Contents. Preface P-5. Syllabus Reference P-7. Flow Chart P-10

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1 Preface P-1 Conens Preface P-5 Syllabus Reference P-7 Flow Char P-10 Chaer 1 Survival Disribuions C Age-a-deah Random Variables C Fuure Lifeime Random Variable C Acuarial Noaion C Curae Fuure Lifeime Random Variable C Force of Moraliy C1-13 Exercise 1 C1-18 Soluions o Exercise 1 C1-23 Chaer 2 Life Tables C Life Table Funcions C Fracional Age Assumions C Selec-and-Ulimae Tables C Momens of Fuure Lifeime Random Variables C Useful Shorcus C2-26 Exercise 2 C2-28 Soluions o Exercise 2 C2-35 Chaer 3 Life Insurances C Coninuous Life Insurances C Discree Life Insurances C mhly Life Insurances C Relaing Differen Policies C Recursions C Relaing Coninuous, Discree and mhly Insurance C Useful Shorcus C3-38 Exercise 3 C3-40 Soluions o Exercise 3 C3-49 Chaer 4 Life Annuiies C Coninuous Life Annuiies C Discree Life Annuiies (Due) C Discree Life Annuiies (Immediae) C mhly Life Annuiies C4-25 Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

2 P-2 Preface 4.5 Relaing Differen Policies C Recursions C Relaing Coninuous, Discree and mhly Life Annuiies C Useful Shorcus C4-38 Exercise 4 C4-39 Soluions o Exercise 4 C4-48 Chaer 5 Premium Calculaion C Tradiional Insurance Policies C Benefi Premium and Equivalence Princile C Benefi Premiums for Secial Policies C The Loss-a-issue Random Variable C Percenile Premium C5-26 Exercise 5 C5-30 Soluions o Exercise 5 C5-44 Chaer 6 Benefi Reserves C The Prosecive Aroach C The Recursive Aroach: Basic Idea C The Recursive Aroach: Furher Alicaions C The Rerosecive Aroach C6-36 Exercise 6 C6-43 Soluions o Exercise 6 C6-62 Chaer 7 Insurance Models Including Exenses C Gross Premium C Gross Premium Reserve C Exense Reserve and FPT Reserve C Basis, Asse Share and Profi C7-19 Exercise 7 C7-31 Soluions o Exercise 7 C7-44 Chaer 8 Mulile Decremen Models: Theory C Mulile Decremen Table C Forces of Decremen C Associaed Single Decremen C Discree Jums C8-23 Exercise 8 C8-28 Soluions o Exercise 8 C8-39 Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

3 Preface P-3 Chaer 9 Mulile Decremen Models: Alicaions C Calculaing Acuarial Presen Values of Cash Flows C Cash Values C Calculaing Asse Share under Mulile Decremen C9-17 Exercise 9 C9-22 Soluions o Exercise 9 C9-32 Chaer 10 Mulile Sae Models C Discree-ime Markov Chain C Coninuous-ime Markov Chain C Kolmogorov s Forward Equaions C Calculaing Acuarial Presen Value of Cash Flows C Calculaing Reserves C10-36 Exercise 10 C10-41 Soluions o Exercise 10 C10-52 Chaer 11 Mulile Life Funcions C Mulile Life Sauses C Insurances and Annuiies C Deenden Life Models C11-29 Exercise 11 C11-36 Soluions o Exercise 11 C11-53 Chaer 12 Ineres Rae Risk C Yield Curves C Ineres Rae Scenario Models C Diversifiable and Non-Diversifiable Risks C12-12 Exercise 12 C12-17 Soluions o Exercise 12 C12-25 Chaer 13 Profi Tesing C Profi Vecor and Profi Signaure C Profi Measures C13-8 Exercise 13 C13-12 Soluions o Exercise 13 C13-15 Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

4 P-4 Preface Chaer 14 Universal Life Insurance C Basic Policy Design C Cos of Insurance and Surrender Value C Oher Policy Feaures C Projecing Accoun Values C Profi Tesing C14-22 Exercise 14 C14-27 Soluions o Exercise 14 C14-37 Chaer 15 Pension Mahemaics C The Salary Scale Funcion C Pension Plans C Seing he DC Conribuion rae C15-10 Exercise 15 C1-16 Soluions o Exercise 15 C15-19 Chaer 0 Numerical Techniques C Numerical Inegraion C Euler s Mehod C0-7 Exercise 1 C1-18 Soluions o Exercise 1 C1-23 Mock Tes 1 T1-1 Soluion T1-18 Mock Tes 2 T2-1 Soluion T2-18 Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

5 Preface P-5 Preface Thank you for choosing ACTEX. Saring in Sring 2012, Exam MLC will be based on a new syllabus, which is significanly differen from he old one. Two new exbooks, Acuarial Mahemaics for Life Coningen Risks (AMLCR) and Models for Quanifying Risk (MQR) are adoed. A number of new oics, for examle, ineres rae risk and universal life insurance, are added o he syllabus. Because of hese major changes, ACTEX decided o bring you his sudy manual, which is wrien anew (raher han develoed from old suff) o fi he curren syllabus. When you ook oher acuarial exams, you migh rely on only one of he alernaive exbooks. This aroach is no going o work for Exam MLC, because no all oics in he syllabus are covered a he same level in each exbook. The exam syllabus says: No all oics lised in he learning objecives are covered a he same level in each source (exbook). The sources differ in heir exosiion, examles, and exercises. The selecion of examinaion quesions is based on coverage of he learning objecives bu he selecion of he examinaion quesions is no necessarily neural wih resec o he sources. Tha is, he sources may no be equivalen in heir coverage of all he secific iems included on an examinaion quesion. For insance, discree-ime Markov chains, which are examined in he new Exam MLC Samle Quesions (#152), are discussed only in MQR. In conras, rofi measures and gain by source, which are also examined in he samle quesions (#300), are menioned only in AMLCR. Using a good sudy manual is herefore crucially imoran for Exam MLC. In his sudy manual, we have fully inegraed he srenghs of he wo exbooks, making sure ha all oics in he syllabus are exlained and raciced in sufficien deh. For your reference, a deailed maing beween his sudy guide and he exbooks are rovided on ages P-7 o P-9. We have aid secial aenion o he newly inroduced oics. Five full-lengh (u o 186 ages) chaers are esecially devoed o hese oics. Insead of reaing he new oics as orhans, we demonsrae, as far as ossible, how hey can be relaed o he old oics in an exam seing. This is very imoran for you, because, as saed exlicily in he exam syllabus, quesions may cover mulile learning oucomes. We oally undersand ha you may be anic, because no as exam quesions are based on he new oics. To hel you bes reare for he exam, we have incororaed all Exam MLC samle quesions ha are relaed o he new oics ino he manual. On o of ha, leny of original exam-ye quesions on he new oics are rovided for you o racice. Besides he oics secified in he exam syllabus, you also need o know a range of numerical echniques in order o succeed. These echniques include, for examle, Euler s mehod, which is involved in Exam MLC Samle Quesion #299. We know ha quie a few of you have no even heard of Euler s mehod before, so we have reared a secial chaer (C0, aended o he end of he sudy manual) o each you all numerical echniques required for his exam. Whenever a Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

6 P-6 Preface numerical echnique is used, we clearly oin ou which echnique i is, leing you follow our examles and exercises more easily. Oher disinguishing feaures of his sudy manual include: We use grahics exensively. Grahical illusraions are robably he mos effecive way o exlain formulas involved in Exam MLC. The exensive use of grahics can also hel you remember various conces and equaions. A sleek layou is used. The fon size and sacing are chosen o le you feel more comforable in reading. Imoran equaions are dislayed in eye-caching boxes. Raher han sliing he manual ino iny unis, each of which ells you a coule of formulas only, we have carefully groued he exam oics ino 15 chaers. Such a grouing allows you o more easily idenify he linkages beween differen conces, which, as we menioned earlier, are essenial for your success. Insead of giving you a long lis of formulas, we oin ou which formulas are he mos imoran. Having read his sudy manual, you will be able o idenify he formulas you mus remember and he formulas ha are jus varians of he key ones. We do no wan o overwhelm you wih verbose exlanaions. Whenever ossible, conces and echniques are demonsraed wih examles and inegraed ino he racice roblems. We wrie he racice roblems and he mock exams in a similar forma as he released exam and samle quesions. This will enable you o comrehend quesions more quickly in he real exam. On age P-10, you will find a flow char showing how differen chaers of his manual are conneced o one anoher. You should firs sudy Chaers 1 o 10 in order. Chaers 1 o 4 will build you a solid foundaion, while he Chaers 5 o 10 will ge you o he core of he exam. You should hen sudy Chaers 11 o 15 in any order you wish. Immediaely afer reading a chaer, do he racice roblems we rovide for ha chaer. Make sure ha you undersand every single racice roblem. Finally, work on he mock exams. Before you begin your sudy, lease download he exam syllabus from he following hyerlink: h:// On age 2 of he df file, you will find a link o Exam MLC Tables, which are frequenly used in he exam. You should kee a coy of he ables, as we are going o refer o hem from ime o ime. On he same age, you will also find links o Exam MLC Samle Quesions and he sulemenary noes for AMLCR. If you find a ossible error in his manual, lease le us know a he Cusomer Feedback link on he ACTEX homeage ( Any confirmed erraa will be osed on he ACTEX websie under he Erraa & Udaes link. Enjoy your sudy! Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

7 Preface P-7 Syllabus Reference Our Manual AMLCR + Sulemen MQR Chaer 1: Survival Disribuions , , , SN Chaer 2: Life Tables , , , 3.8, , , Chaer 3: Life Insurances , 4.4.5, 4.4.7, , , 4.4.6, 4.4.7, , 7.2, , , 7.5.1, Chaer 4: Life Annuiies , 8.2.3, 8.3.3, , 5.4.2, 5.9, , 8.2.2, 8.3.2, , , 8.2.1, 8.3.1, Scaered in Chaer , Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

8 P-8 Preface Our Manual AMLCR + Sulemen MQR Chaer 5: Premium Calculaion , , 9.2, 9.4, Chaer 6: Benefi Reserves , 7.3.1, , , , , , , , SN , Chaer 7: Insurance Models Including Exenses , , SN , 7.3.4, Chaer 8: Mulile Decremen Models: Theory 8.1 SN , SN , 13.2, , SN3.3, SN Chaer 9: Mulile Decremen Models: Alicaions Chaer 10: Mulile Sae Models , 7.6.1, , , 5.5, , , 7.6.2, 8.6, Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

9 Preface P-9 Our Manual AMLCR + Sulemen MQR Chaer 11: Mulile Life Funcions , 12.2, 12.3, , , 12.7 Chaer 12: Ineres Rae Risk , 15.3, , 10.4 Chaer 13: Profi Tesing , 11.2, Chaer 14: Universal Life Insurances 14.1 SN , SN4.3, SN SN SN4.2, SN SN Examle Chaer 15: Pension Mahemaics , Chaer 0: Numerical Techniques Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

10 P-10 Preface Flow Char 1. Survival Disribuions 2. Life Tables 15. Pension Mahemaics 3. Life Insurances 5. Premium Calculaion 12. Ineres Rae Risk 4. Life Annuiies 6. Benefi Reserves 0. Numerical Techniques 7. Insurance Models Including Exenses 8. Mulile Decremen Models: Theory 9. Mulile Decremen Models: Alicaions 10. Mulile Sae Models 13. Profi Tesing 11. Mulile Life Funcions 14. Universal Life Insurance Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

11 Samle Pages from Chaer 10 Chaer 10 conains 72 ages. We are showing ages 4 o 10.

12 C10-4 Chaer 10: Mulile Sae Models The Disabiliy Income Model A mulile sae model models he ransiion raes beween saes. Transiion o any sae can be reversible or irreversible. Consider he following disabiliy income model: 0. Healhy 1. Disabled 2. Dead Model 3: The Disabiliy Income Model In his model, he insured can swich beween sae 0 and sae 1. However, ransiion o sae 2 is irreversible. We can change he above o a ermanen disabiliy model by disallowing ransiion from sae 1 back o sae 0. In his chaer we wan o analyze models wih srucures similar o Model 3. We shall firs discuss a simle version, where sae changes can only occurs a he end of each eriod. Then we shall discuss he more general case when sae changes can only occur any ime Discree-ime Markov Chain In his secion we discuss discree-ime Markov chain, a model for which ransiions can only occur a he end of each eriod. Such a model is useful in modeling healh saus, bonus malus sysem and credi raing in grou healh and non-life insurances. Le he sae a ime be Y. Le he sae sace, which is he se of all ossible saes, be E. For examle, in he disabiliy income model above, we have E {0, 1, 2} and Y 0 0. As ime rogresses, Y swiches beween 0 and 1 and finally ge sruck in sae 2. The following figure shows one of he ossible ahs of Y. Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

13 Chaer 10: Mulile Sae Models C10-5 Y ime Transiion Probabiliy Marix and he Markov Proery For a discree-ime Markov chain, we assume ha when he rocess is in sae i a ime, he robabiliy ha i would be in sae j a ime + 1 is ij. Tha is, Pr(Y +1 j Y i) cusomary o lace all robabiliies ino a ransiion robabiliy marix: P M n M n1 P 0 P 1 P 2 P 1 P M n2 L L L O L 0n 1n 2n M nn ij. I is Y 0 Y 1 Y 2 Y 3 Y 1 Y Y +1 The sum of he robabiliies on each row is 1. Moreover, we assume for any 0 and i, j E. The Markov Proery Pr(Y +1 j Y i, Y 1 i 1, Y 2 i 2, ) Pr(Y +1 j Y i) This above says ha he condiional disribuion of any fuure sae Y +1, given he as saes {Y 0, Y 1,, Y 1 } and he resen sae Y, deends only on he resen sae Y bu no he as saes. Loosely seaking, Markov means given he resen, you can forge abou he as! If P is consan wih, he Markov chain is said o be homogeneous. If P is ime-deenden, he Markov chain is said o be inhomogeneous. Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

14 C10-6 Chaer 10: Mulile Sae Models In many acuarial alicaions, here would be one or more saes ha canno be lef once i is enered (examles include wihdrawal, deah, and bankrucy). If you look a he following 3- sae ransiion robabiliy marix: hen sae 2 canno be lef once i is enered. Such a sae is called an absorbing sae. For an absorbing sae, all elemens in he row, exce he one in he main diagonal, are 0. The one in he main diagonal is 1. Examle 10.1 [Exam M 2005 Nov #4] Kevin and Kira are modeling he fuure lifeime of (60). (i) Kevin uses a double decremen model: (τ ) l x Age (ii) Kira uses a non-homogeneous Markov model: (a) The saes are 0 (alive), 1 (deah due o cause 1), 2 (deah due o cause 2). (b) P 60 is he ransiion marix from age 60 o 61; P 61 is he ransiion marix from age 61 o 62. (iii) The wo models roduce equal robabiliies of decremen. Calculae P (A) (B) (C) (1) d x (2) d x (D) (E) The main diagonal of a square marix is he diagonal which runs from he o lef corner o he boom righ corner. Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

15 Chaer 10: Mulile Sae Models C10-7 Soluion Obviously, boh saes 1 and 2 are absorbing. This means (B) mus be wrong. For a life age (61), he robabiliy ha he/she would sill be alive afer one year is ( τ ) 560 / The robabiliy ha he/she would die due o cause 1 wihin he year is This means (D) is he correc answer. (1) q 160 / [ END ] The following is anoher acuarial alicaion of homogeneous Markov chain. An insurance comany classifies is insureds based on each insured s credi raing, as one of Preferred, Sandard or Poor. Individual ransiion beween classes is modeled by he following ransiion marix: Preferred Sandard Poor Preferred Sandard Poor Very frequenly we are no only ineresed in ransiion robabiliies for a single eriod. For examle, we may wan o find he robabiliy ha an insured who is Sandard a ime 0 would end u in Poor afer 2 years. This is called a 2-se ransiion robabiliy. One way o calculae such robabiliies is o lis ou all ossible ahs: Pah Probabiliy Sandard Preferred Poor Sandard Sandard Poor Sandard Poor Poor Sum Similarly, we can calculae he robabiliy ha an insured who is Sandard a ime 0 would end u in Preferred afer 2 years: Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

16 C10-8 Chaer 10: Mulile Sae Models Pah Probabiliy Sandard Preferred Preferred Sandard Sandard Preferred Sandard Poor Preferred Sum Wha is he robabiliy for he insured o be in he sae of Sandard afer 2 years? I is simly Acually here is a very efficien way o calculae such robabiliies if you know some elemenary marix algebra: In case you have forgoen abou marix mulilicaion, read he following shor examle: If A, B , hen 4 3 (1)(2) + (0)( 1) + (3)( 4) (1)(1) + (0)( 2) + (3)(3) AB. ( 1)(2) + (7)( 1) + ( 5)( 4) ( 1)(1) + (7)( 2) + ( 5)(3) In his chaer we would frequenly calculae he roduc of wo square marices. Remember ha for wo square marices A and B, he roducs AB and BA can be differen! Now le us look a Wha can you find from he marix on he righ? The 2-se ransiion robabiliies are all in he second row of he marix roduc P P P 2. If we wan o find Pr(Y +2 j Y i) ij 2, hen we can simly look a he ij-h elemen of he marix P 2. So, P 2 is he 2-se ransiion marix. More generally, if we wan o find ij Pr(Y x+s j Y x i) s x, we only need o look u he ij-h elemen in s P x P x P x+1... P x+s 1. Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

17 Chaer 10: Mulile Sae Models C10-9 sp x P x P x+1 P x+2 P x+s 1 P x+s This resul can be hough of as a generalizaion of he relaion The above imlies x x + 1 x + 2 x + 3 x + s 1 x + s x + s + 1 Y x Y x+1 Y x+2 Y x+3 Y x+s 1 Y x+s Y x+s+1 s K. x x x+ 1 x+ 2 x+ s 1 The Chaman-Kolmogorov (CK) Equaion + P P P s x x s x+ F O R M U L A The order of mulilicaion on he righ hand side of he CK equaion maers! Examle 10.2 [Samle #151] For a mulile sae model wih hree saes, Healhy (0), Disabled (1), and Dead (2): (i) For k 0, 1: , 0. 2, 0. 1, x+k x+k (ii) There are 100 lives a he sar, all Healhy. Their fuure saes are indeenden. Calculae he variance of he number of he original 100 lives who die wihin he firs wo years. (A) 11 (B) 14 (C) 17 (D) 20 (E) 23 x+k x+k Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

18 C10-10 Chaer 10: Mulile Sae Models Soluion We are given he following informaion: ? P x P x ? 0.25.??? Since sae 2 is an absorbing sae, he las row mus be [0 0 1]. By using he roery ha each row of a ransiion robabiliy marix sums o 1, we ge: P x P x By he CK equaion, 2 Px PxPx ?? ???, where For a single life, he robabiliy ha he would die wihin he firs wo years is If here are 100 indeenden lives, we le 2 D 0 as he number of deahs wihin he firs wo years. Then 2D 0 ~ B(100, 0.22) (B(n, ) here means a binomial disribuion wih number of rials n and robabiliy of success ) and hence he variance is Var( 2 D 0 ) (1 0.22) 17. So he correc answer is (C). [ END ] Calculaing APV of Cash Flows There are mainly wo kinds of cash flows associaed wih discree-ime Markov chain: (1) Cash flow arising from being in a aricular sae. (2) Cash flow arising from a ransiion from one sae o anoher. I is easy o calculae APV of cash flows under a discree-ime Markov chain model. Sudy he following examle: Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

19 Samle Pages from Chaer 14 Chaer 14 conains 46 ages. We are showing ages 2 o 11.

20 C14-2 Chaer 14: Universal Life Insurance Any ineres ha he accoun value earns is credied back o he accoun value and invesed, so essenially, he olicyholder s money is earning money. You may herefore view a universal life olicy as a mixure of life insurance and an invesmen roduc. Le us begin wih a descriion of how he accoun value is accumulaed Basic Policy Design We define he following noaion: P Premium for he h ime eriod, aid a ime 1 EC The exense charge (also called MER, for Managemen Exense Rae) for he h ime eriod, deduced from he accoun value a ime 1 CoI The cos of insurance for he h ime eriod, deduced from he accoun value a ime 1 AV The accoun value a ime, before remium and deducions c i The credied rae of ineres for he h ime eriod (i.e., from ime 1 o ime ) The accumulaion of he accoun value over he h ime eriod (i.e., from ime 1 o ime ) is illusraed in he following diagram: Exense Charge (EC ) Cos of Insurance (CoI ) Deduc Accoun Value (AV 1 ) Accoun Value (AV ) Add Premium (P ) Accumulae a he credied rae of ineres i ) ( c Time 1 Time Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

21 Chaer 14: Universal Life Insurance C14-3 Kee in mind ha he remium (P), exense charge (EC) and he cos of insurance (CoI) are for he h ime eriod and are deduced or aid a ime 1.Assuming he olicy is sill in force a ime, we have he following relaion: Accoun Value Accumulaion F O R M U L A AV (AV 1 + P EC CoI )(1 + i c ) Noe: For a basic universal life olicy, here is no searae invesmen accoun. Policyholders funds are merged o he insurer s general funds. The credied ineres rae is declared by he insurer and is based on he invesmen erformance on he insurer s asses. The amoun of remium P is a he olicyholder s discreion. The exense charge EC is deermined by he insurer. The lengh of each ime se deends on he frequency of remium aymens. In he exam, he accumulaion of accoun value may be calculaed a yearly or monhly ime ses. Examle 14.1 For a basic universal life olicy, you are given: (i) The accoun value a 10 (before remium aymen and deducions) is $10,000. (ii) The olicyholder ays remiums a 10 and 11. Each remium is $500. (iii) The exense charge is 1% of each remium. (iv) The coss of insurance deduced a 10 and 11 are $20 and $15, resecively. (v) The credied ineres rae over he eriod from 10 o 12 is 2% er annum effecive. Assume he olicy is sill in force a 12. (a) Calculae he accoun value a 12 if he lengh of each ime se is one year. (b) Calculae he accoun value a 12 if he lengh of each ime se is one monh. Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

22 C14-4 Chaer 14: Universal Life Insurance Soluion (a) Here, he lengh of each ime se is one year. A 11, he accoun value is given by c AV 11 (AV 10 + P 11 EC 11 CoI 11 )(1 + i 11 ) ( )( ) A 12, he accoun value is given by c AV 12 (AV 11 + P 12 EC 12 CoI 12 )(1 + i 12 ) ( )( ) (b) Here, he lengh of each ime se is one monh. The credied ineres raes are calculaed as follows: 1+ i 1 + i ( ) c c A 11, he accoun value is given by c AV 11 (AV 10 + P 11 EC 11 CoI 11 )(1 + i 11 ) ( ) ( ) A 12, he accoun value is given by c AV 12 (AV 11 + P 12 EC 12 CoI 12 )(1 + i 12 ) ( ) ( ) [ END ] You may wonder how he cos of insurance is calculaed. We are going o address his issue in he nex secion. Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

23 Chaer 14: Universal Life Insurance C Cos of Insurance and Surrender Value Suose ha a universal life insurance olicy is in force a ime 1. A ime, here are hree ossibiliies: The olicy is in force. The olicy is in force. The olicy is surrendered. A surrender value is aid. The olicyholder has died. A deah benefi is aid Time 1 Time Possibiliy (1): The olicy is sill in force A ime, he olicy can sill be in force. In his case, he rocess described in Secion 14.1 reeas. Possibiliy (2): The olicy is surrendered A ime, he olicyholder can choose o surrender he olicy. In his case, he olicyholder will be aid a surrender value. The oal cash available o he olicyholder on surrender is he accoun value (AV ) minus he surrender charge (SC ), which is deermined by he insurer. This amoun of money is called he cash value (CV ) of he conrac. Mahemaically, CV max(av SC, 0) Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

24 C14-6 Chaer 14: Universal Life Insurance Possibiliy (3): The olicyholder has died Deah may occur beween ime 1 and ime. In his case, a deah benefi mus be aid a ime. The amoun of deah benefi deends on he conrac ye. Tye A: Secified Amoun For a secified amoun (Tye A) conrac, he oal deah benefi is level. The oal deah benefi offered by Tye A conrac is called he Face Amoun (FA) of he olicy. Tye B: Secified Amoun lus he Accoun Value For a secified amoun lus he accoun value (Tye B) conrac, he oal deah benefi is he accoun value lus a fixed amoun, which we denoe by X. For boh yes, he deah benefi is subjec o a corridor facor requiremen, which siulaes he minimum amoun of insurance coverage required by he ax laws o reven he conrac from becoming one of invesmen only (which has a less favorable ax saus). The minimum oal deah benefi is defined by he corridor facor (γ ) imes he accoun value (AV ) a deah. Taking he corridor facor requiremen ino accoun, he oal deah benefi a ime can be exressed as follows. F O R M U L A Toal Deah Benefi Secified Amoun (Tye A): max(fa, γ AV ) Secified Amoun lus he Accoun Value (Tye B): max(av + X, γ AV ) The difference beween he oal deah benefi and he accoun value is known as he Addiional Deah Benefi (ADB). If he corridor facor requiremen is saisfied, hen he ADB for a secified amoun (Tye A) conrac, which equals FA AV, decreases as he accoun value increases, bu he ADB for a secified amoun lus he accoun value (Tye B) conrac is a Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

25 Chaer 14: Universal Life Insurance C14-7 consan X. If he corridor facor requiremen is no saisfied, hen he ADB equals (γ 1)AV for boh yes. Hence, we have he following exressions for he ADB a ime. Addiional Deah Benefi Secified Amoun (Tye A): ADB max(fa AV, (γ 1)AV )) Secified Amoun lus he Accoun Value (Tye B): ADB max(x, (γ 1)AV )) F O R M U L A Examle 14.2 For a secified amoun universal life insurance olicy, you are given: (i) The face amoun of he olicy is $100,000 (ii) The accoun value on December 31, 2011 was $50,000. (iii) On January 1, 2012, a remium of $2000 was made. No oher remiums were made in (iv) The exense charge and he cos of insurance deduced on January 1, 2012 were $150 and $200, resecively. (v) The credied ineres rae in 2012 was 6% er annum effecive. (vi) The corridor facor alicable in 2012 was 2.5. (vii) The surrender charge alicable in 2012 was $10 er $1,000 face amoun. Calculae he following: (a) The cash value of he olicy on December 31, (b) Suose ha deah occurred in 2012 and ha he deah benefi was ayable a he end of he year of deah. Calculae he amoun of deah benefi. Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

26 C14-8 Chaer 14: Universal Life Insurance Soluion (a) Firs, we need o rojec he accoun value on December 31, For convenience, we refer December 31, 2011 o 1 and December 31, 2012 o. We have AV (AV 1 + P EC CoI )(1 + i c ) ( )(1.06) The cash value of he olicy on December 31, 2012 is given by CV max(av SC, 0) max( , 0) (b) This is a Tye A olicy. The oal deah benefi is given by max(fa, γ AV ) max(100000, ) [ END ] Suose ha a he ime of deah, he olicyholder s accoun value is AV. The amoun of AV will cover ar of he oal deah benefi, and he shorfall (i.e., ADB ) will be oed u by he insurer. I is now clear ha he cos of insurance colleced a he beginning of a ime eriod is used o suor he execed ADB over ha ime eriod. In general, he cos of insurance can be calculaed by he following formula. General Formula for he Cos of Insurance CoI * q v q ADB, F O R M U L A where CoI is he cos of insurance for he h ime eriod, deduced from he accoun value a ime 1, * q is he deah robabiliy (for he h ime eriod) used o calculae he cos of insurance, v q is he discoun facor for discouning he cos of insurance o ime 1, ADB is he addiional deah benefi a ime. Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

27 Chaer 14: Universal Life Insurance C14-9 Noe: (1) The credied ineres rae and he ineres rae for calculaing he cos of insurance are no necessarily he same. For his reason, he noaion v q is inroduced. (2) If remiums are aid annually, hen he cos of insurance is jus he single remium for a 1- year erm life insurance wih sum insured equal o he ADB. The acual calculaion of he cos of insurance is a lo more comlicaed han i seems. This is because of he following circular relaionshi: CoI AV ADB The cos of insurance CoI is a funcion of he addiional deah benefi ADB, which is a funcion of he accoun value AV. However, AV deends on CoI, which is deduced a he beginning of he ime eriod. We now discuss how his roblem can be solved. Firs, le us consider a secified amoun (Tye A) olicy. If he corridor requiremen is saisfied, hen ADB FA AV FA (AV 1 + P EC CoI )(1 + i c ). Le CoI f be he cos of insurance when he corridor facor requiremen is saisfied. We have which imlies CoI f * q v q ADB * q v q [FA (AV 1 + P EC CoI f )(1 + i c )], CoI f qv(fa (AV + P EC )(1 + i)). * q 1 c * c 1 qv q(1 + i ) Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

28 C14-10 Chaer 14: Universal Life Insurance If he corridor facor requiremen is no saisfied, hen ADB (γ 1)AV (γ 1) (AV 1 + P EC CoI )(1 + i c ). Le CoI c be he cos of insurance when he corridor facor requiremen is no saisfied. We have + + c * c c CoI qv q( γ 1)(AV 1 P EC CoI )(1 i ) qv(1 + i)( γ 1)(AV + P EC ). * q c 1 * c 1 + qv q(1 + i )( γ 1) Finally, he insurer should charge CoI max( CoI f, CoI c ) for he cos of insurance. Cos of Insurance for a Secified Amoun (Tye A) Policy F O R M U L A where CoI f CoI c CoI max( CoI f, CoI c ) qv(fa (AV + P EC )(1 + i)), * q 1 c * c 1 qv q(1 + i ) qv(1 + i)( γ 1)(AV + P EC ). * q c 1 * c 1 + qv q(1 + i )( γ 1) We now consider a secified amoun lus he accoun value (Tye B) olicy. If he corridor facor requiremen is saisfied, hen he addiional deah benefi is ADB X. The cos of insurance is simly CoI f * q v q X. The same as a Tye A conrac, he addiional deah benefi is ADB (γ 1)AV if he corridor facor requiremen is no saisfied. Hence, for Tye B, olicy, we also have CoI c qv(1 + i)( γ 1)(AV + P EC ). * q c 1 * c 1 + qv q(1 + i )( γ 1) Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC

29 Chaer 14: Universal Life Insurance C14-11 F O R M U L A Cos of Insurance for a Secified Amoun lus Accoun Value (Tye B) Policy CoI max( CoI f, CoI c ) where CoI f * q v q X and CoI c qv(1 + i)( γ 1)(AV + P EC ). * q c 1 * c 1 + qv q(1 + i )( γ 1) Examle 14.3 For a universal life insurance olicy, you are given: (i) The deah benefi is $100,000 lus he accoun value a he end of he year of deah. The benefi is ayable a he end of he year of deah. (ii) The accoun value on December 31, 2011 was $20,000. (iii) On January 1, 2012, a remium of $1,000 was made. No oher remiums were made in (iv) The exense charge deduced on January 1, 2012 was $100. (v) The credied ineres rae in 2012 was 1% er annum effecive. (vi) The surrender charge alicable in 2012 was $10 er $1,000 face amoun. (vii) There is no corridor facor requiremen for his olicy. (viii) In calculaing he cos of insurance, i was assumed ha he robabiliy of deah in 2012 was The ineres rae used was 3% er annum effecive. (iv) The olicyholder is alive on December 31, Calculae he accoun value on December 31, Soluion For convenience, we refer December 31, 2011 o 1 and December 31, 2012 o. The required value, AV, can be calculaed wih he following relaion: Acex 2012 Johnny Li and Andrew Ng SoA Exam MLC