Actuarial Models 1: solutions example sheet 4

Transcription

1 Actuarial Moel 1: olution example heet 4 (a) Anwer to Exercie 4.1 Q ( e u ) e σ σ u η η. (b) The forwar equation correponing to the backwar tate e are t p ee(t) σp ee (t) + ηp eu (t) t p eu(t) σp ee (t) ηp eu (t), with bounary conition p ee (0) 1 an p eu (0) 0. (c) Working with the firt equation an replacing p eu (t) by 1 p ee (t) (a p ee (t)+p eu (t) 1) we get t p ee(t) (σ + η)p ee (t) + η. T i a firt orer, linear, non-homogeneou ODE that we nee to olve. There are everal way to o t, but here we follow the metho of firt etermining the general olution of the homogeneou verion of the ODE an then fining a particular olution to the non-homogeneou ODE 1. Let firt olve the homogeneou part of above ODE, i.e. (t) (σ + η)p hom (t). We ee that t phom ee ee t log ( p hom ee (t) ) t phom ee p hom ee (t) (σ + η) (t) an o by integrating on both ie we ee that the general olution to the homogeneou part i given by p hom ee (t) Ce (σ+η)t for ome contant C. Secon, we try to fin a particular olution to the non-homogeneou ODE (note that for uch a particular olution the bounary conition oe not have to be atifie). For t we ue the metho of unetermine coefficient. A the non-homogeneou part of the ODE i η wch i contant in t, we gue that a particular olution might be alo a contant in t. So we try p part ee (t) A. Then by the ODE we mut have 0 (σ + η)a + η, i.e. A η. Hence ppart σ+η ee (t) η i a particular olution to the non-homogeneou ODE. σ+η Hence (by a theorem from the theory of ODE) the general olution to the original ODE i given by p ee (t) p hom ee (t) + p part ee (t) Ce (σ+η)t + η σ + η. 1 Alternative metho are uing an integrating factor or uing the general formula (3). 1

2 Invoking the bounary conition p ee (0) 1 lea to C σ. Hence in the en we σ+η have p ee (t) σ σ + η e (σ+η)t + η σ + η, p eu (t) 1 p ee (t) σ ( ) 1 e (σ+η)t. σ + η Remark In general for the non-homogeneou, firt orer, linear ODE, f (t) g(t)f(t) + h(t), (1) where g an h are given function (that are (locally) boune an continuou) an with bounary conition f(t 0 ) a, (2) with t 0, a R given, we have that it olution i given by ( t t ) t g(u)u f(t) e 0 h()e t g(u)u 0 + a. (3) t 0 Inee, it i eay to check, via the prouct rule of ifferentation an the funamental theorem of calculu, that (3) i a - an thu the unique - olution to the ODE (1) t with bounary conition (2). Note that Ce g(u)u 0 with C a contant i the general olution of the homogeneou part of the ODE (i.e. f (t) g(t)f(t)), wherea t g(u)u e 0 t 0 h()e t g(u)u 0 i a particular olution to the ODE (1) (but oe not atify the bounary conition (2) unle a 0). Anwer to Exercie 4.2 The Q-matrix, wch epen on age t, i given by (µ(t) + ν(t)) µ(t) ν(t) Q So for each 0, the Kolmogorov forwar equation for the tranition probabilitie with backwar tate 1 are given by t p 11(, t) (µ(t) + ν(t))p 11 (, t), t p 12(, t) µ(t)p 11 (, t), t p 13(, t) ν(t)p 11 (, t), where t > with bounary conition p 11 (, ) 1, p 12 (, ) 0 an p 13 (, ) 0. The olution of the firt equation i eaily een to be given by p 11 (, t) e (µ(u)+ν(u))u. (4) 2

3 In orer to ee t, note that by the firt ODE we have for all u >, u log (p 11(, u)) p u 11(, u) p 11 (, u) (µ(u) + ν(u)) an thu integrating both ie with repect to u from to t an uing the bounary conition, log(p 11 (, t)) log(p 11 (, t)) log(p 11 (, )) u log (p 11(, u)) u (µ(u)+ν(u))u. Hence (4) follow. For p 12 (, t), we have by uing repectively the bounary conition p 12 (, ) 0, the funamental theorem of calculu, the econ ODE an (4), p 12 (, t) p 12 (, t) p 12 (, ) u p 12(, u)u µ(u)p 11 (, u)u µ(u)e u (µ(v)+ν(v))v u. Similarly, we get p 12 (, t) ν(u)e u (µ(v)+ν(v))v u. Anwer to Exercie 4.3 The Q-matrix for t moel i given by (λ + µ) λ µ Q 0 ν ν, whereby h i the firt tate, i i the econ an the tr. (a) The 3-imenional ytem of Kolmogorov forwar equation correponing to the backwar tate i are: t p ih(t) (λ + µ)p ih (t) t p ii(t) λp ih (t) νp ii (t) t p i(t) µp ih (t) + νp ii (t). We ee that the firt ODE involve only p ih (t). Hence we can olve it by eparation of variable to get p ih (t) Ae (λ+µ)t, t 0. 3

4 Since p ih (0) 0, we mut have A 0 an o p ih (t) 0 for all t 0. Now the econ ODE become t p ii(t) νp ii (t) an ince p ii (0) 1, we euce p ii (t) e νt. We can fill t in into the tr ODE an olve for p i (t), or intea we ue the fact that p ih (t)+p ii (t)+p i (t) 1 to euce that p i (t) 1 p ii (t) 1 e νt. (b) The 3-imenional ytem of Kolmogorov forwar equation correponing to the backwar tate h are: t p hh(t) (λ + µ)p hh (t) t p (t) λp hh (t) νp (t) t p h(t) µp hh (t) + νp (t). Note that they are baically the ame a in part (a); only the backwar tate change from i to h. However, the olution will be ifferent becaue the bounary conition are not the ame. The firt ODE together with the bounary conition p hh (0) 1 give p hh (t) e (λ+µ)t, t 0. The econ ODE now become t p (t) λe (λ+µ)t νp (t). The olution to t ODE i of the form p (t) p hom (t) + p part (t), where p hom (t) i the general olution to the homogeneou problem: t phom (t) νp hom (t) an p part (t) i a particular olution to the original ODE. We can eaily euce that p hom (t) Ae νt, where A i a contant. Further, a the non-homogeneou part i equal to e (λ+µ)t it i a goo iea to tart looking for a particular olution of the form p part (t) Be (λ+µ)t, with B another contant uch that the original ODE i atifie. We eaily euce that the contant B mut be uch that B(λ + µ) λ νb. T equation only ha a olution if ν λ + µ in wch cae B λ/(λ + µ ν). Hence if ν λ + µ, then p (t) Ae νt 4 λ λ + µ ν e (λ+µ)t.

5 The contant A can now be euce from p (0) 0. T give A λ. λ+µ ν If ν λ + µ we have to fin a particular olution of a ifferent form. The next bet gue i to try p part (t) Bte νt. We ee that p part (t) atifie the ODE if B λ. In t cae the contant A will then be equal to zero. Hence we have { λ ( λ+µ ν e νt e (λ+µ)t) if ν λ + µ, p (t) λte νt if ν λ + µ. Note that by l Hôpital rule lim (λ+µ) ν λ ( e νt e (λ+µ)t) λte νt λ + µ ν an t coul alo have been ue to gue the olution of the ODE in the cae where ν λ + µ. (Alternatively, one can ue the general formula (3).) Uing p hh (t) + p (t) + p h (t) 1, we euce that { ( 1 1 λ+µ ν λe νt + (µ ν)e (λ+µ)t) if ν λ + µ, p h (t) 1 (1 + λt)e νt if ν λ + µ. Anwer to Exercie 4.4 We nee to prove p ik (, t) p ij(, u)p jk (u, t) for all 0 < u < t an i, k S. We have p ik (, t) Pr(X t k X i) Pr(X t k, X u j X i) Pr(X t k X u j, X i)pr(x u j X i) Pr(X t k X u j)pr(x u j X i) p ij (, u)p jk (u, t), where we have ue the aitivity property of (conitional) probability meaure in the econ equality (note that j S {X t k, X u j} {X t k} an that {X t k, X u j} {X t k, X u i} for i j), the efinition of conitional probability in the tr equality an the Markov property in the fourth equality. 5

6 We have that Hence Anwer to Exercie 4.5 p il (, t) 1 ( t p il(, t) 1 t j1 p ij (, t) j1 p ik (, t). p ik (, t) t p ik(, t) ) p ij (, t)µ jk (t) p ij (, t)µ jl (t), j1 µ jk (t) where we have ue that ( ) i atifie for k l in the tr equality an the fact that the rowum of Q(t) are 0 (becaue Q(t) i a Q-matrix) in the lat equality. Hence ( ) i atifie for k l a well. 6