Math 6710 Homework 13: Solutions Fall, 2016 Grading: jerison/math6710/hw13-sol.pdf · Math 6710 Homework…

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Math 6710 Homework 13: Solutions Fall, 2016Grading: TBDProblem 1: First, P L. If J P then J (X1, . . . , Xk) for some k. Take In = J forn k; then each In (X1, . . . , Xn) and P (InJ) = 0 for n k, so J L. To show that Lis a -system, we must check that L (which is already implied by P L) and that L isclosed under subset differences and countable increasing unions.Subset differences: Suppose A,B L with A B. Find sequences {In}, {Jn} such thateach In, Jn (X1, . . . , Xn) and P (InA), P (JnB) 0 as n. To show that B \A Lit suffices to show that P ((Jn \In)(B \A)) 0 as n, since each Jn \In (X1, . . . , Xn).(Jn \ In)(B \ A) = [(Jn Icn) (B Ac)c] [(Jn Icn)c (B Ac)]= [(Jn Icn) (Bc A)] [(J cn In) (B Ac)]= (Jn Icn Bc) (Jn Icn A) (J cn B Ac) (In B Ac) (Jn Bc) (Icn A) (J cn B) (In Ac)= (JnB) (InA).Therefore, P ((Jn \ In)(B \ A)) P (JnB) + P (InA) 0 as n.Countable increasing unions: Suppose Am A and each Am L. For each m, find setsIm,n (X1, . . . , Xn) such that P (Im,nAm) 0 as n . Let n(m) be the least integergreater than or equal to m such that P (Im,n(m)Am) 1m . In particular, n(1) = 1.Since P (Im,n(m)Am) 0 and P (A \ Am) 0 as m , it is not hard to showthat P (Im,n(m)A) 0 as m . This would suffice except that Im,n(m) is containedin (X1, . . . , Xn(m)) rather than (X1, . . . , Xm). The solution is to slow down the sequence{Im,n(m)} by repeating terms. For example, if the sequence begins (I1,1, I2,4, I3,6, . . .) we woulddefine (J1, J2, . . .) = (I1,1, I1,1, I1,1, I2,4, I2,4, I3,6, . . .) so that each Jk (X1, . . . , Xk). The fol-lowing argument makes this rigorous.For each k 1, let m(k) be the greatest integer such that n(m(k)) k. Since n(1) kand n(m) m > k for all m > k, m(k) is well-defined and at most k. Set Jk = Im(k),n(m(k)) (X1, . . . , Xk). We show that P (JkA) 0 as k . First note that for all M 1, ifk n(M) then m(k) M . Thus m(k) as k . We haveP (JkA) = P (Jk \ A) + P (Am(k) \ Jk) + P ((A \ Am(k)) \ Jk) P (Jk \ Am(k)) + P (Am(k) \ Jk) + P (A \ Am(k))= P (JkAm(k)) + P (A \ Am(k)).By construction, P (JkAm(k)) = P (Im(k),n(m(k))Am(k)) 1m(k) 0 as k . Also, sinceAm A, P (A \ Am(k)) 0 as k . Hence P (JkA) 0 as k , so A L.1Math 6710 Homework 13: Solutions Fall, 2016Problem 2: Since {Yn} is a Cauchy sequence in the complete metric space L2, it has a limitwhich we call Z. So Yn Z in L2 and therefore in probability, and there is a subsequenceYnm that converges to Z almost surely. However, we know that Yn (and therefore also Ynm)converges to Y almost surely. Hence Y = Z except on a set of measure zero and Yn Y inL2. By the triangle inequality in L2,Y 2 Yn2 Y Yn2, Yn2 Y 2 Yn Y 2which meansYn2 Y 2 Yn Y 2 0 as n and so Yn2 Y 2. It followsthat E[Y 2n ] = Yn22 Y 22 = E[Y 2].Problem 3: Case (i) is only possible when X1 = 0 almost surely, which contradicts E[X21 ] > 0.To rule out case (ii), assume for contradiction that Sn a.s. Let T = max{n 0 : Sn 0},so that T < a.s., and choose N such that P (T > N) 14. By the CLT, Sn/(n) Z N(0, 1) and soP (Sn 0) = P(Snn 0) P (Z 0) = 12as n.Choose n > N large enough that P (Sn 0) 13 . However, if Sn 0 then T n > N , soalso P (T > N) 13, which is a contradiction. This rules out case (ii), and case (iii) is alsoimpossible by symmetry, so case (iv) is all that remains.Problem 4: (a) By Proposition 17.1 in the notes, {S n}, {T n} (X1, . . . , Xn) for eachn. Therefore,{S T n} = {S n} {T n} (X1, . . . , Xn),{S T n} = {S n} {T n} (X1, . . . , Xn)and so S T , S T are stopping times, again by Proposition 17.1.(b) Yes, S + T is a stopping time: {S + T = n} = nk=1({S = k} {T = n k}), and each{S = k} {T = n k} (X1, . . . , Xk(nk)) (X1, . . . , Xn). No, T S is not necessarilya stopping time. For a counterexample, let the Xn be iid, P (Xn = 0) = P (Xn = 1) =12. SetS = 10 and T = 11 if X11 = 0, T = 12 if X11 = 1. Both S and T are stopping times, but T Sis 1 or 2 depending on the value of X11, so for example {T S = 1} / (X1).Problem 5: Use the notation of Example 4.1.4 in the textbook. First suppose P ( Math 6710 Homework 13: Solutions Fall, 2016textbook implies that the increments Yk := Sk Sk1 are iid, and since each Yk > 0 we have0 < E[Yk] . In case E[Yk] = Math 6710 Homework 13: Solutions Fall, 2016if Y = infm Sm then Y Sn 1. By part (ii),E[|Y |] =k=1P (|Y | k) =k=1P (Y k) =k=1P (

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