The Experts In Actuarial Career Advancement. Product Preview. For More Information: or call 1(800)

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1 P U B L I C A T I O N S The Eers In Acuarial Career Advancemen Produc Preview For More Informaion: Suor@AceMadRiver.com or call (8) 8-839

2 Preface P- Conens Preface P-7 Syllabus Reference P- Flow Char P-3 Chaer Some Facual Informaion C-. Tradiional Life Insurance Conracs C-. Modern Life Insurance Conracs C-3.3 Underwriing C-3.4 Life Annuiies C-4.5 Pensions C-6 Chaer Survival Disribuions C-. Age-a-deah Random Variables C-. Fuure Lifeime Random Variable C-4.3 Acuarial Noaion C-6.4 Curae Fuure Lifeime Random Variable C-.5 Force of Moraliy C- Eercise C- Soluions o Eercise C-7 Chaer Life Tables C-. Life Table Funcions C-. Fracional Age Assumions C-6.3 Selec-and-Ulimae Tables C-8.4 Momens of Fuure Lifeime Random Variables C-9.5 Useful Shorcus C-39 Eercise C-43 Soluions o Eercise C-5 Chaer 3 Life Insurances C3-3. Coninuous Life Insurances C3-3. Discree Life Insurances C mhly Life Insurances C Relaing Differen Policies C Recursions C Relaing Coninuous, Discree and mhly Insurance C Useful Shorcus C3-46 Eercise 3 C3-48 Soluions o Eercise 3 C3-6 Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

3 P- Preface Chaer 4 Life Annuiies C4-4. Coninuous Life Annuiies C4-4. Discree Life Annuiies (Due) C Discree Life Annuiies (Immediae) C mhly Life Annuiies C Relaing Differen Policies C Recursions C Relaing Coninuous, Discree and mhly Life Annuiies C Useful Shorcus C4-44 Eercise 4 C4-47 Soluions o Eercise 4 C4-6 Chaer 5 Premium Calculaion C5-5. Tradiional Insurance Policies C5-5. Ne Premium and Equivalence Princile C Ne Premiums for Secial Policies C5-5.4 The Loss-a-issue Random Variable C Percenile Premium and Profi C5-7 Eercise 5 C5-38 Soluions o Eercise 5 C5-55 Chaer 6 Ne Premium Reserves C6-6. The Prosecive Aroach C6-6. The Recursive Aroach: Basic Idea C The Recursive Aroach: Furher Alicaions C The Rerosecive Aroach C6-33 Eercise 6 C6-4 Soluions o Eercise 6 C6-65 Chaer 7 Insurance Models Including Eenses C7-7. Gross Premium C7-7. Gross Premium Reserve C Eense Reserve and Modified Reserve C Basis, Asse Share and Profi C7-3 Eercise 7 C7-39 Soluions o Eercise 7 C7-57 Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

4 Preface P-3 Chaer 8 Mulile Decremen Models: Theory C8-8. Mulile Decremen Table C8-8. Forces of Decremen C Associaed Single Decremen C8-8.4 Discree Jums C8-3 Eercise 8 C8-9 Soluions o Eercise 8 C8-4 Chaer 9 Mulile Decremen Models: Alicaions C9-9. Calculaing Acuarial Presen Values of Cash Flows C9-9. Calculaing Reserve and Profi C Cash Values C Calculaing Asse Shares under Mulile Decremen C9- Eercise 9 C9-6 Soluions o Eercise 9 C9-37 Chaer Mulile Sae Models C-. Discree-ime Markov Chain C-4. Coninuous-ime Markov Chain C-3.3 Kolmogorov s Forward Equaions C-9.4 Calculaing Acuarial Presen Value of Cash Flows C-3.5 Calculaing Reserves C-39 Eercise C-44 Soluions o Eercise C-6 Chaer Mulile Life Funcions C-. Mulile Life Sauses C-. Insurances and Annuiies C-7.3 Deenden Life Models C-3 Eercise C-4 Soluions o Eercise C-6 Chaer Ineres Rae Risk C-. Yield Curves C-. Ineres Rae Scenario Models C-3.3 Diversifiable and Non-Diversifiable Risks C-7 Eercise C-4 Soluions o Eercise C-35 Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

5 P-4 Preface Chaer 3 Profi Tesing C3-3. Profi Vecor and Profi Signaure C3-3. Profi Measures C3-3.3 Using Profi Tes o Comue Premiums and Reserves C3-6 Eercise 3 C3-4 Soluions o Eercise 3 C3-3 Chaer 4 Universal Life Insurance C4-4. Basic Policy Design C4-4. Cos of Insurance and Surrender Value C Oher Policy Feaures C Projecing Accoun Values C4-4.5 Profi Tesing C Asse Shares for Universal Life Policies C4-4 Eercise 4 C4-43 Soluions o Eercise 4 C4-53 Chaer 5 Pariciaing Insurance C5-5. Dividends C5-5. Bonuses C5- Eercise 5 C5-4 Soluions o Eercise 5 C5-8 Chaer 6 Pension Mahemaics C6-6. The Salary Scale Funcion C6-6. Pension Plans C6-6.3 Seing he DC Conribuion Rae C DB Plans and Service Table C6-9 Eercise 6 C6-39 Soluions o Eercise 6 C6-49 Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

6 Preface P-5 Aendi Numerical Techniques A-. Numerical Inegraion A-. Euler s Mehod A-7.3 Solving Sysems of ODEs wih Euler s Mehod A- Aendi Review of Probabiliy A-. Probabiliy Laws A-. Random Variables and Eecaions A-.3 Secial Univariae Probabiliy Disribuions A-6.4 Join Disribuion A-9.5 Condiional and Double Eecaion A-.6 The Cenral Limi Theorem A- Eam MLC: General Informaion T- Mock Tes T- Soluion T-9 Mock Tes T- Soluion T-8 Mock Tes 3 T3- Soluion T3-9 Mock Tes 4 T4- Soluion T4-9 Mock Tes 5 T5- Soluion T5-9 Mock Tes 6 T6- Soluion T6-9 Mock Tes 7 T7- Soluion T7-9 Mock Tes 8 T8- Soluion T8-8 Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

7 P-6 Preface Suggesed Soluions o MLC May S- Suggesed Soluions o MLC Nov S-7 Suggesed Soluions o MLC May 3 S-9 Suggesed Soluions o MLC Nov 3 S-45 Suggesed Soluions o MLC Aril 4 S-55 Suggesed Soluions o MLC Oc 4 S-69 Suggesed Soluions o MLC Aril 5 S-8 Suggesed Soluions o Samle Srucural Quesions S-97 Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

8 Chaer : Survival Disribuions C- Chaer Survival Disribuions OBJECTIVES. To define fuure lifeime random variables. To secify survival funcions for fuure lifeime random variables 3. To define acuarial symbols for deah and survival robabiliies and develo relaionshis beween hem 4. To define he force of moraliy In Eam FM, you valued cash flows ha are aid a some known fuure imes. In Eam MLC, by conras, you are going o value cash flows ha are aid a some unknown fuure imes. Secifically, he imings of he cash flows are deenden on he fuure lifeime of he underlying individual. These cash flows are called life coningen cash flows, and he sudy of hese cash flows is called life coningencies. I is obvious ha an imoran ar of life coningencies is he modeling of fuure lifeimes. In his chaer, we are going o sudy how we can model fuure lifeimes as random variables. A few simle robabiliy conces you learn in Eam P will be used.. Age-a-deah Random Variable Le us begin wih he age-a-deah random variable, which is denoed by T. The definiion of T can be easily seen from he diagram below. Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

9 C- Chaer : Survival Disribuions Deah occurs T Age The age-a-deah random variable can ake any value wihin [, ). Someimes, we assume ha no individual can live beyond a cerain very high age. We call ha age he limiing age, and denoe i by ω. If a limiing age is assumed, hen T can only ake a value wihin [, ω]. We regard T as a coninuous random variable, because i can, in rincile, ake any value on he inerval [, ) if here is no limiing age or [, ω] if a limiing age is assumed. Of course, o model T, we need a robabiliy disribuion. The following noaion is used hroughou his sudy guide (and in he eaminaion). F () Pr(T ) is he (cumulaive) disribuion funcion of T. d f () () d F is he robabiliy densiy funcion of T. For a small inerval Δ, he roduc f ()Δ is he (aroimae) robabiliy ha he age a deah is in beween and + Δ. In life coningencies, we ofen need o calculae he robabiliy ha an individual will survive o a cerain age. This moivaes us o define he survival funcion: S () Pr(T > ) F (). Noe ha he subscri indicaes ha hese funcions are secified for he age-a-deah random variable (or equivalenly, he fuure lifeime of a erson age now). No all funcions can be regarded as survival funcions. A survival funcion mus saisfy he following requiremens:. S (). This means every individual can live a leas years.. S (ω) or lim S ( ). This means ha every individual mus die evenually. 3. S () is monoonically decreasing. This means ha, for eamle, he robabiliy of surviving o age 8 canno be greaer han ha of surviving o age 7. Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

10 Chaer : Survival Disribuions C-3 Summing u, f (), F () and S () are relaed o one anoher as follows. F O R M U L A Relaions beween f (), F () and S () f d d () () () d F d S, (.) S ( ) f ( u)du f ( u)du F ( ), (.) b Pr(a < T b) f ( u)du F ( b) F ( a) S ( a) S ( b). (.3) a Noe ha because T is a coninuous random variable, Pr(T c) for any consan c. Now, le us consider he following eamle. Eamle. [Srucural Quesion] You are given ha S () / for. (a) Verify ha S () is a valid survival funcion. (b) Find eressions for F () and f (). (c) Calculae he robabiliy ha T is greaer han 3 and smaller han 6. Soluion (a) Firs, we have S () /. Second, we have S () /. Third, he firs derivaive of S () is /, indicaing ha S () is non-increasing. Hence, S () is a valid survival funcion. (b) We have F () S () /, for. Also, we have and f () d d F () /, for. (c) Pr(3 < T < 6) S (3) S (6) ( 3/) ( 6/).3. [ END ] Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

11 C-4 Chaer : Survival Disribuions. Fuure Lifeime Random Variable Consider an individual who is age now. Throughou his e, we use () o reresen such an individual. Insead of he enire lifeime of (), we are ofen more ineresed in he fuure lifeime of (). We use T o denoe he fuure lifeime random variable for (). The definiion of T can be easily seen from he diagram below. + T Age Now Deah occurs T Time from now [Noe: For breviy, we may only dislay he orion saring from age (i.e., ime ) in fuure illusraions.] If here is no limiing age, T can ake any value wihin [, ). If a limiing age is assumed, hen T can only ake a value wihin [, ω ]. We have o subrac because he individual has aained age a ime already. We le S () be he survival funcion for he fuure lifeime random variable. The subscri here indicaes ha he survival funcion is defined for a life who is age now. I is imoran o undersand ha when modeling he fuure lifeime of (), we always know ha he individual is alive a age. Thus, we may evaluae S () as a condiional robabiliy: S ( ) Pr( T > ) Pr( T > + T > ) Pr( T > + T > ) Pr( T > + ) S( + ). Pr( T > ) Pr( T > ) S ( ) The hird se above follows from he equaion Eam P. Pr( A B) Pr( A B), which you learn in Pr( B) Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

12 Chaer : Survival Disribuions C-5 Survival Funcion for he Fuure Lifeime Random Variable S () S ( + ) S ( ) F O R M U L A (.4) Wih S (), we can obain F () and f () by using resecively. F () S () and f () d F (), d Eamle. [Srucural Quesion] You are given ha S () / for. (a) Find eressions for S (), F () and f (). (b) Calculae he robabiliy ha an individual age now can survive o age 5. (c) Calculae he robabiliy ha an individual age now will die wihin 5 years. Soluion (a) In his ar, we are asked o calculae funcions for an individual age now (i.e., ). Here, ω and herefore hese funcions are defined for 9 only. S( + ) ( + ) / Firs, we have S(), for 9. S () / 9 Second, we have F () S () /9, for 9. Finally, we have f d () F (). d 9 (b) The robabiliy ha an individual age now can survive o age 5 is given by 5 5 Pr(T > 5) S (5). 9 6 (c) The robabiliy ha an individual age now will die wihin 5 years is given by Pr(T 5) F (5) S (5) 6. [ END ] Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

13 Chaer : Life Tables C- Chaer Life Tables OBJECTIVES. To aly life ables. To undersand wo assumions for fracional ages: uniform disribuion of deah and consan force of moraliy 3. To calculae momens for fuure lifeime random variables 4. To undersand and model he effec of selecion Acuaries use sreadshees eensively in racice. I would be very helful if we could eress survival disribuions in a abular form. Such ables, which are known as life ables, are he focus of his chaer.. Life Table Funcions Below is an ecer of a (hyoheical) life able. In wha follows, we are going o define he funcions l and d, and elain how hey are alied. l d Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

14 C- Chaer : Life Tables In his hyoheical life able, he value of l is,. This saring value is called he radi of he life able. For,,., he funcion l sands for he eeced number of ersons who can survive o age. Given an assumed value of l, we can eress any survival funcion S () in a abular form by using he relaion l l S (). In he oher way around, given he life able funcion l, we can easily obain values of S () for inegral values of using he relaion Furhermore, we have l S( ). l + + S(), S( ) l / l l S ( + ) l / l l which means ha we can calculae for all inegral values of and from he life able funcion l. The difference l l + is he eeced number of deahs over he age inerval of [, + ). We denoe his by d. I immediaely follows ha d l l +. We can hen calculae q and m n q by he following wo relaions: q d l l+ l+, l l l d n + m + m + m+ n m n q. l l When, we can omi he subscri and wrie d as d. By definiion, we have l l Grahically, d d + d d +. d + d + + d + + d d + d l l age l l + Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

15 Chaer : Life Tables C-3 Also, when, we have he following relaions: l+ d d l l +,, and q. l l Summing u, wih he life able funcions l and d, we can recover survival robabiliies and deah robabiliies q for all inegral values of and easily. Life Table Funcions l F O R M U L A + (.) l d l l + d + d d + (.) d l l+ l+ q (.3) l l l Eam quesions are ofen based on he Illusraive Life Table, which is, of course, rovided in he eaminaion. To obain a coy of his able, download he mos udaed Eam MLC syllabus. On he las age of he syllabus, you will find a link o Eam MLC Tables, which encomass he Illusraive Life Table. You may also download he able direcly from h:// The Illusraive Life Table conains a lo of informaion. For now, you only need o know and use he firs hree columns:, l, and q. For eamle, o obain he value q 6, simly use he column labeled q. You should obain q , which means q I is also ossible, bu more edious, o calculae q 6 using he column labeled l ; we have q 6 l 6 / l To ge values of and q for >, you should always use he column labeled l. For eamle, we have 5 6 l 65 / l / and 5 q Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

16 C-4 Chaer : Life Tables Here, you should no base your calculaions on he column labeled q, arly because ha would be a lo more edious, and arly because ha may lead o a huge rounding error. Eamle. You are given he following ecer of a life able: Calculae he following: (a) 5 (b) q 4 (c) 4 q l d Soluion l (a) l d4.7 (b) q4.77. l d 4.7 (c) 4 q. 7. l [ END ] Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

17 Chaer : Life Tables C-5 Eamle. [Srucural Quesion] You are given: (i) S (), (ii) l (a) Find an eression for l for. (b) Calculae q. (c) Calculae 3 q. Soluion (a) l l S (). l l (b) q. l l l (c) 3q. l [ END ] In Eam MLC, you may need o deal wih a miure of wo oulaions. As illusraed in he following eamle, he calculaion is a lo more edious when wo oulaions are involved. Eamle.3 For a cerain oulaion of years old, you are given: (i) /3 of he oulaion are nonsmokers, and /3 of he oulaion are smokers. (ii) The fuure lifeime of a nonsmoker is uniformly disribued over [, 8). (iii) The fuure lifeime of a smoker is uniformly disribued over [, 5). Calculae 5 4 for a life randomly seleced from hose surviving o age 4. Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

18 C-6 Chaer : Life Tables Soluion The calculaion of he required robabiliy involves wo ses. Firs, we need o know he comosiion of he oulaion a age. Suose ha here are l ersons in he enire oulaion iniially. A ime (i.e., a age ), here are 3 l nonsmokers and 3 l smokers. For nonsmokers, he roorion of individuals who can survive o age 4 is /8 3/4. For smokers, he roorion of individuals who can survive o age 4 is /5 3/5. A 3 age 4, here are 43 l.5l 3 nonsmokers and 53 l.l smokers. Hence, among hose who can survive o age 4, 5/7 are nonsmokers and /7 are smokers. Second, we need o calculae he robabiliies of surviving from age 4 o age 45 for boh smokers and nonsmokers. For a nonsmoker a age 4, he remaining lifeime is uniformly disribued over [, 6). This means ha he robabiliy for a nonsmoker o survive from age 4 o age 45 is 5/6 /. For a smoker a age 4, he remaining lifeime is uniformly disribued over [, 3). This means ha he robabiliy for a smoker o survive from age 4 o age 45 is 5/3 5/6. Finally, for he whole oulaion, we have [ END ]. Fracional Age Assumions We have demonsraed ha given a life able, we can calculae values of and q when boh and are inegers. Bu wha if and/or are no inegers? In his case, we need o make an assumion abou how he survival funcion behaves beween wo inegral ages. We call such an assumion a fracional age assumion. Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

19 Chaer 5: Pariciaing Insurance C5- Chaer 5 Pariciaing Insurance OBJECTIVES. To undersand he key feaures of ar olicies. To describe how dividends on a ar olicy are comued 3. To describe how bonuses on a ar olicy are comued, and how hey can be eressed as bonus raes 4. To calculae he rofi vecor of a ar olicy Pariciaing insurances (also known as ar olicies in he US and wih rofi olicies in he UK) are insurance olicies ha ay dividends. The dividends are a orion of he insurance comany s rofis and are aid o he olicyholder as if he or she were a sockholder. When claims are low and he comany s invesmens erform well, dividends rise. However, in many cases dividends are no guaraneed and hey may no be aid if he insurance comany canno make a rofi. In his shor chaer we briefly discuss ariciaing insurances and we shall sress on he following asecs: () he calculaion of dividends and bonus, and () he oions ha are available o olicyholders. Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

20 C5- Chaer 5: Pariciaing Insurance 5. Dividends In Chaer 3 we have discussed he calculaion of annual rofi in grea deail. Recall ha for a single decremen model, in olicy year h +, for a olicy ha is in force, Toal reserve h V a ime h Deah benefi q +h b h+ Selemen eense q +h E h+ Toal reserve needed ( +h ) h+v ime h h + + Conrac remium G h fracion of remium aid for eense c h G h er olicy eense e h The eeced rofi ha emerges a he end of he (h + )h olicy year er olicy in force a he sar of he year is hen Pr h + [ h V + G h ( c h ) e h ]( + i h+ ) (b h+ + E h+ )q +h +h h+ V. The acual rofi can be comued using realized ineres rae, moraliy and eenses. For non-ariciaing olicies, he acual rofis belong o he insurance comany. The rofis from a ariciaing olicy, however, are shared beween he olicyholder and he insurance comany, usually in re-secified roorions. In he official ebook Acuarial Mahemaics for Life Coningen Risks, he following definiions are used: The erm dividend is used when he rofi share is disribued in he form of cash (or cash equivalen, such as a reducion of remium). The erm bonus is used when he rofi share is disribued in he form of addiional insurance, such as increased deah benefis, or era erm life insurance. Bonus in his case is he (addiional) face amoun of insurance ha can be urchased. In he US, i is common for olicyholders of ariciaing insurances o be given choices abou he disribuion, wih remium reducion being a sandard oion. For single-remium olicy, he olicyholder may receive he cash dividend. Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

21 Chaer 5: Pariciaing Insurance C5-3 Since he case for cash dividend or remium refund is simler han addiional deah benefi, le us discuss dividends firs. To begin wih, we look a an eamle wih no surrender. Eamle 5. [Srucural Quesion] A life aged 5 urchases a fully discree ariciaing whole life conrac. The sum insured is,. You are given: (i) Moraliy follows he Illusraive Life Table. (ii) The remium is 5. (iii) The insurer holds ne remium reserves. (iv) Ne remium reserves are based on an ineres rae of 6% and moraliy following he Illusraive Life Table is used for comuing he reserve. The reserves and deah robabiliies in he firs 4 years are: (v) k q 49+k kv Pre-conrac eense is, assumed incurred a, and remium eenses of 5 are incurred wih each remium aymen, ecluding he firs. Calculae he eeced emerging rofi Pr k, k,,, 3 and 4, assuming an invesmen reurn of 8% under he conracual arrangemens in (a), (b) and (c): (a) The ariciaing conrac ays no dividends during he firs 4 years. (b) From he second olicy year onwards, 8% of rofi ha emerges each year is disribued o all olicyholders as a cash dividend. Surviving olicyholders would receive he dividend as cash, and beneficiaries of he olicyholders who die during he year would receive he dividend as addiional deah benefi. No dividend is ayable in he firs olicy year. No dividend is declared if he emerging rofi is negaive. (c) From he second olicy year onwards, surviving olicyholders receive 8% of he rofi er olicy as cash dividends. The beneficiaries of he olicyholders who die during he year would no receive he dividend as addiional deah benefi. Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

22 C5-4 Chaer 5: Pariciaing Insurance (d) Consider he arrangemen in (b). How much remiums during he firs 5 olicy years would a surviving olicyholders need o ay if he dividends are no disribued as cash bu as a remium refund? (e) Consider he arrangemen in (c). How much of he rofi ha emerges from he hird year is disribued as cash? Soluion (a) When he conrac ays no dividends during he firs 4 years, he rofi would no be shared, and hence we can use he same mehodology as illusraed in Chaer 3: Pr Pr Pr ( ) Pr 3 ( ) Pr 4 ( ) (b) When 8% of he emerging rofi (ece for he firs olicy year) are disribued, he rofis calculaed in (a) would be denoed by Pr k, and he dividends are Div k.8pr k, for k >. Since he dividends have o be disribued no maer if he olicyholder survives (disribued as dividends) or dies (disribued as addiional deah benefi ayable a he end of he year) Pr k Pr k Div k.pr k, for k >. k Div k Pr k (c) The eeced rofis generaed er olicy are he Pr k s in (a). The dividend is disribued only o surviving olicyholders a he end of he year. So Pr k Pr k 49+k Div k, for k >. k Pr k Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

23 Chaer 5: Pariciaing Insurance C5-5 (d) In his case he remium ayable a ime k would be deduced by he dividend. So, k Premium (e) The ercenage is / % 79.48%, which is less han 8%. This is because here are no cash disribued for olicies who erminae during year 3 and he whole rofi would be grabbed by he insurer. [ END ] Eamle 5. Consider he olicy in Eamle 5. again, bu wih he following rofi sharing mechanism. From he second olicy year onwards, he insurer disribues a cash dividend o holders of olicies which are in force a he year end. The cash dividend is deermined using 8% of he emerging surlus a he year end, if he surlus is osiive. Projec he cash dividend for his olicy a he end of year 3 for surviving olicies. Soluion The goal of he insurer is o disribue 8% of he emerging surlus. Since only surviving olicies would ge cash dividends and olicies ha erminae ge nohing, he surviving olicies would ge more han 8% of he emerging surlus. Le he cash dividend o be aid be Div 3. The rofi afer disribuion of cash dividend is We wan and hence Div Div Div Div This is / % of he rofis before disribuion of he cash dividends. [ END ] Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

24 C5-6 Chaer 5: Pariciaing Insurance Noe ha in he above eamle, he dividend for a olice in force a he end of he year is calculaed from Div 3.8Pr 3 / 5, while in Eamle 5.(c), we have Div 3.8Pr 3 for surviving olicies and hence he eeced dividends o be aid for a olicy ha is in olice a he beginning of he year is.8pr A sligh change in wording can give quie differen answer. You can coninue wih he calculaion of Eamle 5. and 5. for k 5, 6, and so on, and find all he Pr k s for k being sufficienly large so ha he remaining rofis would be negligibly small. If you do so for Eamle 5., you can verify ha he NPV in (b) is under a risk adjused rae of %. However, wihou he hel of a sreadshee rogram i is ne o imossible o comue all Pr k s and hen obain he rofi signaure and he associaed rofi measures. In Eam MLC, i is more likely ha hey give you he reserves, eenses and moraliy for wo o hree years and ask you o comue he dividend, jus like he following eamle: Eamle 5.3 A life aged 6 urchases a fully discree ariciaing whole life conrac. The sum insured is,. You are given: (i) Moraliy follows he Illusraive Life Table. (ii) The remium is 4. (iii) The insurance comany holds full reliminary erm reserves. (iv) Full reliminary erm reserves are based on an ineres rae of 6% and moraliy following he Illusraive Life Table is used for comuing he reserve. (v) Premium eenses of 5 are incurred wih each remium aymen, ecluding he firs. (vi) Annual rofis are comued assuming reserves and remiums earn 8% ineres er year. Calculae he dividend ayable years afer he issuance of he olicy for he following conracual agreemens: (a) The insurance comany would disribue 9% of rofi ha emerges a he end of each year from he fifeenh olicy year onwards o olicyholders (or his / her beneficiaries) as a cash dividend no maer if he olicyholder survives or no. Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

25 Chaer 5: Pariciaing Insurance C5-7 (b) The insurance comany would disribue 9% of rofi ha emerges a he end of each year from he fifeenh olicy year onwards is disribued o surviving olicyholders as a cash dividend. Soluion We firs comue V and V. Because a full reliminary erm reserve is used, a&& V V a&& a&& V V a&& Also, q So, Pr ( ) (a) The dividend ayable is.9pr (b) Le he dividend be Div. We need. Pr 8 Div and hence Div.9 Pr [ END ] The eamles above are no realisic because hey do no consider he ossibiliy of lases. However, lase is very imoran because he remiums for such olicies are high (see he end of his secion for he elanaion.) To model lases, we use he framework inroduced in Chaer 8: deahs (d) occur coninuously hroughou he year, and lases (w) may haen immediaely before a remium aymen. Le and (d ) q + h (w) q + h be he indeenden raes of decremen. Then assuming annual remiums are ayable a he beginning of each year, we have (d ) q + h, (d ) q + h and he rofi before dividend is ( w) ( d ) ( w) q h h q h ( τ ) ( d ) ( w), ( q ), + h + h + h Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

26 Mock Tes 7 T7- Eam MLC Mock Tes 7 SECTION A Mulile Choice Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

27 T7- Mock Tes 7 **BEGINNING OF EXAMINATION**. You are given: (i) The force of moraliy follows Gomerz s law wih B.5 and c.. (ii) f (, + k) is he forward ineres rae, conraced a ime, effecive from ime o + k. (iii) The following forward ineres raes: f(, 3) Calculae & s& 6:3. (A) 3.4 (B) 3.4 (C) 3.4 (D) 3.43 (E) 3.44 Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

28 Mock Tes 7 T7-3. You are given: (i) A.3 (ii) A :n.55 (iii) A.35 (iv) A :n.56 (v) Deahs are uniformly disribued over each year of age. Find A. :n (A).35 (B).38 (C).4 (D).44 (E).47 Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

29 T7-4 Mock Tes 7 3. A cerain olicy roviding a deah benefi of, a he end of he year of deah is issued wih a gross remium of 5. Le AS be he asse share a ime. You are given: (i) i.3 (ii) AS 7 (iii) Percen of remium eense for he h remium is.. (iv) Eense er, incurred a he beginning of he h year is.5. (v) The robabiliy of decremen by deah in he h year is.3 (vi) The robabiliy of decremen by wihdrawal in he h year is. (vii) The cash value in he h year is CV 7. Find AS. (A) 69 (B) 7 (C) 85 (D) 9 (E) 96 Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

30 Mock Tes 7 T You are given: (i) i.6 (ii) a&& 6.5 (iii) μ 6.5 (iv) Using Woolhouse s formula wih hree erms, he value of () a&& 6 is γ. (v) Assuming uniform disribuion of deahs over each year of age, he value of a&& is γ. () 6 Find (γ γ ). (A) 6 (B) 3 (C) (D) 3 (E) 6 Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

31 T7-6 Mock Tes 7 5. For a fully discree whole life insurance of on (3), you are given: (i) kv is he ne remium reserve a he end of year k. (ii) V 8 (iii) A (iv) v.95 (v) q 4., q 4.3 Calculae V. (A) (B) 8 (C) (D) 8 (E) 4 Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

32 T7- Mock Tes 7 Eam MLC Mock Tes 7 SECTION B Wrien Answer Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

33 Mock Tes 7 T7-3. (8 oins) Le T be a coninuous random variable whose ossible values are in (, ). (a) (3 oins) Suose ha T has a unique mode a >. Show ha a, μ. μ (b) (5 oins) Consider he force of moraliy Bc μ A + for >. + Dc Here A, B, D >, and c >. (i) Find S (). (ii) Wha is he mode of he disribuion of T when A? Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

34 T7-4 Mock Tes 7. (8 oins) For wo indeenden fuure lifeime random variables T and T y, you are given: q, y q y, q.8 and q y.4 (a) ( oins) For < <, derive an eression for f y () in erms of. (b) (c) ( oins) Calculae he robabiliy ha (y) will die wihin 6 monhs and will be redeceased by (). (4 oins) Le he force of ineres be 8%. Calculae ( I A), y: he eeced resen value of $ dollars ayable when dies a ime, if he deah of haens wihin year, and if dies afer y. Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

35 Mock Tes 7 T (8 oins) Consider a 3-year erm universal life insurance on (5). The deah benefi is $8, lus he accoun value a he end of he year of deah. The deah benefi is ayable a he end of he year of deah. There is no corridor facor requiremen. You rofi es he conrac using he following basis: Premiums of $3, each are aid a he beginning of years, and 3. The rojeced accoun values a he end of years, and 3 are $,95, $4, and $6,4, resecively. Incurred eenses are $ a inceion, $5 lus % of remium a renewal, $ on surrender, and $ on deah. The surrender charge is $8 for all duraions. The insurer s earned rae of ineres is % for all hree years. Moraliy eerience is % of he Illusraive Life Table. Surrenders occur a year ends only. The surrender raes for years, and 3 are %, % and %, resecively. The insurer holds he full accoun value as reserve for he conrac. Find he rofi vecor. Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

36 Mock Tes 7 T7-9 Soluions o Mock Tes 7 Secion A. A. C. A. B 3. E 3. E 4. E 4. D 5. B 5. D 6. D 6. C 7. B 7. D 8. C 8. B 9. E 9. B. A. C. [Chaer ] Answer: (A) We need o comue law, We have & s 6:3 + f (, 3) 6 ( + + f (, 3)) 6 ( + + f (, 3)) + u c c e ( Bc )du e B. ln c e ln e ln e ln So he answer is For Gomerz s. [Chaer 3] Answer: (A) i A We need o find n E. From (i), (iii) and (v), we ge δ A On he oher hand, A A A A : n i : n : n : n. So, δ n n E E n n E E Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

37 T7-3 Mock Tes 7 3. [Chaer 9] Answer: (E) By he recursive relaion for asse shares,.55(.35).56(.3) (.35.3) n E E.35 ( ) AS [Chaer 4] Answer: (E) Woolhouse s formula for whole life annuiies is given by n ( m) m m a ( + ) a&& && m m δ μ. Therefore, γ.5 (ln(.6) +.5).454. Assuming uniform disribuion of deahs over each year of age, we have a&& α( m) a&& β( m). ( m) From Eam MLC Tables, we obain α()., β() Therefore, γ Finally, (γ γ ) ( ) 6. Hence, he answer is (E). 5. [Chaer 6] Answer: (B) From saemens (iii) and (iv), we can back ou he ne annual remium: d v.5, By Fackler s accumulaion formula, ( da P 3 7. A.3565 V + π )( + i) q 3 (8 + 7) / V 4 ( V + π )( + i) q ( ) / V 4 6. [Chaer ] Answer: (D) We need o comue has a densiy of n q y.9334, Since he number is aached o (), we condiion on T, which Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

38 Mock Tes 7 T7-3 As a resul, n q y n < < f ( ). f ( )d y e μ y d + d ( e ( ) μ n μ ( ) ()d y ). 7. [Chaer ] Answer: (B) We calculae 6 q 5, 6 q 6, and 6 q 5:6. Firsly, 6 q Secondly, 6 q Thirdly 6 q 5:6 6 5:6 q 56:66 (.9.8) (.97.95).565. Finally, by symmeric relaion, he answer is [Chaer 8] Answer: (C) () () q From (iii), decremen is SUDD. So μ +.75 holds. ().75q From (i), () 4 q () 97.75q () () 97q 4 3 q () q. Now we can aly he formula for SUDD (Equaion (8.6)) o obain.75 3 () ().75 () (3).75 () (3) q + + q.75 ( q q ) q q [Chaer 7 + 9] Answer: (E) ( τ ) The aniciaed robabiliy of survival + is The eeced rofi for a olicy ha is in force a ime is ( )(.4) The acual robabiliy of survival is The acual rofi for he olicy is ( )(.) Since here are 355 / olicy in force a he beginning of he year, he oal gain is 3 ( ) Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

39 T7-3 Mock Tes 7. [Chaer ] Answer: (A) (I) d d The saemen is incorrec. The ei rae for sae is By Equaion (.),.6h o( h) h + +. (II) Saes and 3 are absorbing saes. The Q mari of he CTMC is d KFE says ( P d As a resul, d d ) P Q +.. For (III) The differenial equaion saisfied by , we look a , and hence saemen (ii) is also incorrec. is d.. d.6 e If is he soluion, we can u i ino he differenial equaion above 3 o obain.6.e. which would gives us from sae,.6 e. However, since sae can be re-enered > and his means ha soluion. Thus, saemen (III) is incorrec.. [Chaer 4] Answer: (C) We firs calculae he mean of Y, which is acually a. : Under consan force of moraliy (over age o + ), ( e μ + δ e ). ( μ + δ) a :.6 e is no he correc Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

40 Mock Tes 7 T7-33 To calculae robabiliies involving Y, we firs wrie down he recise definiion of Y as a funcion of T : Y.8T e.8.8 e The even Y is equivalen o.8 T e This means he required robabiliy is. [Chaers 9 + 6] Answer: (B) or T T T > Pr(T 5.968) e The salary over age (63, 64) is 75, / ,8.45. The salary over age (64, 65) is 75, / , The deah benefi ayable a age 6.5 is 75, 3 5,. The robabiliy for his o occur is 3 / 468. The deah benefi ayable a age 63.5 is 76, , The robabiliy for his o occur is 84 / 486. The deah benefi ayable a age 64.5 is 77, , The robabiliy for his o occur is 5 / 486. So he APV of he deah benefi is 5, / 3 / 5 / [Chaer 9] Answer: (E) Le he ne remium be π and he conrac remium be G. Decremen due o deah is denoed by () and decremen due o wihdrawal is denoed by (). Since V, he recursion relaion for ne remium reserves says.π ( τ ) () () 36 + q + 36(.4) q Since AS, he recursion relaion for asse shares says or ( τ ) ().(.5G) 8 + q + 36(.4) q ( τ ) ().G 36 + q + 7(.4) q Uon subracion, we ge (). () Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

41 T7-34 Mock Tes 7 On solving, we ge ().(G π) q + 36(.4) q () q.9. () ().(9.36) q + 4.4q () 4. [Chaer 3] Answer: (D) The benefi funcion is e.5 during he firs year ( < < ), e.5 during he second year ( < ), and 3e.5 during he hird year ( < 3), and so on, u o he enh year. The APV of his insurance is.5.4. bv f ( )d + e e.e d. d +.( K+ ) [Chaer ] Answer: (D) Le Z be he resen value random variable for a discree whole life insurance of $ on (). I is known ha Var(Z) ( A 5 (A 5 ) ). We can eress he random variable U as follows: U N N i MZ i, where Z i s are muually indeenden and are idenically disribued as Z. I follows ha N Var( U ) Var MZi Var N i N i N N Var( MZ ) M i N i N i M NVar( Z) M Var( Z). N N Hence, Var(U ) M ( A 5 (A 5 ) )/N. 6. [Chaer ] Answer: (C) The EPV of he benefi is e ( N MZ Var( Z ) μ + μ )d. The firs ar of he EPV comes from a direc ransiion from sae o, wihou going hrough sae. The second ar of he EPV comes from a ransiion from sae o sae and hen o sae. Since + v , e (.5 +.3s)ds + i e(.5.65 i ) Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

42 Mock Tes 7 T7-35 So, g(.5) 5e.45 (e ) [Chaer + 4] Answer: (D) The accoun value a he end of monh is.8 AV ( ) So, he cash surrender value is The insured is age 6 when he surrenders. The eeced resen value of he annuiy is.4pa&& 6:5 +.6Pa&& 6:5 P(.4a&& 6:5 +.6a&& 5 +.6a&& 6.6a&& 6:5) P( ) So he annuiy has an annual aymen of P [Chaer 3, 4 and 5] Answer: (B) For he sandard oulaion, A da5 So,. P A5 For his aricular life, A 5 * ,.5.5 A5 * a& 5 *.97 d So, E( L) ( ) [Chaer 7] Answer: (B) This is like he se u of FPT reserve, jus ha he valuaion remiums during he firs wo years comleely offse he cos of insurances during he firs wo years. As a resul, mod a&& V 5V3 5 5 a&& [Chaer ] Answer: (C) From (iv), a 4% ineres, 87 a& & : Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

43 T7-36 Mock Tes A : Now we add back he APV of he cash flows on or afer ime 4: a& & 8: / A / / :5 The ne annual remium is / Secion B. [Chaer ] (a) f () S ()μ. By roduc rule, d d d f ( ) μ S ( ) + S ( ) μ d d d μ [ S ( ) μ ] + S ( ) μ S ( )( μ μ ) S () > for any < <. A oin, f ( ) mus have o be zero. This gives he resul desired. Bc B B + Dc (b) (i) μ d A + d A + [ln( + Dc )] A + ln + Dc D ln c D ln c + D B + Dc So, S () e A ln. D ln c + D Bc ln c (ii) I is easy o see ha μ and hence ( + Dc ). [Chaer ] Bc ln c B c μ μ ( + Dc ) ( + Dc ) ( B ln c) c B c ( + Dc ) Bc (ln c Bc ) ( + Dc ) d For f ( ), ln c Bc ln(lnc) ln B, and hence. d ln c (a) F y () q y y q + q y 4 q q y , and by differeniaion, f y () Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

44 Mock Tes 7 T (b) f ( )d d( ).8 d(.4 ).64 q y q q y d..5 (c) e q yf ( )d e q y qd k k k e k e e k e k k k e d k e 4 3 k d e e e k d( ) + k k d k k d( ) k k k 4e k e 4e k + e e k k d + k k k d( ) 3 k e k k k k k e 4e e 4 k e 4e e 4 + e + k k k k d k k k k k (where we have used so ha e.8 4 e k d ( Ia) k a k e k k in he las se) d.873, and hence he EPV is.975. k 3. [Chaer 4] A ime Iniial eenses: Pr : A ime AV : AV : 95 P : 3 E : (since iniial eenses are incororaed in Pr ) EDB :..59 ( ) [Noe: This is a secified amoun lus he accoun value (Tye B) olicy. In he absence of any corridor facor requiremen, he deah benefi is $8 lus he ime- accoun value of $95. The value.59 is obained from he Illusraive Life Table.] ESB : (..59). ( ) 4.. [Noe: The cash value CV is he accoun value AV 95 less he surrender charge SC 8.] EAV : (..59)(.)(95) Pr : ( + 3 )(.) The answer is hus (A). Of course you can coninue o calculae Pr and Pr 3 : A ime AV : 95 AV : 4 P : 3 E : EDB :..64 ( ) Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

45 T7-38 Mock Tes 7 ESB : (..64). (4 8 + ) EAV : (..64)(.)(4) Pr : ( )(.) A ime 3 AV : 4 AV 3 : 64 P 3 : 3 E 3 : EDB 3 :..697 ( ) ESB 3 : (..697) ( ) EAV 3 : (..697)( )(64) [Noe: The surrender rae for 3 is %, which means all olicyholders surrender a he end of he erm. Hence, EAV 3 mus be zero.] Pr 3 : ( )(.) Therefore, he rofi vecor is given by (, 85, 779, 346). 4. [Chaer ] We are given: q [] q []+ q q By using he informaion from saemens (i) and (ii), we obain q [] q []+ q q q q The rick o solve his quesion is o make use of he fac ha of l [86] and l [87] would evolve o l 89. Firsly, Similarly, l 89 l [87] [87] [87]+ l [87] (.5q 87 )(.768). l 89 l [86] [86] [86]+ 88 (.55)(.5q 87 )(.3536) As a resul, (.55)(.5q 87 )(.3536) l [87] (.5q 87 )(.768) (.55)(.3536) On solving, we ge l [87] Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

46 Suggesed Soluions o MLC Aril 5 S-8 Suggesed Soluions o MLC Aril 5 Secion A: Mulile Choice Quesions. [Chaer ] Answer: (B). B. D. C. B 3. B 3. B 4. D 4. A 5. A 5. E 6. C 6. D 7. C 7. A 8. E 8. D 9. C 9. E. D. C Le I i if he ih members who are age 35 now is alive 3 years afer he clue is esablished, and oherwise. Similarly, le J i if he ih members who are age 45 now is alive 3 years afer he clue is esablished, and oherwise. Then i N ( I i + J i ). Since he I i s and J i s are indeenden, and follow B(, 3 35 ) and B(, 3 45 ), resecively, E( N) [E( I i [E( I ( i ) + E( J 35 )] ) + E( J + i 45 )] ) Var( N) [Var( I i [Var( I ( Since 3 35 l 65 / l / and 3 45 l 75 / l / , i ) + Var( J 35 E(N) and Var(N) 4.7. By normal aroimaion (wihou he coninuiy correcion), n So n and he leas ineger value of n is 4. )] ) + Var( J q +. [Chaer ] Answer: (C) The robabiliy of he even A s ha he life healhy a 5 would be disabled firs ime (couning from age 5) for a eriod of a leas year, saring age 5 + s, is 35 i )] q ) Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

47 S-8 Suggesed Soluions o MLC Aril 5 Pr(A s ) s 5.5s. ( μ ds) e.e ds. 5+ s 5+ s Noing ha he evens A s above are muually eclusive for differen values of s (because of he condiion disabled firs ime ), he required robabiliy is obained by inegraing (acually he same as summing if ds is no involved in he robabiliy above) s from o 4 (no 5, because we need he era one year o fulfil he year disabiliy): 4 e.5s 3. [Chaer ] Answer: (B).e. ds.e. e Noing ha ransiion from sae o sae or sae are irreversible, To his end, he ei rae for sae is v :3+ μ 3+ :3+ + μ3+ : :3 3: :3 d. + ( , ln.75 v3+ + ) and hence he answer required is e [Chaer 3] Answer: (D) T v Rewrie he random variable as Z + Z Z Z T T + 3. v v T T T v v v (If you wonder why we use his aricular decomosiion, he hin is ha in he oions given, mos of hem, ece (C), sars wih A or is equivalen. So we firs creae he resen value random variable ha corresonds o his APV, and hen build u he in he layer < T < 3, and finally subrac he era ar for T 3.) Hence, he correc eression for E(Z) is A A 3 + A. (A): I misses he coefficien for he las erm. (B): The eression for (B) is equivalen o A + A 3 A. Hence he firs erm is no correc. (C): The firs erm is no correc. (D): This is equivalen o he eression above. (E): This eression can be rearranged as E A E A and hence he las erm is no correc. E E A A A E A, Ace 5 Johnny Li and Andrew Ng SoA Eam MLC

48 S-88 Suggesed Soluions o MLC Aril 5 Secion B: Wrien Answer. [Chaer ] (a) Le v() be he resen value (a ime ) of dollar ayable a ime. Le I(A) be if even A occurs, and oherwise. Then E(I(A) B) Pr(A B). For any sae j, j a E ( ) ( ( ) )d ( ) : v I Y + j Y v( )E[ I( Y ( + ) j) Y ( ) ]d v( ) j d. The sum required is j j j a v( ) d v( ) d v( )d a. : j j j e (b) Using (a), we ge a a a a : : :. EPV of all benefis EPV of coninuous disabiliy benefis + EPV of deah benefi a + A : : The ne remium rae is (c) By Thiele s differenial equaion for reserves, (d) d d V () () ( j) ( j) () π + δ V ( b + V V ) μ. j A ime 3, when he sae is, he remium rae is 967. The ransiion benefi o sae is, while he ransiion benefi o sae is he deah benefi, which is. So, d () d V ( ).4 ( ) e + d e (.4. )d + v e e Le P* be he new ne remium rae. To ge he refund of remium, no any ransiion can occur wihin years. This means EPV of all benefis is he original EPV of benefis lus he EPV of refund of remiums P* e P* By he equivalence rincile, P* 4.49P* j + Ace 5 Johnny Li and Andrew Ng SoA Eam MLC