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Solution to Selected Problems Solution to selected problems. Chapter 1. Preliminaries 1. 8A 2 FS, 8t ¸ 0, A \ fT · tg = (A \ fS · tg) \ fT · tg, since fT · tg ½ fS · tg. Since A \ fS · tg 2 Ft and fT · tg 2 Ft, A \ fT · tg 2 Ft. Thus FS ½ FT . 2. Let Ω = N and F = P(N) be the power set of the natural numbers. Let Fn = σ(f2g; f3g;:::; fn+ 1g), 8n. Then (Fn)n¸1 is a filtration. Let S = 3 ¢ 13 and T = 4. Then S · T and ( f3g if n = 3 f! : S(!) = ng = ; otherwise ( Ω if n = 4 f! : T (!) = ng = ; otherwise Hence fS = ng 2 Fn, fT = ng 2 Fn, 8n and S; T are both stopping time. However f! : T ¡ S = 1g = f! : 1f3g(!) = 1g = f3g 2= F1. Therefore T ¡ S is not a stopping time. 3. Observe that fTn · tg 2 Ft and fTn < tg 2 Ft for all n 2 N , t 2 R+, since Tn is stopping time and we assume usual hypothesis. Then (1) supn Tn is a stopping time since 8t ¸ 0, fsupn Tn · tg = \nfTn · tg 2 Ft. (2) infn Tn is a stopping time since finfn Tn < tg = [fTn < tg 2 Ft (3) lim supn!1 is a stopping time since lim supn!1 = infm supn¸m Tn (and (1), (2).) (4) lim infn!1 is a stopping time since lim infn!1 = supm infn¸m Tn (and (1), (2).). 4. T is clearly a stopping time by exercise 3, since T = lim infn Tn. FT ½ FTn , 8n since T · Tn, and FT ½ \nFTn by exercise 1. Pick a set A 2 \nFTn .8n ¸ 1, A 2 FTn and A \ fTn · tg 2 Ft. Therefore A \ fT · tg = A \ (\nfTn · tg) = \n (A \ fTn · tg) 2 Ft. This implies \nFTn ½ FT and completes the proof. p P p p 5. (a) By completeness of L space, X 2 L . By Jensen’s inequality, EjMtj = EjE(XjFt)j · p p E [E(jXj jFt)] = EjXj < 1 for p > 1. p 1 (b) By (a), Mt 2 L ½ L . For t ¸ s ¸ 0, E(MtjFs) = E(E(XjFt)jFs) = E(XjFs) = Ms a.s. fMtg is a martingale. Next, we show that fMtg is continuous. By Jensen’s inequality, for p > 1, n p n p n p EjMt ¡ Mtj = EjE(M1 ¡ XjFt)j · EjM1 ¡ Xj ; 8t ¸ 0: (1) n p n p It follows that supt EjMt ¡ Mtj · EjM1 ¡ Xj ! 0 as n ! 1. Fix arbitrary " > 0. By Chebychev’s and Doob’s inequality, µ ¶ µ ¶p n p n 1 n p p supt EjMt ¡ Mtj P sup jMt ¡ Mtj > " · p E(sup jMt ¡ Mtj ) · p ! 0: (2) t " t p ¡ 1 " n Therefore M converges to M uniformly in probability. There exists a subsequence fnkg such that Mnk converges uniformly to M with probability 1. Then M is continuous since for almost all !, it is a limit of uniformly convergent continuous paths. 1 6. Let p(n) denote a probability mass function of Poisson distribution with parameter ¸t. Assume ¸t is integer as given. X¸t ¡ ¡ EjNt ¡ ¸tj =E(Nt ¡ ¸t) + 2E(Nt ¡ ¸t) = 2E(Nt ¡ ¸t) = 2 (¸t ¡ n)p(n) n=0 à ! X¸t (¸t)n X¸t (¸t)n ¸tX¡1 (¸t)n =2e¡¸t (¸t ¡ n) = 2¸te¡¸t ¡ (3) n! n! n! n=0 n=0 n=0 (¸t)¸t =2e¡¸t (¸t ¡ 1)! 7. Since N has stationary increments, for t ¸ s ¸ 0, 2 2 2 E(Nt ¡ Ns) = ENt¡s = V ar(Nt¡s) + (ENt¡s) = ¸(t ¡ s)[1 + ¸(t ¡ s)]: (4) 2 2 As t # s (or s " t), E(Nt ¡ Ns) ! 0. N is continuous in L and therefore in probability. 8. Suppose ¿® is a stopping time. A 3-dimensional Brownian motion is a strong Markov process we know that P (¿® < 1) = 1. Let’s define Wt := B¿®+t ¡B¿® . W is also a 3-dimensional Brownian W motion and its argumented filtration Ft is independent of F¿®+ = F¿® . Observe kW0k = kB¿® k = W W ®. Let S = inftft > 0 : kWtk · ®g. Then S is a Ft stopping time and fS · sg 2 Fs . So fS · sg has to be independent of any sets in F¿® . However f¿® · tg \ fS · sg = ;, which implies that f¿® · tg and fS · sg are dependent. Sine f¿® · tg 2 F¿® , this contradicts the fact that F¿® and W Ft are independent. Hence ¿® is not a stopping time. 2 n Pn i 9. (a) Since L -space is complete, it suffices to show that St = i=1 Mt is a Cauchy sequence w.r.t. n. For m ¸ n, by independence of fM ig, à ! m 2 m m X X ¡ ¢2 X 1 E(Sn ¡ Sm)2 = E M i = E M i = t : (5) t t t t i2 i=n+1 i=n+1 i=n+1 P1 2 n m 2 n , Since i=1 1=i < 1, as n; m ! 1, E(St ¡ St ) ! 1. fSt gn is Cauchy and its limit Mt is well defined for all t ¸ 0. (b)First, recall Kolmogorov’sP convergence criterion:P Suppose fXigi¸1 is a sequence of independent random variables. If i V ar(Xi) < 1, then i(Xi ¡ EXi) converges a.s. i i i For all i and t, 4Mt = (1=i)4Nt and 4Mt > 0. By Fubini’s theorem and monotone convergence theorem, X X X1 X1 X X1 X 4N i X1 N i 4M = 4M i = 4M i = s = t : (6) s s s i i s·t s·t i=1 i=1 s·t i=1 s·t i=1 i Let Xi = (Nt ¡t)=i. Then fXigi is a sequence of independent random variables such that EXi = 0, 2 P1 V ar(Xi) = 1=i and hence i=1 V ar(Xi) < 1. Therefore Kolmogorov’s criterion implies that P1 P1 P i=1 Xi converges almost surely. On the other hand, i=1 t=i = 1. Therefore, s·t 4Ms = P1 i i=1 Nt =i = 1. 2 P 1 i P 1 i 10. (a) Let Nt = i i (Nt ¡ t) and Lt = i i (Lt ¡ t). As we show in exercise 9(a), N, M are well defined in L2 sense. Then by linearity of L2 space M is also well defined in L2 sense since X 1 £ ¤ X 1 X 1 M = (N i ¡ t) ¡ (Li ¡ t) = (N i ¡ t) ¡ (Li ¡ t) = N ¡ L : (7) t i t t i t i t t t i i i Both terms in right size are martingales change only by jumps as shown in exercise 9(b). Hence Mt is a martingale which changes only by jumps. P (b) First show that given two independent Poisson processes N and L, s>0 4Ns4Ls = 0 a.s., i.e. N and L almostP surely don’t jumpP simultaneously. Let fTngn¸1 be aP sequence of jump times of a process L. Then s>0 4Ns4Ls = n 4NTn . We want to show that n 4NTn = 0 a.s. Since 4NTn ¸ 0, it is enough to show that E4NTn = 0 for 8n ¸ 1. Fix n ¸ 1 and let ¹Tn be a induced probability measure on R+ of Tn. By conditioning, Z 1 Z 1 E(4NTn ) = E [E (4NTn jTn)] = E (4NTn jTn = t) ¹Tn (dt) = E (4Nt) ¹Tn (dt); (8) 0 0 where last equality is by independence of N and Tn. It follows that E4NTn = E4Nt. Since 1 4Nt 2 L and P (4Nt = 0) = 1 by problem 25, E4Nt = 0, hence E4NTn = 0. Next we show that the previous claim holds even when there are countably many Poisson processes. i assume that there exist countably many independent Poisson processes fN gi¸1. Let A ½ Ω be a i set on which more than two processes of fN gi¸1 jump simultaneously. Let Ωij denotes a set on i j which N and N don’t jump simultaneously. Then P (Ωij) = 1 for i 6= j by previous assertion. c P c Since A ½ [i>jΩij, P (A) · i>j P (Ωij) = 0. Therefore jumps don’t happen simultaneously almost surely. Going back to the main proof, by (a) and the fact that N and L don’t jump simultaneously, 8t > 0, X X X j4Msj = j4Nsj + j4Lsj = 1 a:s: (9) s·t s·t s·t 11. Continuity: We use notations adopted in Example 2 in section 4 (P33). Assume EjU1j < 1. By independence of Ui, elementary inequality, Markov inequality, and the property of Poisson process, we observe X lim P (jZt ¡ Zsj > ²) = lim P (jZt ¡ Zsj > ² jNt ¡ Ns = k)P (Nt ¡ Ns = k) s!t s!t k X Xk X h ² i · lim P ( jUij > ²)P (Nt ¡ Ns = k) · lim kP (jU1j > ) P (Nt ¡ Ns = k) (10) s!t s!t k k i=1 k X EjU1j 2 EjU1j · lim k P (Nt ¡ Ns = k) = lim f¸(t ¡ s)g = 0 s!t ² s!t ² k 3 Independent Increment: Let F be a distribution function of U. By using independence of fUkgk and strong Markov property of N, for arbitrary t; s : t ¸ s ¸ 0, ³ ´ ³ PN P ´ iu(Z ¡Z )+ivZ iu t U +iv Ns U E e t s s = E e k=Ns+1 k k=1 k ³ ³ ´´ PNt PNs iu Uk+iv Uk =E E e k=Ns+1 k=1 jFs ³ ³ P ´´ ³ ³ P ´´ (11) PNs Nt PNs Nt iv Uk iu Uk iv Uk iu Uk =E e k=1 E e k=Ns+1 jFs = E e k=1 E e k=Ns+1 ³ P ´ ³ PN ´ ³ ´ iv Ns U iu t U ¡ ivZ ¢ iu(Z ¡Z ) =E e k=1 k E e k=Ns+1 k = E e s E e t s : This shows that Z has independent increments.
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