustrative Life Table we get , so that . Answer: D

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8. Decrement 2 (withdrawal) is a discrete decrement occurring at the end of the year of death.In this case the relationships between and are

and .From , if we divide by , it follows that .Therefore, .Using the relationships above, we get , and .Therefore, , and . Answer: A

9. The information about the single table behavior gives us values."Age" 0 corresponds to the start of the training program.Decrement 1 (insufficient progress): Decrement 2 (resign): Decrement 3 (other reasons): We are given , and we are asked to find .This can be formulated as , where .From the given information, we have ,and . Then, .Under the Uniform Distribution of Decrement assumption in multiple decrement table, the

relationship between multiple decrement probabilities and absolute rates is ,

so that .

From the given information , we have

and . Then .

Finally, . Answer: D

10. The total profit on a the applications accepted in a given day can be modeled as acompound Poisson random variable with mean frequency rate , and with severitydistribution . represents the profit on an arriving application. can be modeled as amixture of three component distributions. Good risks will always be accepted no matter whatthe result of the coin flip. Since good risks arrive of the time, it follows that of the acceptedrisks are "good". Bad risks arrive of the time, but half of the arriving bad risks will be rejectedsince the coin flip is "tails" half of the time, and in that case the bad risk is rejected.

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10. continuedTherefore, of the accepted risks are "bad". The other of the accepted risks are 0 (arejected bad risk). The accepted risks are a mixture of(i) "good" risks (say ) with mixing weight , (ii) "bad" risks (say ), with mixing weight ,and (iii) risks of amount 0 (say ) with mixing weight .Since has a compound Poisson distribution, the variance of can be formulated as

.Since is a mixture distribution, .

,and , and Then .Finally, . Answer: C

11. .

From the Illustrative Life Table from the column we have and . Also, .

Then, . Answer: B

12. Suppose that the annual benefit premium (this is the equivalence principle premium) is .The actuarial present value of benefit is .The equivalence principle equation is .We use the relationship to get .Then . Answer: A

13.

. .

. Answer: C

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14. The number of express trains arriving forms a Poisson process with mean rate per hour. The amount of time between successive express trains has an

exponential distribution with mean hours 12 min. The expected time it will take your co-worker to get to work is minutes (expected time until the express train arrivesplus travel time to work). The number of local trains arriving per hour forms a Poisson processwith a mean rate of per hour. Since you take the first available train, your timeto work is a mixture of the two travel times. The expected time until the next train is hours 3 min . There is a .25 chance that the next train is express and a .75 chance that the nexttrain is local. The expected travel time to work for the next train is .The expected time for you to get to work is min. The expected times are the same. Answer: C

15. The APV of the last survivor insurance is .Suppose that the premium rate is . Since premiums are payable until the first death, the APVof premium is . Since and are independent, we have

. With the constant forces of mortality given,this becomes (note that this is , which is

the joint-life form of life annuity APV when lives are independent and the forces of mortality areconstant). From constant force of mortality we get , and

(or ).

The benefit premium equation is . Answer: C

16. We use the recursive relationship for benefit reserves: .For , this equation becomes

.We solve for the interest rate, .The recursive relationship for becomes

.We can now solve for , which is equal to . Answer: E

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17. The annual benefit premium will be the solution of the equation. Since there is an extra risk at age 60 we cannot use the insurance and annuity

values from the Illustrative Life Table at age 60. We can use the values at age 61,and then use the recursive relationship to get .

. .

Finally, . Answer: E

18. The one-step (one-year) transition probability matrix is Preferred Standard Poor Preferred 0.95 0.04 0.01 Standard 0.15 0.80 0.05 Poor 0.00 0.25 0.75

The two-step transition probability matrix is

Of the 300 Preferred policies now, we expect to still bepreferred in 2 years. Of the 600 standard policies now, we expect

to be preferred in 2 years, and of the 100 poor policiesright now, we expect to be preferred in 2 years.The total expected preferred policies in 2 years is , or43% of the total of 1000 policies. Answer: B

19. We can outline the possible sequences of events leading to the event in question in thefollowing table. 00-01 01-02 02-03 03-04 D (given) D (given) I D D (given) D (given) D D

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19 continuedThe probability of the first sequence is ; this is row 2 of the table of probabilities,followed by the complement of row 4.The probability of the first sequence is ; this is the complement of row 2 of thetable of probabilities, followed by the complement of row 2.The overall probability in question is .

A more tedious approach uses formal probability rules.We use the following probability rule

,where form a partition of the probability space.30. continued

is the event that interest rates decrease for 2003 to 2004 . is the event that interest rates decreased from 2000 to 2001 and from 2001 to 2002 ,

is the event that interest rates increase for 2001 to 2002 and 2002 to 2003 ,is the event that interest rates decrease from 01 to 02 and from 02 to 03,

is the event that interest rates increase from 01 to 02and decrease from 02 to 03, and

is the event that interest rates decrease from 01 to 02and increase from 02 to 03.Then is given, as is

, and .

We must now find , , and .2 3

Since is the event that interest rates decrease from 00 to 01 and from 01 to 02, onlyand can occur given . Given that rates decreased from 00 to 01 and from 01 to 02, rates

did not increase from 01 to 02 and therefore it is impossible for or to occur. Therefore, .

Since includes the event that interest rates decrease from 01 to 02, in order for to occurafter , rates have to decrease from 02 to 03 as well; therefore,

03 will inc from 02 dec from 00 to 01 and dec from 01 to 02 .

Also, 03 will inc from 02 dec from 00 to 01 and dec from 01 to 02 .Finally, . Answer: B

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20. The APV is ¨2 |

.sum of payments exceeds 13.67 at least 14 payments .

The 14-th payment is made at age 78 (1st payment is at age 65), so thatat least 14 payments survival to at least age 78 .

Answer: B

21.

. Then . Answer: C

22. Tool A has , and tool B has . We are asked to find .°

We use the relationship .° ° ° °

Since A follows DeMoivre's Law with , we have , and°

similarly .°° ( we use the independence assumption to get

). From DeMoivre's Law we have for ,and for Therefore,° (the upper limit is 7 because for all

). This becomes .°

Then . Answer: E° ° ° °

23. The force of decrement for both causes combined is constant at , and thesurvival probability is .The APV of the decrement 1 benefit is

.The APV of the decrement 2 benefit is

.Total APV is . Answer: A

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24. .We are given

[ . Answer: A

25. The prospective form of the reserve at the end of the 3rd year is .

Under the DeMoivre model, ,

and .

Then, . Answer: D

26. Based on the original force of interest and force of mortality, the continuous whole lifeannuity value is . The current payment form of the annuity is

Based on the new force of mortality, the survival probability is .

Based on the new force of mortality and the new force of interest, the annuity value is ;the new annuity value is the same as the old annuity value (the increase in force of mortality isexactly offset by the decrease in the force of interest).The new insurance value is .Answer: D

27. We calculate the expected number of bulb failures at the end of each year for three years.Each year some of the bulbs are replaced by new bulbs (age 0) and some bulbs remain (at anincreasing age). We keep track of how many bulbs of each age there are each year, since thistells us how many failures to expect for that year. For the first year, all 10,000 bulbs are new,and we expect .1 of them to fail; i.e., we expect 1000 failures in the first year. At the start of thesecond year there are 1000 new and 9000 one-year-old bulbs. We expect 100 of the 1000 newbulbs to fail ( ) and 2700 of the 9000 one-year-old bulbs to fail ( ). At the start ofthe 3rd year, there are 2800 new bulbs (replacing the that stopped working), and900 one-year-old bulbs (remaining from the 1000 new ones at the start of the second year) and

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27 continued6300 two-year-old bulbs (remaining from the 9000 on-year-old bulbs at the start of the secondyear). The number of replacements needed at the end of the third year is

.This can be summarized in the following table.

Time 0 Time 1 Time 2 Time 310,000 1000 replacednew bulbs 1000 new bulbs, 100 repl 9000 one year old 2700 repl 2800 new, 280 repl 900 one year old 270 repl 6300 two years old 3150 repl

Total replaced 1000 2800 3700

The present value of the cost of replacement bulbs is . Answer: A

28. Expected lifetime of system X is °

years. Expected lifetime of system Y is .° ° ° °° , so that the expected lifetime of system Y is

. The difference in expected lifetimes of systems X and Y is . Answer: B

29. If the purchase price is , the extended warranty pays if failure is in the 4th or 5th year.The APV of this benefit is . ,

,and . Under UDD, .

. The APV of the extended warranty is . Answer: C

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30. The 2-step transition probability matrix from Jan. 1, 2006 to Jan.1, 2008 is

Expected health care costs for someone living independently on Jan. 1, 2006 will bein 2006, in 2007, and

in 2008, for a total of 1209 .

Expected health care costs for someone living with assistance on Jan. 1, 2006 will be700 in 2006, in 2007, and

in 2008, for a total of 1879 .

Expected health care costs for someone disabled on Jan. 1, 2006 will be2000 in 2006, in 2007, and

in 2008, for a total of .

Total expected costs for all residents are . Answer: C

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 7

1. You are given:

Calculate .°

A) 14.0 B) 14.4 C) 14.8 D) 15.2 E) 15.6

2. For a select-and-ultimate mortality table with a 3-year select period:(i)

60 0.09 0.11 0.13 0.15 6361 .010 0.12 0.14 0.16 6462 0.11 0.13 0.15 0.17 6563 0.12 0.14 0.16 0.18 6664 .013 0.15 0.17 0.19 67

(ii) White was a newly selected life on 01/01/2000.(iii) White’s age on 01/01/2001 is 61.(iv) is the probability on 01 /01 /2001 that White will be alive on 01 /01 /2006.Calculate P.A) B) C) D) E)

3. For a continuous whole life annuity of 1 on :(i) is the future lifetime random variable for .(ii) The force of interest and force of mortality are constant and equal.(iii)Calculate the standard deviation of

A) 1.67 B) 2.50 C) 2.89 D) 6.25 E) 7.22

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4. For a special fully discrete whole life insurance on :(i) The death benefit is 0 in the first year and 5000 thereafter.(ii) Level benefit premiums are payable for life.(iii)(iv)(v) a(vi)(vii) is the benefit reserve at the end of year 10 for this insurance.CalculateA) 795 B) 1000 C) 1090 D) 1180 E) 1225

5. For a fully discrete 2-year term insurance of 1 on :(i) 0.95 is the lowest premium such that there is a 0% chance of loss in year 1(ii)(iii)(iv) is the random variable for the present value at issue of future benefits.Calculate VarA) 0.15 B) 0.17 C) 0.19 D) 0.21 E) 0.23

6. Each of 100 independent lives purchase a single premium 5-year deferred whole lifeinsurance of 10 payable at the moment of death. You are given: (i) (ii) (iii) is the aggregate amount the insurer receives from the 100 lives.Using the normal approximation, calculate such that the probability the insurer has sufficientfunds to pay all claims is 0.95.A) 280 B) 390 C) 500 D) 610 E) 720

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7. For a select-and-ultimate table with a 2-year select period:

.

Keith and Clive are independent lives, both age 50. Keith was selected at age 45 and Clive wasselected at age 50. Calculate the probability that exactly one will be alive at the end of threeyears.A) Less than 0.115 B) At least 0.115, but less than 0.125C) At least 0.125, but less than 0.135 D) At least 0.135, but less than 0.145E) At least 0.145

8. A fund is established by collecting an amount P from a life age 70. The fund will pay thefollowing benefits: • 10, payable at the end of the year of death, for death before age 72, or

• , payable at age 72, if the individual survives survive.You are given:(i) Mortality follows the Illustrative Life Table.(ii)Calculate P, using the equivalence principle.A) 2.33 B) 2.38 C) 3.02 D) 3.07 E) 3.55

9. You are given:(i) (ii) (iii)

Calculate

A) 0.008 B) 0.024 C) 0.040 D) 0.065 E) 0.085

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10. For a fully discrete 5-payment 10-year deferred 20-year term insurance of 1000 on (30), youare given the following schedule of expenses:

Year 1 Years 2 - 10Percent of Premium Per Policy Percent of Premium Per Policy

Taxes 5% 0 5% 0Sales Commission 25% 0 10% 0

Policy Maintenance 0 20 0 10Expenses are paid at the beginning of each policy year.Which of the following is a correct expression for the expense-loaded premium?

A) 4

B)

C) 4

D) 4

E)

11. You are given:(i) Mortality follows De Moivre's law with (ii)(iii) The following annuity-certain values:

CalculateA) 0.075 B) 0.077 C) 0.079 D) 0.081 E) 0.083

12. Don, age 50, is an actuarial science professor. His career is subject to two decrements:(i) Decrement 1 is mortality. The associated single decrement table follows De Moivre’s lawwith .(ii) Decrement 2 is leaving academic employment, with Calculate the probability that Don remains an actuarial science professor for at least five but lessthan ten years.A) 0.22 B) 0.25 C) 0.28 D) 0.31 E) 0.34

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13. For a double decrement model:(i) In the single decrement table associated with cause (1), and decrements areuniformly distributed over the year.(ii) In the single decrement table associated with cause (2), and all decrements2

occur at time 0.7.Calculate .A) 0.114 B) 0.115 C) 0.116 D) 0.117 E) 0.118

14. A taxi driver provides service in city R and city S only. If the taxi driver is in city R, theprobability that he has to drive passengers to city S is 0.8. If he is in city S, the probability thathe has to drive passengers to city R is 0.3. The expected profit for each trip is as follows: a trip within city R: 1.00 a trip within city S: 1.20 a trip between city R and city S: 2.00The driver is currently in city S. Find the expected profit to the nearest .25 for his next two rides.A) 2.00 B) 2.25 C) 2.50 D) 2.75 E) 3.00

15. For a special 2-payment whole life insurance on (80):(i) Premiums of are paid at the beginning of years 1 and 3.(ii) The death benefit is paid at the end of the year of death.(iii) There is a partial refund of premium feature: If (80) dies in either year 1 or year 3, the death benefit is .

Otherwise, the death benefit is 1000.(iv) Mortality follows the Illustrative Life Table.(v)Calculate , using the equivalence principle.A) 369 B) 381 C) 397 D) 409 E) 425

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16. For a special fully continuous whole life insurance on (65):(i) The death benefit at time is , t 0.(ii) Level benefit premiums are payable for life.(iii)(iv) 0.04Calculate , the benefit reserve at the end of year 2.2

A) 0 B) 29 () 37 () 61 E) 83

17. You are given:(i) (ii) 40 (iii) (iv) +20

Calculate .:

A) 11.0 B) 11.2 C) 11.7 D) 12.0 E) 12.3

18. A machine is in one of four states (F, G, H, I) and migrates annually among them accordingto a Markov process with transition matrix:

F G H IFGHI

0.20 0.80 0.00 0.000.50 0.00 0.50 0.000.75 0.00 0.00 0.251.00 0.00 0.00 0.00

At time 0, the machine is in State F. A salvage company will pay 500 at the end of 3 years if themachine is in State F. Assuming calculate the actuarial present value at time 0 of thispayment.A) 150 B) 155 C) 160 D) 165 E) 170

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19. You are given: (i) The survival function for males is , .

(ii) Female mortality follows De Moivre’s law. (iii) At age 60, the female force of mortality is 60% of the male force of mortality.For two independent lives, a male age 65 and a female age 60, calculate the expected time untilthe second death.A) 4.33 B) 5.63 C) 7.23 D) 11.88 E) 13.17

20. For a fully discrete insurance of 10,000 on , you are given(i) (ii) (iii) (iv) is the fraction of gross premium paid at time 9 for expenses(v) is the amount of per policy expense paid at time 9(vi) Death and withdrawal are the only decrements(vii) (viii) (ix) Calculate .A) 1316 B) 1390 C) 1472 D) 1524 E) 1565

21. For a fully continuous whole life insurance of 1 (i) (ii) (iii) is the loss-at-issue random variable based on the benefit premium.Calculate Var .A) B) C) D) E) 10 4 3 25

22. A fully discrete 4-year term insurance policy to has level annual benefit premiums andhas the following schedule of death benefits: for .The annual effective interest rate is 6% and for .Find .A) Less than 50 B) At least 50 but less than 60 C) At least 60 but less than 70D) At least 70 but less than 80 E) At least 80

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23. For a special 3-year deferred whole life annuity-due on : (i) (ii) The first annual payment is 1000. (iii) Payments in the following years increase by 4% per year. (iv) There is no death benefit during the three year deferral period. (v) Level benefit premiums are payable at the beginning of each of the first three years. (vi) is the curtate expectation of life for .

(vii)

Calculate the annual benefit premium.A) 2625 B) 2825 C) 3025 D) 3225 E) 3425

24. For a special fully discrete 10-payment whole life insurance on (30) with level annualbenefit premium :(i) The death benefit is equal to 1000 plus the refund, without interest, of the benefit premiumspaid.(ii)(iii) 0(iv)

(v) ¨

CalculateA) 14.9 B) 15.0 C) 15.1 D) 15.2 E) 15.3

25. For two independent lives (30) and (40), you are given:• Mortality follows DeMoivre's Law with • is the probability that the first death occurs between 10 and 20 years from now• is the probability that the last death occurs between 10 and 20 years from nowCalculate .A) Less than 0.155 B) At least 0.155, but less than 0.175C) At least 0.175, but less than 0.195 D) At least 0.195, but less than 0.215E) At least 0.215

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26. You are given the following 2-decrement table: Calculate the difference between and .A) Less than 0.24 B) At least 0.24, but less than 0.26 C) At least 0.26, but less than 0.28D) At least 0.28, but less than 0.30 E) At least 0.30

27. For a 20-year term life insurance on , you are given:•• , the force of mortality due to accident• , the force of mortality due to other causes• The benefit is paid at the moment of death.• The benefit of 2 is paid if death occurs by accident, and a benefit of 1 is paid if death occurs byother causes.Calculate the actuarial present value of this insurance.A) B) C) D) E)

28. 24-hour Lube-n-Tune auto shop does oil changes and tune-ups for automobiles.Cars arrive at Lube-n-Tune according to a Poisson process at a rate of 4 per hour(during business hours): 50% of cars arriving are for an oil change only, 25% are for a tune-uponly, and 25% are for an oil change and tune-up.Find the probability that between 10AM and 11 AM at least two cars arrive requesting an oilchange and at least one car arrives requesting a tune up.A) .54 B) .60 C) .66 D) .72 E) .78

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29. Lucky Tom finds coins on his way to work at a Poisson rate of 0.5 coins/minute.The denominations are randomly distributed:(i) 60% of the coins are worth 1;(ii) 20% of the coins are worth 5; and(ii) 20% of the coins are worth 10.Calculate the conditional expected value of the coins Tom found during his one-hour walk today,given that among the coins he found exactly ten were worth 5 each.A) 108 B) 115 C) 128 D) 165 E) 180

30. For a certain mortality table, you are given:(i) (80.5) 0.0202(ii) (81.5) 0.0408(iii) (82.5) 0.0619(iv) Deaths are uniformly distributed between integral ages.Calculate the probability that a person age 80.5 will die within two years.A) 0.0782 () 0.0785 () 0.0790 D) 0.0796 E) 0.0800

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 7 SOLUTIONS

1. . °

° .

Alternatively, ° ° °

, same as above. Answer: E

2. .Answer: C

3. With constant force of morality and force of interest also equal to , we have . Therefore, .

Then,

The standard deviation is . Answer: E

4. The APV of the benefit is .

The annual benefit premium is . .

We are given

A . Then . Answer: D

5. With annual benefit premium , if death occurs in the first year the loss is , and theprobability of death in the first year is . In order for there to 0% chance of a positive loss inthe first year, we must have , or equivalently . The minimum premium forwhich there is a 0% probability of positive loss in the first year is , which we are told is .95.Therefore, . The variance of the two-year term insurance present value random variable

is

. Answer: D

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6. For one policy, the PVRV is with mean

and variance .

For 100 policies, the PVRV of total claims has mean , andvariance . We wish to find so that .Using the normal approximation, we get

. Answer: A

7. exactly one will be alive in 3 years . .

.Prob . Answer: D

8. Answer: C

9. . Answer: E

10.

Answer: E

11. Under DeMoivre's Law, .

With , we have , and

Then . Answer: A

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12. We wish to find .Under DeMoivre's Law with we have ,and .Under constant force of decrement, , and .Then . Answer: A

13. . .

(this last equality is true since density is constant within associated table 1 because of the UDDassumption for single table 1).Since cause 2 occurs at time 7, we have

.Then, and .Answer: C

14. The one-step (one ride) transition probability matrix is

There are 4 possible combinations for the next two trips. The combinations with the probabilitiesand profits are Next Two Trips Probability Profit Expected profit is . Answer: E

15.

. Answer: C

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16. The annual benefit premium is the solution of .

. Answer: E

17.

. Answer: B

18. The one-step transition probability matrix is

The two-step transition probability matrix is

The 3-step transition probability from F to F is (the other probabilities in the 3-step transition matrix are not needed). The present value at time 0 is .Answer: E

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19. The male survival distribution is DeMoivre's Law with . The force of mortality fora male at age 60 is . Female mortality follows DeMoivre's Law with upper agelimit . Female force of mortality at age 60 is .° ° ° °

. °

Answer: E

20. We use the relationship

to get

from which we get . Answer: E

21. For fully continuous whole life insurance with premiums for life at rate . Since , this becomes

,

. Answer: B

22. We use the relationship .

Since this is a 4-year term insurance, we have , so that

, which becomes

.

Solving for results in .Using the net-amount-at-risk form of the recursive relationship for reserves, we have

. Answer: D

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23.

.Using the relation , we get

.Then .

Using , we get .

Answer: B

24.

. Answer: B

25.

. Answer: B

26.(for instance ).

.The difference is . Answer: E

27. .APV of accidental death benefit is Using the substitution, , , and the integral becomes

, so the APV is .

The APV of the benefit for death due to other causes is . We see that this integral is twice as large as

, so .The total APV is . Answer: E

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28. We have the following independent sub-processes:- cars arrive requesting an oil change only at a rate of 2 per hour,- cars arrive requesting a tune-up only at a rate of 1 per hour,- cars arrive requesting an oil change and tune-up at a rate of 1 per hour.

is the event of 0 or 1 cars arriving requesting an oil change between 10 and 11, is the event of 0 cars arriving requesting a tune up between 10 and 11.

The probability in question is .

The number of cars arriving between 10 and 11 requesting an oil change is Poisson with a meanof , and the number of cars arriving between 10 and 11 requesting a tune-up is Poissonwith a mean of .

, .no cars arriving between 10 and 11

1 car arriving between 10 and 11 requesting an oil change no cars arriving between 10 and 11 requesting a tune-up .

no cars arriving between 10 and 11 , and

1 car arriving between 10 and 11 requesting an oil change no cars arriving between 10 and 11 requesting a tune-up

1 car arriving between 10 and 11 requesting an oil change only no cars arriving between 10 and 11 requesting a tune-up

(the two events in the intersection are independent).

Then , and the probability in question is .

The probability can also be formulated asoil tune oil tune oil only

oil tune oil only tune only

. Answer: D

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29. The number of coins found worth 1 follows a Poisson process with a rate of per minute. The expected number of coins worth 1 found by Lucky Tom in one hour is

. The number of coins found worth 10 follows a Poisson process with a rate of per minute. The expected number of coins worth 10 found by Lucky Tom in one

hour is . Given that Lucky Tom found 10 coins worth 5 each, the conditionalexpected value of the coins he found is (expected 18 coinsworth 1 each, 6 coins worth 10 each, and the 10 coins worth 5 each that we are told he found). Answer: C

30. We are to find .The reason for writing the probability in this way is that we must reflect the mortality withineach year of age. Under the UDD assumption, .

Also, , for integer age . Thus,

,

, and

, so that

, , and

.Then, . Answer: A

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 8

1. For a given life age 30, it is estimated that an impact of a medical breakthrough will be anincrease of 4 years in , the complete expectation of life. Prior to the medical breakthrough,°

followed de Moivre's law with as the limiting age. Assuming de Moivre’s lawstill applies after the medical breakthrough, calculate the new limiting age.A) 104 B) 105 C) 106 D) 107 E) 108

2. On January 1, 2002, Pat, age 40, purchases a 5-payment, 10-year term insurance of 100,000:(i) Death benefits are payable at the moment of death.(ii) Contract premiums of 4000 are payable annually at the beginning of each year for 5 years.(iii)(iv) is the loss random variable at time of issue.Calculate the value of if Pat dies on June 30, 2004.A) 77,100 B) 80,700 C) 82,700 D) 85,900 E) 88,000

3. Lucky Tom finds coins on his way to work at a Poisson rate of 0.5 coins per minute.The denominations are randomly distributed:(i) 60% of the coins are worth 1;(ii) 20% of the coins are worth 5;(iii) 20% of the coins are worth 10.Calculate the variance of the value of the coins Tom finds during his one-hour walk to work.A) 379 B) 487 C) 566 D) 670 E) 768

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4. For a fully discrete 20-payment whole life insurance of 1000 on , you are given:(i)(ii)(iii) The level annual benefit premium is 13.72.(iv) The benefit reserve at the end of year 19 is 342.03.Calculate , the level annual benefit premium for a fully discrete whole life insuranceof 1000 on .A) 27 B) 29 C) 31 D) 33 E) 35

5. For a multiple decrement model on (60):(i) , , follows the Illustrative Life Table.(ii) , Calculate , the probability that decrement occurs during the 11 year.A) 0.03 B) 0.04 C) 0.05 D) 0.06 E) 0.07

6. A coach can give two types of training, “light” or “heavy,” to his sports team before a game.If the team wins the prior game, the next training is equally likely to be light or heavy. But, if theteam loses the prior game, the next training is always heavy. The probability that the team willwin the game is 0.4 after light training and 0.8 after heavy training. The team lost its last game.Find the probability of winning at least one of the next two games.A) 0.90 B) 0.92 C) 0.94 D) 0.96 E) 0.98

7. and are two lives with identical expected mortality. You are given: , , where is the annual benefit premium for a fully discrete insurance of 1 on ,

Calculate the premium , the annual benefit premium for a fully discrete insurance of 1 on.

A) 0.14 B) 0.16 C) 0.18 D) 0.20 E) 0.22

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8. A benefit scheme allows retirements to occur starting at age 65. Prior to retirement there aretwo other decrements, those being death and withdrawal, and these two decrements are assumedto be uniformly distributed in their associated single decrement tables. The employer isconsidering allowing early retirement at age 64. Under the current scheme,

(death decrement), (withdrawal decrement), and (retirement decrement). If the early retirement plan is adopted, the pension

actuary estimates a net probability of retirement at age 64 of , with the other netprobabilities remaining unchanged. Find the revised value of if early retirement is adoptedand the retirement decrement is assumed to be uniformly distributed in its associated singledecrement table.A) .180 B) .197 C) .214 D) .231 E) .250

9. Justin takes a train to work each day. It takes 10 minutes for Justin to walk from home to thetrain station. In order to get to work on time, Justin must board the train by 7:50AM. Trainsarrive at the station at a Poisson rate of 1 every 8 minutes. What is the latest time he must leavehome each morning so that he is on time for work at least 90% of the time?A) 7:21AM B) 7:22AM C) 7:31AM D) 7:32AM E) 7:41AM

10. Mr. Ucci has only 3 hairs left on his head and he won’t be growing any more.(i) The future mortality of each hair follows and is Mr. Ucci’s age(ii) Hair loss follows the hyperbolic assumption at fractional ages.(iii) The future lifetimes of the 3 hairs are independent.Calculate the probability that Mr. Ucci is bald (has no hair left) at age A) 0.098 B) 0.103 C) 0.108 D) 0.113 E) 0.118

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11. The following graph is related to current human mortality:

Which of the following functions of age does the graph most likely show?A) B) C) D) E)

12. For a special 3-year term insurance on , you are given:(i) is the present-value random variable for the death benefits.(ii) q (iii) The following death benefits, payable at the end of the year of death:

0 300,0001 350,0002 400,000

(iv)Calculate .A) 36,800 B) 39,100 C) 41,400 D) 43,700 E) 46,000

13. For a special fully discrete 20-year endowment insurance on (55):(i) Death benefits in year are given by (ii) The maturity benefit is 1. (iii) Annual benefit premiums are level. (iv) denotes the benefit reserve at the end of year (v) (vi) (vii) (viii) Calculate .A) 4.5 B) 4.6 C) 4.8 D) 5.1 E) 5.3

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14. A whole life insurance has annual premiums payable at the beginning of the year and deathbenefits payable at the moment of death. The following expenses are allocated to this policy atthe beginning of each year: % of Premium Per 1000 of Insurance Per PolicyFirst year 30% 3.00 150Renewal 10% 0.00 50You are given: (i) (ii) A level policy fee is used to recognize per policy expenses in the expense-loaded premiumformula. Calculate the minimum face value such that the policy fee does not exceed 50% of theexpense-loaded premium.A) 2,650 B) 3,000 C) 3,450 D) 5,300 E) 6,000

15. Each day, traffic passing through the Washington Tunnel increases during the morning andafternoon rush hours, and decreases at other times as follows: (i) From 12 a.m. to 8 a.m., the numbers of cars follows a Poisson distribution with an increasing hourly rate of for . (ii) From 8 a.m. to 12 p.m., the numbers of cars follows a Poisson distribution with an decreasing hourly rate of for . (iii) From 12 p.m. to 6 p.m., the numbers of cars follows a Poisson distribution with an increasing hourly rate of for . (iv) From 6 p.m. to 12 a.m., the numbers of cars follows a Poisson distribution with an decreasing hourly rate of for .What is the probability that exactly 25 cars pass through the tunnel between 11:30 a.m. and12:30 p.m.?A) 0.0187 B) 0.0273 C) 0.0357 D) 0.0432 E) 0.0511

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16. The Simple Insurance Company starts at time with a surplus of . At thebeginning of every year, it collects a premium of . Every year, it pays a random claimamount:

Claim Amount Probability of Claim Amount0 0.151 0.252 0.504 0.10

Claim amounts are mutually independent. If, at the end of the year, Simple’s surplus is morethan 3, it pays a dividend equal to the amount of surplus in excess of 3. If Simple is unable to payits claims, or if its surplus drops to 0, it goes out of business. Simple has no administrativeexpenses and its interest income is 0.Calculate the expected dividend at the end of the third year.A) 0.115 B) 0.350 C) 0.414 D) 0.458 E) 0.550

17. A continuous two-life annuity pays: 100 while both (30) and (40) are alive; 70 while (30) is alive but (40) is dead; and 50 while (40) is alive but (30) is dead.The actuarial present value of this annuity is 1180. Continuous single life annuities paying 100per year are available for (30) and (40) with actuarial present values of 1200 and 1000,respectively.Calculate the actuarial present value of a two-life continuous annuity that pays 100 while at leastone of them is alive.A) 1400 B) 1500 C) 1600 D) 1700 E) 1800

18. If is an increasing function of for 50, then which of the following must be true?I. II. III. A) I only B) II only C) III only D) None E) None of A, B, C or D

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19. For a special fully discrete 3-year term insurance on :(i) Level benefit premiums are paid at the beginning of each year.(ii)

0 200,000 0.031 150,000 0.062 100,000 0.09

(iii)Calculate the initial benefit reserve for year 2.A) 6,500 B) 7,500 C) 8,100 D) 9,400 E) 10,300

20. For a special fully continuous whole life insurance on :(i) The level premium is determined using the equivalence principle.(ii) Death benefits are given by where is the interest rate.(iii) is the loss random variable at for the insurance.(iv) is the future lifetime random variable of Which of the following expressions is equal to ?

A) B) C)

D) E)

21. For a 4-year college, you are given the following probabilities for dropout from all causes: Dropouts are uniformly distributed over each year.Compute the temporary 1.5-year complete expected college lifetime of a student entering thesecond year, .°

A) 1.25 B) 1.30 C) 1.35 D) 1.40 E) 1.45

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22. Lee, age 63, considers the purchase of a single premium whole life insurance of 10,000 withdeath benefit payable at the end of the year of death.The company calculates benefit premiums using:(i) Mortality based on the Illustrative Life Table,(ii)The company calculates contract premiums as 112% of benefit premiums. The single contractpremium at age 63 is 5233. Lee decides to delay the purchase for two years and invests the5233. Calculate the minimum annual rate of return that the investment must earn to accumulateto an amount equal to the single contract premium at age 65.A) 0.030 B) 0.035 () 0.040 D) 0.045 E) 0.050

23. You have calculated the actuarial present value of a last-survivor whole life insurance of 1on and . You assumed:(i) The death benefit is payable at the moment of death.(ii) The future lifetimes of and are independent, and each life has a constant force ofmortality with .(iii) .Your supervisor points out that these are not independent future lifetimes. Each mortalityassumption is correct, but each includes a common shock component with constant force 0.02.Calculate the increase in the actuarial present value over what you originally calculated.A) 0.020 B) 0.039 C) 0.093 D) 0.109 E) 0.163

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24. For a claims process, you are given:(i) The number of claims is a nonhomogeneous Poisson process with intensityfunction:

(ii) Claims amounts are independently and identically distributed random variables that arealso independent of .(iii) Each is uniformly distributed on (iv) The random variable is the number of claims with claim amount less than 500by time (v) The random variable is the number of claims with claim amount greater than 500by time .(vi) is the conditional expected value of , given .CalculateA) 2.0 B) 2.5 C) 3.0 D) 3.5 E) 4.0

25. A government creates a fund to pay this year’s lottery winners.You are given:(i) There are 100 winners each age 40.(ii) Each winner receives payments of 10 per year for life, payable annually, beginningimmediately.(iii) Mortality follows the Illustrative Life Table.(iv) The lifetimes are independent.(v)The amount of the fund is determined, using the normal approximation, such that the probabilitythat the fund is sufficient to make all payments is 95%.Calculate the initial amount of the fund.A) 14,800 B) 14,900 C) 15,050 D) 15,150 E) 15,250

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26. For a special fully discrete 35-payment whole life insurance on (30):(i) The death benefit is 1 for the first 20 years and is 5 thereafter.(ii) The initial benefit premium paid during the each of the first 20 years is one fifth of thebenefit premium paid during each of the 15 subsequent years.(iii) Mortality follows the Illustrative Life Table.(iv)(v) 0.32307

(vi) 14.835¨

Calculate the initial annual benefit premium.A) 0.010 B) 0.015 C) 0.020 D) 0.025 E) 0.030

27. For a single premium discrete 15-year term insurance of $100,000 on a person age (45), youare given: % Mortality follows DeMoivre with .After 5 years, it is discovered that the insurance should have been calculated using .At that time, a one-time premium adjustment is made only for the remaining 10 years of theinsurance. Calculate the one-time adjustment.A) Refund of $7,091 B) Refund of $4,412 C) Refund of $2,678D) Additional premium of $4,412 E) Additional premium of $7,091

28. In a double decrement table:(i) 100030

( )

(ii) .10030(1)

(iii) .30030(2)

(iv) | 0.0751 30(1)

(v) 47232( )

Calculate .31(2)

A) 0.11 B) 0.13 C) 0.14 D) 0.15 E) 0.17

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29. A company offers two special discrete life policies to a life age (30).The first policy has a death benefit that starts at 5 and increases 5 per year until age (50).It then decreases at 5 per year until age (70). From age (70) onward it is level at 10.The second policy has a death benefit that starts at 10 and increases 5 per year until age (50).It then decreases at 5 per year until age (70). From age (70) onward it is level at 20. Mortality follows the Illustrative Table. %Compute the difference in the actuarial present value of these two death benefits at issue.A) Less than 0.500 B) At least 0.500, but less than 0.625C) At least 0.625, but less than 0.750 D) At least 0.750, but less than 0.825E) At least 0.825

30. For a last-survivor insurance of 10,000 on independent lives (70) and (80), you are given:(i) The benefit, payable at the end of the year of death, is paid only if the second death occursduring year 5.(ii) Mortality follows the Illustrative Life Table.(iii) 0.03Calculate the actuarial present value of this insurance.A) 235 () 245 C) 255 D) 265 E) 275

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 8 SOLUTIONS

1. Under deMoivre's Law with upper age limit , complete expectation of life at age is° . Prior to the medical breakthrough, with limiting age in deMoivre's law, thecomplete expectation of life at age 30 is .After the breakthrough, the complete expectation of life at age 30 will be 39 (an increase of 4years over 35). DeMoivre's law still applies after the breakthrough. Let us denote by thelimiting age after the breakthrough. Then . Answer: E°

2. The loss-at-issue is introduced in Chapter 6 of "Actuarial Mathematics" and is reviewed inthe notes section of this study material. June 30, 2004 is 2.5 years after the issue date of January1, 2002. There will have been a premium paid on January 1, 2002, 2003 and 2004, and the deathbenefit will be paid on June 30, 2004 (the moment of death). Based on the given time of death,the value of (loss random variable at time of issue) is

PV at issue of benefit paid PV at issue of premium received .

The PV is taken as of January 1, 2002, the issue date. Answer: A

3. The overall rate at which coins are found is 30 per hour (.5 per minute). The rate at whichcoins worth 1 are found is per hour,for coins worth 5 the rate is per hour, and for coins worth 10 the rate is

per hour. Since the overall way in which coins are found is a Poisson process, andsince the denominations are randomly distributed (and therefore independent) it is also true thatfor each denomination of coin amount, coins are found according to a Poisson process. Supposethat is the number of coins worth 1 found in one hour, is the number of coins worth found in one hour, is the number of coins worth 10 found in one hour.

Then and , and and are mutually

independent. The total value of all coins found in the hour is , and the

variance is . Answer: E

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4. This problem makes use of the recursive relationship for accumulating benefit reserves. Sincethe policy is a 20-payment whole life policy, the 20-th year terminal benefit reserve is (for faceamount 1) APV future benefit APV future premium .However, since we are given the 19-th year terminal reserve, the level annual benefit premium,and other factors, we can use the recursive relationship for reserves to find . The recursiverelationship is

. With this becomes

, so that

, and then solving for results

in .

Therefore, and

(since ) . Then, .

Answer: D

5. This problem relies on the following relationship .Since we are told that follows the Illustrative Life Table, it follows that

(from the Illustrative Life Table).

Then, since , we get

(from the Illustrative Life Table)..

Then, using the relationship , we get

. Answer: B

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6. The event of winning or losing a game can be put in the context of a homogeneous Markovchain, where the probability of winning or losing depends on whether the previous game waswon or lost. The probability that the next game is a win if the last game was a win is

Heavy training last game was a win Win next game heavy trainingLight training last game was a win Win next game light training

. Then .In a similar way, the probability of winning the next game given that the last game was a loss is

Heavy training last game was a loss Win next game heavy trainingLight training last game was a loss Win next game light training

. Then .Since the team just lost a game, the possible sequences of wins/losses and their probabilities forthe next two games is Sequence Probability L-W-W L-W-L L-L-W L-L-L Probability of winning at least one of the next two games is .96 . Answer: D

7. We wish to find , so if we can find , we can find

. From , we get , and also

. Also, , so that

.

However, since , we get ,

so that . Then, . Answer: C

8. . Answer: C

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9. Let denote the number of minutes Justin is at the train station prior to 7:50AM.The number of trains arriving in minutes has a Poisson distribution with a mean of (1 train every 8 minutes is the same as .125 per minute, or in minutes).The probability of at least one train arriving while he is at the station is the complement of notrains arriving in minutes, which is . We want this probability to be at least .9(we want a 90% probability that a train arrives while he is waiting). We want

, which is the same as , which is the same as

, which is the same as .He must arrive at least 19 full minutes (18.4 rounded up to the next minute) before 7:50AM.Therefore, he must leave the house at least 29 minutes before 7:50AM, which is 7:21AM.Answer: A

10. For a particular hair, the probability that this hair has been lost by time 2.5 is . Theprobability of being bald by time 2.5 is (lose hair 1) (lose hair 2) (lose hair 3) .Since the hairs are independent of one another, this probability is

(lose hair 1) (lose hair 2) (lose hair 3)(we use the rule for independent events and that says ).From the given information we have

.

Then so that .

.From the hyperbolic assumption, we have .

Then, , and .

Answer: E

11. This graph is shown on page 65 of "Actuarial Mathematics". It is the graph of . Answer: B

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12. If the death benefit is for death between ages and , then actuarialpresent value of the death benefit (or single benefit premium) is

. In this case, that will be

. Answer: A

13. Since we are given we use the relationship

,where is the annual benefit premium and is the death benefit payable at the end of the 11-th year for death in the 11-th year. The insurance is a 20-year decreasing insurance, so that

. Also, the issue age is , so that the equation becomes

, which is .In order to find we need to know . We can find by using the recursive relationship for

the final (20-th) year: .When a policy matures, the maturity amount is the reserve at the end of the policy term.Therefore, , since we are told that this is a 20-year policy with maturity value 1.The death benefit is also in the 20-th year. The recursive relationship becomes

, which becomes , since

for any . We then get , and then . Answer: E

14. With face amount , the expense-loaded premium equation is

, which can be written as

.

Solving for results in .

The policy fee is . In order for this to not exceed 50% of the premium, it must be true

that , which is equivalent to . Answer: B

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15. The number of cars passing through the tunnel forms a non-homogeneous Poisson process.The time period from 11:30 a.m. to 12:30 p.m. corresponds to The expectednumber of cars passing through the tunnel between 11:30 a.m. and 12:30 p.m. is

The probability that exactly 25 cars pass through the tunnel between 11:30 a.m. and 12:30 p.m.

is Answer: D

16. The amount of surplus at the end of a year after any dividends are paid can be formulated asa Markov Chain. Let denote the event that at the end of the -th year, the surplus afterdividends are paid is , where refers to the event that the company has

gone out of business). The one-step transition probability matrix is .

The entries in the matrix are calculated as follows:If then the company is already out of business and will stay out of business in futureyears so that .

since if then after receipt of the premium of 2, the surplus is3 and this surplus will fall to 0 in year is the claim amount is 4 (prob .1)in year . Other entries are found the same way.

The 2-step transition matrix is

The expected dividend at the end of the third year can be conditioned over the surplusamount at the end of the second year (after dividend):

initial surplus is 3

Since the initial surplus is 3, the surplus at the end of the second year is described by theprobabilities in the bottom row of the matrix, so that , etc.We note that (since the company stays out of business once surplus reaches0), and (since the premium brings surplus to 3, so even if there are no claimsthere will be no dividend).

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16 continuedAlso, (if there are no claims then surplus at the end of the yearwill be , from which a dividend of 1 will be payable), and

.Overall expected dividend at the end of the third year is

initial surplus is 3 . Answer: D

17. A review of the general form of a last-survivor annuity in which payments vary according towhich of the individuals is alive is found in the notes section of this study material. The annuityspecified has actuarial present valueand we are told this actuarial present value is 1180.

We are also given and .

We are asked to find .

Since and =10 we have40

. It follows that .

Then, . Answer: A

18. I. . True II. False.

III. Suppose that .

Then False. Answer: A

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19. This problem can be solved with the accumulation relationship for benefit reserves.The initial benefit reserve for a particular is the terminal reserve for the previous year just endedplus the annual benefit premium:year initial benefit reserve benefit premium .

, and the initial benefit reserve for year 2 is. It is always the case that , and for this problem,

and . We find from the equivalence principle premium (equivalence principleis always used for benefit reserves).

.

Then, and .Finally, the initial benefit reserve for year 2 is . Answer: D

20. This problem involves the loss-at-issue random variable. The loss-at-issue ispresent value random variable of benefit

present value random value random variable of premiums .The PVRV of benefit is and the PVRV of premium is , where

is the equivalence principle premium. is found from the equivalence principle equation APVbenefit APV premium . This equation is , so that .

The loss random variable becomes

. Answer: A

21. This problem involves complete expectation of life. We use the following identity forexpectation: . Therefore, .° ° ° ° ° °1 1.5| 1 1

Under the UDD assumption, we use for ° .1 1

° .

Then, . Answer: D°1 1.5|

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22. This problem involves single benefit premium for life insurance with death benefit payableat the end of the year of death, and it also involves the recursive relationship for insurancevaluation. The accumulated value of the 5233 invested for 2 years is at annual rateof return . The single benefit premium for the insurance at age 65 (2 years after age 63) is

, and the contract premium is . We cannot use the values inthe Illustrative Life Table, since they are based on an interest rate of 6%. We are given thesingle benefit premium at age 63, so that at interest of 5%, . We use therelationship . From the Illustrative Life Table, we have

. The single benefit premium at age 63 isTherefore,

. The contract premium for the insurance at age 65 will beIn order for the investment to be enough to purchase the

insurance at age 65, we must have . Answer: A

23. This problem involves the common shock model for dependence of two lives. The originalcalculation assumes independent lives. With constant force of mortality for each life, we have

and since the force of failure for the joint life status is

for independent lives, we have and then . The last survivor insurance value originally calculated is

then .Recognizing that 's and 's forces of mortality include the common shock factor .02, wehave and 's force of mortality is .04 before common shock. Then, the insurance for and for with common shock is (same as before) but for the

joint status, . Under the common shock assumption, the insurance

present value is . The difference between new and new valuesis Answer: B

24. This problem involves a nonhomogeneous Poisson process. and will be independent,since each successive claim is independent of previous claims, and therefore, the number ofclaims greater than 500 does not depend on the number of claims that have been less than 500.

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24 continuedThe probability of any particular claim being less than 500 is .5, since claim amounts areuniformly distributed between 200 and 800. The number of claims by time 3 has a Poissondistribution with expected value .The expected number of claims by time 3 which are less than 500 is

a given claim is less than 500 Answer: C

25. This problem involves the variance of a life annuity present value random variable and thenormal approximation. The expected present value of payment to the winners is

(for 100 annuitants). Since the prize winners areindependent, the combined variance of the present value random variables is the sum of thevariances of the separate present value random variables. For one prize winner, the variance ofthe present value random variable is

.The variance of the total present value paid to the winners isWe assume that is approximately normal to find the initial fund amount needed so that

. We standardize the probability to get .

Since the 95-th percentile of the standard normal distribution is 1.645, we get , and then . Answer: E

26. This problem involves the equivalence principle for finding a benefit premium.For the first 20 years, the benefit premium is , and for the next 15 years it is . The deathbenefit is 1 for the first 20 years and then 5 after that. The equivalence principle equation is

5 .

We use the relationships , and

, and .

From the Illustrated Life Table, we get .Then,

and

.

Then, is the initial annual benefit premium. Answer: B

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27. Under DeMoivre with upper age limit , term insurance valuation is

After 5 years, the actuarial present value of the remaining benefit is .

Based on the original , the actuarial present value of the remaining benefit is

Based on the original , the actuarial present value of the remaining benefit is

The one-time adjustment is a refund of Answer: B

28. Since there are two decrements, we use and to find

.

From and we get .

Then, since , we get , and

.

Since , we get .

Finally, since , we get . Answer: B

29. The schedule of benefits payable for the two insurance policies is:Age is at the end of the year of death.1st PolicyAge 31 32 33 50 51 52 53 69 70 71 72Benefit 5 10 15 100 95 90 95 5 0 10 10

2nd PolicyAge 31 32 33 50 51 52 53 69 70 71 72Benefit 10 15 20 105 100 95 90 10 5 20 20

2nd minus 1st PolicyAge 31 32 33 50 51 52 53 69 70 71 72Benefit 5 5 5 5 5 5 5 5 5 10 10

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29 continuedThe difference in actuarial present value is the present value, at issue age 65, of a 20 year terminsurance of amount 5 plus a 20 year deferred insurance of amount 10.The difference in actuarial present values is

and , so that the difference is

Answer: C

30. The probability that the second death occurs during year 5 is

Note that follows from independence of the future lifetimes of and .

According to the Illustrative Life Table, ,

, so that

. The actuarial present value of the insurance paid at the end of the 5th year is

. Answer: A

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 9

1. For independent lives ( ) and ( ):(i) .05(ii) .10(iii) The force of mortality is constant over each year of age.(iv) The force of interest isCalculate (one year contingent term insurance).

A) Less than .045 B) At least .045 but less than .046 C) At least .046 but less than .047D) At least .047 but less than .048 E) At least .048

2. A person age 40 wins 10,000 in the actuarial lottery. Rather than receiving the money at once,the winner is offered the actuarially equivalent option of receiving an annual payment of (atthe beginning of each year) guaranteed for 10 years and continuing thereafter for life.You are given: (i) 0.04 (ii) 0.3040

(iii) 0.3550

(iv) .0940:10|1

Calculate .A) 538 B) 541 C) 545 D) 548 E) 551

3. Mortality for Audra, age 25, follows De Moivre’s law with 100. If she takes up hot airballooning for the coming year, her assumed mortality will be adjusted so that for the comingyear only, she will have a mortality probability of 0.1 based on uniform distribution of deathwith that year. Calculate the decrease in the 11-year temporary complete life expectancy forAudra if she takes up hot air ballooning.A) 0.55 B) 0.65 C) 0.75 D) 0.85 E) .95

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4. For a multiple decrement table, you are given:(i) Decrement (1) is death, decrement (2) is disability, and decrement (3) is withdrawal.(ii) 0.01060

(1)

(iii) 0.05060(2)

(iv) 0.10060(3)

(v) Withdrawals occur only at the end of the year.(vi) Mortality and disability are uniformly distributed over each year of age in the associatedsingle decrement tables.Calculate .60

(3)

A) 0.088 B) 0.091 C) 0.094 D) 0.097 E) 0.100

5. For a special whole life insurance of 100,000 on ( ), you are given:(i) 0.06(ii) The death benefit is payable at the moment of death.(iii) If death occurs by accident during the first 30 years, the death benefit is doubled.(iv) ( ) 0.008, 0( )

(v) ( ) 0.001, 0 where is the force of decrement due to death by accident.(1) (1)

Calculate the single benefit premium for this insurance.A) 11,765 B) 12,195 C) 12,622 D) 13,044 E) 13,235

6. You are given the following extract from a select-and-ultimate mortality table with a 2-yearselect period: 2

60 80,625 79,954 78,839 6261 79,137 78,402 77,252 6362 77,575 76,770 75,578 64

[ ] [ ] 1 [ ] 2

Assume that deaths are uniformly distributed between integral ages. Calculate ..9 [60] 6

A) 0.0102 B) 0.0103 C) 0.0104 D) 0.0105 E) 0.0106

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7. For a 5-year fully continuous term insurance on ( ):(i) 0.10(ii) All the graphs below are to the same scale.(iii) All the graphs show ) on the vertical axis and on the horizontal axis.(

Which of the following mortality assumptions would produce the highest benefit reserve at theend of year 2?

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8. The expense-loaded premium, , for a fully discrete 3-year endowment insurance of 1000issued to is calculated using the equivalence principle. Expenses are paid at the beginning ofeach year. You are given:(i) (ii)

(iii) (iv) (v) (vi) Expenses Percentage of Premium Per Policy First Year 30% 8 Renewal 10% 4Calculate the expense reserve at the end of the first year.A) B) C) D) E)

Questions 9 and 10 are based on the following information.A 3-year term insurance with death benefit 1000 is issued to (90).You are given for and .

9. is the smallest annual premium such that , where is the issue date loss.In what range is .A) Less than 200 B) At least 200 but less than 250 C) At least 250 but less than 300D) At least 300 but less than 350 E) At least 350

10. Using an annual premium of 300, find the standard deviation of , the first year terminalprospective loss, given that the policy is still in force at the end of the first year.A) Less than 400 B) At least 400 but less than 450 C) At least 450 but less than 500D) At least 500 but less than 550 E) At least 550

11. For a double decrement table with 2000:40( )

(1) (2) (1) (2)

40 .24 .10 .2541 -- -- .20 2

Calculate .42( )

A) 800 B) 820 C) 840 D) 860 E) 880

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12. A non-homogeneous Markov Chain model with annual transitions is used for a ContinuingCare Retirement Home with four states: Independent Living (I) , Temporarily in Assisted LivingCenter (T), Permanently in Assisted Living Center (P), and No Longer in the Home (N). Newlyadmitted residents are always initially assigned to Independent Living. For a newly admittedresident in the home, the one-year transition probabilities

.The one-year transition probability matrices for the 2nd and 3rd year of residence are (rows andcolumns states in the order I , T , P , N)

and

Find the probability that a newly admitted resident is temporarily in the assisted living centersometime in the next 3 years.A) .300 B) .324 C) .348 D) .372 E) .396

13. For a fully discrete whole life insurance of 10,000 on (30):(i) denotes the annual premium and ( ) denotes the loss-at-issue random variable for thisinsurance.(ii) Mortality follows the Illustrative Life Table. (iii) 0.06Calculate the lowest premium, , such that the probability is less than 0.5 that the loss ( ) ispositive.A) 34.6 B) 36.6 C) 36.8 D) 39.0 E) 39.1

14. is the present-value random variable for a special 3-year temporary life annuity-due on( ). You are given:(i) .9 , 0(ii) is the curtate-future-lifetime random variable for ( ).

(iii)1.00, 01.87, 12.72, 2, 3,

Calculate ( ).A) 0.19 B) 0.30 C) 0. 37 D) 0.46 E) 0.55

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15. For a fully continuous whole life insurance of 1 on ( ):(i) is the benefit premium.(ii) is the loss-at-issue random variable with the premium equal to .(iii) is the loss-at-issue random variable with the premium equal to 1.25 .(iv) 5.0 (v) 0.08 (vi) ( ) 0.5625Calculate the sum of the expected value and the standard deviation of .A) 0.59 B) 0.71 C) 0.86 D) 0.89 E) 1.01

16. Workers’ compensation claims are reported according to a Poisson process with mean 100per month. The number of claims reported and the claim amounts are independently distributed.2% of the claims exceed 30,000. Calculate the number of complete months of data that must begathered to have at least a 90% chance of observing at least 3 claims each exceeding 30,000.A) 1 B) 2 C) 3 D) 4 E) 5

17. For students entering a three-year law school, you are given:(i) The following double decrement table:

probability ofFor a student at the beginning of that academic year,

Year Failure All Other ReasonsAcademic Academic Withdrawal for

Academic Year Survival Through

1 .40 .20 ---2 --- .30 ---3 -- --- .60

(ii) Ten times as many students survive year 2 as fail during year 3.(iii) The number of students who fail during year 2 is 40% of the number of students whosurvive year 2.

Calculate the probability that a student entering the school will withdraw for reasons other thanacademic failure before graduation.A) Less than 0.35 B) At least 0.35, but less than 0.40C) At least 0.40, but less than 0.45 D) At least 0.45, but less than 0.50E) At least 0.50

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18. Given:(i) Superscripts and identify two forces of mortality and the curtate expectations of lifecalculated from them.

(ii) ( ) ( ) .1 (1 ) 0 1( ) 125

25

25

(iii) 10.025

Calculate .25

A) 9.2 B) 9.3 C) 9.4 D) 9.5 E) 9.6

19. A fund is established to pay annuities to 100 independent lives age . Each annuitant willreceive 10,000 per year continuously until death. You are given:(i) .06(ii) 0.40(iii) 0.252

Calculate the amount (in millions) needed in the fund so that the probability, using the normalapproximation, is 0.90 that the fund will be sufficient to provide the payments.A) 9.74 B) 9.96 C) 10.30 D) 10.64 E) 11.10

20. A decreasing term life insurance on (80) pays (20 ) at the end of the year of death if (80)dies in year 1, for 0, 1, 2, , 19.You are given:(i) 0.06(ii) For a certain mortality table with 0.2, the single benefit premium for this insurance is80

13.(iii) For this same mortality table, except that 0.1, the single benefit premium for this80

insurance is .Calculate .A) 11.1 B) 11.4 C) 11.7 D) 12.0 E) 12.3

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21. For independent lives (50) and (60):( ) , 0 1001

100Calculate .50:60

A) 30 B) 31 C) 32 D) 33 E) 34

22. For an industry-wide study of patients admitted to hospitals for treatment of cardiovascularillness in 1998, you are given:

(i) Duration in Days Remaining HospitalizedNumber of Patients

0 4,386,000 5 1,461,55410 486,73915 161,80120 53,48825 17,38430 5,34935 1,33740 0

(ii) Discharges from the hospital are uniformly distributed between the durations shown in thetable.Calculate the expected time remaining hospitalized, in days, for a patient who has beenhospitalized for 21 days.A) 4.4 B) 4.9 C) 53 D) 5.8 E) 6.3

23. For a fully discrete three-year endowment insurance of 10,000 on (50), you are given(i) (ii) (iii) (iv) (v)

(vi) is the prospective loss random variable at issue, based on the benefit premium.Calculate the variance of (nearest 50,000) .A) 250,000 B) 300,000 C) 350,000 D) 400,000 E) 450,000

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24. A customer service department receives 0 or 1 complaint each day, depending on the numberof complaints on the previous 2 days, as follows: (i) If there were no complaints in the past 2 days, then there will be no complaints today with probability 0.75 . (ii) If there were no complaints 2 days ago but 1 complaint yesterday, then there will be no complaints today with probability 0.40 . (iii) If there was 1 complaint 2 days ago but no complaints yesterday, then there will be no complaints today with probability 0.55 . (iv) If there was 1 complaint on each of the past 2 days, then there will be no complaints today with probability 0.10 .Suppose there were no complaints 2 days ago and 1 complaint yesterday. Calculate theprobability that there will be at least 1 complaint over the next 2 days.A) 0.4375 B) 0.5700 C) 0.6975 D) 0.7800 E) 0.8400

25. For a special 30-year deferred annual whole life annuity-due of 1 on (35):(i) If death occurs during the deferral period, the single benefit premium is refunded withoutinterest at the end of the year of death.(ii) 9.90ä65

(iii) 0.2135:30|

(iv) 0.0735:30|1

Calculate the single benefit premium for this special deferred annuity.A) 1.3 B) 1.4 C) 1.5 D) 1.6 E) 1.7

26. Given: (i) ( ) , 02

(ii) .500.4 0

Calculate .A) .20 B) .09 C) .00 D) .09 E) .20

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27. Loans transition through five states (Current, 30, 60, 90, and Foreclosed) based on thefollowing matrix: Current 30 60 90 ForeclosedCurrent 0.80 0.20 0.00 0.00 0.0030 0.50 0.00 0.50 0.00 0.0060 0.25 0.00 0.00 0.75 0.0090 0.10 0.00 0.00 0.00 0.90Foreclosed 0.00 0.00 0.00 0.00 1.00

The transitions happen monthly. Out of 100,000 Current loans how many are expected to beForeclosed in six months?A) Less than 16,500 B) At least 16,500, but less than 16,750C) At least 16,750, but less than 17,000 D) At least 17,000, but less than 17,250E) At least 17,250

28. For a fully discrete insurance of 1000 on , you are given:(i) (ii) (iii) (iv) (v) is the fraction of gross premium paid at time 4 for expenses(vi) is the amount of per policy expenses paid at time 4(vii) is the probability of decrement by death(viii) 0.26 is the probability of decrement by withdrawal.Calculate .A) 0.050 B) 0.055 C) 0.060 D) 0.065 E) 0.070

29. A 20-payment, 30-year fully discrete endowment insurance policy issued to has anendowment amount of $1000 for survivors at age . If death occurs before age , the deathbenefit is equal to the benefit reserve at the end of the year of death. Mortality followsDeMoivre's law with and the annual effective rate of interest is 10%. Find .Premiums are level and are based on the equivalence premium.A) 6.13 B) 6.33 C) 6.53 D) 6.73 E) 6.93

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30. For a fully discrete 3-year endowment insurance of 1000 on ( ):(i) 0.201

(ii) 0.06(iii) 1000 373.63:3|

Calculate 1000 ( ).2 1:3| :3|

A) 320 B) 325 C) 330 D) 335 E) 340

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 9 SOLUTIONS

1.

. Answer: B

2. .

With we have .

Also, from , we get .

Then, .

Finally, . Answer: A

3. If she does not take up hot-air ballooning, Audra's 11-year temporary complete lifeexpectancy will be , which can also be°

written as

° ° °

.If Audra does take up hot-air ballooning for one year (year of age 25), her survival probabilityfor year of age 25 will be for (UDD for the year).Her revised 11-year temporary complete life expectancy will be

° ° °

.Her expectation would decrease . Answer: D

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4. Since decrement 3 (withdrawal) is discrete, we cannot find using the relationship

since does not exist for a discrete decrement.

We find .

(this relationship is valid whether decrements are continuous or discrete).1 1 1 .

The assumption of UDD in associated single decrement tables for decrements 1 and 2 means that1 and for . Also, since decrement 3 occurs

only at the end of the year, for . Then,1 1

.

In a similar way

.

Then, .

An alternative solution that follows from "general reasoning" is as follows. Since decrement 3occurs only at the end of the year, the probability of surviving to just before the end of the year is

. This proportion of people is subject to decrement 3 only at

the end of the year, so that . Answer: C

5. The SBP for the basic benefit of 100,000 is

.The SBP for the additional accidental benefit in the first 30 years is

. Total SBP is . Answer: D

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6. .

From UDD, .

and .

(since the table has a 2-year select period, ).

Then, , and . Answer: B

7. We use the following principles:if for then ,

and for .

Constant force of mortality results in whole life and term reserves of 0 at all durations.Therefore, the reserve for E is 0 at the end of year 2 so that .

Force A Force B, for the first 2 years but Force A Force B after that. Therefore, . Then using the retrospective reserve at time 2, , we

see that and are the same for A and B for the first 2 years (they depend only on

behavior up to time 2) so that .Force D Force B for the first 2 years but Force D Force B after that. Therefore .Then using the prospective reserve at time 2, we see that

and are the same for D and B after time 2, so that D

For Force C, we see that Force C Force B prior to time 2 and Force C Force B after time 2.If we imagine a Force C which is the same as Force C below time 2 and which is the same asForce B above time 2, then by the reasoning above for D, we have reserve at time 2 for C is lessthan reserve for B. But by the same reasoning as for A, we see that reserve for Force C is belowthat for Force C , and therefore .Answer: B

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8. The level expense loading is .

We can use the accumulation relationship for expense reserve , where is the total expense at the

start of year and is the settlement expense at the end of year . In this problemthere is no settlement expense. Also, from the Equivalence Principle, we have . Theexpenses at the start of the first year are

. Answer: C

9. The issue date loss is

prob.prob.prob.prob.

If , then no matter when death occurs.If , then if death is in the first year, which has probability .1, and

otherwise. So if then .If , then if death is in the first year

(since ), and if death occurs after the first year. In this case .Note that is equivalent to

If , then if death is in the first or second year,

which has probability .If , then if death occurs in the first three years, which has

probability .Therefore, is the smallest annual premium for which .Answer: C

10.prob.prob.prob.

,

.Standard deviation is . Answer: B

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11. .Then .Finally, .Answer: A

12. The sequences of transitions over the next three years that have the newly admitted residentin state T at some point are Sequence Probability I-I-I-T I-I-T I-T Total probability is . Answer: B

13. The premium is found in the following way. We first find the smallest for which . We then solve for from . The premium found this way will

satisfy the property that . Inorder to have , we must have , so that

. We are trying to find the smallest which satisfies this. From theIllustrative Life Table (handed out with the exam) we have

Therefore, the smallest for which is . We find from . Answer: C

14. is a 3-point random variable:

.

. . Answer: B

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15. .

.We are told that is the benefit premium, so that .

.We are given that , so that .

Then, , and .

Then . Answer B

16. Proposition 5.2 on page 296 of the Ross book (8th Ed.) shows the following result. Supposethat (i) is a Poisson process with rate , and(ii) the events can be classified into 2 distinct types, and(iii) is the probability that an event is of type 1 when an event occurs, and(iv) is the probability that an event is of type 1 when an event occurs.Then if is the number of events of type 1 that have occurred up to time , then

is a Poisson process with rate .In this question, is the number of claims in months, with a mean rate of permonth. Type 1 claims are those of amount over 30,000 . The probability of an event being type1 is . The number of claims exceeding 30,000 in months is , a Poisson processwith a mean rate of . In months, the number of claims exceeding 30,000has a Poisson distribution with mean . The probability of observing at least 3 claimsexceeding 30,000 in months is23. continued

.For month, ,for months, ,for 3 months, .In 3 complete months, the probability of observing at least 3 claims exceeding 30,000 is greaterthan 90%, but not for 2 months or 1 month. Answer: C

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17. (ii) says that , from which it follows that . We are given that , and therefore, .

(iii) says that , from which it follows that . Then, since ,we get ,and .The probability that a student entering the school will withdraw for reasons other than academicfailure before graduation is

. Answer: B

18. , .Since for , we have .

Therefore,

.Then, . Answer: D

19. Let denote the present value random variable for one annuity. Then , and

.Let be the sum of the pvrv's for all 100 annuities.Then , and since the 's are mutually independent,

.

The amount needed so that there is a .9 probability that will be sufficient to provide forpayments must satisfy . Using the normal approximation for , this becomes

is the 90-th percentile of the standard normal distribution, so that

, or 10.641 million . Answer: D

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20. The APV at age 80 is .

We are given that , and .

Therefore, .

If is changed to .1, then 12.225 remains unchanged, and the revised value of

is . Answer: E

21. . The force of mortality is for DeMoivre's Law with° ° ° ° . Therefore, , so that and .° ° °

Under DeMoivre's Law, .°

. The reason that the upper limit of the integral is 40 isthat for , and therefore, .Then, . Answer: A°

22. The expected lifetime for someone who is at time is the same as the complete expectation° ° . For , , since there are no survivors

(patients remaining hospitalized) past 40 days. Since we are assuming uniform distribution ofdischarges between successive durations in the table, we have

, and

( 8) ( )( ) ( ) .

° . Answer: A

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23.

.

.Answer: E

24. This a four state Markov chain, in which each state is the complaint experience on twoconsecutive days. We will label the states asNN - no complaints on the first day and no complaints on the second day,NC - no complaints on the first day and 1 complaint on the second day,CN and CC are defined in a similar way.We are told to assume that there were no complaints 2 days ago and 1 complaint yesterday. Thatis state NC. We wish to find the probability of at least one complaint in the next two days. Thatis the complement of the probability that there are no complaints in the next two days. Nocomplaints in the next 2 days is state NN. A one-step transition is from the two-day period

to the two day period . If we label two days ago as day , then we wish tofind the two-step transition from state NC on days to state NN on days .This can only happen if day has no complaints, so the one-step transition from to

must be from NC to CN. This probability is .4 (this is transition (ii) in thestatement of the problem).Then the one-step transition from to must be from CN to NN, whichhas probability .55 (statement (iii)). The combined probability from NC to NN in two steps is

. This is the probability of no complaints in the next two days. The probabilityof at least one complaint in the next two days is .Answer: D

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25. .

We are given .

. We are given .We also have .

Then, . Answer: C

26. . Answer: E

27. We first find the probability , the probability of transferring from Current toforeclosed in 6 months. Since Foreclosed is an absorbing state, includes those loansthat transferred from Current to Foreclosed in less than 6 months. The expected number of loansthat are foreclosed at the end of six months is .

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27 continued

. The number of Foreclosed loans is Note that it is not necessary to compute the full matrices or . Answer: D

28. We use the recursive relationship for asset shares,

for .

.

Solving for results in . Answer: A

29. Since benefit reserve, this is a savings account and mortality is irrelevant. The premium satisfies , so that . The reserve at then of the first year is the first

year premium accumulated with interest, so that . Answer: D

30. Using the recursive relationship for benefit reserve , and the starting value ,

we have and .

. Answer: B

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 10

1. A select-and-ultimate life table has a select period of 10 years. The force of mortality is during the select period, and the force of mortality is for someone at

age who is past the select period.Find an expression for the complete expectation for .°

A) B) C) D) E)

2. An insurer creates a discrete single premium insurance policy with issue age (35) for whichthe death benefit will be 200,000 if death occurs before age 65 and 100,000 if death occurs afterage 65. Mortality follows the Exam M Illustrative Life Table at a 6% rate of interest. The insurerissues this policy to 1000 independent individuals and charges a premium per policy that resultsin a 90% probability that premiums will be enough to cover total cost of benefits based on anormal approximation to the distribution of the present value of total cost of benefits.Find the premium per policy.A) Less than 24,000 B) At least 24,000 but less than 24,250C) At least 24,250 but less than 24,500 D) At least 24,500 but less than 24,750E) At least 24,750

Questions 3 and 4 refer to the following information.You are given the following for every integer age :(i) and (ii)

3. Calculate .A) .400 B) .408 C) .416 D) .424 E) .432

4. Assuming UDD over each year of age, calculate .

A) 6.0 B) 6.1 C) 6.2 D) 6.3 E) 6.4

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5. Smith, who has just turned 62, takes a 3-year mortgage loan of $100,000 on his home. Theloan has an annual effective rate of interest of 6% and has scheduled level annual payments atthe end of year for 3 years (the day before Smith's 63rd, 64th and 65th birthdays). The lenderoffers Smith mortgage insurance. The insurance policy will pay the mortgage balance owing atthe end of the year of death if Smith dies before the final (3rd) mortgage payment is made. Theinsurance policy is based on an annual effective interest rate of 4% and the following life table.

The insurance policy has a level annual premium payable at the start of the year. The premium iscalculated so that the expected value of the issue date loss to the insurer is 1000. Find thelevel annual premium for the insurance (nearest 100).A) 6,500 B) 6,600 C) 6,700 D) 6,800 E) 6,900

Questions 6 and 7 are based on the following information.A 4-year fully discrete term insurance with face amount 1000 is issued at age (premiums arescheduled for the lifetime of the policy). The effective annual interest rate is 25%, and themortality probabilities are .The equivalence principle annual premium is .

6. Which of the following is the retrospective form of the 2nd year terminal reserve?

A) B)

C) D)

E)

7. Find the 2nd year terminal benefit reserve (nearest $1).A) 181 B) 184 C) 187 D) 190 E) 193

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8. Smith will be retiring at age 65 and is eligible to receive a pension annuity starting at age 65.The annuity will pay $50,000 at the beginning of each year as long as Smith survives (firstpayment at age 65). Smith is married and his wife is 5 years younger than he is. Smithconsiders some alternative arrangements for his pension benefit.You are given that , .The following assumes that Smith and his wife are both alive when Smith turns 65 when hisretirement benefit is to start. Equivalent benefits are based on actuarial equivalence.Smith considers a retirement annuity-due with payment of per year while both he and his wifeare alive, and with the payment reducing to after the first death, and continuing at that leveluntil the second death. Find .A) 32,000 B) 32,250 C) 32,500 D) 37,500 E) 40,000

9. You are given the following table of probabilities in a 3-decrement model.

Using an annual effective interest rate of 6%, find the actuarial present value at age 30 for a 3-year discrete term benefit that pays 1000 at the end of the year of decrement if due todecrement 1.A) 48.0 B) 48.1 C) 48.2 D) 48.3 E) 48.4

10. An year fully discrete endowment insurance for $100,000 issued to (x) has a benefitpremium of $2061. Expenses on the policy per year are $25 plus 20% of the expense-loadedpremium G , along with a settlement expense of $A . Given that ( L ) ( L) = 1.0005 , whate

is G (nearest 100)?A) 2,500 B) 2,600 C) 2,700 D) 2,800 E) 2,900

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11. A Poisson process has a rate of per unit time.Events are classified as Type 1 and Type 2.When an event occurs, there is a probability that it is a Type 1 event and a probability that it is aType 2 event. Event types are independent of one another.

is the process of Type 1 events.Find .A) 1 B) 2 C) 3 D) 4 E) 5

Question 12 and 13 are based on the following information.An auto insurance company has a rating system which rates drivers as Low, Medium, or High risk. Arating is assigned to the policyholder at the time an auto insurance policy is issued. The rating is updatedat the end of each year. The transition probability matrix of risk rating in a policyholder's first year is

.

For all years after the first year, the transition probability matrix of risk rating is

.

12. Suppose that a new policyholder is rated as low risk. Find the probability that thispolicyholder will never be rated as high risk by the end of (and including) the third policy year.A) .60 B) .62 C) .64 D) .66 E) .68

13. Based on a policyholder's rating at the start of a year, the insurance company estimates annualclaims for that year as follows (assumed to be paid at the end of each year)Low risk: $100 ; Medium Risk: $200 ; High Risk: $1000 .For a new policyholder rated Low risk, find the total expected claims for the first three years.A) 624 B) 634 C) 644 D) 654 E) 664

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14. A late-night television infomercial claims that the "Extend Your Life"rejuvenation cocktail has the following benefits for an 80-year old male:(i) doubles the probability of surviving 10 years, and(ii) increases expected future lifetime by 7.5 years.A late-night actuary viewing this infomercial models survival from birth based on DeMoivresLaw with different for before and after the rejuvenation cocktail is taken. Find the value of for survival after the rejuvenation cocktail is taken.A) 102 B) 104 C) 106 D) 108 E) 110

15. Actuary A uses the following mortality model is used for life insurance valuation:

.

Actuary B uses the following mortality model is used for life insurance valuation:

.

The force of interest is .06. Each actuary calculates the single benefit premium for a continuouswhole life insurance of 1 at age 50. Find .A) .06 B) .07 C) .08 D) .09 E) .10

16. An insurer issues a continuous 10-year certain and life annuity of 1 per year 100independent lives all at age B. The insurer charges a single contract premium that is 1% largerthan the single benefit premium. The mortality model used has constant force of mortality of .01at all ages, and force of interest .05. Using the normal approximation, find the probability thatthe insurer's aggregate issue date loss for all 100 policies is greater than 0.A) .22 B) .26 C) .30 D) .34 E) .38

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17. The following information is given for a fully discrete whole life insurance issued to withface amount , and level benefit premium payable for life:

Find the face amount for this policy.A) 150 B) 200 C) 250 D) 300 E) 350

18. Two 60-year old individuals have independent future lifetimes, but both have survival basedon DeMoivre's Law with . Annual effective interest is 6%. Find the actuarial presentvalue of a continuous whole life annuity that pays at a rate of 3 per year until the first death, andafter the first death continues at a rate of 1 per year until the second death.A) 28.9 B) 29.2 C) 29.5 D) 29.8 E) 30.1

19. A 3-decrement model for mortality for individuals with a particular genetic makeupidentifies three causes of death. The three causes are Disease A, Disease B and all other causes.According to the model, at age the following absolute rates of decrement are known:

(Disease A) , (Disease B) , (Disease C) .As a result of ongoing research, Disease B is reclassified as two separate diseases, Disease and Disease . The multiple decrement probabilities for original Disease B and the reclassifiedDiseases and are consistent in that B in that in the original 3 decrement model isequal to the sum of and in the reclassification model. Furthermore, it is determinedthat Disease is twice as prevalent as Disease for individuals with the particular geneticmakeup being studied. Find the value of for the reclassification model if UDD in themultiple decrement model is assumed after reclassification.A) .17 B) .19 C) .21 D) .23 E) .25

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20. A fully discrete whole life insurance is issued at age 35 with a face amount of $1,000,000.Premiums are payable for life, and calculations and mortality are based on the illustrative ExamM table, at an annual effective interest rate of 6%. The policy has the following expenses:- 25% of premium in the 1st year, 10% of premium in subsequent years,- per policy expense of $1000 in the first year and $200 per year in subsequent years,- face amount expense of % of face amount in the first year and % of face amount insubsequent years

Calculate the second year expense augmented reserve, .A) Less than 7,400 B) At least 7,400 but less than 7,500C) At least 7,500 but less than 7,600 D) At least 7,600 but less than 7,700E) At least 7,700

21. A 2-year fully discrete endowment insurance policy with level premiums issued at age hasa death benefit of 2000 and an endowment benefit of 1000. The policy expenses are as follows: 1st Year 2nd YearPercent of Premium 50% 20%Per Policy 100 20The policy is based on a two-decrement model, with decrement 1 being death and decrement 2being policy cancellation. Cancellation can only occur at the end of the first year. Interest is at arate of , and mortality probabilities are , and the policycancellation probability is .

With a premium of and a first year cash value of , find theexpected second year-end asset share, .A) 800 B) 900 C) 1000 D) 1100 E) 1200

22. A two-state homogeneous Markov chain is being used to model the transitions between dayswith rain (R) and without rain (N). You are given .If it is raining today, find the expected number of non-rainy days until the next rainy day.A) 1.0 B) 1.5 C) 2.0 D) 2.5 E) 3.0

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23. A 3-year college names some students to the dean's honor list at the end of each year.The student might or might not be named to the list at the end of the 2nd and the end of the 3rdyear. Some students also drop out each year. The college models student behavior according to anon-homogeneous Markov Chain. At the start of each year, each student is classified as being inone of three states:1 - not on the hon our list at the end of the previous year,2 - on the hon our list at the end of the previous year, and3 - dropped out before the end of the previous year.Every new student starts out in state 1.It is assumed that once a student drops out, the student does not return.

For the 3 years that the student is in college, we have the following one-step transition matrices:

1st year , 2nd year ,

3rd year .

The Dean's Honour List award pays $1000 to each student that is named to the list at the endof each year. Find the total expected amount of scholarship that will be awarded to a new studentover his 3 year college career from the Dean's Hon our List Award.A) 680 B) 681.25 C) 682.50 D) 683.75 E) 685.00

24. Customers arrive at a service facility at a rate of 20 per hour. The customer arrival process ismodeled as a Poisson process. On average, 25% of the arriving customers require specialattention, and the other 75% do not require special attention. The attention requirements ofsuccessive customers are independent of one another. The 4th customer requiring specialattention arrives at time 1. What is the expected number of customers not requiring specialattention that have arrived by that time.A) 11 B) 12 C) 13 D) 14 E) 15

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25. For a fully discrete whole life insurance on with a benefit of 1, you are given:• 5%• for Calculate the benefit reserve at time 2.A) Less than .32 B) At least .32, but less than .36 C) At least .36, but less than .40D) At least .40, but less than .44 E) At least .44

26. An insurance company has agreed to make payments to a worker age who was injured atwork.(i) The payments are 150,000 per year, paid annually, starting immediately and continuing forthe remainder of the worker’s life.(ii) After the first 500,000 is paid by the insurance company, the remainder will be paid by areinsurance company.

(iii) (.7) , 0 5.50, 5.5

(iv) 0.05Calculate the actuarial present value of the payments to be made by the reinsurer.A) Less than 50,000 B) At least 50,000, but less than 100,000C) At least 100,000, but less than 150,000 D) At least 150,000, but less than 200,000E) At least 200,000

27. The pricing actuary at Company XYZ sets the premium for a fully continuous whole lifeinsurance of 1000 on (80) using the equivalence principle and the following assumptions:(i) The force of mortality is 0.15.(ii) .06The pricing actuary’s supervisor believes that the Illustrative Life Table with deaths uniformlydistributed over each year of age is a better mortality assumption. Calculate the insurer’sexpected loss at issue if the premium is not changed and the supervisor is right.A) 124 B) 26 C) 0 D) 37 E) 220

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28. For a last-survivor whole life insurance of 1 on ( ) and ( ):(i) The death benefit is payable at the moment of the second death.(ii) The independent random variables *( ), *( ), and are the components of a commonshock model.(iii) *( ) has an exponential distribution with ( ) 0.03, 0.*( )

(iv) *( ) has an exponential distribution with ( ) 0.05, 0.*

(v) , the common shock random variable, has an exponential distributionwith ( ) 0.02, 0.(vi) 0.06Calculate the actuarial present value of this insurance.A) 0.216 B) 0.271 C) 0.326 D) 0.368 E) 0.423

29. A continuous annuity just issued to (35) will pay 1 per year until the earlier of(i) 10 years past the death of (35) and (ii) 30 years from now .What is the actuarial present value the annuity?

A) a B) a + a C) a + v a_ _ _ _ _

: | | : | | : |

D) a + v a E) a + v a_ _ _ _

| : | : | : |

30. Company ABC issued a fully discrete three-year term insurance of 1000 on Pat whose statedage at issue was 30. You are given:(i)

30 .0131 .0232 .0333 .04

(ii) 0.04(iii) Premiums are determined using the equivalence principle.During year 3, Company ABC discovers that Pat was really age 31 when the insurance wasissued. Using the equivalence principle, Company ABC adjusts the death benefit to the leveldeath benefit it should have been at issue, given the premium charged.Calculate the adjusted death benefit.A) 646 B) 664 C) 712 D) 750 E) 963

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 10 SOLUTIONS

1. We first note that for .°

° ° °for and for . for .

For , we can write .° °

It was noted above that , and .

And °

Then for , .°

Answer: D

2. We will denote by the present value random variable for one policy. Then

(we can also express this as , but the first expression is

usually more convenient for calculation).

From the Illustrative Table at 6% we have and .

Then .

In order to solve the problem, we need to know .Since is found by squaring the benefit and the present value factor, we have

, which can also be written as

(because ).

From the Illustrative Table, , and .

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2. continuedThen,

.

Then, .

The total present value of benefit costs is for the 1000 policies,so and

.

We first find so that . If is the total premium for all 1000 policies, thenthere will be a 90% probability that premium is enough to cover benefit costs.

We can write the probability in the form .

According to the normal approximation applied to , has a standard normal

distribution, and therefore is the 90th percentile of the standard normal, which is

1.282 (from interpolation in the normal table from the Exam M tables).

Therefore from which we get .

This is the premium for 1000 policies, so the premium per policy is . Answer: A

3. Using the relationship we get , from which itfollows that . Then and .

Then using the relationship we get from which it follows that . This is valid for any integer .

Then, ,and . Answer: E

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4. We use the relationship : .

From UDD we have for all ,and then for all .Therefore

and then .

Note that since and , we have , so that . Answer: C

5. The annual loan payment is

The outstanding balances just before the payment is made at the end of years 1, 2 and 3 are

and .

We make these calculations because these are the death benefits at the ends of years 1, 2 and 3.For instance, if death occurs during the first year, before the loan payments is made at the end ofthe year, then the amount owing at the end of the year (the death benefit) is the present value ofthe payment due at the end of the year along with the two future payments.The APV of the death benefits is

,where and .

The APV of the death benefits is .With annual premium of amount , the APV of premiums is

.

The expected issue date loss for the insurance policy is benefit premium .

In order for this to be , we must have . Answer: B

6. Answer: B

7. .

, .

. Answer: C

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8. .

Answer: E

9. APV

. Answer: C

10. G a = (100,000 + A) A + (.2 G + 25) a G = (100,000 + A) P + .2 G + 25¨ ¨x:n| x:n| x:n|_ _ _

.8 G 25 = (100,000 + A) P (where P is the net annual premium) .L = (100,000 + A) Z + (.2 G + 25 G) (where Z is the present value randome

1 Zd

variable for an n year endowment insurance) = (100,000 + A) Z P (100,000 + A) 1 Zd

= (100,000 + A) 1 + Z (100,00 + A)P Pd d .

L = 100,000 Z 100,000 P = 100,000 1 + Z 100,0001 Z Pd d

Pd

= = 1 + = 1.0005 A = 50( L )( L) 100,000

(100,000 + A) 1 +

100,000 1 +Ae

Pd

Pd

G = + 31.25 = + 31.25 = 2609 . Answer: B(100,050) P.8 .8

(1.0005) (2061)x:n|_

11. since and are independent, and

is a Poisson process with rate . Answer: C

12. We convert state H to an absorbing state. The transition matrices become

for the first year,

and for the second and subsequent years ( ).

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12 continued

Then

Then the probability in question is(the probability of being rated L or M at the end of 3 years). Answer: E

13. For a policyholder rated Low risk at time 0, expected first year claim is 100.The rating probabilities at the start of the second year for this policyholder areL: M: H: .The expected claim in the 2nd year is .The two-step transition probabilities from time 0 are in the matrix

.

The rating probabilities at the start of the third year for this policyholder areL: M: H: .The expected claim in the 3rd year is .Three-year total expected claim is . Answer: B

14. Suppose that the DeMoivre age limit before the cocktail is taken is , and after the cocktail is takenit is .Before rejuvenation cocktail: , .°

After rejuvenation cocktail: , .°

We are given that and .

From the final equation, we get , and then the previous equation can be written as.

This equation can be written in the form ,which becomes the quadratic equation .There are two roots, and . We discard as infeasible (since the modelconsiders someone who is still alive at age 80). Then . Answer: E

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15. We use the rule .

With constant force of mortality and constant force of interest ,

, and .

With force of mortality , mortality follows DeMoivre's Law with ,and .

For Actuary A, we have

Therefore, .

For Actuary B, we have

Therefore, .

The absolute difference is .0634 . Answer: A

16. The single benefit premium for the annuity is

.The contract premium charged for each annuity is .Since the present value random variable of the 10-year certain and life annuity differs from a 10-year deferred annuity only by the constant , it follows that the variance of the present value

random variable for the 10-year certain and life annuity is the same as the variance of a 10-yeardeferred annuity. The variance of the present value random variable of an -year deferredcontinuous annuity of 1 per year is .For constant force of mortality , we have .

For this 10-year deferred annuity, this is

The issue date loss random variable for a single annuity is , with an expectedvalue of , and variance of .The aggregate issue date loss for 100 independent annuities is ,with mean and with variance

.Using the normal approximation, the probability that the aggregate issue date loss is greater than0 is .

Answer: D

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17. This is an exercise in combining the recursive relationship for benefit reserves with therecursive relationship for the variance of the prospective loss random variable.The recursive relationship for benefit reserves is

,and the recursive relationship for loss variable variance is

From the first equation, we get

so that

Substituting this into the second equation, we get , which becomes

After multiplying through by we get the quadratic equation .

The equation has two roots, one negative, which we can ignore. The positive root is .

From the relationship , we get . Answer: B

18. We wish to find

For DeMoivre's Law with upper age limit , we have ,so that , and .

, and under DeMoivre's Law, with independent lives both of age , ,

since for independent lives, .This integral for can be written as , which is .Therefore, in this case, ,and .

The APV of the annuity is . Answer: B

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19. For the 3 decrement model, we have .

For the reclassification model, we have .

The survival probability is not affected by the reclassification, so that .

It follows that .

We are given that .Under UDD in the multiple table assumption, and .

It follows that , so that .

Then, from , we get , and ,and . Finally, . Answer: C

20. The equivalence principle equation for the expense-loaded premium is..

.. .... is the equivalence principle expense

loaded premium... ..

is the 2nd year terminal expense augmented reserve. Answer: D

21. . .

Answer: C

22. Let be the expected number of non-rainy days until the next rainy day if it is raining today,and let be the expected number of future non-rainy days until the next rainy day if it is not rainingtoday. Conditioning over the weather tomorrow, we have

.We get an equation for in the same way

. From this equation we get .Then from the earlier equation we get . Answer: C

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23. There is a payment of 1000 at time 1 with probability .There is a payment of 1000 at time 2 with probability .There is a payment of 1000 at time 3 with probability .

. .

Alternatively, is the -entry in .

is the -entry in .

.This can also be found by identifying paths from state 1 at time 0 to state 2 at time 3.The expected amount of scholarship received by a student is

. Answer: D

24. Since the customer types are independent, the number of customers not requiring specialattention forms a Poisson process with rate per hour. This is the expected numberof customers not requiring special attention that have arrived by that time. Answer: E

25. For the fully discrete whole life insurance, the reserve is ... (the summation ends because for ).

(note that ), .

Then, ...

.. .

. Answer: E

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26. The reinsurer's first potential payment is 100,000 and will be made 3 years from now (theprimary insurer would pay 150,000 now and for the next 2 years, and 50,000 in three years toreach the limit of 500,000). Starting 4 years from now, the reinsurer would 150,000 per year, aslong as the worker survives. The actuarial present value of the reinsurer's payments is

Since for , and for , this becomes

Answer: B

27. With force of mortality constant at .15 for all ages, we have

.

The expected loss at issue is .If the Illustrative Table with UDD is used, then , and

.

The expected loss at issue using the original fully continuous premium of .15 is . For face amount 1000, the expected loss is . Answer: A

28. . In the common shock model,

's force of mortality is

's force of mortality is and the joint status force of mortality is

.

Then,

and

.

The last survivor insurance has APV . Answer: D

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29. Using the current payment method APV = v dt + v p dt0 1010 30t t

t 10 35

(let s = t 10 in the second integral) = a + v p ds = a + v a ._ _ _

10| 10| 35:20|020 s+10 10

s 35

Answer: C.

30. The original premium is

Using this premium from age 31, the level benefit from issue is , where

.

Answer: B

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 11

1. For a 10-year deferred whole life annuity of 1 on (35) payable continuously:(i) Mortality follows De Moivre’s law with 85.(ii) 0(iii) Level benefit premiums are payable continuously for 10 years.Calculate the benefit reserve at the end of five years.A) 9.38 B) 9.67 C) 10.00 D) 10.36 E) 10.54

2. An investment fund is established to provide benefits on 400 independent lives age .(i) On January 1, 2001, each life is issued a 10-year deferred whole life insurance of 1000,payable at the moment of death.(ii) Each life is subject to a constant force of mortality of 0.05.(iii) The force of interest is 0.07.Calculate the amount needed in the investment fund on January 1, 2001, so that the probability,as determined by the normal approximation, is 0.95 that the fund will be sufficient to providethese benefits.A) 55,300 B) 56,400 C) 58,500 D) 59,300 E) 60,100

3. A risky investment with a constant rate of default will pay:(i) principal and accumulated interest at 16% compounded annually at the end of 20 years if itdoes not default; and(ii) zero if it defaults.A risk-free investment will pay principal and accumulated interest at 10% compounded annuallyat the end of 20 years. The principal amounts of the two investments are equal. The actuarialpresent values of the two investments are equal at time zero. Calculate the median time untildefault or maturity of the risky investment.A) 9 B) 10 () 11 D) 12 E) 13

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4. For a fully discrete whole life insurance with non-level benefits on (70):(i) The level benefit premium for this insurance is equal to .50

(ii) 0.01, 0, 1, , 19 (iii) 0.0136870 50 60

(iv) , 0, 1, , 19 (v) 0.1663750 11 50

Calculate , the death benefit in year 11.11

A) 0.482 B) 0.624 C) 0.636 D) 0.648 E) 0.834

5. You are given:(i) is the benefit reserve at the end of year k for type A insurance, which is a fully discrete10-payment whole life insurance of 1000 on .(ii) is the benefit reserve at the end of year k for type B insurance, which is a fully discretewhole life insurance of 1000 on .(iii)(iv) The annual benefit premium for type B is 8.36.(v)(vi)CalculateA) 91 B) 93 C) 95 D) 97 E) 99

6. For a whole life annuity-due of 1 on ( ), payable annually:(i) 0.01(ii) 0.051

(iii) 0.05(iv) 6.951ä 1

Calculate the change in the actuarial present value of this annuity-due if is increased1

by 0.03.A) 0.16 B) 0.17 C) 0.18 D) 0.19 E) 0.20

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7. In the state of Elbonia all adults are drivers. It is illegal to drive drunk. If you are caught, yourdriver’s license is suspended for the following year. Driver’s licenses are suspended only fordrunk driving. If you are caught driving with a suspended license, your license is revoked andyou are imprisoned for one year. Licenses are reinstated upon release from prison.Every year, 5% of adults with an active license have their license suspended for drunk driving.Every year, 40% of drivers with suspended licenses are caught driving.Assume that all changes in driving status take place on January 1, all drivers act independently,and the adult population does not change.Calculate the percentage of currently active drivers that will be jailed some time in the next 3years.A) 0.031 B) 0.033 C) 0.035 D) 0.037 E) 0.039

8. For a special 3-year temporary life annuity-due on , you are given:(i) Annuity Payment 0 15 0.95 1 20 0.90 2 25 0.85(ii)Calculate the variance of the present value random variable for this annuity.A) 91 B) 102 C) 114 D) 127 E) 139

9. For a special 3-year fully discrete term insurance on :

(i) forfor

(ii) 0 0.20 1 0.10 2 0.097(iii)Calculate the level annual benefit premium for this insurance.A) 518 B) 549 C) 638 D) 732 E) 799

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10. For Shoestring Swim Club, with three possible financial states at the end of each year:(i) State 0 means cash of 1500. If in state 0, aggregate member charges for the next year are setequal to operating expenses.(ii) State 1 means cash of 500. If in state 1, aggregate member charges for the next year are setequal to operating expenses plus 1000, hoping to return the club to state 0.(iii) State 2 means cash less than 0. If in state 2, the club is bankrupt and remains in state 2.(iv) The club is subject to four risks each year. These risks are independent. Each of the fourrisks occurs at most once per year, but may recur in a subsequent year.(v) Three of the four risks each have a cost of 1000 and a probability of occurrence 0.25 peryear.(vi) The fourth risk has a cost of 2000 and a probability of occurrence 0.10 per year.(vii) Aggregate member charges are received at the beginning of the year.(viii) 0Calculate the probability that the club is in state 2 at the end of three years, given that it is instate 0 at time 0.A) 0.24 B) 0.27 C) 0.30 D) 0.37 E) 0.56

11. For a continuous whole life annuity of 1 on ( ):(i) ( ), the future lifetime of ( ), follows a constant force of mortality 0.06.(ii) The force of interest is 0.04.Calculate ( ).( )|

A) 0.40 B) 0.44 C) 0.46 D) 0.48 E) 0.50

12. You are given the following: .Which of the following relationships are correct?I. II. III. A) I only B) II only C) III only D) I and II only E) I and III only

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13. The force of mortality is given to be for .

median future lifetime for someone at age , andmean future lifetime for someone at age .

Find .A) B) C) D) E)

14. A select and ultimate life table with a one year select period satisfies UDD, with and . Suppose that is changed to .05, but all other mortality probabilities areunchanged. Find the change in if .A) Increase of .06 B) Increase of .03 C) 0 D) Decrease of .03 E) Decrease of .06

15. An insurer is calculating the single premium to charge for continuous whole life insurance to with face amount 100,000. The insurer calculates the premium in the following way.

Using and DeMoivre's Law with a continuous single premium whole lifeinsurance policy with face amount 100,000 is issued to each of 100 individuals at age 81, and thesingle premium charged results in a probability of .1587 (based on the normal approximation) ofthe insurer experiencing a positive loss on all policies combined. That is the premium charged to

by the insurer. Using the same interest and mortality assumptions, suppose that thisinsurance policy is issued to 100 independent individuals at age . Using the normalapproximation, find the probability of a positive loss on all policies combined.A) Less than .01 B) At least .01, but less than .03 C) At least .03, but less than .05D) At least .05, but less than .07 E) At least .07

16. A 4-payment, 5-year fully discrete term insurance issued to has a death benefit of$100,000 if death occurs within the first 3 years, and $50,000 if death occurs in the 4th year or5th year. The annual benefit premium payable for 4 years is $2,379.30 .Given that and for find , the 2nd year terminal benefitreserve for this policy.A) 65 B) 70 C) 75 D) 80 E) 85

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Questions 17 and 18 refer to the following information. is the present value random variable for a continuous -year certain and life annuity of 1 per

year issued to . is the present value random variable for a continuous whole life annuityof 1 per year issued to . The force of mortality is constant at for all , and theforce of interest is . Find each of the following in terms of and or annuity notation:

17. FindA) B) C) D) E)

18. Find , where and are the distribution functions of and ,respectively.A) E) B) C) D)

19. A fully discrete 20-year deferred life annuity-due is issued to .The annuity will pay $100,000 every year starting at age 60.The annuity will have level annual benefit premiums payable for 10 years starting at age 40.There is no benefit during the deferral period. Mortality is based on the Illustrative Table(download from the SOA website at the Exam M webpage) with effective annual interest of 6%Find the maximum age so that of death occurs after age but before age , the issuedate loss is negative.A) 71 B) 72 C) 73 D) 74 E) 75

20. A fully discrete whole life insurance with premiums for life with face amount $100,000 isissued to The annual benefit premium is $2,779.01. A fully discrete one-year deferredinsurance with the same face amount issued to , would have annual benefit premium of$2,577.34 payable for life (starting at age ). These premiums are based on a particular lifetable and effective annual interest rate of 7.5% Suppose that the mortality probability at age is increased by .04, but all other mortality probabilities are unchanged. Find the premium for the$100,000 fully discrete whole life policy issued to .A) Less than 3000 B) At least 3000, but less than 3100C) At least 3100, but less than 3200 D) At least 3200, but less than 3300E) At least 3300

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21. You are given the following: .

Find .

A) .085 B) .089 C) .093 D) .097 E) .101

22. The following information is given about a fully discrete whole life insurance of $100,000issued to with level benefit premiums for life: • mortality follows DeMoivre's Law • • • • Find the DeMoivre upper age limit .A) 105 B) 106 C) 107 D) 108 E) 109

23. A company insures the lives of two senior executives. The following assumptions are made. • both executives have the same constant force of mortality • mortality of the two executives is modeled with a common shock model • the force of interest is 7.2% • single benefit premium for $1,000,000 continuous insurance on one of the lives is $200,000 • single benefit premium for $1,000,000 continuous insurance payable on the first death of the two lives is $294,118Find the expected time until the second death.A) 77.2 B) 77.4 C) 77.6 D) 77.8 E) 78.0

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24. The progression of a disease is being studied. Patients with the disease will either be cured(decrement 1), or they die before being cured (decrement 2). A study of the disease is madeusing two multiple decrement model. It is assumed that all individuals in the study begin with amild form and each year, an individual either continues next year with the mild form, or theindividual is cured (decrement 1), or the individual dies during the year before being cured(decrement 2). For the purpose of the study, if an individual is cured, that individual will notreturn to the decrement group. The following table summarizes the behavior of a group at age who all have the mild form.Age

In the analysis, it is assumed that the decrements are uniformly distributed in the multipledecrement table. A new medical procedure is discovered that will increase the force ofdecrement due to being cured (decrement 1) by 50% and will reduce the force of decrementdue to dying before being cured by 25%. With , find based on therevised forces of decrement.A) 210 B) 220 C) 230 D) 240 E) 250

25. and are lives subject to a common shock mortality model, with common shockparameter . has a constant force of mortality of , and has a constant force of mortalityof . Find .

A) B) C) 1 D) E) 2

26. A fully discrete whole life insurance policy with face amount and premiums for life hasthe following expenses:• level annual percentage of premium expense is %• level annual per policy expense is $• settlement expense is $The expense-loaded premium based on the equivalence principle is .Of which of the variables and is a linear function?A) All but B) All but C) All but D) All but E) All

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27. The expense-loaded premium, , for a fully discrete 3-year endowment insurance of 1000issued to is calculated using the equivalence principle. Expenses are paid at the beginning ofeach year. You are given:(i) (ii) Expense reserve at the end of the first year is

(iii) (iv) (v) (vi) Expenses Percentage of Premium Per Policy First Year 30% 8 Renewal 10% 4Calculate .A) Less than 400 B) At least 400, but less than 410C) At least 410, but less than 420 D) At least 420, but less than 430E) At least 430

28. A casino creates a gambling game in which the chance of winning the game depends uponthe outcome of the last two games. There are 3 wheels that can be spun with the followingprobabilities of winning and losing:Wheel 1: win , loseWheel 2: win , loseWheel 3: win , loseThe game starts out with the gambler spinning Wheel 1. The gambler can choose to continue thegame and spin Wheel 1 again. If the gambler wants to continue after that, the wheel that he spinsdepends on the outcome of the previous two spins. If he won both of the previous two spins, hemust spin Wheel 3, if he lost both of the previous two spins, he must spin Wheel 2, and if wonexactly 1 of his previous two spins, he spins Wheel 1. The game consists of 4 consecutive spinsand pays $1 for a win on spin 3 and $2 for a win on spin 4. Find the expected payout on a play ofthe game.A) Less than 1.00 B) At least 1.00, but less than 1.25 C) At least 1.25, but less than 1.50D) At least 1.50, but less than 1.75 E) At least 1.75

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Questions 29 and 30 relate to a Poisson process with rate per period.

29. The process is a count of the number of events occurring by time . The events are of twotypes. Type 1 events occur twice as frequently as Type 2 events. Event types are independent ofone another. Given that at least two events have occurred by time 1, find the probability that atleast one of them is of Type 2.A) .69 B) .71 C) .73 D) .75 E) .77

30. Suppose that 6 events have occurred by time 1. Find the expected number of events of Type2 that have occurred between time 0 and time 2.A) 1 B) 2 C) 3 D) 4 E) 5

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 11 SOLUTIONS

1. The prospective form of the reserve is , and

the retrospective form is , where and .

With , and Demoivre's Law with , we have and

, ,

, .

. Answer: A

2. The actuarial present value of the benefit for one life is , and the variance of thepresent value random variable of the benefit for one life is .

With constant force of mortality at all ages, and force of interest , the APV ofa continuous whole life insurance at age is , and at a doubled force of

interest, . The APV of the deferred insurance is

.

Also, ,

so that the variance is .

If denotes the aggregate present value random variable for 400 independent policies, then and .

Using the normal approximation for , we wish to find so that .After standardizing, this becomes , so that

, and . Answer: A

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3. Suppose the constant rate (force of mortality) of default for the risky investment is per year.The probability that the risk investment does not default within 20 years is (like ).The risky investment pays , the risk free investment pays (with 100%probability of survival). The APV's of the two investments arerisky: risk-free: Since the two APV's are equal, we have so that .The median time until default for the risky investment is , where , sothat . Answer: E

4. The accumulation of reserve formula gives us , and from the relationships in the question .Then,

. Answer: D

5. From the recursive relationship for reserves, for policy A from time 10 to time 11 we have (since policy A is a 10-payment policy, there is

no premium at the start of the 11-th year).For policy B, the recursive relationship is .Subtracting the second equation from the first results in

.Therefore, , so that

. Answer: E

6.

The revised values (superscript ) are found as follows:

, , (only change is at age )

.Change is . Answer: C

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7. The one-year transition matrix for the states 1 - active ; 2 - suspended ; 3 - jail is

. Suppose a driver is currently active. The sequences of transitions that

will result in the driver being jailed some time in the next 3 years (3 transitions) is Sequence Probability 1-1-2-3 1-2-3 Total probability is .039 . 3.9% of currently active drivers will be jailed within the next threeyears. Answer: E

8. The present value random variable is a 3-point random variable which depends upon whendeath takes place. If death is in the first year (prob. ), the pv is 15 (one paymentreceived). If death is in the second year (prob. ) the pv is (pv of 2

payments). If survives two years, then the pv is . The pvrv can be

described as .prob.prob.prob.

Then , and

. Then

. Answer: C

9. The annual premium is the solution the equivalence principle equation:

. Therefore

, which becomes

. Then .

Answer: A

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10. 1 big loss no big loss and 2 or more small losses .State 2 is absorbing, so that the probability of not entering state 2 in 3 years is

, and the probability of entering state 2 within 3 years is .

Alternatively, (no losses) , (exactly one small loss),

(as above).(no losses) ,

(exactly one small loss). .

The transition probability matrix is .

The two-step transition probability matrix is ,

and the three-step transition probability matrix is .

Answer: E

11. With constant force of mortality , .

. Answer: C

12. I. . True.II. . False.III. . False.

Answer: A

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13. If , then .

The median future lifetime is the 50-th percentile of the survival time random variable, so that . Therefore, , from which we get .

The mean future lifetime is .°

. Answer: B

14. Using the original value of , we have .

Under UDD, we have ,

and .Using the original value of , we have

.Using the new value , we have ,of

and as before .Using the new value of , we have

.The change in is a decrease of .0288 .Alternatively,

New Old New New Old OldNew Old

New Old New Old New Old) ) (since ) .

. Answer: D

15. Under DeMoivre's Law, ,_ _

and ._ _

With , we have ,

andThe loss on a single policy with premium is , where is thepresent value random variable of a continuous insurance of 100,000 issued to .We have , and

.

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15 continued

The combined loss on 100 policies is , where each has the same distribution as

but the 's are independent of one another. We wish to find so that .We have and it follows from independence of the 'sthat .

.

Applying the normal approximation to , it follows that ,

so that . This is the premium charged to each of 100 individuals at age .

Now the loss on a single policy with premium is , where is thepresent value random variable of a continuous insurance of 100,000 issued to .We have , and

.

The combined loss on 100 policies is , with

and .

Then

. Answer: C

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16. At the end of the 2nd year there are 3 years remaining on the policy.One of 4 events must occur to (i) death occurs before age ,(ii) death occurs between and ,(iii) death occurs between and ,(iv) survival to age .

can be represented as the 4-point random variable

prob.prob.prob.prob.

Since for , it follows that , and . Then, since for , it follows that

for .

Then, , and and .

alive at age .

Answer: E

17. If lives beyond years, then , since both annuities will pay up to the time of's death. If dies at time , then and , so that .

Therefore, dies before time . Answer: B

18. Since and are non-negative random variables with finite means, it is true that and .

Then,

.

Answer: E

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19. The annual premium is , which is the solution of the equation , which can be written as

.Using values from the Illustrative Table, solving for results in .The issue date loss is PV of benefit PV of premium .If death occurs before age 60, there is no benefit, so the loss is PV of premium .If death occurs between age and , there will be annuity payments.The PV at age 40 of the annuity payments will be .

The present value at age 40 of the premiums (if all 10 premiums are paid) is .

If death occurs between age and , the issue date loss is .

This will be negative if .

By trial and error, or by solving for from ,we see that and .

Therefore, if death occurs between ages 73 and 74, there will be 14 annuity payments, and theissue date loss will be ,

but if death occurs after age 74, there will be at least 15 annuity payments, and the issue date losswill be at least . The age is 73. Answer: C

20. .

.

.The new value of is , and the new value of is

( is unchanged since only was changed).But, we know that

.Then , and

Answer: C

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21. From the prospective form of the reserve we have .

Then .

Using we get .

Then, . Answer: B

22. We use the recursive reserve relationship .

For we get .

For we get .

Subtracting the first equation from the second equation results in

.This reduces to the equation , or equivalently

.This is a quadratic equation in . The two roots are and .We ignore the negative root. Answer: A

23. , and are the common shock components, and we are given that , whichwe will denote . The overall force of mortality for is and same for .The force of mortality for the joint status of and is .The expected time until the second death is° ° ° ° .

The single premium for a continuous whole life insurance of 1 for is

and the same for . The single premium for a continuous whole life insurance of 1 for thejoints status is .

We are given that based on common shock parameter , we have and . Writing these two equations as

andgives us two equations in the two unknown quantities and .Solving the equations results in and .Expected time until second death is . Answer: D

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24 If is multiplied by 1.5, then the new is old ,since .Similarly, if is reduced by 25%, then new is old Under UDD, and .With the original decrement table, we have and . Then and .Under the decrement table based on the new medical procedure, we have

, ,and .Based on these revised absolute rates, the multiple decrement probabilities will be

, , and

The revised table isAge

Answer: E

25. is defined as , however this definition applies

only if there is no common shock. When there is a common shock, the definition is , where is the force of mortality for excluding the common shock

component. , and in a similar way, .

Then, , and . , and

.

. Answer: B

26.Solving for results in

is a linear function of and , but not of . Answer: B

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27. The level expense loading is .

We can use the accumulation relationship for expense reserve , where is the total expense at the

start of year and is the settlement expense at the end of year . In this problemthere is no settlement expense. Also, from the Equivalence Principle, we have . Theexpenses at the start of the first year are

. Answer: B

28. The outcome of the 3rd spin depends on what happened in the first two spins.The following table summarizes the probabilities of the possible outcomes of the first two spins. Spin 1 Spin 2 Probability W W .25 W L .25 L W .25 L L .25 Coning in to the 3rd spin, there is .25 chance that the first two spins were , in which caseWheel 3 is spun. There is a .25 chance that the first two spins were and a .25 chance thatthey were , in which case Wheel 1 is spun. There is a .25 chance that the first two spinswere , in which case Wheel 2 is spun. We can represent the probabilities of the possibleresults of the first 2 spins in vector form

.The successive results of the spins can be formulated as a Markov Chain with states

, with one-step transition matrix

If we multiply the vector by the matrix , we get

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28 continuedThis is the matrix of probabilities of the states for (2nd spin, 3rd spin):

. The probability of winning on the third spin is the sum ofthe probabilities of states and on (2nd spin,3rd spin).This sum is .If we multiply we get

This is the matrix of probabilities of the states for (3rd spin, 4th spin).The probability of winning on the fourth spin is the sum of the probabilities of states and on (3rd spin,4th spin). This sum is .The expected payout is . Answer: D

29. When an event occurs, there is a probability that it is Type 1 and a probability that it isType 2. We will denote by and the numbers of Type 1 and Type 2 events occurringby time , respectively; these will be independent processes with rates and ,respectively. We wish to find .This probability is .

.

(the last equality follows from the independence of and . , and

.

Then, . Answer: C

30. We wish to find

. , since is independent of .

. .

. Answer: C

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 12

1. The following annuities immediate are actuarially equivalent: I. Straight life annuity of $100 payable to a life age 65II. $90 payable during the joint life of two independent lives both age 65, reducing to $54 on the first deathIII. $70 payable during the joint life of two independent lives both age 65, reducing to P on the first deathIn what range is P?A) Less than 57 B) 57 but less than 59 C) 59 but less than 61D) 61 but less than 63 E) 63 or more

2. Which of the following is equal to the expression° °

under the assumption of UDD?A) B) C) D) E)

3. Smith (age 65) purchases a special single premium life annuity with an annual payment of$1,000 at the end of each year. The annuity continues for 5 years after Smith's death. Theinterest rate is 5% annual effective. You are given 10.17548In what range is the single premium paid by Smith?A) Less than $12,150 B) $12,150 but less than $12,250C) $12,250 but less than $12,350 D) $12,350 but less than $12,450E) $12,450 or more

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4. For a population comprised of smokers and nonsmokers, you are given:(i) Mortality of nonsmokers is based on DeMoivre's Law with .(ii) The force of mortality for smokers is times as large as that for nonsmokers at all ages.For a pair of individuals with independent lifetimes, each at age , one a nonsmoker and one asmoker, the complete expected time until the second death is 30.71 years. Find .A) Less than 1 B) At least 1, but less than 1.2 C) At least 1.2, but less than 1.4D) At least 1.4, but less than 1.6 D) At least 1.6

5. Smith (age 65) purchases an annuity that pays $1,000 at the end of each year. Payment ceasesat the earlier of Smith's death or 25 years from purchase date. The interest rate is 5% annualeffective. For all , we have , and mortality rates at ages 75 and greater aredouble those under age 75. Find the present value of this annuity.A) Less than $6,500 B) $6,500 but less than $7,500 C) $7,500 but less than $8,500D) $8,500 but less than $9,500 E) $9,500 or more

6. Customers enter a supermarket at a Poisson rate of 3 customers per minute. Each personspends $50 on average but the amount spent by any one individual follows a Normal distributionwith coefficient of variation, , of 25%. The amount spent by a customer is revenue to the

supermarket. Calculate the standard deviation of the revenue generated by customers who enterthe supermarket in a single hour.A) Less than $500 B) At least $500, but less than $550C) At least $550, but less than $600 D) At least $600, but less than $650 E) At least $650

7. has the following mortality distribution: .

A 4-year discrete increasing insurance has a death benefit of if for .

When the policy is issued with a single premium of 1900, there is a 70%probability that the issue date loss will be positive. Which of the following annual effectiveinterest rates are consistent with this probability?I. 5% II. 10% III. 15% IV. 20%A) All but I B) All but II C) All but III D) All but IV E) All

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8. Smith and Jones are independent lives of the same age. They pay a single premium to purchase anannuity immediate of $1,000, payable annually as long as at least one of them is alive.The interest rate is 5% annual effective. Mortality follows the model for all with In what range is the single premium? A) Less than $19,250 B) $19,250 but less than $19,750C) $19,750 but less than $20,250 D) $20,250 but less than $20,750E) $20,750 or more

9. Independent lives Smith and Jones, both age 40, purchase an insurance policy with a death benefit of$100,000 payable if and only if both die in the same year. The death benefit is payable at the end of thepolicy year. The interest rate is 5% annual effective. Mortality follows the model .In what range is the single premium for this policy?A) Less than $500 B) $500 but less than $520 C) $520 but less than $540D) $540 but less than $560 E) $560 or more

10. A gambler is playing a game in which he either wins 1 dollar or loses 1 dollar on each play of thegame. The probability of winning 1 on any play of the game is .4. If he ever reaches 0 dollars, he stopsgambling and stays at 0 dollars. Suppose that he currently has 2 dollars. Find the minimum number ofplays so that the probability of reaching 0 by time is at least .5.A) 2 B) 3 C) 4 D) 5 E) 6

11. Data from a double-decrement table:

What is ?A) 115 B) 116 C) 117 D) 118 E) 119

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12. You are given the following information: • The number of wild fires per day in a state follows a Poisson distribution. • The expected number of wild fires in a thirty-day time period is 15.Calculate the probability that the time between the eighth and ninth fire will be greater than threedays.A) Less than 20% B) At least 20%, but less than 25% C) At least 25%, but less than 30%D) At least 30%, but less than 35% E) At least 35%

13. You are given the following information about a fully continuous whole life insurance of 1.• If the premium is increased by 10% the variance of the issue date loss increases by 4.04%.• If the premium is increased by .003 the variance of the issue date loss increases by 10.25%.Find the percentage increase in the variance of the issue date loss if the premium is increasedby 20%A) Less than 7.5% B) At least 7.5%, but less than 8.5%C) At least 8.5%, but less than 9.5%D) At least 9.5%, but less than 10.5% E) At least 10.5%

14. Mortality follows the model , for .The probability that two independent lives age 30 and 50 will die within 10 years of each other.

What is ?A) B) C) D) E)

15. A pricing actuary is reviewing cash values on fully discrete whole life insurance policy.The estimated mortality probability for the 20th year is .006 and the estimated withdrawalprobability for the 20th year is .03. Find the reduction in the 20th year-end asset share if the 20thyear cash value is increased by $100.A) Less than 2.00 B) At least 2.00, but less than 2.25C) At least 2.25, but less than 2.50D) At least 2.50, but less than 2.75 E) At least 2.75

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16. You are given••• deaths have constant force of mortality over each year of age.Find °A) Less than 1.450 B) 1.450 but less than 1.460 C) 1.460 but less than 1.470D) 1.470 but less than 1.480 E) 1.480 or more

17. A fully discrete whole life insurance is issued to with face amount 100,000.Mortality is based on the Exam MLC Illustrative Life Table (available at the SOA website)with an annual effective interest rate of 6%. The annual contract premium is 20% larger than theannual benefit premium. Assuming that the policy is still in force at the end of 10 years, find theprobability that the 10-th year terminal prospective loss is positive.A) Less than .45 B) At least .045 but less than .46 C) At least .46 but less than .47D) At least .47 but less than .48 E) At least .48

18. Drivers are classified as either Preferred, (1), or Standard, (2), at the beginning of each year.The movement between drivers' classifications is modeled as a non-homogeneous Markov chainusing the following transition matrices. There is an administrative cost at the a driver transitionsbetween states.

• • •

• The administrative cost of moving from Preferred to Standard is 20 units. • The administrative cost of moving from Standard to Preferred is 10 units. • The annual interest rate is 5%.For a driver classified as Preferred, calculate the present value of the cash flows for theadministrative cost for the first three years.A) Less than 10.25 B) At least 10.25, but less than 10.35C) At least 10.35, but less than 10.45D) At least 10.45, but less than 10.55 E) At least 10.55

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19. A 2-decrement model has absolute rates and .Actuary A uses the following model for decrement behavior:• decrement 1 is uniformly distributed in its associated single decrement table• decrement 2 is discrete and occurs at time .5Actuary B uses the following model for decrement behavior:• decrement 1 has a constant force of decrement• decrement 2 is discrete and occurs at time .5Find , the ratio of Actuary B's calculated value of to that of Actuary A.

A) Less than 1.00 B) At least 1.00, but less than 1.05C) At least 1.05, but less than 1.10D) At least 1.10, but less than 1.15 E) At least 1.15

20. Smith (age 55) is entitled to an annual payment of X at the beginning of each yearguaranteed for 10 years and continuing as long as Smith is alive. Instead, Smith elects anactuarially equivalent annuity that pays the following as long as Smith is alive:(1) $10,000 at the beginning of each year for the first 5 years,(2) $7,500 at the beginning of each year for the next 5 years; and(3) $5,000 at the beginning of each year thereafter.The interest rate is 7% annual effective.You are given:

In what range is X?A) Less than $7,375 B) $7,375 but less than $7,425C) $7,425 but less than $7,475 D) $7,475 but less than $7,525 E) $7,525 or more

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21. The following matrices with states 1, 2 and 3, , show the probability of transitioningfrom State to State at time :

For a subject in State 2 at time 0, calculate the probability of transitioning from State 3 to State 1at time 3.A) Less than 0.06 B) At least 0.06, but less than 0.07C) At least 0.07, but less than 0.008D) At least 0.08, but less than 0.09 E) At least 0.09

22. Smith and Jones are independent lives.Smith (age 65) and Jones (age 64) pay $100,000 for an annuity with the following payments: (1) Annual payments of at the beginning of each year and ending on the first death.(2) Annual payments of ( - $3,000) beginning at the end of the year of the first death andcontinuing for the lifetime of the survivor.The interest rate is 7% annual effective.You are given: In what range is ?A) Less than $8,522 B) $8,522 but less than $8,622 C) $8,622 but less than $8,722D) $8,722 but less than $8,822 E) $8,822 or more

23. A fully discrete whole life policy with death benefit of $100,000 is issued to .The annual benefit premium is $1 00, the single benefit premium is $20,000,and the 30th year terminal reserve is $40,000.Based on the same mortality table and interest rate as the first policy, another fully discretepolicy has a death benefit of $100,000 up to age 65 and $ 0,000 after age 65. The annualbenefit premium is $1 00 payable for the first 30 years, and the benefit premium is payablefrom age 65 on. Find .A) 6400 B) 6500 C) 6600 D) 6700 E) 6800

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24. Historical data from the Actuarial School of Hard Knocks (ASHK) shows the followinginformation about when a graduating student gets their first job during the three years aftergraduation. Decrement 1 refers to the first job being an actuarial job and decrement 2 refers tothe first job being a non-actuarial job. Graduation occurs at time . For a group of 80 graduates from ASHK, find the expected number whose first job is an actuarialjob within three years after graduating (nearest integer).A) 44 B) 46 C) 48 D) 50 E) 62

25. You are given the following annuity values :

Find the value of .1

A) .010 B) .015 C) .020 D) .025 E) .030

26. A fully discrete whole life policy with death benefit of $100,000 is issued to .The annual benefit premium is $1 00, the 30th year terminal benefit reserve is $40,000and the 31st year terminal reserve is $42,405.Based on the same mortality table and interest rate as the first policy, another fully discretepolicy has a death benefit of $100,000 up to age 65 and $ 0,000 after age 65. The annualbenefit premium is $1 00 payable for the first 30 years, and the benefit premium is 6400 payablefrom age 65 on. The 31st year terminal benefit reserve is $45,791. Find .A) .020 B) .021 C) .022 D) .023 E) .024

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27. and are lives subject to a common shock. The usual assumptions for the common shockmodel are made:• represents 's future lifetime in the absence of common shock• represents 's future lifetime in the absence of common shock• represents the time until the common shock occurs• and are independent

In addition, the following assumptions are made:• has a constant intensity of • has a constant force of mortality• has a constant force of mortality

For insurance calculations, there is a constant force of interest .

You are given the following insurance values:• the single benefit premium for an insurance of 1 paid at the instant of 's death is .072464• the single benefit premium for an insurance of 1 paid at the instant of 's death if dies before is .045699

Find the single benefit premium for an insurance of 1 paid at the instant that the common shockoccurs if and are still both alive then.A) Less than .020 B) At least .020 but less than .025 C) At least .025 but less than .030D) At least .030 but less than .035 E) At least .035

28. It is January 1, 2008 and and are two independent lives. has a constant force of mortality of .03 until his death,

and has a constant force of mortality of .02 until his death.Find the probability that dies either in the same calendar year as dies,or in the calendar year following the calendar year of 's death.A) Less than .025 B) At least .025 but less than .030 C) At least .030 but less than .035D) At least .035 but less than .040 E) At least .040

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29. A special fully discrete 20-year endowment insurance policy issued to has a deathbenefit of if dies in the -th year for . If survives to age

, the endowment benefit is 1000. The annual effective rate of interest is 8% and mortalityprobability is for . Find the variance of the issue date loss randomvariable.A) Less than 100,000 B) At least 100,000 but less than 120,000C) At least 120,000 but less than 140,000D) At least 140,000 but less than 160,000 E) At least 160,000

30. 1,300 lives, all age 50. are issued a: 3-year term insurance, paying $10,000 at end of year ofdeath. Each insured pays a single premium based on 5% interest and mortality which follows thelaw . All premiums are deposited into a fund out of which all benefits are paid.Over the 3-year period, the fund earns interest of 5¼% per annum and mortality follows

.Find the amount remaining in the fund immediately after third year claims are paid (ignoringexpenses)A) Less than $0 B) $0 but less than $10,000 C) $10,000 but less than $20,000D) $20,000 but less than $30,000 E) $30,000 or more

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S. BROVERMAN MLC STUDY GUIDEPRACTICE EXAM 12 SOLUTIONS

1. The APV's of these annuities areI.II.III.

Since all three APV's are equal, by setting I equal to II we get so that .

Then setting I equal to III, we get .

The factor cancels out, resulting in ,Solving for results in . Answer: D

2. ° ° ° ° . Answer: C

3. The annuity pays during the first 5 years even if Smith dies immediately, so the APV of the first 5years of payments is .

After 5 years, the annuity pays if Smith was alive 5 years earlier, so the APV of the payments starting at

time 6 are (substitution of ),

which becomes .The total APV is .

An alternative solution is (annuity for life plus 5 more payments after death).

We can see that can be written as

. Answer: C

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4. Nonsmokers have force of mortality , and smokers have force of

mortality . Under DeMoivre's Law, survival probability is

and force of mortality is .

Since smokers have force of mortality , it follows that smoker survival is

, with . Both nonsmokers and smokers have upper age limit 90.Last survivor expectation for two independent lives aged 40, one a nonsmoker and one a smokeris ° ° ° °

.

From this equation, we get . This results in the quadratic

equation . The two roots are and .We ignore the negative root. Answer: D

5. This is a 25-year life annuity-immediate. Since the mortality rate changes 10 years after the annuitybegins, we write the APV as .

The survival probability is .95 each year up to age 74, sofor .

The mortality probability at ages less than 75 is .05, so the mortality probability at each age from 75 onis .10. Then for , so that .

..

.APV We have used the finite geometric series formula

. Answer: C

6. The total revenue generated in one hour , has a compound Poisson distribution with Poissonmean of customers per hour. The severity (revenue per customer) isNormal with mean and coefficient of variation , so that and

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. The variance of the total revenue per hour for this compound Poissondistribution is .But .Then, , and the standard deviation of is . Answer: E

7. The issue date losses are if (prob. .1), if (prob. .2), if (prob. .3), and if (prob. .4) .

The 70% chance of positive loss indicates that there is a positiveloss if or , and a non-positive loss if or .Therefore the following conditions must all be satisfied:(a)(b)(c)(d) .Interest rates I, II and III satisfy all conditions, but interest rate IV violatesthe condition required by (c). Answer: D

8. This is a last survivor annuity-immediate. From the mortality model, we see that .

Suppose that both individuals have the same age .Since the two are independent, the joint life status survival probability is

.The last survivor annuity-immediate has APV .

.

.

. Answer: A

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9. Mortality follows DeMoivre's Law with ,so that for .The probability that Smith and Jones both die between ages 40 and 41 is

.The probability that Smith and Jones both die between ages 41 and 42 is

.In general, the probability that Smith and Jones both die between ages and

is .The APV of the insurance policy is

. Answer: C

10. The general one-step transition probability matrix for this transition process has the form

State State

The process starts in state 2, so after one transition, row 2 of the two step transition matrix is therow resulting from multiplying the row vector by the one-step transitionmatrix above. This results in the row vector.State State These are the two step transition probabilities from initial state 2. We see that the probability ofreaching 0 by time 2 is .36 . We multiply this row vector by the one-step transition matrix to getthe following row vector.State State

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These are the three step transition probabilities from initial state 2. We see that the probability ofreaching 0 by time 3 is .36 . We multiply this row vector by the one-step transition matrix to getthe following row vector.State State These are the four step transition probabilities from initial state 2. We see that the probability ofreaching 0 by time 4 is .5328 which is greater than .5 . Time is the first for which theprobability of reaching state 0 by time is at least .5. Answer: C

11. Since , it must be true that.

.

Then . Answer: E

12. The number of fires in a 3-day period is Poisson with a mean of . The number offires in the 3-day period following the 8-th fire is Poisson with a mean of 1.5, so the probabilityof no fires in that 3-day period is . Answer: B

13. We denote the annual premium by and the force of interest is .The variance of the issue date loss is• based on premium • based on premium 1.11.1

• based on premium +.003

We are given , and

.

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We get the following two equations: and .

Solving for and results in and .If the premium is increased by 20%, the issue date variance is increased by a factor of

, an increase of 8.16% .

Answer: B

14. The two individuals will die within 10 years of each other if either the 30-year old dies first and the50- year old dies within 10 years after that or if the 50-year old dies first and the 30-year old dies within10 years after that. The first probability is

.Since the force of mortality is constant, when we set up the second probability integral, all the factorswill be identical to the first integral. The probability is .Answer: E

15. The accumulation relationship for the asset share for the 20th year

.

The 20th year asset share is .

If is increased by 100 , then the asset share is decreased by

. Answer: E

16. ° ° °

. . Answer: D

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17. The benefit premium is , and the contract premium is

. , where is the curtate future lifetime of .

We first solve for to make :

.

Solving for results in . Alternatively, we can use the unknown timefunction on a calculator to solve for .If survives 32 complete years and dies between ages 78 and 79, then

,

but if survives 31 complete years and dies between ages 77 and 78, then .

The conditional probability that is positive given that the policy is still in force at age 45 isthe probability that survives at most 31 complete years, or equivalently, dies within 32years. This probability is .

18. The driver starts out as Preferred. The only possible cost at the end of the first year is 20 andthis occurs if the driver transfers from Preferred to Standard. That probability is .3, found in .The actuarial present value of the first year end transition cost is .

The possible transition costs at the end of the second year are 20 and 10.20 occurs if the transition is from P to S at the end of the 2nd year, which has probability

(path P-P-S).10 occurs if the transition is from S to P at the end of the 2nd year, which has probability

(path P-S-P).The APV of 2nd year end transition costs is .

The possible transition costs at the end of the third year are 20 and 10.20 occurs if the transition is from P to S at the end of the 3rd year, which has probability

(paths P-P-P-S and P-S-P-S).10 occurs if the transition is from S to P at the end of the 3rd year, which has probability

(paths P-P-S-P and P-S-S-P) .The APV of 3rd year end transition costs is .

The total APV is . Answer: D

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19. For both Actuaries, .For Actuary A, the value of is The value of is .For Actuary B, the value of is The value of is .The ratio is . Answer: B

20. The APV of Smith's original benefit is.

The APV of the actuarial equivalent annuity is

We can find from the relationship ,

so that and .In a similar way, we can find from , so that

. Then, .Since the two annuities are actuarially equivalent, we have

,from which we get . Answer: A

21. The subject must transition from State 2 at time 0 to State 3 at time 2 and then transition toState 1 at time 3. This is . The matrix is the transition matrix from time

to time (which is not the same as the notation used in the study note; the study notewould denote this matrix ). is found from the matrix .

is the 2-3 entry in the matrix product . This will be .

The probability in question is . Answer: B

22. The APV is

We find from .Then solving for results in . Answer: C

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23. We are given the following: , .

From , we get ,

and from , we get .

The second policy has the same benefit and premium for the first 30 years, soit must have the same reserves as the first policy during the first 30 years, so

for the second policy. The prospective form of the reserve forpolicy 2 at time 30 is .We get , so that

.Solving for results in . Answer: A

24. The expected number is .

The expected number is . Answer: C

25. We use the following identities.1

,

From (5) we have .

From (3) we have From (4) we have From (2) we have From (1) we have . Answer: C1

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26. Using the accumulation relationship for benefit reserves from time 30 to time 31, we have

for the first policy, and

for the second policy.Solving these two equations for and results in and . Answer: C

27. 's force of mortality in absence of common shock is ,and similarly for we have .We are given .

It follows that .Since forces of mortality and common shock are constant, the single benefit premium for aninsurance of 1 paid at the moment of 's death if dies before (not the same time as) is

.

It follows that .The single benefit premium for an insurance of 1 paid at the instant that the common shockoccurs if and are still both alive then is . Answer: B

28. For we have for all , andfor we have for all .Therefore,and .The probability that and die in the same calendar year is

(because of independence of and )

.

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The probability that dies in the calendar year following the calendar year of 's death is

.

The total probability is .Answer: A

29. We use the recursive variance relationship .

For we have.

For this 20-year policy, endowment amount , so .Therefore, since it follows that

.Then, since , we have

. .

Then, since , we have.

, where .Continuing in this way, we see that

.Answer: B

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30. Assumed mortality follows DeMoivre's Law with . The single premium for a 3-year terminsurance at age 50 is .

Actual mortality is DeMoivre with so the mortality probability at age 50 is and the number of deaths at age 50 is

This will also be the number of deaths at age 51 and 52.At time 0 the insurer collects This earns 5.25% , so that at the end of the year before claims are paid, the fund is

. The insurer pays 25 claims for a total of 250,000 ,so the fund at the end of the first year is .After interest and claims in the 2nd year, the fund is

.After interest and claims in the 3rd year, the fund is

. Answer: E

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MAY 2007 SOA EXAM M

1. You are given(i) (ii) (iii) Calculate .A) 0.85 B) 0.86 C) 0.87 D) 0.88 E) 0.89

2. You are given:(i) (ii) (iii) (iv) is the future lifetime random variable for .Calculate .

A) 12 B) 14 C) 16 D) 18 E) 20

3. For a fully discrete whole life insurance of 1000 on the select life [60]:(i) Ultimate mortality follows the Illustrative Life Table.(ii) The select period is 3 years.(iii)(iv)Calculate , the benefit reserve at the end of year 5 for this insurance.A) 112 B) 116 C) 121 D) 126 E) 130

4. For a fully discrete whole life insurance of 150,000 on , you are given:(i) (ii) (iii) The annual premium is determined using the equivalence principle.(iv) is the loss-at-issue random variable.Calculate the standard deviation of .A) 14,000 B) 14,500 C) 15,100 D) 15,600 E) 16,100

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5. Heart/Lung transplant claims in 2007 have interarrival times that are independent with acommon distribution which is exponential with mean one month. As of the end of January, 2007no transplant claims have arrived. Calculate the probability that at least three Heart/Lungtransplant claims will have arrived by the end of March, 2007.A) 0.18 B) 0.25 C) 0.32 D) 0.39 E) 0.45

6. People arrive at a food bank at a Poisson rate of 10 per day. 80% of them donatenonperishable units of food and 20% withdraw units of food. Individual food donations aredistributed with mean 15 and variance 75 and individual food withdrawals are distributedwith mean 40 and variance 533. The number arriving and the amounts of donations andwithdrawals are independent. Using the normal approximation, calculate the probability that theamount of food units at the end of seven days will be at least 600 more than at the beginning ofthe week.A) 0.07 B) 0.09 C) 0.11 D) 0.13 E) 0.15

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7. For five special fully continuous whole life insurances on (35):(i) The death benefit on each insurance is 1000.(ii) Benefit premium rates for the first five years are shown in the graphs below.(iii) Benefit premium rates after five years for each insurance are level, not necessarily at thesame rate as during year 5.(iv)Which of these benefit premium patterns results in the largest benefit reserve at the end ofyear 3?

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8. Kevin and Kira excel at the newest video game at the local arcade, “Reversion”. The arcadehas only one station for it. Kevin is playing. Kira is next in line. You are given:(i) Kevin will play until his parents call him to come home.(ii) Kira will leave when her parents call her. She will start playing as soon as Kevin leaves if heis called first.(iii) Each child is subject to a constant force of being called: 0.7 per hour for Kevin; 0.6 per hourfor Kira.(iv) Calls are independent.(v) If Kira gets to play, she will score points at a rate of 100,000 per hour.Calculate the expected number of points Kira will score before she leaves.A) 77,000 B) 80,000 C) 84,000 D) 87,000 E) 90,000

9. For a double decrement table, you are given:(i) In the single decrement table associated with cause (1), decrements are uniformly distributedover the year.(ii) In the single decrement table associated with cause (2), decrements occur at only two pointsduring the year. Three-quarters of the decrements occur at time 1/5 and the remaining one-quarter occur at time 3/5.(iii) and Calculate .A) 0.108 B) 0.110 C) 0.112 D) 0.114 E) 0.116

10. For whole life insurances of 1000 on (65) and (66):(i) Death benefits are payable at the end of the year of death.(ii) The interest rate is 0.10 for 2008 and 0.06 for 2009 and thereafter.(iii) and (iv) The actuarial present value on December 31st 2007 of the insurance on (66) is 300.Calculate the actuarial present value on December 31st 2007 of the insurance on (65).A) 279 B) 284 C) 289 D) 293 E) 298

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11. For a 10-payment, fully discrete, 20-year term insurance of 1000 on (40), you are given:(i)(ii) Mortality follows the Illustrative Life Table.(iii) The following expenses, which are incurred at the beginning of each policy year: Year 1 Years 2+ % of premium constant % of premium constantTaxes 4% 0 4% 0Sales Commission 25% 0 5% 0Policy Maintenance 0% 10 0% 5Calculate the expense-loaded premium using the equivalence principle.A) 18.21 B) 18.35 C) 18.53 D) 18.71 E) 18.95

12. For a special fully discrete 3-year term insurance on (55), whose mortality follows a doubledecrement model:(i) Decrement 1 is accidental death; decrement 2 is all other causes of death.(ii) 55 0.002 0.020 56 0.005 0.040 57 0.008 0.060(iii)(iv) The death benefit is 2000 for accidental deaths and 1000 for deaths from all other causes.(v) The level annual contract premium is 50.(vi) is the prospective loss random variable at time 1, based on the contract premium.(vii) is the curtate future lifetime of (55).Calculate the smallest value of such that .A) 743 B) 793 C) 843 D) 893 E) 943

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13. For a fully discrete 3-year term insurance of 1000 on , you are given:(i)(ii) The mortality rates and terminal reserves are given by: 0 0.3 95.833 1 0.4 120.833 2 0.5 0 (iii) is the prospective loss random variable at time 1, based on the benefit premium.(iv) is the full number of years lived by prior to death; i.e. the curtate future lifetimerandom variable for .Calculate .A) 238,000 B) 247,000 C) 256,000 D) 265,000 E) 274,000

14. The probability that (30) will survive five years is . The probability that (30) will die before(35) is . (30) and (35) are independent lives both based on the same life table.Determine the probability that (30) will die second and within five years of the death of (35).A) B) C) D) E)

15. Residents in a Continuing Care Retirement Community (CCRC) can be in one of threestatuses: Independent Living (#1), Health Center (#2) and Dead (#3).

(i) ,

,

(ii) Transitions occur at the end of each year.(iii) The CCRC incurs a cost of 1000 at the end of year for transition from IndependentLiving at the start of that year to Health Center at the start of the next year, for all .(iv) The CCRC wishes to charge a fee at the start of each of the first 3 years for eachresident then in Independent Living.(v) Nathan enters Independent Living at time 0.(vi)Calculate for Nathan using the equivalence principle.A) 239 B) 242 C) 245 D) 248 E) 251

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16. The number of coins Lucky Tom finds in successive blocks as he walks to work follows ahomogeneous Markov model:(i) States 0, 1, 2 correspond to 0, 1, or 2 coins found in a block.

(ii) The transition matrix is:

(iii) Tom found one coin in the first block today.Calculate the probability that Tom will find at least 3 more coins in the next two blockstoday.A) 0.43 B) 0.45 C) 0.47 D) 0.49 E) 0.51

17. Your son has been driving for one year without an accident. To encourage continued safedriving you offer him a choice now of one of the following:(i) For the next three years, 100 at the end of each year in which he has no accidents.(ii) at the end of Year 3 if he has no accidents for the next three years.You assume that the probability of an accident-free year is 0.8 if the previous year was accident-free and 0.7 if there was an accident in the previous year. Using 4% annual interest, calculate Rso that the two choices are actuarially equivalent.A) 315 B) 378 C) 426 D) 479 E) 513

18. You are given the following extract from a 2-year select-and-ultimate mortality table: 65 - - 8200 67 66 - - 8000 68 67 - - 7700 69The following relationships hold for all :(i) (ii) Calculate .A) 7940 B) 8000 C) 8060 D) 8130 E) 8200

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19. For a whole life insurance of 1 on (40), you are given:(i)(ii)(iii)(iv)Calculate ..A) 0.942 B) 0.946 C) 0.950 D) 0.954 E) 0.958

20. For professional athletes Derek and A-Rod:(i) Professional athletes are subject to a constant total force of mortality .(ii) Professional athletes are subject to a constant force of mortality due to crashes of theteam airplane .(iii) Mortality of athletes on the same team follows a common shock model, where allteam members die if the team plane crashes.(iv) Future lifetimes of athletes on different teams are independent.(v) Derek and A-Rod are on the same team now, but after one year will play for differentteams.Calculate the probability that both Derek and A-Rod survive two years.A) 0.9958 B) 0.9960 C) 0.9962 D) 0.9964 E) 0.9966

21. You are given the following information about a new model for buildings with limitingage(i) The expected number of buildings surviving at age x will be , .(ii) The new model predicts a % higher complete life expectancy (over the previousDeMoivre model with the same ) for buildings aged 30.(iii) The complete life expectancy for buildings aged 60 under the new model is 20 years.Calculate the complete life expectancy under the previous DeMoivre model for buildingsaged 70.A) 8 B) 10 C) 12 D) 14 E) 16

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22. For a special whole life insurance on (40), you are given:(i) The death benefit is 1000 for the first 10 years and 2500 thereafter.(ii) Death benefits are payable at the moment of death.(iii) is the present-value random variable.(iv) Mortality follows DeMoivre’s law with = 100 .(v)Calculate .A) 0.059 B) 0.079 C) 0.105 D) 0.169 E) 0.212

23. Assuming constant forces of decrement in each year of age, you are given:(i) (ii) (iii) (iv) Calculate for a double decrement table.A) 0.169 B) 0.172 C) 0.175 D) 0.178 E) 0.181

24. For a three-year temporary life annuity due of 100 on (75), you are given:(i)(ii)Calculate the actuarial present value of this annuity.A) 264 B) 266 C) 268 D) 270 E) 272

25. Subway trains arrive at a certain station according to a nonhomogeneous Poisson process., the intensity function (trains per minute), varies with , the time in minutes after 7:00 AM:

(i)(ii)(iii)Calculate the probability that exactly four trains arrive between 7:00 AM and 7:25 AM.A) 0.05 B) 0.07 C) 0.09 D) 0.11 E) 0.13

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26. A certain scientific theory supposes that mistakes in cell division occur according to aPoisson process with rate 4 per day, and that a specimen fails at the time of the 289th suchmistake. This theory explains the only cause of failure. T is the time-of-failure random variablein days for a newborn specimen. Using the normal approximation, calculate the probabilitythat .A) 0.84 B) 0.86 C) 0.88 D) 0.90 E) 0.92

27. For a special whole life insurance, you are given:(i)(ii) is constant.(iii)(iv) , where is the future lifetime random variable.(v)Calculate .A) 0.017 B) 0.021 C) 0.025 D) 0.029 E) 0.033

28. You are the pricing actuary reviewing cash values on fully discrete whole life insurances of10,000 on (40). A desired asset share pattern has been chosen. You are to determine cash valuesthat will produce those asset shares. You are given:(i) The gross or contract premium is 90.(ii) Renewal expenses, payable at the start of the year, are 5% of premium.(iii) (iv) (v)death withdrawal

(vi) and are the asset shares at the ends of years 15 and 16.Calculate , the cash value payable upon withdrawal at the end of year 16.A) 810 B) 860 C) 910 D) 960 E) 1010

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29. For a special fully discrete, 30-year deferred, annual life annuity-due of 200 on (30), you aregiven:(i) The single benefit premium is refunded without interest at the end of the year of deathif death occurs during the deferral period.(ii) Mortality follows the Illustrative Life Table.(iii)Calculate the single benefit premium for this annuity.A) 350 B) 360 C) 370 D) 380 E) 390

30. For a special fully continuous whole life insurance of 1 on , you are given:(i) Mortality follows a double decrement model.(ii) The death benefit for death due to cause 1 is 3.(iii) The death benefit for death due to cause 2 is 1.(iv)(v)(vi) The force of interest, , is a positive constant.Calculate the benefit premium for this insurance.A) 0.07 B) 0.08 C) 0.09 D) 0.10 E) 0.11

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MAY 2007 SOA EXAM MLC SOLUTIONS

1. . .

. Answer: E

2.

Since the force of mortality is constant at , we have and .

Therefore, from , we get ,

and then .

. Answer: B

3. (since we are past the select period of 3 years, the insurance annuityreverts to ultimate values). We can find from

.

From the Illustrative Table we have and , so that the reserve forface amount 1 is .Multiplying by 1000 gives the reserve for the face amount 1000. Answer: D

4. Since this is a fully discrete whole life insurance, for face amount 1, the variance of is , and the standard deviation is .

For face amount 150,000, the standard deviation is scaled up by a factor of to . Answer: E

5. The exponential interarrival times with mean time between arrivals is equivalent to arrivalsfollowing a Poisson process with a mean of per unit time. We are given that the average

interarrival time is 1 month, so the average number of arrivals per month is 1. Because of theindependence of arrivals in disjoint intervals of time, the fact that there have been no arrivals bythe end of January has no effect on how many arrivals will occur in February and March. Thenumber of arrivals in Feb. and Mar. is Poisson with a mean of 2. The probability of at least 3arrivals in Feb. and Mar. is the complement of the probability of at most 2 arrivals. This is

. Answer: C

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6. The units donated and the units withdrawn are independent of one another. The units donatedfollows a compound Poisson process and so do the units withdrawn. The mean of a compoundPoisson distribution is and the variance is , where is thePoisson mean, and is the amount of an individual deposit (or withdrawal for the withdrawalprocess). For the deposits in one week, has a mean of (since 80% of foodbank visitors make a deposit) and has mean 15 and variance 75. For the withdrawals in oneweek, has a mean of (since 20% of food bank visitors make a withdrawal)and has mean 40 and variance 533. The expected amount deposited in one week is

and the variance of the amount deposited is

(since ).Similarly, the expected amount withdrawn in one week is

and the variance of the amount withdrawn is.

The net amount deposit in the week is , which has a mean of and a variance of (because of independence of

and ). The probability that the amount of food units at the end of 7 days will be at least600 more than at the beginning of the week is .Using the normal approximation, this

.

Answer: A

7. The earlier premium is paid, the higher the reserve will be. This can be seen retrospectively,since the accumulated cost of insurance is the same in all cases (level benefit of 1000), so thereserves differ because of different premium payment patterns. Earlier premium payment resultsin greater accumulation to time 5. Pattern E has the most premium paid earliest. E has the sametotal in the first 3 years as A and C and the same premium in years 4 and 5, so E's accumulatedpremium will be greater than that of A and C. The difference between E and D is that E haspremium of 1 more than D in the first year and 1 less than D in the 3rd year, but D has one morethan E in the 4th year and 1 less than E in the fifth year. Since E's excess differential with Doccurs earlier (years 1 and 3, vs years 4 and 5), the accumulation of E's premium is greater thanthat of D. From the diagram, it can be seen that D's accumulated premium is greater than thatof B. Answer: E

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8. The expected number of points that Kira will score isProb. that Kira gets to play Expected number of points Kira scores given that she starts to playIf Kira gets to play, the expected time until she will be called is hours.

The expected number of points she would score in that time is .The probability that Kira will get to play is the probability that Kevin gets called first. This is

, where is Kevin and is Kira. This probability is

.

The expected number of points Kira will score before she leaves is . Answer: E

9. We first find , the decrement probability for the continuous decrement. .

The last inequality follows from UDD in associated single tables for decrement 1. for , since decrement 2 does not occur until time .

Then for , because of decrement 2 occurs at time and no more of decrement 2 occurs until time Then for , because the rest of decrement 2 occurs attime .Then .We know that for a 2-decrement model

,and we also know that , so that ,from which we get . Answer: E

10. We are given that , where "08" refers to valuation at the start of 2008.We wish to find . Using the recursive insurance relationship,

, so we need to find .Again using the recursive relationship, we have ,so if we can find then we can get , and then get .

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10. continuedApplying the recursive relationship to , we get

, so that ,and we get . Since the interest remains at 6% for 2009 and thereafter, it followsthat is the same as , so .10

Then .Finally, . Answer: C

11. .

Using the relationship , from the Illustrative Table, we get

, so that .

Using the relationship , from the Illustrative Table, we get

, so that .

We also use , so from the Illustrative Table we get

, so that .

Then, and .

Substituting these values into the original equation results in , and solving for results in

. Answer: A

12. Given that (means that (55) is still alive at age 56) is a 5-point randomvariable as of age 56. The 5 possible values for are

if death is acc. at age 56, prob. if death is not acc. at age 56, prob. if death is acc. at age 57, prob. if death is not acc. at age 57, prob. 2

if (56) survives to age 58, prob. From this table, we see that (only on survival toage 58),and (still not ),and . Answer: D

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13. We can use the recursive relationship

to find .Since the policy terminates at time 3, .Then

(since and ).Then,

. Answer: C

14. In order for (30) to die second and within 5 years the death of (35) it must be true that (35)dies first and (30) dies within 5 years after that. This probability is .The integral is set up based on the density of (35)'s death at time , and (30) being alive at thetime but dying in the next 5 years. We can write in the form

The integral becomes

.The first integral is the probability that 35 will die before 30, which is .The second integral is , because it is the probability that one of two people of equal age 35 willbe the first to die. We are also given .The overall probability is . Answer: E

15. The probability of transform from state 1 to state 1 in the first year is .3 .The probability of transferring from state 1 to state 2 in the second year is

.The probability of transferring from state 1 to state 2 in the third year is

2 [ (we can see from that since , the only wayto still be in state 1 at the start of the third year is to stay in state 1 from year 1 to year 2 and thestay from year 2 to year 3).The actuarial present value of the payments made because of transfer from state 1 to state 2 is

.

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15. continuedThe fee is paid if in state 1. The fee will be paid at the star of the first year.The fee will be paid with probability at the start of the second year.The fee will be paid with probability at the start of the third year.The APV of fees is .According to the equivalence principle, we have , so .Answer: D

16. In order for Tom to find at least 3 coins in the next two blocks, Tom must find either1,2 or 2,1 or 2,2 in the next two blocks. The probabilities of these combinations are

, , and . The total probability is .

Answer: B

17. The probability of getting 100 at the end of the first year is .8.The probability of getting 100 at the end of the second year is(these are the 2 combinations of and , where and denote the events of noaccident and accident).The probability of getting 100 at the end of the third year is

.(these are the probabilities of the combinations of ).The actuarial present value of the payments is .

The probability of getting at the end of 3 years is . The actuarial present value ofthat payment is . The two choices are actuarially equivalent if they havethe same actuarial present value. Solving for from results in .Answer: D

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18. From the ultimate column, we have , so that .Then from (ii), we get , so that ,from which we get , and then .Then from , we get .

We continue in a similar way. From (ii) again we get .But from the ultimate table, we have .Then, so that .Then from (i), we get , so that ,

and . Since ,

we get . Answer: C

19. For fully discrete whole life reserves, we have .

and .

Since , it follows that .Using the relationship , we get

, where is the common value of , and .By trial and error we try each of the possible values of . The value satisfies theequation. Answer: D

20. In order for Derek and A-Rod to survive two years, they must both survive the first year inwhich the are subject to a single common shock, and they must both survive the second year asindependent lives. The probability is , where is the probability that they bothsurvive the first year, is the probability that Derek survives the 2nd year, and is theprobability that A-Rod survives the 2nd year. and are both (they are each subject tothe total force of mortality of .001). is (this is the probability that Derek doesnot die to causes other than the common shock, and A-Rod also doesn't die to causes other thancommon shock, and the common shock doesn't occur in the first year). Derek' force of mortalitydue to causes other than common shock is , and same for A-Rod.Therefore, . The total probability we are looking for is

. Answer: C

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21. The original model is a DeMoivre model. Survival under the new model is based on aGeneralized DeMoivre model. The new model has a new , but the same .(ii) tells us that . It follows that .

(iii) tells us that , so that , and .

Under the original DeMoivre model, we have Answer: B°

22. if death occurs with years, and if . if , or equivalently, .

if , or equivalently, , but in this case, wemust also have (for the benefit to be 2500).The total probability is the combination of and .This is . Answer: C

23.For a 2 decrement table, .We are given , so that .From constant force , we get , so that .From we get , and then from constant force of decrement we get

, and .

Then, also from constant force of decrement, .

From constant force ,we have . so that ,and then .Finally, . Answer: B

24. .

From (i) we get .Therefore, and

.The APV of the annuity is . Answer: A

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25. The expected number of trains that will arrive between 7:00 AM and 7:25AM. is .

The probability of exactly four trains in the time interval is the Poisson probability

. Answer: B

26. For a Poisson process with rate per unit time, the time of the -th event, , has a gammadistribution with mean and variance . In this problem, and , so hasmean and variance . Applying the normal approximation to ,we get . Answer: A

27. .

Since , it follows that .Then, .

, Answer: A

28. The recursive relationship for assets shares is

.

Using this, we have

,which becomes

.Solving for results in . Answer: C

29. The equivalence principle equation is .

. , so that

.

Solving for results in . Answer: A

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30. The benefit premium is . The equivalence principle equation is .

Since and , it follows that ,and . The equation becomes

which becomes . Answer: D

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TEXTBOOK REFERENCES FOR SOA EXAM MLC QUESTIONS

QUESTION NUMBERSM07

ACTUARIAL MATHEMATICS (Bowers, etc.)

Chapter 3 1,18,21

Chapter 4 10,22,27,29

Chapter 5 2,24

Chapter 6 4

Chapter 7 3,7

Chapter 8 12,19

Chapter 9 14,20

Chapter 10-11 9,23,30

Chapter 15 11,28

PROBABILITY

Multi-State Transition 15,16,17

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