Multiple Life Models

1 Multiple Life ModelsLecture: Weeks 9-10 Lecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez1 / 38 Chapter summaryChapter summaryApproaches to studying Multiple life Models :define Multiple statestraditional approach (use joint random variables)Statuses:joint life statuslast-survivor statusInsurances and annuities involving Multiple livesevaluation using special mortality lawsSimple reversionary annuitiesContingent probability functionsDependent lifetime modelsChapter 9 (Dickson et al.)Lecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez2 / 38 Approachesmultiple statesStates in a joint life and last survivor model 02x+t:y+t 13x+t 01x+t:y+t 23y+txaliveyalive(0)xaliveydead(1)xdeady alive(2)xdeadydead(3)Lecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez3 / 38 Approachesjoint future lifetimesJoint distribution of future lifetimesConsider the case of two lives currently agesxandywith respectivefuture cumulative dist.

2 Function:FTxTy(s,t) =Pr[Tx s,Ty t] independence :FTxTy(s,t) =Pr[Tx s] Pr[Ty t] =Fx(s) Fy(t)Joint density function:fTxTy(s,t) = 2 FTxTy(s,t) s tindependence:fTxTy(s,t) =fx(s) fy(t)Joint survival dist. function:STxTy(s,t) =Pr[Tx> s,Ty> t] independence :STxTy(s,t) =Pr[Tx> s] Pr[Ty> t] =Sx(s) Sy(t)Lecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez4 / 38 ApproachesillustrationIllustrative example 1 Consider the joint density expressed byfTxTy(s,t) =164(s+t),for0< s <4,0< t < thatTxandTyare not the covariance the probability(x)outlives(y)by at least one to be discussed in : Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez5 / 38 Statusesjoint life statusThe joint life statusThis is a status that survives so long as all members are alive, andtherefore fails upon the first :(xy)for two lives(x)and(y)For two lives.

3 Txy= min(Tx,Ty)Cumulative distribution function:FTxy(t) =qt xy=Pr[min(Tx,Ty) t]= 1 Pr[min(Tx,Ty)> t]= 1 Pr[Tx> t,Ty> t]= 1 STxTy(t,t)= 1 pt xywherept xy=Pr[Tx> t,Ty> t] =STxy(t)is the probability that bothlives(x)and(y)survive : Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez6 / 38 Statusesjoint life statusThe case of independenceAlternative expression for the distribution function:FTxy(t) =Fx(t) +Fy(t) FTxTy(t,t)In the case whereTxandTyare independent:pt xy=Pr[Tx> t,Ty> t]=Pr[Tx> t] Pr[Ty> t]=pt x pt yandqt xy=qt x+qt y qt x qt yRemember this (even in the case of independence ):qt xy6=qt x qt yLecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez7 / 38 Statuseslast-survivor statusThe last-survivor statusThis is a status that survives so long as there is at least one member alive,and therefore fails upon the last :(xy)For two lives:Txy= max(Tx,Ty)General relationship amongTxy,Txy,Tx, andTy:Txy+Txy=Tx+TyTxy Txy=Tx TyaTxy+aTxy=aTx+aTyfor any constanta > each outcome, note thatTxyis equal eitherTxorTy, andtherefore,Txyequals the : Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez8 / 38 StatusesdistributionDistribution ofTxyRecall method of inclusion-exclusion of probability:Pr[A B] +Pr[A B] =Pr[A] +Pr[B].

4 Choose eventsA={Tx t}andB={Ty t}so thatA B={Txy t}andA B={Txy t}.This leads us to the following useful relationships:FTxy(t) +FTxy(t) =Fx(t) +Fy(t)STxy(t) +STxy(t) =Sx(t) +Sy(t)pt xy+ptxy=pt x+pt yfTxy(t) +fTxy(t) =fx(t) +fy(t)These relationships lead us to finding distributions ofTxy, (t) =Fx(t) +Fy(t) FTxy(t) =FTxTy(t,t)which is obvious fromFTxy(t) =Pr[Tx t Ty t].Lecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez9 / 38 StatusesdistributionInterpretation of probabilitiesNote that:pt xyis the probability that both lives(x)and(y)will be alive the probability that at least one of lives(x)and(y)will be contrast:qt xyis the probability that at least one of lives(x)and(y)will be the probability that both lives(x)and(y)will be dead : Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez10 / 38 StatusesillustrationIllustrative example 2 For independent lives(x)and(y), you are given.

5 Qx= ,andqx+1= +1= are assumed to be uniformly distributed over each year of and interpret the following to be discussed in : Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez11 / 38 Force of mortalityjoint lifeForce of mortality ofTxyDefine the force of mortality (similar manner to any random variable): x+t:y+t=fTxy(t)1 FTxy(t)=fTxy(t)STxy(t)=fTxy(t)pt can then write the density ofTxyasfTxy(t) =pt xy x+t:y+tIn the case of independence , we have: x+t:y+t=pt x pt y( x+t+ y+t)pt x pt y= x+t+ y+ force of mortality of the joint life status is the sum of theindividuals force of mortality, when lives are : Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez12 / 38 Force of mortalitylast-survivorForce of mortality forTxyThe force of mortality forTxyis defined as x+t:y+t=fTxy(t)1 FTxy(t)=fTxy(t)STxy(t)=fx(t) +fy(t) fTxy(t)ptxy=pt x x+t+pt y y+t pt xy x+t:y+tptxyIndeed we have the density ofTxyexpressed asfTxy(t) =ptxy x+t:y+ what this formula gives in the case of.

6 Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez13 / 38 Insurance benefitsdiscreteInsurance benefits - discreteConsider an insurance under which the benefit of $1 is paid at theEOY of ending (failure) of be any joint life or last survivor status , time at which the benefit is paid:Ku+ 1the present value (at issue) of the benefit:Z=vKu+1 APV of benefits: E[Z] =Au= k=0vk+1 Pr[Ku=k]variance: Var[Z] =A2u (Au)2 Lecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez14 / 38 Insurance benefitscontinuousInsurance benefits - continuousConsider an insurance under which the benefit of $1 is paidimmediately of ending (failure) of be any joint life or last survivor status , time at which the benefit is paid:Tuthe present value (at issue) of the benefit:Z=vTuAPV of benefits: E[Z] = Au= 0vt pt u u+tdtvariance: Var[Z] = A2u ( Au)2 Lecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez15 / 38 Insurance benefitscontinuousSome illustrationsFor a joint life status(xy), consider whole life insurance providingbenefits at the first death.

7 Axy= k=0vk+1 qk|xy= k=0vk+1 pk xy qx+k:y+k Axy= 0vt pt xy x+t:y+tdtFor a last-survivor status(xy), consider whole life insurance providingbenefits upon the last death:Axy= k=0vk+1 qk|xy= k=0vk+1 (qk|x+qk|y qk|xy) Axy= 0vt ptxy x+t:y+tdt= 0vt(pt x x+t+pt y y+t pt xy x+t:y+t)dtLecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez16 / 38 Insurance benefitscontinuous- continuedUseful relationships:Axy+Axy=Ax+Ay Axy+ Axy= Ax+ AyLecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez17 / 38 Annuity benefitsdiscreteAnnuity benefits - discreteConsider ann-year temporary life annuity-due on present value (at issue) of the benefit:Y={ aKu+1, Ku< n an,Ku nAPV of benefits: E[Y] = au:n= n 1k=0 ak+1 qk|u+ an pn uvariance: Var[Y] =1d2[A2u:n (Au:n)2]Other ways to write APV: au:n=n 1 k=0vk pk u=1d(1 Au:n).}

8 Lecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez18 / 38 Annuity benefitscontinuousAnnuity benefits - continuousConsider an annuity for which the benefit of $1 is paid each yearcontinuously for years so long as a present value (at issue) of the benefit:Y= aTuAPV of benefits: E[Y] = au= 0 at pt u u+tdt= 0vtpt udtvariance: Var[Y] =1 2[ A2u ( Au)2]Note that the identity aTu+vTu= 1provides the connectionbetween insurances and : Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez19 / 38 Annuity benefitscontinuousSome illustrationsFor joint life status(xy), consider a whole life annuity providingbenefits until the first death: axy= k=0vk pk xyand axy= 0vt pt xydtFor last survivor status(xy), consider a whole life insurance providingbenefits upon the last death: axy= k=0vk pkxyand axy= 0vt ptxydtUseful relationships: axy+ axy= ax+ ay axy+ axy= ax+ ayLecture.

9 Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez20 / 38 Annuity benefitscontinuousComparing benefits - annuitiesType of life annuitySingle lifexJoint life statusxyLast survivor statusxyWhole life a-due ax axy axyWhole life a-immediateaxaxyaxyTemporary life a-due ax:n axy:n axy:nTemporary life a-immediateax:naxy:naxy:nWhole life a-continuous ax axy axyTemporary life a-continuous ax:n axy:n axy:nLecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez21 / 38 Annuity benefitscontinuousComparing benefits - insurancesType of life insuranceSingle lifexJoint life statusxyLast survivor statusxyWhole life - discreteAxAxyAxyWhole life - continuous Ax Axy AxyTerm - discreteA1x:nA1 xy8:nA1xy:nTerm - continuous A1x:n A1 xy8:n A1xy:nEndowment - discreteAx:nAxy:nAxy:nEndowment - continuous Ax:n Axy:n Axy:nPure endowmentA1x:norEn xA1xy:norEn xyA1xy:norEnxyLecture: Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez22 / 38 Annuity benefitscontinuousIllustrative example 3 You are given.

10 (45)and(65)have independent future for either life follows deMoivre s law with = 105. = 5%Calculate A45 : Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez23 / 38 Contingent functionsContingent functionsIt is possible to compute probabilities, insurances and annuities basedon the failure of the status that is contingent on the order of thedeaths of the members in the group, (x)dies before(y).These are called contingent the probability that(x)fails before(y)- assumingindependence:Pr[Tx< Ty] = 0fTx(t) STy(t)dt= 0pt x x+t pt ydt= 0pt xy x+tdtThe actuarial symbol for this isq1 xy. It should be obvious this is thesame asq2 : Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez24 / 38 Contingent functions- continuedThe probability that(x)dies before(y)and withinnyears is given byq1n xy= n0pt xy x+ , we have the probability that(y)dies before(x)and withinnyears:q1n xy= n0pt xy y+ is easy to show thatq1n xy+q1n xy=qn can similarly define and interpret the following:q2n xyandq2n xy,and show thatq2n xy+q2n xy= : Weeks 9-10 (STT 456) Multiple Life ModelsSpring 2015 - Valdez25