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I'm working on this probability problem and am quite stuck on how to approach the following proof:

$ X_n, n = 1, 2, ... \text{and } Z \text{ are random variables defined on} (\Omega, \mathcal{F}, P)$

We may assume that:

  • $E|Z| < \infty$
  • $P(|X_n|\leq c) = 1$ for some constant $c > 0$
  • $EX_n=0$

The following sigma-algebras are defined as: $\mathcal{A}_t = \sigma(X_n, n \leq t)$
$\mathcal{A}_\infty = \sigma(X_n, n \geq 1)$

The question asks to prove that if $Z$ is $\mathcal{A}_m$-measurable then:
$\lim_{n \rightarrow \infty} E(X_n Z)=0$

Note we are also given that $\lim_{t \rightarrow \infty} E(Z|\mathcal{A}_t) = E(Z|\mathcal{A}_\infty) a.s.$ (1)

I've tried approaching it by substituting $Z$ as $X_nZ$ into the given limit (1) and then applying dominated convergence theorem but I can't seem to figure out how get the entire expression to converge to 0. Intuitively, it feels like I have to find an equality/inequality which involves $EX_n$ but I can't figure out how to get to this result.

I wish I could've progressed further prior to asking for input here but I'm rather new to formally learning probability, so appreciate any advice.

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1 Answer 1

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The statement is false. Let $X$ have uniform distrbution on $(-1,1)$, $X_n=X=Z$ for all $n$. Then $EX_nZ=\frac 1 3$ for all $n$.

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