Homework 5 Solutions

Transcription

1 Stat 310B/Math 230B Theory of Probabiity Homework 5 Soutions Andrea Montanari Due on 2/19/2014 Exercise [5.3.20] 1. We caim that n 2 [ E[h F n ] = 2 n i=1 A i,n h(u)du ] I Ai,n (t). (1) Indeed, integrabiity and F n -measurabiity of the RHS of (1) are obvious. Further, denoting the RHS of (1) by f n, ceary E[f n I Ai,n ] = E[hI Ai,n ] for each vaue of i and n (as E(I Ai,n ) = P(A i,n ) = 2 n ), which suffices since {A i,n : i = 1,..., 2 n } is a π-system. 2. Each function X n (t) is piecewise constant on the intervas generating F n, having the vaue h i,n on the interva A i,n. Thus, X n (t) is F n measurabe and further integrabe with n 2 c n = E X n = 2 n h i,n finite. It is easy to check that h i,n = (h 2i 1,n+1 + h 2i,n+1 )/2, namey, the constant vaue of X n (t) on each interva A i,n is the average of the vaues that X n+1 (t) take on the two adjacent intervas A 2i 1,n+1 and A 2i,n+1 into which A i,n spit. By part (a) E[X n+1 F n ] = 2 n i=1 i=1 g i,n I Ai,n (t), where g i,n is the expected vaue of X n+1 (t) with respect to the uniform measure on A i,n. Since A i,n is the disjoint union of the intervas A 2i 1,n+1 and A 2i,n+1 of same ength on each of which X n+1 (t) is constant, it foows that g i,n = h i,n and consequenty, that X n is a martingae. 3. Let c n = E X n and c = sup n c n. If c = then there exist n k k such that c nk 2 k. Taking for each k the coection of intervas A i,nk with the m = m (k) = 2 nk k argest vaues of h i,nk then yieds some s (k) 1 < t (k) 1 s (k) 2 < t (k) 2 t (k) m for which m =1 t (k) s (k) 2 k k 0 whie m =1 x(t (k) ) x(s (k) ) 1 contradicting the absoute continuity of x( ). Therefore, c < and hence for a ρ > 0 by Markov s inequaity n 2 2 n I hj,n >ρ = P( X n > ρ) E X n /ρ c/ρ (2) Further, note that j=1 E[ X n I Xn >ρ] = {j: h j,n >ρ} 2 n h j,n. (3) 1

2 Let ɛ > 0 be fixed and δ = δ(ɛ, x) > 0 be determined as in the definition of absoute continuity. Taking ρ = c/δ observe that (2) and (3) impy by the absoute continuity of x( ) that hence {X n } is U.I. E[ X n I Xn >ρ] ɛ for a n, 4. Assuming hereafter that x( ) is absoutey continuous, hence {X n } is a U.I. martingae, by Coroary , X n = E[h F n ] for some h = X L 1. Hence, by (1) we see that x(i2 n ) x((i 1)2 n ) = i2 n h(u)du for n = 0, 1, and i = 1,, 2 n. (i 1)2 n By inearity of the integra we thus have x(j2 n ) x(i2 n ) = j2 n i2 n h(u)du for a j i, and a n vaues. (4) Consider now arbitrary 1 > t s 0 and et j m i m, n m be such that j m 2 nm t and i m 2 nm s as m. By continuity of x( ), the LHS of (4) for this sequence converges to x(t) x(s), whie the RHS of (4) is 1 0 h(u)i [i m2 nm,jm2 nm ] (u)du, with the integrand converging a.e. to h(u)i [s,t] (u) and dominated by the integrabe function h. Thus, by dominated convergence these integras converge to t s h(u)du. 5. Consider now So that Hence h(t) = dx dt Exercise [5.3.39] s+ 1 [x(s + ) x(s)] h(s) = 1 [h(u) h(s)]du. im 0 1 (x(s + ) x(s)) h(s) im 1 0 a.e. [0, 1) as caimed. Set λ > 0 and et ψ = e λ λ 1. s s+ s h(u) h(s) du = 0 a.e. [0, 1). 1. Obviousy, N n is measurabe on F n. By our assumptions about the L 2 martingae (M n, F n ), part (a) of Exercise appies for the aw of Y = λ(m n+1 M n ) conditiona on F n, taking there κ = λ and λ 2 E[Y 2 F n ] = E[(M n+1 M n ) 2 F n ] = M n+1 M n. With M n+1 mf n, we consequenty have that E[N n+1 F n ] = N n exp( ψ( M n+1 M n ))E[e Y F n ] N n, impying that (N n, F n ) is a non-negative sup-mg. 2. Since λ > 0 the event {M τ u, M τ r} impies that N τ a for a = exp(λu ψr). Hence, by Markov s inequaity P(M τ u, M τ r) P(N τ a) a 1 EN τ. Appying Doob s convergence theorem for the non-negative sup-mg {N n τ } whose a.s. imit is N τ (by the a.s. finiteness of τ), you thus deduce that EN τ EN 0 = 1 as needed to compete the proof. 2

3 3. Reca that the L 2 bounded martingae S n of Exampe converge a.s. (and in L 2 ) to a finite imit S. As the martingae S n has independent increments, we deduce that S n = E(Sn) 2 is a non-random sequence which converges to the finite constant S = k Eξ2 k. By part (a) and our assumption that ξ k 1 we further have that N n = exp(λs n ψ S n ) is a non-negative sup-mg for any λ > 0. Thus, by Doob s convergence theorem N n N amost surey and EN EN 0 = 1. Since necessariy N = exp(λs ψ S ), it foows that E[exp(λS )] exp(ψ S ) is finite. This concusion extends to a λ R since the same argument appies aso for the martingae S n. Exercise [5.4.10] 1. Since τ is a stopping time for Fn, ξ we know that I k τ = 1 I τ k 1 is measurabe on F ξ k 1, and hence independent of ξ k. Consequenty, with ξ k identicay distributed, Eξ k I k τ = Eξ k P(k τ) = Eξ 1 P(τ k). The representation S τ = ξ ki k τ appies when τ < a.s. (hence when Eτ < as assumed). Thus, by Fubini s theorem with respect to the product of the probabiity measure P and the counting measure on k {1, 2,...}, we find that ES τ = E[ ξ k I k τ ] = E[ξ k I k τ ] = E[ξ 1 ] P(τ k) = E[ξ 1 ]E[τ], where the integrabiity condition for Fubini s theorem is merey that E[ ξ k I k τ ] = E[ ξ 1 ]P(τ k) = E[ ξ 1 ]E[τ] is finite. As the atter foows from the assumed finiteness of Eτ, we are done. 2. Without oss of generaity assume that Eξ 1 = 0, for otherwise, we can aways work with {ξ i Eξ i } which are i.i.d. and have the same variance as {ξ i }. Setting v := Var(ξ 1 ), reca that X n = S 2 n vn is a martingae with X 0 = 0. Since EX n τ = EX 0 = 0 and τ < a.s., we have by monotone convergence that as n ES 2 n τ = ve[n τ] veτ <. This shows that the martingae {S n τ } is L 2 -bounded and by Doob s L 2 -martingae convergence theorem, S n τ S τ in L 2, resuting with ES 2 τ = im n ES2 n τ = veτ. 3. When estabishing Wad s identity in part (a) we used the condition Eτ < ony for justifying the representation S τ = ξ ki k τ and for estabishing Fubini s theorem integrabiity condition when interchanging the order of summation (over k) and expectation (with respect to P). For a non-negative sequence ξ k we have a non-negative integrand, in which case Fubini s theorem requires no integrabiity assumption (under the convention that 0 = 0), and the representation for S τ is then vaid even when τ(ω) =. Exercise [5.4.12] This is the stricty positive product martingae M n of Exampe , for the positive i.i.d. variabes Y k = e λξ k /M(λ) of mean one. 3

4 1. For p = 1 q 1/2 we know from parts (c) and (d) of Exercise that τ b is finite a.s. Further, by definition S n τb b for a n and as M(λ) = pe λ + qe λ 1 whenever λ 0, in this case M n τb = exp(λs n τb (n τ b ) og M(λ)) exp(λb). Thus, {M n τb } is a uniformy bounded, hence U.I. martingae. With S τb Doob s optiona stopping theorem that = b, it then foows from 1 = EM 0 = EM τb = e λb E[M(λ) τ b ]. 2. Setting 0 < s < 1 there exists for p 1/2 a unique λ > 0 such that M(λ) = pe λ + qe λ = 1/s. Indeed, soving qsx 2 x + ps = 0 for x = e λ in (0, 1), we find from part (a) that E[s τ b ] = x b and E[s τ1 ] = x = 1 1 4pqs 2 2qs when q (0, 1/2], whereas x = s in the trivia case q = Since τ a,b = min(τ b, τ a ) is finite a.s. and τ a a we have that P(τ b < a) P(τ b < τ a ) P(τ b < ). Consequenty, P(τ b < τ a ) P(τ b < ) as a. To compete the proof reca that from Coroary we have that P(τ b < τ a ) = 1 r = 1 e λ a e λ b e λ b e λ a when a (as λ > 0). 4. Ceary, {τ b < } if and ony if {Z b + 1}. Hence, for any positive integer b, Exercise [5.5.8] 1. First, we have that P(Z = b) = P(Z b) P(Z b + 1) = (1 e λ )e λ (b 1). L(s) = E(s N ) = s k p(1 p) k = p (s(1 p)) k p = 1 s(1 p). k=0 Then, note that for s = 1/m we have k=0 L( 1 m ) = p 1 (1 p)/m = p 1 p = 1 m impying that ρ, being the unique soution of s = L(s) in [0, 1) must equa 1/m. Reca that Proposition gives the recursive formua L n (s) = L[L n 1 (s)] starting at L 0 (s) = s. The stated formua we have for the non-critica case is of the form, L n (s) = (pa n (s) + b(s))/((1 p)a n (s) + b(s)) ( ) with a n (s) = m n (1 s). Moreover, b(s) = (1 p)s p is such that the identity pa 0 (s) + b(s) = s[(1 p)a 0 (s) + b(s)] hods, or equivaenty L 0 (s) = s. Having estabished the case n = 0, we proceed 4

5 to verify this formua by induction over n. To this end, suppressing the argument of a n and b, note that by the induction hypothesis L[L n (s)] = p((1 p)a n + b) (1 p)a n + b (1 p)(pa n + b) = pa n+1 + b (1 p)a n+1 + b, where the second identity foows from our choice of a n+1 = (1 p)a n /p and so we are done. In the critica case, L(s) = 1/(2 s) and our formua L n (s) = [na(s) + s]/[na(s) + 1] ceary starts at L 0 (s) = s. In the induction step, by the induction hypothesis (and suppressing the argument of a), L[L n (s)] = na + 1 (n + 1)a + s = 2(na + 1) (na + s) (n + 1)a + 1 where the second identity foows from our choice of a = (1 s) and we are once more done. 2. Reca Lemma that X n = m n Z n is a non-negative MG, so by Doob s convergence theorem X n D X 0. Further, f(x) = exp( λx) is continuous and bounded on [0, ) when λ 0. Thus, L (e λ ) = E[e λx ] is the imit of L n (s n ) as n, where s n = exp( λ/m n ) 1 in the supercritica case m > 1. More precisey, a n (s n ) = m n (1 s n ) λ and b(s n ) = (1 p)s n p 1 2p. Hence, from the expression in part (a) we deduce that L n (s n ) = pa n (s n ) + b(s n ) pλ + 1 2p ρλ + (1 ρ) = (1 p)a n (s n ) + b(s n ) (1 p)λ + 1 2p λ + (1 ρ). We obtained the right most identity by substituting (1 p)(1 ρ) = (1 2p) and (1 p)ρ = p (reca part (a) that ρ = 1/m), and it is easy to verify the right most expression of the preceding dispay matches the formua we gave for L (e λ ). Since P(X = 0) = L (0) = ρ, ceary E[e λx X > 0] = E[e λx ] E[e λx I {X =0}] 1 P(X = 0) = L (e λ ) L (0) 1 L (0) = 1 ρ λ + (1 ρ) = E[e λt ], where T has the exponentia distribution of parameter 1 ρ. Reca part (a) of Exercise that the aw of a non-negative random variabe is uniquey determined by its Lapace transform, hence the aw of X conditiona on X > 0 is an exponentia distribution of parameter (1 ρ). Further, the event X > 0 impies non-extinction and though the converse is in genera not obvious, here P(X = 0) = ρ is aso the probabiity of extinction, hence if there is a difference between these two events, it is of probabiity zero (and hence can be ignored). 3. In the sub-critica case, p > 1 p and thus m < 1. Then, a n (s) 0 when n and hence L n (s) 1 for a s [0, 1]. More precisey, using the formua (*) for L n (s) we find that m n (1 L n (s)) = (1 2p)(1 s) (1 2p)(1 s)/b(s) (1 p)a n (s) + b(s) as n. It is easy to check that b(s) = b(0)(1 ms) and hence 1 L n (s) im = (1 s)b(0) n 1 L n (0) b(s) = 1 s 1 ms. 5

6 Moreover, foowing our treatment in part (b) of the conditioning on being positive, we thus get here that Setting s = e λ we get that E[s Zn Z n 0] = L n(s) L n (0) 1 L n (0) E[e λzn Z n 0] = 1 1 L n(s) (1 m)s 1 L n (0) 1 ms. (1 m)e λ 1 me λ which is precisey the Lapace transform of a Geometric(1 m) random variabe. Reca part (b) of Exercise that such convergence of Lapace transforms of non-negative random variabes impies the corresponding weak convergence, which eads to our thesis. 4. For the critica case, taking s n = e λ/n 1 yieds L n (s n ) 1. More precisey, from our formua L n (s) = (na(s) + s)/(na(s) + 1) we find that n(1 L n (s)) = na(s)/(na(s) + 1) since a(s) = 1 s, and with na(s n ) λ it foows that n(1 L n (s n )) λ/(1+λ). Simiary, n(1 L n (0)) 1. Consequenty, simiary to part (c), here E[e λzn/n Z n 0] = L n(s n ) L n (0) 1 L n (0) 1 λ 1 + λ = λ. = 1 1 L n(s n ) 1 L n (0) The ast expression is the Lapace transform of the standard exponentia distribution (with parameter one), from which we deduce that conditiona to non-extinction, n 1 Z n converges weaky to an exponentia distribution. Exercise [5.3.32] 1. Since (S n, F ξ n) is a martingae, the same appies for the stopped martingae X n = S n τ. With n τ n, we further have that ES 2 n τ ES 2 n is finite, so {X n } is an L 2 -martingae. Ceary, X 0 = 0 and X k X k 1 = ξ k I k τ. Further, τ is a stopping time for F ξ n so the event {k τ} = {τ k 1} c is in F ξ k 1 for each k 1. With {ξ k} i.i.d. the predictabe compensator of {X n } is thus with X = τeξ 2 1. X n = n E[ξkI 2 k τ F ξ k 1 ] = n I k τ E[ξ 2 k F ξ k 1 ] = (n τ)eξ2 1, 2. Since E[ τ] is finite, we have from part (a) that so is E[ X 1/2 ]. It then foows from Proposition that sup k X k is integrabe, hence {X n } is a U.I. martingae. In particuar, EX n = EX 0 = 0 for a n. Further, with τ a.s. finite, X n = S n τ S τ as n which by uniform integrabiity of {X n } impies that ES τ = 0. 6