6 Insurances on Joint Lives

Transcription

1 6 Insurances on Joint Lives 6.1 Introduction Itiscommonforlifeinsurancepoliciesandannuitiestodependonthedeathorsurvivalof more than one life. For example: (i) Apolicywhichpaysamonthlybenefittoawifeorotherdependentsafterthedeathof the husband(widow s or dependent s pension). (ii) Apolicywhichpaysalump-sumontheseconddeathofacouple(oftentomeet inheritance tax liability). We will confine attention to policies involving two livesbut the same approaches can be extendedtoanynumberoflives.weassumethatpoliciesaresoldtoalifeage xandalife age y,denoted (x)and (y). The basic contract types we will consider are: Assurances paying out on first death, and annuities payable until first death. 131

2 Assurances paying out on second death, and annuities payable until second death. Assurances and annuities whose payment depends on the order of deaths. 6.2 Multiple-state model for two lives The following multiple-state model represents the joint mortality of two lives, (x) and (y). State 1 (x) Alive (y) Alive µ 12 t State 2 (x) Dead (y) Alive µ 13 t µ 24 t State 3 (x) Alive (y) Dead µ 34 t State 4 (x) Dead (y) Dead 132

3 Notes: We just index the transition intensities by time t. Other notations are possible. We implicitly assume that the simultaneous death of (x) and (y) is impossible: there isnodirecttransitionfromstate1tostate4. HerewelistthemainEPVsmetinpractice,givingtheirsymbolsinthestandardactuarial notation. In the first place, we assume all insurance contracts(assurance- or annuity-type) tobeforalloflifeandtobeofunitamount,i.e.$1sumassuredor$1annuityper annum. Joint-life assurances ĀxyistheEPVof$1paidimmediatelyonthefirstdeathof (x)or (y). ĀxyistheEPVof$1paidimmediatelyontheseconddeathof (x)and (y). 133

4 Contingent assurances Ā1 xy istheepvof$1paidimmediatelyonthedeathof (x)provided (y)isthenalive, i.e.provided (x)isthefirsttodie. Ā 1 xy istheepvof$1paidimmediatelyonthedeathof (y)provided (x)isthenalive, i.e.provided (y)isthefirsttodie. Ā2 xy istheepvof$1paidimmediatelyonthedeathof (x)provided (y)isthendead, i.e.provided (x)isthesecondtodie. Ā 2 xy istheepvof$1paidimmediatelyonthedeathof (y)provided (x)isthendead, i.e.provided (y)isthesecondtodie. Joint-life annuities ā xy istheepvofanannuityof$1perannum,payablecontinuously,untilthefirstof (x)or (y)dies. 134

5 ā xy istheepvofanannuityof$1perannum,payablecontinuously,untilthesecond of (x)or (y)dies. Reversionary annuities ā x y istheepvofanannuityof$1perannum,payablecontinuously,to (y)aslongas (y)isaliveand (x)isdead. ā y x istheepvofanannuityof$1perannum,payablecontinuously,to (x)aslongas (x)isaliveand (y)isdead. Thenotationforreversionaryannuitieshelpfullysuggests(taking ā x y asanexample)an annuitypayableto (y)deferreduntil (x)isdead.thinkof m ā n fromfinancial Mathematics. 135

6 Limited terms Theabovecontractsmayallbewrittenforalimitedtermof nyears,inwhichcasethe usual n is appended to the subscript. Evaluation of EPVs In the multiple-state, continuous-time model, we can compute all the above EPVs simply by appropriatechoicesofassurancebenefits b ij orannuitybenefits b i inthiele s differential equations. The following table lists these choices for whole-life contracts. 136

7 EPV b 1 b 2 b 3 b 12 b 13 b 24 b 34 Ā xy Ā xy Ā 1 xy Ā 1 xy Ā 2 xy Ā 2 xy ā xy ā x y ā y x ā xy Example:Consider Āxy:n.Thiele sequationsare: 137

8 d dt V 1 (t) = V 1 (t)δ (µ 12 t + µ 13 t )(1 V 1 (t)) d dt V 2 (t) = d dt V 3 (t) = d dt V 4 (t) = 0 withboundaryconditions V 1 (n) = 1andallother V i (n) = 0. Composite benefits ManyjointlifecontractscanbebuiltupoutoftheaboveEPVsandtheEPVsofsinglelife benefits. This is generally the simplest way to compute them, especially if using tables rather than a spreadsheet. Example: Consider a pension of $10,000 per annum, payable continuously as long as (x) and (y)arealive,reducingbyhalfonthedeathof (x)if (x)diesbefore (y).(thiswould be a typical retirement pension with a spouse s benefit.)its EPV, denoted ā say, is most easilycomputedbynotingthat$10,000p.a.ispayableaslongas (x)isalive,andin 138

9 addition,$5,000p.a.ispayableif (y)isalivebut (x)isdead,hence: ā = 10,000 ā x + 5,000 ā x y. Bysimilarreasoning,anannuityof$1p.a.payableto (y)forlifecanbedecomposedinto anannuityof$1p.a.payableuntilthefirstdeathof (x)and (y),plusareversionary annuityof$1p.a.payableto (y)afterthepriordeathof (x);hencetheuseful: ā x y = ā y ā xy. Computing contingent assurance EPVs TheonetypeofjointlifebenefitwhoseEPVisnoteasilywrittenintermsoffirst-death, second-death and single-life EPVs is the contingent assurance. We defer discussion until 139

10 Section Random joint lifetimes Itisalsopossibletospecifyajointlivesmodelviarandomfuturelifetimes.Wehavethe random variables: T x = thefuturelifetimeof (x) T y = thefuturelifetimeof (y). Now define the random variables: T min = min(t x,t y )and T max = max(t x,t y ). T min istherandomtimeuntilthefirstdeathoccurs,and T max istherandomtimeuntilthe 140

11 second death occurs. Thedistributionof T min Define: tq xy = P {T min t} (c.d.f.) tp xy = P {T min > t} Sothat: t q xy + t p xy = 1. Nowif T x and T y areindependent: tp xy = P {T min > t} = P {T x > tand T y > t} = P {T x > t} P {T y > t} = t p x t p y 141

12 and(usefulresult)thejointlifesurvivalfunction t p xy istheproductofthesinglelife survival functions. If T x and T y arenotindependentthenwecannotwrite t p xy = t p x t p y. If T x and T y areindependentthenalso: tq xy = 1 t p xy = 1 t p x t p y = 1 {(1 t q x )(1 t q y )} = t q x + t q y t q x t q y. Example:Given n q x = 0.2and n q y = 0.4 calculate n q xy and n p xy. Solution: nq xy = n q x + n q y n q x n q y 142

13 = = 0.52 np xy = n p x n p y = = 0.48 Tosimplifytheevaluationofprobabilities,like t p xy,wecandevelopalifetablefunction, l xy,associatedwith T min. If T x and T y areindependentthen: tp xy = t p x t p y = l x+t l x l y+t l y = l x+t:y+t l x:y where: l xy = l x l y. 143

14 Weobtainthep.d.f.of T min,denoted f xy (t),bydifferentiationwhen T x and T y are independent: f xy (t) = d dt t q xy = d dt t p xy = d dt t p x tp y [ ] d d = tp x dt t p y + t p y dt t p x = [ t p x ( t p y µ y+t ) + t p y ( t p x µ x+t )] = t p xy (µ x+t + µ y+t ). 144

15 Compare this to the single life case: f x (t) = t p x µ x+t. Now,define µ xy (t)asthe forceofmortality associatedwith T min.wecanshowdirectly that µ xy (t) = µ x+t + µ y+t when T x and T y areindependentbecause: µ xy (t) = f xy(t) tp xy = µ x+t + µ y+t. However,usingthemultiple-stateformulationofthemodel,weseethat T min isjustthe 145

16 time when state 1 is left.the survival function associated with this event is, by definition, tp 11 (wherethebulletrepresentspolicyinception)andweknowthat: ( t ) tp 11 = exp µ 12 s + µ 13 s ds. 0 By differentiating this, the force of mortality associated with leaving state 1 is the sum of theintensitiesoutofstate1withoutassumingthat T x and T y areindependent. Thedistributionof T max Define: tq xy = P {T max t} (c.d.f.) tp xy = P {T max > t} Sothat t q xy + t p xy = 1.Wehave: 146

17 tp xy = P{T max > t)} = P{T x > t or T y > t} = P{T x > t} +P{T y > t} P{T x > tand T y > t} = t p x + t p y t p xy whichdoesnotrequire T x and T y tobeindependent.iftheyareindependentthen: tp xy = t p x + t p y t p x t p y and also: tq xy = P{T max t} 147

18 = P {T x tand T y t} = P{T x t}p{t y t} = t q x t q y. Thep.d.f.of T max andthe forceofmortality µ xy (t)associatedwith T max areleftas tutorial questions. Expectations of joint lifetimes Define: e xy =E[T min ] and e xy =E[T max ]. Now, consider the following identities: Identity1 T min + T max = T x + T y. Identity2 T min T max = T x T y. 148

19 These can be verified by considering the three exhaustive and exclusive cases: T x > T y, T x < T y and T x = T y. Taking expectations of the first identity gives: E [T min + T max ] =E [T x + T y ] = E [T min ] +E [T max ] =E [T x ] +E [T y ] = e xy + e xy = e x + e y. From the second identity we have: E [T min T max ] =E [T x T y ] andif T x and T y areindependentwehave: E [T min T max ] =E [T x ] E [T y ]. 149

20 Evaluation of EPVs WecanevaluateEPVsusingthedistributionsof T min and T max,justasforasinglelife, as an alternative to solving Thiele s equations. For example, consider an assurance with a sumassuredof$1payableimmediatelyonthefirstdeathof (x)and (y). ThePVofbenefits = v T min withp.d.f. = t p xy µ xy (t) sotheexpectedpresentvalue Āxyis: Ā xy =E [ ] v T min = 0 v t tp xy µ xy (t)dt. Note that evaluating an integral numerically is not significantly easier than solving Thiele s equations numerically. 150

21 6.4 Curtate Future Joint Lifetimes Basic definitions We now introduce discrete random variables. Let: K min = Integerpartof T min K max = Integerpartof T max. Wecanthendefinethecurtateexpectationoflifeas: e xy = E [K min ] e xy = E [K max ]. Formorepropertiesof e xy and e xy seetutorial. Theassociateddeferredprobabilities k qxy and k qxy (i.e.thedistributionfunctionsof 151

22 K min and K max aregivenby: qxy = P {k T min < k + 1} k = P {K min = k} k q xy = P {k T max < k + 1} = P {K max = k}. Example: Given: tq non-smoker x = 70 x + t 80 x validfor 0 x 70and 0 t 10,andthattheforceofmortalityforsmokersistwice thatfornon-smokers,calculatetheexpectedtimetofirstdeathofa(70)smokeranda(70) non-smoker. You are given that the lives are independent. 152

23 Solution: We want: e s n s 70:70 = 10 0 tp s n s 70:70 dt = 10 Nowlet µ x betheforceofmortalityfornon-smokers,then: tp s x = exp = ( [ ( exp ) 2 t 0 t 0 = ( tp n s x ( ) 2 10 t = ) 2 µ x+r dr tp s 70 tp n s 70 dt. )] 2 µ x+r dr Since: tp n s x = 1 ( ) t = 10 t

24 therefore: e s n s 70:70 = 10 0 [ ] 3 10 t dt = 2.5years. 10 Discrete joint life annuities Anyannuityorassurancewecandefineasafunctionofthesinglelifetime K x,wecan defineusing K min,thereforedependingonthefirstdeathor K max,therefore depending on the second death. Example: Consider an annuity of $1 per annum, payable yearly in advance in advanceas longasboth(x)and(y)arealive(i.e.payableuntilthefirstdeath).thesamereasoning asinthesingle-lifecaseshowsthatthepresentvalueofthisis: PV = ä Kmin

25 WedenotetheEPVofthisbenefit ä xy so: ä xy = E[ä Kmin +1 ] = ä k+1 k q xy = k=0 v k kp xy k=0 Sincethedistributionofthepresentvalueisknown,thevariance,Var[ä Kmin +1 ],and other moments can be calculated. Example: Consider an annuity of $1 per annum, payable yearly in advance in advanceas longasatleastoneof(x)and(y)isalive(i.e.payableuntiltheseconddeath).thesame reasoning as before leads to: PV = ä K max

26 WedenotetheEPVofthisbenefit ä xy so: Tohelpevaluate ä xy notethat: ä xy = E[ä Kmax +1 ] = ä k+1 k q xy = k=0 v k kp xy. k=0 ä xy = = v k kp xy k=0 v k ( k p x + k p y k p xy ) k=0 156

27 = v k kp x + v k kp y k=0 = ä x + ä y ä xy. k=0 v k kp xy k=0 Hence also: ä xy + ä xy = ä x + ä y. Computing ä xy etc.usingtables ItiseasytocomputeEPVsofdiscretebenefitsusingaspreadsheetif t p xy isknown, whichisparticularlysimpleif T x and T y areindependentso t p xy = t p x t p y. However,othermethodscanbeusedifjointlifetablesareavailable,astheywillbein examinations. (i) Giventabulatedvaluesof ä xy directly. 157

28 e.g.a(55)tables,butnotethatthevaluesof a xy aregivennot ä xy,andtheyareonly givenforevenages mayneedtouselinearinterpolation. For example: a 66: (a 66:60 + a 66:62 ). (ii) Given commutation functions. e.g. A tables, but note that they are only given for x = yandat4%interest. Note that(assuming independence): ä xy = = v t tp xy = t=0 t=0 v t tp x t p y t=0 { v t l x+t l y+t l x l y } 158

29 = t=0 { v 1 2 (x+y)+t l x+t l y+t v 1 2 (x+y) l x l y } = N xy D xy where: D xy = v 1 2 (x+y) l x l y N x+t:y+t = D x+t+r: y+t+r. r=0 159

30 Warning: ä xy:n = ä xy n ä xy = ä xy v n np x n p y ä x+n:y+n = ä xy 1 v n D x+n D y+n D x D y ä x+n:y+n. Example: Using a(55) mortality and 6% interest calculate the following: (i) ä 80:75 (ii) 10 ä 70:65. Solution: ä 80:75 = 0.5(ä 80:74 + ä 80:76 ) = 0.5(1 + a 80: a 80:76 ) = 0.5( ) =

31 10 ä 70:65 = v 10 10p p 65 ä 80:75 = 1 ( v 10 v p 70 v 10 ) 10p 65 ä80:75 ( ) ( ) D80 D75 = 1 v 10 D 70 ( =(1.06) = D 65 ä 80:75 ) ( ) We can extend many of the formulations for single life annuities to joint life annuities. This applies to: Deferred annuities(e.g. previous example) Temporaryannuities, ä xy:n etc. 161

32 Increasingannuities, (Iä) xy etc. Annuities paid monthly. For example: ä (m) xy ä xy m 1 2m. Approximations for annuities paid continuously. For example: a xy ä xy 0.5 = a xy Example(Question No 8 Diploma Paper June 1997): A certain life office issues a last survivor annuity of $2,000 per annum, payable annually in arrear,toamanaged68andawomanaged65. (a) Using the a(55) ultimate table(male/female as appropriate) and a rate of interest of 4% per annum, estimate the expected present value of this benefit. (b) Usingthebasisof(a)above,deriveanexpression(whichyouneedNOTevaluate) 162

33 forthestandarddeviationofthepresentvalueofthisbenefit,intermsofsinglelife and joint life annuity functions. Solution: (a)wewant: 2000 a m f 68:65 = ( a m 68 + a f 65 am f 68:65 ( a m 68 + a f = 2000( = 25, ) ( a m f 68:64 + am f 68: ( ) ) ) ) 163

34 (b)presentvalue = 2000 a K max Sowewant:Var [ ] 2000 a K max ] = Var [ä K max +1 1 ] = Var [ä K max [ +1 1 v = K max +1 ] Var d = = ( ) [ Var v K max +1] d ( 2000 d ) 2 ( ( ) ) 2 A m f A m f 68:65 68:65 164

35 where* means evaluated at: j = i 2 + 2i = 8.16%. Wenowuse: A xy = 1 d ä xy and: ä xy = ä x + ä y ä xy and: to get: 2000 d { 1 d SD[PV ] = V ar[pv ] ) ( ä m 68 + ä f 65 ä m f 68:65 [ ( )] } d ä m 68 + ä f 65 äm f 2 68:

36 Joint life assurances Consideranassurancewithsumassured$1,payableattheendoftheyearofdeathofthe firstof(x)and(y)todie.thebenefitispayableattime K min + 1sohaspresentvalue: v K min +1. Theexpectedpresentvalueisdenoted A xy,and: A xy =E [ v K min +1] = k=0 v k+1 k qxy. Relationships: Recall from single life: A x = 1 dä x. 166

37 Itcanbeshownbysimilarreasoningthat: A xy = 1 dä xy Ā xy = 1 δ ā xy A xy:n = 1 dä xy:n Ā xy:n = 1 δ ā xy:n andsoon. Associated with the second death we have: A xy =E[v K max+1 ] = v k+1 k q xy k=0 167

38 and similar relationships can be derived, e.g: A xy = 1 dä xy A xy:n = 1 dä xy:n. Reversionary Annuities We noted before that: ā x y = ā y ā xy just by noticing that the following define identical cashflows no matter when (x) and(y) should die: (1) a reversionary annuity of $1 per annum payable continuously while (y) is alive, following the death of(x). (2) anannuityof$1perannumpayablecontinuouslyto (y)forlife,lessanannuityof$1 168

39 per annum payable continuously until the first death of (x) and (y). Hence the cashflows have identical present values, and the same EPVs. We can apply similar reasoning to other variants of reversionary annuities, for example: ä x y = ä y ä xy = v k kp y (1 k p x ) k=0 a x y = a y a xy = v k kp y (1 k p x ) k=1 ä (m) x y = ä y (m) = 1 m ä (m) xy k=0 v k m k p y (1 k p x ) m m 169

40 a (m) x y = a (m) y = 1 m a (m) xy k=1 v k m k p y (1 k p x ). m m We can take advantage of some relationships to simplify calculations. For example: ä x y = ä y ä xy = (1 + a y ) (1 + a xy ) = a y a xy. Hence: ä x y = a x y. Similarly, we can derive other relationships like: 170

41 (a) ä x y ā x y (b) ä (m) x y ä x y. Example: Calculate the annual premium(payable while both lives are alive) for the followingcontractforamanaged70andwomanaged64. Benefits: SAof$10,000payableimmediatelyonfirstdeath;and a reversionary annuity of $5,000 payable continuously to the survivor for the remainder of their lifetime. Basis: Renewal expenses 10% of all premiums. a(55) Ultimate, male/female as appropriate Interestat8%. 171

42 Solution: Let P = Annual Premium. EPV of premiums less expenses EPV of assurance = 0.9 P ä m f 70:64 = 0.9 P ( ) = P. = 10,000Ām f 70:64 ( ) = 10, δ ā m f 70:64 ( ) 10, (a m f 70: ) = 10,000 ( ( )) = 5,

43 EPV of reversionary annuity ( ) = 5, 000 ā m f āf m ( ) = 5, 000 a m f af m ( ) = 5, 000 a f 64 am f 70:64 + am 70 a f m 64:70 = 5,000 ( ) = 18, Now set: EPV[Income] = EPV[Outgo] = P = 5, ,450.0 = P = 4,

44 Contingent assurances Consider a contingent assurance with sum assured $1 payable immediately on the death of(x),if(y)isstillthenalive. Theprobabilitythatlife(x)dieswithin tyearswithlife(y)beingalivewhenlife(x)diesis denoted t qxy 1 and: tqxy 1 = P[(x)dieswithin tyearsandbefore (y)] = P[T x < tand T x < T y ]. This is an example of a contingent probability,so called because they are associated with thedeathofalifecontingentonthesurvivalordeathofanotherlife. Nowlet f x (r)bethedensityof T x and f y (r)bethedensityof T y,then: t tq 1 xy = r=0 s=r f x (r)f y (s)ds dr 174

45 = t r=0 [ f x (r) s=r ] f y (s)ds dr. Since: then: tq 1 xy = s=r t r=0 f y (s)ds = r p y rp x µ x+r r p y dr. Illustration: (x) survives to time r and then dies and (y) survives beyond r, giving: rp x µ x+r r p y (x)dies = µ x+r Age (x)survives = r p (y)survives x { }} { = r p y x x + r y 175y + r

46 Notethatsinceoneof (x)and (y)hastodiefirst: tq 1 xy + t q 1 xy = t q xy. Wedefine t q 2 xy astheprobabilitythat (x)dieswithin tyearsbutafter (y)hasdied. Therefore: tq x = t q 1 xy + t q 2 xy from which: tq 2 xy = t 0 rp x µ x+r (1 r p y ) dr. WiththeseprobabilitieswecannowevaluatetheEPV Ā1 xy. Ā 1 xy = v T x if T x < T y 0 if T x T y. 176

47 Fromthisitcanbeshownthat: Ā 1 xy = 0 v r rp x µ x+r r p y dr. Similarly,wecanderiveanexpressionfortheEPVofabenefitof$1payableimmediately onthedeathof (x)ifitisafterthatof (y),denoted Ā2 xy : Ā 2 xy = 0 v r rp x µ x+r (1 r p y ) dr. We see, what is intuitively clear, that: Ā x = Ā1 xy + Ā2 xy. Other important relationships: (a) Ā xy = Ā1 xy + Ā 1 xy and Āxy = Ā2 xy + Ā 2 xy. 177

48 (b) A 1 xy = t=0 v t+1 tp xt p y q 1 x+t:y+t. (c) A xy = A 1 xy + A 1 xy and A xy = A 2 xy + A 2 xy. (d) A 1 xy (1 + i) 1 2 Ā 1 xy. Ifthetwolivesarethesameage,andthesamemortalitytableisappliestoboth,then: Ā 1 xx = 1 2Āxx A 1 xx = 1 2 A xx Ā 2 xx = 1 2Āxx A 2 xx = 1 2 A xx. 178

49 6.5 Examples Example 1 Find the expected present value of an annuity of $10,000 p.a. payable annually in advance toamanaged65,reducingto$5,000p.a.continuinginpaymenttohiswife,nowaged61, if she survives him. Basis: Mortality a(55) ultimate, male/female as appropriate Interest: 8% p.a. Ignore the possibility of divorce. Solution EPV = 10,000 ä m ,000 ä m f = 10,000 (a m ) + 5,000 a m f

50 = 10,000 ( ) + 5,000 ( ) a f 61 am f 65:61 = 10, 000 (8.4) + 5, 000 ( ) = 96, 525. Since: a m f 65: ( ) a m f 66:62 + am f 66:60 + am f 64:62 + am f 64:60 = 1 4 ( ) = Example 2 (a) Express n q 2 xy intermsofsinglelifeprobabilitiesandcontingentprobabilities referring to the first death. (b) Suppose µ x = 1 80 x for 0 x 80, 180

51 evaluate 20 q 2 40:50. Solution (a) t q 2 xy = = t r=0 t r=0 = t q y t q 1 xy. rp y µ y+r (1 r p x ) dr rp y µ y+r dr t r=0 rp y µ y+r r p x dr (b) n q x = 1 n p x = 1 exp { n 0 } µ x+t dt 181

52 { n } 1 = 1 exp 0 80 x t dt = 1 exp {[log (80 x t)] n 0 { ( )} } 80 x n = 1 exp log = n 80 x. 80 x andwehave n q 1 xy = = n 0 tp x t p y µ y+t dt 1 (80 x)(80 y) = n (80 x) 1 2 n2 (80 x)(80 y) n 0 (80 x t) dt 182

53 2 20 q40:50 = 20 q q 1 40:50 = = 1 6. Example 3 Consider a reversionary annuity of $1 p.a. payable quarterly in advance during the lifetime of (y)followingthedeathof (x).showthattheexpectedpresentvalueofthisbenefitis approximately equal to: a x y + 1 8Ā1 xy. Solution Hint:Fixatime ttorepresentthedeathof(x),asbelow: 183

54 (x)dies = µ x+t Age (x)survives = t p (y)survives x { }} { = t p y x x + t y y + t andgets ä (4) y+t Hence: EPV = = v t tp x µ x+t t p y ä (4) y+t dt v t tp x µ x+t t p y ( ā y+t v t tp x µ x+t t p y ā y+t dt ) dt 184

55 v t tp x µ x+t t p y dt = ā x y Ā1 xy a x y Ā1 xy. 185