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Understanding Multiple State Models in Life Contingencies

School
York University**We aren't endorsed by this school
Course
MTH 3280
Subject
Mathematics
Date
Dec 12, 2024
Pages
25
Uploaded by ChiefOxide15741
Multiple state modelMathematics of life contingencies 1Multiple state modelJinghui ChenJinghui ChenMathematics of life contingencies 11/24
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An introduction and basic quantities of interestDefinition 5.1 (Stochastic process)Family of random variables{Tx:x∈ T }is called a stochasticprocess. It is called discrete if the setTis finite or countable,and it is called continuous otherwise. Also, the setSsuch thatTx∈ Sis called the state-space of the stochastic process{Tx:x∈ T }. Note thatTx: Ωx× T → S ⊆[0,∞).Definition 5.2 (Non-homegenious Markov chain)Stochastic process{Tx:x∈ T }is a Markov chain if (a) it isdiscrete, (b) has finite number of states, and (c)P(Tx+1=s|Tx=sx,Tx-1=sx-1, . . . ,T0=s0)=P(Tx+1=s|Tx=sx)for allx=1,2,3, . . .and alls0,s1, . . . ,st,s∈ S.JinghuiChenMaths of Life contingenices MATH 32802 / 24
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An introduction and basic quantities of interestiWe will use the notationpx,j=P(Tx+1=j|Tx=i)for allx=1,2,3, . . .and alls0,s1, . . . ,st,s∈ S.Definition 5.3 (Transition probability matrix)i jLetm= #(S), then matrixPn∈Matm×m([0,1])that haspx,+nas itsi-th row andj-th column entry is called transitionprobability matrix, and it gathers probabilities that(x)‘jumps’from thei-th state at timento thej-th state at timen+1.NoteWe must havemj=1i jmj=1Xpx,+n=XP(Tx+n+1=j|Tx+n=i) =1.This does not have to be so for the sum of columns of thematrixPn.JinghuiChenMaths of Life contingenices MATH 32803 / 24
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An introduction and basic quantities of interestExample 5.1 (Simplest alive-dead situation)SetS={1,2}, where{1}denotes “alive” and{2}denotes“dead”. ThenPn=px+nqx01+n,where obviously(Pn)1,1+ (Pn)1,2=px+n+qx+n=1 and also(Pn)2,1+ (Pn)2,2=0+1=1.JinghuiChenMaths of Life contingenices MATH 32804 / 24
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An introduction and basic quantities of interestExample 5.2 (Multiple-decrement situation)SetS={0,1,2, . . . ,m}, where{0}denotes “alive” and{1,2, . . . ,m}denote various reasons of decrement. Then thetransition matrixPnis(m+1)×(m+1)as followingPn=p(τ)x+nq(1)x+nq(2)x+n· · ·q(m)x+n010· · ·0001· · ·0· · ·· · ·· · ·· · ·· · ·000· · ·1,where obviously the sum of row elements is equal to one.JinghuiChenMaths of Life contingenices MATH 32805 / 24
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An introduction and basic quantities of interestExample 5.3 (Multiple-life situation)SetS={0,1,2,3}, where{0}denotes “both are alive”,{1}denotes “(x)is alive and(y)is dead”,{2}denotes “(x)is deadand(y)is alive”,{3}denotes “both are dead”. Then thetransition matrixPnis as following under the assumption ofindependence of the future life-time random variablesPn=px+npy+npx+nqy+nqx+npy+nqx+nqy+n0px+n0qx+n00py+nqy+n0001,where obviously the sum of row elements is equal to one.JinghuiChenMaths of Life contingenices MATH 32806 / 24
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An introduction and basic quantities of interestExample 5.4 (Disability situation)SetS={0,1,2,3}, where{0}denotes “(x)is active”,{1}denotes “(x)is temporarily disabled”,{2}denotes “(x)ispermanently disabled”,{3}denotes “(x)is dead”. Then thetransition matrixPnis such that(Pn)2,0= (Pn)2,1=0.(Pn)3,3=1 and so(Pn)3,0= (Pn)3,1= (Pn)3,2=0, whereasother probabilities are chosen so that they reflect observations.ther popular examples are driving ratings and continuing careretirement communities.JinghuiChenMaths of Life contingenices MATH 32807 / 24
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An introduction and basic quantities of interestProposition 5.1Let Txbe a non-homogeneous Markov Chain withS={1, . . . ,m}, thenk(Pn)i,j:=P(Tx+n+k=j|Tx+n=i) = (PnPn+1· · ·Pn+k-1)i,jfor non-negative and integer x,k,n and i,j∈ S.Proof.We will prove by induction. To start off, letn=0 for simplicity ofthe exposition and without loss of generality. So what is theprobabilityP(T2=jkT0=i)?JinghuiChenMaths of Life contingenices MATH 32808 / 24
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An introduction and basic quantities of interestProof.2(P0)i,j=P(T2=j|T0=i)=mXl=1P(T2=j|T1=l,T0=i)P(T1=l|T0=i)=mXl=1P(T2=j|T1=l)P(T1=l|T0=i)=mXl=1(P1)l,j(P0)i,l=mXl=1(P0)i,l(P1)l,j=(P0P1)i,j(?)= (P2)i,jfor alli,j∈ Sand the last equality holds if the Markov Chain ishomogeneous.JinghuiChenMaths of Life contingenices MATH 32809 / 24
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An introduction and basic quantities of interestProof.Further assume thatP(Tk-1=j|T0=i) = (P0P1· · ·Pk-2)i,jforalli,j∈ S, thenk(P0)i,j=P(Tk=j|T0=i)=mXl=1P(Tk=j|Tk-1=l,T0=i)P(Tk-1=l|T0=i)=mXl=1P(Tk=j|Tk-1=l)(P0P1· · ·Pk-2)i,l=mXl=1(Pk-1)l,j(P0P1· · ·Pk-2)i,l=mXl=1(P0P1· · ·Pk-2)i,l(Pk-1)l,j=(P0P1· · ·Pk-1)i,j(?)= (Pk)i,jfor alli,j∈ Sand the last equality holds if the Markov Chain ishomogeneous.JinghuiChenMaths of Life contingenices MATH 328010 / 24
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An introduction and basic quantities of interestCorollary 5.1LetT0,T1, . . .be a non-homogeneous Markov Chain withS={1, . . . ,m}and transition probability matricesP0,P1, . . .,thenk(Pn) =Pn×Pn+1× · · · ×Pn+k-1for non-negative and integerk,n. Also, if the Markov Chain ishomogeneous, thenk(Pn) = (Pkn)for non-negative and integerk,n.JinghuiChenMaths of Life contingenices MATH 328011 / 24
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An introduction and basic quantities of interestDenote byπthe distribution of the random variableTx, and letπk=P(Tx=k),k∈ S.Corollary 5.2We have that the distribution of the random variableTx+kisπ0(P0P1× · · · ×Pk-1).Proof.We haveP(Tx+k=j) =mXi=1P(Tx+k=j,Tx=i)=mXi=1P(Tx+k=j|Tx=i)P(Tx=i)=mXi=1(P0P1× · · ·Pk-1)i,jπi.JinghuiChenMaths of Life contingenices MATH 328012 / 24
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An introduction and basic quantities of interestProof.Last line yieldsP(Tx+k=j) =mXi=1πi(P0P1× · · ·Pk-1)i,j,which establishes the assertion.Proposition 5.2Consider again a non-homogeneous Markov Chain{Tx,x∈ T }, thenP(Tx+n+1=Tx+n+2=· · ·=Tx+n+k=i|Tx+n=i)=(Pn)i,i(Pn+1)i,i× · · · ×(Pn+k-1)i,i,for non-negative and integer n,k and i∈S.JinghuiChenMaths of Life contingenices MATH 328013 / 24
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An introduction and basic quantities of interestProof.As we have, by conditioning and evoking the Markovianproperty,P(Tx+n+k=i,Tx+n+k-1=i, . . . ,Tx+n+1=i,Tx+n=i)=P(Tx+n+k=i|Tx+n+k-1=i, . . . ,Tx+n+1=i,Tx+n=i)×P(Tx+n+k-1=i|Tx+n+k-2=i, . . . ,Tx+n+1=i,Tx+n=i)× · · · ×P(Tx+n=i)=P(Tx+n+k=i|Tx+n+k-1=i)×P(Tx+n+k-1=i|Tx+n+k-2=i)× · · · ×P(Tx+n+1=i|Tx+n=i)×P(Tx+n=i),then the conditional probability in the assertion follows.JinghuiChenMaths of Life contingenices MATH 328014 / 24
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An introduction and basic quantities of interestFrom now and on consider a continuous stochastic process{Yx(t)}t≥0with the state spaceS={0,1, . . . ,m},m∈NandT= [0,∞). We assume that:(a)For anys≥0 andi,j∈ S, the conditional probabilityP(Yx(t+s) =j|Yx(t) =i)is independent on the history of the process for all timesbeforet∈[0,∞).(b)For any time lengthh>0,P(2 or more transitions occure within h) =o(h),where we say that the functionf(h)iso(h)iflimh→0f(h)/h=0.JinghuiChenMaths of Life contingenices MATH 328015 / 24
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An introduction and basic quantities of interestWe will use the notationtpi,jx=P(Yx(t) =j|Yx=i),i,j∈ S,t≥0andtpi,ix=P(Yx(s) =j∀s∈[0,t]|Yx=i),i∈ S,t≥0(c)The functiont7→tpi,jxis differentiable for allt∈(0,∞).Note that now we can define the force of transition as followingμi,jx=limh↓0hpi,jxh,fori6=j∈ S.JinghuiChenMaths of Life contingenices MATH 328016 / 24
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An introduction and basic quantities of interestNote that we can say equivalently thathpi,jx=h×μi,jx+o(h),fori6=j∈ Sor in other wordshpi,jx≈h×μi,jx,fori6=j∈ S.The latter expression should remind you the simple alive-deadframework.Proposition 5.3For a general multiple-state model, we have for h>0,x≥0and i∈ S,hpi,ix=hpi,ix+o(h)JinghuiChenMaths of Life contingenices MATH 328017 / 24
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An introduction and basic quantities of interestProof.The right hand side is obtained by the law of total probability:hpi,ix=P(Yx(h) =i|Yx=i)=P(Yx(h) =i|Yx=i,∃t∈[0,h) :Yx(t)6=i)×P(∃t∈[0,h) :Yx(t)6=i|Yx=i)+P(Yx(h) =i|Yx=i,∀t∈[0,h) :Yx(t) =i)×P(∀t∈[0,h) :Yx(t) =i|Yx=i)=P(Yx(h) =i,Yx=i,∃t∈[0,h) :Yx(t)6=i)/P(Yx=i)+P(Yx(h) =i,Yx=i,∀t∈[0,h) :Yx(t) =i)/P(Yx=i)=P(Yx(h) =i,∃t∈[0,h) :Yx(t)6=i|Yx=i)+P(Yx(h) =i,∀t∈[0,h) :Yx(t) =i|Yx=i)=o(h) +hpi,ix,which completes the proof.JinghuiChenMaths of Life contingenices MATH 328018 / 24
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An introduction and basic quantities of interestProposition 5.4For any multiple-state model and for h>0,x≥0and i,j∈ S,we havehpi,ix=1-hmXj=0,j6=iμi,jx+o(h)Proof.We have1=hpi,ix+mXj=0,j6=ihpi,jx+o(h),or1=hpi,ix+hmXj=0,j6=iμi,jx+m×o(h),which completes the proof sincem×o(h) =o(h).JinghuiChenMaths of Life contingenices MATH 328018 / 24
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An introduction and basic quantities of interestProposition 5.5For any multiple-state model, we have for h≥0and i∈ S,hpi,ix=exp-Zh0mXj=0,j6=iμi,jx(s)dsProof.Start with the observationh+Δhpi,ix=hpi,ix×Δhpi,ix+h,which is true for allh≥0 because{Yx}h≥0is a Markovprocess.JinghuiChenMaths of Life contingenices MATH 328019 / 24
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An introduction and basic quantities of interestProof.Further evoking Proposition 1.4, we obtainh+Δhpi,ix=hpi,ix×1-ΔhmXj=0,j6=iμi,jx(h) +o(Δh)orh+Δhpi,ix-hpi,ix=-hpi,ix×ΔhmXj=0,j6=iμi,jx(h) +o(Δh)or forΔh>0h+Δhpi,ix-hpi,ixΔh=-hpi,ixmXj=0,j6=iμi,jx(h) +o(Δh)Δh.Then takeΔh↓0, and obtainJinghuiChenMaths of Life contingenices MATH 328020 / 24
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An introduction and basic quantities of interestProof.ddhhpi,ix=-hpi,ixmXj=0,j6=iμi,jx(h).This is an ODE we have already seen, and its solution isexactly the assertion of this proposition..Think of the ODEf0(h) =g(f(h),h),that has the initial conditionf(0) =c>0; setf(h) =hpx,g(f(h),h) =-hpx×μx(h)such that0px=1. Thesolution ishpx=exp(-Zh0μx(s)ds).JinghuiChenMaths of Life contingenices MATH 328021 / 24
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An introduction and basic quantities of interestProposition 5.6For any multiple-state model withS={0,1, . . . ,m∈N},i,j∈ S, we haveddhhpi,jx=Xk6=jhpi,kxμk,jx(h)-hpi,jxXk6=jμj,kx(h).Proof.Start with the expressionh+Δhpi,jx=Xk∈Shpi,kx×Δhpk,jx+h=Xk6=jhpi,kx×Δhpk,jx+h+hpi,jx×Δhpj,jx+h,which hold by conditioning.JinghuiChenMaths of Life contingenices MATH 328022 / 24
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An introduction and basic quantities of interestProof.Further we haveh+Δhpi,jx=Xk6=jhpi,kx×(Δhμk,jx(h) +o(Δh)) +hpi,jx×1-Xk6=jΔhpj,kx+h,which yieldsh+Δhpi,jx-hpi,jx=Xk6=jhpi,kx×(Δhμk,jx(h) +o(Δh))-hpi,jx×Xk6=jΔhμj,kx(h) +o(Δh).Now divide byΔh>0 throughout and getJinghuiChenMaths of Life contingenices MATH 328023 / 24
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An introduction and basic quantities of interestProof.h+Δhpi,jx-hpi,jxΔh=Xk6=jhpi,kx×μk,jx(h)-hpi,jx×Xk6=jμj,kx(h) +o(Δh).Finally, take the limitΔh↓0, and the assertion follows.JinghuiChenMaths of Life contingenices MATH 328024 / 24
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