6 Insurances on Joint Lives

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

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

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

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

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

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

EPV b 1 b 2 b 3 b 12 b 13 b 24 b 34 Ā xy 0 0 0 1 1 0 0 Ā xy 0 0 0 0 0 1 1 Ā 1 xy 0 0 0 1 0 0 0 Ā 1 xy 0 0 0 0 1 0 0 Ā 2 xy 0 0 0 0 0 1 0 Ā 2 xy 0 0 0 0 0 0 1 ā xy 1 0 0 0 0 0 0 ā x y 0 1 0 0 0 0 0 ā y x 0 0 1 0 0 0 0 ā xy 1 1 1 0 0 0 0 Example:Consider Āxy:n.Thiele sequationsare: 137

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

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

Section 6.4. 6.3 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

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

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

= 0.2 + 0.4 0.2 0.4 = 0.52 np xy = n p x n p y = 0.8 0.6 = 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

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

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

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

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

= 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

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

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

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

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

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 =. 10 0 ) 2 µ x+r dr tp s 70 tp n s 70 dt. )] 2 µ x+r dr Since: tp n s x = 1 ( ) 70 70 + t 80 70 = 10 t 10 153

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 +1. 154

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 +1. 155

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

= 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

e.g.a(55)tables,butnotethatthevaluesof a xy aregivennot ä xy,andtheyareonly givenforevenages mayneedtouselinearinterpolation. For example: a 66:61 1 2 (a 66:60 + a 66:62 ). (ii) Given commutation functions. e.g. A1967 70 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

= 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

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 80:74 + 1 + a 80:76 ) = 0.5(4.395 + 4.538) = 4.467. 160

10 ä 70:65 = v 10 10p 70 10 p 65 ä 80:75 = 1 ( v 10 v 10 10 p 70 v 10 ) 10p 65 ä80:75 ( ) ( ) D80 D75 = 1 v 10 D 70 ( 3440.5 =(1.06) 10 11559 = 1.054. D 65 ä 80:75 ) ( 8536.2 19281 ) 4.467 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

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 + 0.5. 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

forthestandarddeviationofthepresentvalueofthisbenefit,intermsofsinglelife and joint life annuity functions. Solution: (a)wewant: 2000 a m f 68:65 = 2000 2000 ( a m 68 + a f 65 am f 68:65 ( a m 68 + a f 65 1 2 = 2000(8.688 + 11.497 = 25, 766.00. ) ( a m f 68:64 + am f 68:66 1 2 (7.418 + 7.186) ) ) ) 163

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

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 [ ( )] } 1 2 1 d ä m 68 + ä f 65 äm f 2 68:65. 165

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

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

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

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

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

(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

Solution: Let P = Annual Premium. EPV of premiums less expenses EPV of assurance = 0.9 P ä m f 70:64 = 0.9 P (1 + 5.576) = 5.9184 P. = 10,000Ām f 70:64 ( ) = 10, 000 1 δ ā m f 70:64 ( ) 10, 000 1 0.076961(a m f 70:64 + 0.5) = 10,000 (1 0.076961(5.576 + 0.5)) = 5, 323.85. 172

EPV of reversionary annuity ( ) = 5, 000 ā m f 70 64 + āf m 64 70 ( ) = 5, 000 a m f 70 64 + af m 64 70 ( ) = 5, 000 a f 64 am f 70:64 + am 70 a f m 64:70 = 5,000 (8.574 + 6.268 2 5.576) = 18, 450.0. Now set: EPV[Income] = EPV[Outgo] = 5.9184 P = 5,323.85 + 18,450.0 = P = 4, 016.94. 173

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

= 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

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

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

(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

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 65 + 5,000 ä m f 65 61 = 10,000 (a m 65 + 1) + 5,000 a m f 65 61 179

= 10,000 (7.4 + 1) + 5,000 ( ) a f 61 am f 65:61 = 10, 000 (8.4) + 5, 000 (9.124 6.619) = 96, 525. Since: a m f 65:61 1 4 ( ) a m f 66:62 + am f 66:60 + am f 64:62 + am f 64:60 = 1 4 (6.392 + 6.519 + 6.709 + 6.854) = 6.619. 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

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

{ 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

2 20 q40:50 = 20 q 50 20 q 1 40:50 = 20 30 20 40 1 2 202 40 30 = 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

(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 = = 0 0 0 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 + 1 8 v t tp x µ x+t t p y ā y+t dt ) dt 184

+ 1 8 0 v t tp x µ x+t t p y dt = ā x y + 1 8 Ā1 xy a x y + 1 8 Ā1 xy. 185