Lx du fdx jo lxo uo 1 1 f fdx du f fudu tft uo lxu jo

Info icon This preview shows pages 242–246. Sign up to view the full content.

[lx du] F(dx) Jo lx=O u=O = 1 1 [ f' F(dx)] du = f' F(u)du - tF(t) u=O lx=u Jo ;::: lot F(u)du. From (7.21), given() > 0, there exists xo such that x 2: xo implies .F{x) ;::: (1 + ())kx-a =: k1x-a. Thus from (7 .26) 1 xo 1t 1' m(t);::: + ;:::c+kt u-adu, 0 xo xo t 2: XQ. Now for a> 1, E(IZII) < oo so that L iPiim(c/IPiD:::: L IPjiE(IZtD < oo j j by (7.23). For a = 1, we find from (7.28) that m(t) :::: c' + kzlogt, t::: xo (7.26) (7.27) (7.28) for positive constants c', kz. Now choose 1J > 0 so small that 1 - 1J > 8, and for another constant c" > 0 IPilm(c/IPiD:::: c" IPil + kz IPillog c:i,) 1 1 1 :::: c"L IPil +k3 L 1Pi1 1 - 11 < oo j j
Image of page 242

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

230 7. Laws of Large Numbers and Sums of Independent Random Variables where we have used (7.23) and Finally for a < 1, t > xo so that L IPjlm(l/IPjD :S Cz L IPjl +k4 L IPjll+a- 1 < oo j j j from (7.23). 0 7.6.1 Necessity of the Kolmogorov Three Series Theorem We now examine a proof of the necessity of the Kolmogorov three series theo- rem. Two lemmas pave the way. The first is a partial converse to the Kolmogorov convergence criterion. Lemma 7.6.1 Suppose {Xn, n ::: 1} are independent random variables which are uniformly bounded, so that for some a > 0 and all w e Q we have IX n (w) I ::: a. If"f:.n(Xn - E(Xn)) converges almost surely, then Var(Xn) < oo. Proof. Without loss of generality, we suppose E (X n) = 0 for all n and we prove the statement: If {Xn, n ::: 1} are independent, E(Xn) = 0, IXn I :Sa, then Ln Xn almost surely convergent implies Ln E (X;) < oo. We set Sn = "'£7= 1 X;, n ::: 1 and begin by estimating Var(SN) = "'f:.f: 1 E (X f) for a positive integer N. To help with this, fix a constant A > 0, and define the first passage time out of [-A, A] r := inf{n::: 1: ISnl >A}. Set r = oo on the set I ::: A]. We then have N LE(XI) = = + (7.29) i=1 =I+ II. Note on r > N, we have vf: 1 IS; I ::: A, so that in particular, ::: A 2 . Hence, (7.30) For I we have N I= L j=1
Image of page 243
For j < N Note while and thus 7.6 The Kolmogorov Three Series Theorem 231 N = E((Sj + L X;) 2 )1[T=j]). i=j+l j-1 [r = j] = lV IS; I::: A, ISjl >A] E a(Xt. ... , Xj) i=1 N L X; E a(Xj+t. ... ,XN), i=j+l N l[T=j) JL L X;. i=j+1 Hence, for j < N, = E((SJ + 2Sj t X;+ ( t X;) 2 )l[T=j)) i=j+1 i=j+1 N = E(SJl[T=j]) + 2E(Sj1[T=jJ)E( L X;) i=j+1 N + E( L X;) 2 P[r = j] i=j+l N = E(SJl[T=j]) + 0 + E( L X;) 2 P[r = j] i=j+l N ::: E((ISj-11 + IXji) 2 1[T=j]) + L E(X;) 2 P[r = j] i=j+1 N _:::(A+a) 2 P[r=j]+ L E(X;) 2 P[r=j]. i=j+l Summarizing, we conclude for j < N that E(X;) 2 )P[r=j], (7.31) 1=]+1
Image of page 244

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

232 7. Laws of Large Numbers and Sums of Independent Random Variables and defining E(X;) 2 = 0, we find that (7.31) holds for 1 j N. Adding over j yields I= E(S11[r:::NJ) (o. +a) 2 + tE(Xf))P[r N] (7.32) Thus combining (7.30) and (7.32) N ECS1) = 'L,E<Xf) =I+ II i=l (o. + a) 2 + ECS1)) P[r N] +(A +a) 2 P[r > N] and solving for E (S1) yields Let N-+ oo. We get oo (A+ a)2 LE(Xf) ::S _ , i=l P[r- oo] which is helpful and gives the desired result only if P [ r = oo] > 0. However, note that we assume Ln X n is almost surely convergent, and hence for almost all w, we have {Sn(w), n :::: 1} is a bounded sequence of numbers. So VniSnl is almost surely a finite random variable, and there exists A > 0 such that 00 P[r = oo] = P[V ISnl 0, n=l else I < oo] = 0, which is a contradiction. This completes the proof of the lemma.
Image of page 245
Image of page 246
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern