A Probability Path.pdf

# Assume the random variables x xn n 1 are all defined

• Test Prep
• 464

This preview shows pages 203–208. Sign up to view the full content.

Assume the random variables {X, Xn, n 1} are all defined on (Q, B, P).

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

190 6. Convergence Concepts 1. A form of Scheffe 's lemma: We have the following equivalence for L 1 con- vergence: As n -+ oo iff sup I { XndP- { XdPI-+ 0. AeB )A )A (6.16) Note that if we replace A by Q in (6.16) that we get IE(Xn) - E(X)I EIXn- XI-+ 0 so that first moments converge. This, of course, also follows by the modulus inequality. To verify (6 . 16), suppose first that Xn X. Then we have sup I { XndP- { XdPI A )A )A =sup I r (Xn- X)dPI A )A sup [ IXn - XldP A )A IXn -XIdP =E(IXn -XI)-+ 0. For the converse, suppose (6 .16) holds. Then 2. If then EiXn-XI= { (Xn-X)dP+ { (X-Xn)dP lrxn>X] lrxn!:X] = (frXn>X] Xn -lXn>X] X) + ( { X- { Xn) lrxn!:X) lrxn!:X] { Xn- {XI . A )A )A L Xn4X 0
6.6 More on L p Convergence 191 or equivalently For this verification, write X=Xn+X-Xn and Minkowski 's inequality implies liXIip ::: liXn lip+ liX- Xn lip · Interchange the roles of X n and X in (6. 17) to get liXn lip ::: liXIip + liX- Xn lip · So combining (6.17)and (6.18) we get lliXnllp -IIXIIpl::: IIX -Xnllp 0, as was to be proved . (6 . 17) (6.18) (6 . 19) 0 Towards a resolution of the problem of when moments of a sequence of random variables converge , we present the following result which deals with the case p=l. Theorem 6.6.1 Suppose for n 2: 1 that Xn e Lt. The following statements are equivalent: (a) {X n } is L 1-convergent. (b) {X n} is L 1-cauchy ; that is, asn,m oo . (c) {Xn} is uniformly integrable and {Xn} converges in probability. So if X n X or X n .!:. X and {X n} is ui, then the first moments converge: IE(Xn)- E(X)I ::: E(IXn- 0. Later, we will see that convergence i.p . of {Xn} can be replaced by convergence in distribution to X . Proof. L 1 convergence implies Cauchy convergence because of the tri- angle inequality. Given (b) we first show that {Xn} is ui. Given > 0, there exists N such that if m, n ::: N then J IXn- XmldP < E/2. (6.20)

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

192 6. Convergence Concepts To show {Xn} is ui, we use Theorem 6.5.1. For any A E B i IXnldP::: i IXn -XN. +XN.IdP ::: iiXN.IdP+ fiXn-XN.IdP. For any n ::: Nf that is, sup { IXnldP::: { iXN.IdP + /2. n'?;N• }A }A and thus sup { IXnldP::: sup { IXmldP + /2. n }A If A = n, we conclude supE(IXnD::: sup EOXmD + /2 < oo. n Furthermore, since finite families of L 1 rv's are ui, {Xm, m ::: Nf} is ui and given > 0, there exists 8 > 0 such that if P(A) < 8, then sup { IXmldP < /2 }A so we may conclude that whenever P(A) < 8, Hence {X n} is ui. sup { IXnl::: /2 + /2 = . n JA To finish the proof that (b) implies (c), we need to check that {Xn} converges in probability. But so {Xn} is Cauchy i.p. and hence convergent in probability. If Xn X, then there exists a subsequence {nk} such that X a.s. X nk '
6.6 More on L p Convergence 193 and so by Fatou's lemma since {Xn} is ui. So X E L Also, for any > 0 f IXn- XldP:;: { IXn- XldP + { IXnldP luxn-XI:SE) luxn-XI>E) + { IXIdP J[IXn-XI>E) :;: +A +B. SinceXn P[IXn -XI> ] 0 and hence B 0 as n oo by Exercise 6 of Chapter 5. To verify A 0, note that since {Xnl is ui, given > 0, there exists 8 > 0 such that sup11XkldP < A if P(A) < 8. Choose n so large that P[IXn -XI> ] < 8 and then A < .

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

This is the end of the preview. Sign up to access the rest of the document.
• Fall '08
• Staff

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern