2
$\begingroup$

Let $(X_n)_n$ be a sequence of random variables on the probability space $(\Omega, \mathcal{F}, P)$, and let $(\mathcal{F}_n)_n$ be a filtration that increases to $\mathcal{F}$. We can assume $(X_n)_n$ is uniformly integrable, but I'm also interested in the general case if anyone wants to comment on that.

Is it true that $\int |X_n - E(X_n \mid \mathcal{F}_n)|dP \to 0$ as $n \to \infty$?

I haven't made any real progress on this and am just looking for some hints so I can try to prove or disprove it myself.

I know that if $X_n$ is held fixed and $\mathcal{F}_n$ is allowed to increase, then the result holds. This is just a textbook martingale convergence result. But I don't know how to generalize this to a whole sequence of random variables and puttering around with Fatou's lemma and the like hasn't gotten me anywhere.

Again, I'm just looking for some hints or suggestions so I can try to get it myself.

$\endgroup$
1
  • $\begingroup$ It's definitely not true in the general case (i.e. without uniform integrability), because you can take $\xi_i$ i.i.d. with $P(\xi_i = 1) = P(\xi_i = -1) = 1/2$ and $S_n = \sum_{i = 1}^n \xi_i$ and $\mathcal{F}_n = \sigma(\{\xi_i\}_{i=1}^{n-1}$ then $|X_n - E(X_n | \mathcal{F_n})| = 1$. $\endgroup$
    – Marcus M
    Jun 1 '17 at 16:12
2
$\begingroup$

It's not true, and a counterexample would be one similar to my comment. Take $\{\xi_i\}$ i.i.d. with $P(\xi_i = 1) = P(\xi_i = - 1) = 1/2$. Set $X_k = \prod\limits_{i = 1}^k \xi_k$ and $\mathcal{F}_n = \sigma(\{\xi_i\}_{i = 1}^{n-1})$. Then $$|X_n - E(X_n | \mathcal{F_{n}})| = |X_n - 0| = 1.$$

This means that $E|X_n - E(X_n | \mathcal{F}_n)| = 1$.

$\endgroup$
6
  • $\begingroup$ Thanks, that's helpful. I always (stupidly) forget to check simple iid examples like this. $\endgroup$
    – aduh
    Jun 1 '17 at 16:40
  • 3
    $\begingroup$ Even $X_n=\xi_{n+1}$ works like a charm. $\endgroup$
    – Did
    Jun 1 '17 at 16:45
  • $\begingroup$ @MarcusM Doesn't $E(X_n | \mathcal{F}_n) = 0$ because $$E(X_n|\mathcal{F}_n) = E(\xi_n X_{n-1}|\mathcal{F}_n)=X_{n-1}E(\xi_n | \mathcal{F}_n) = X_{n-1}E(\xi_n)=0,$$ or am I messing something up? In any event, the counterexample still works. $\endgroup$
    – aduh
    Jun 2 '17 at 14:52
  • $\begingroup$ @aduh, you're right, yeah, I'm not sure what I was thinking. I'll update it. $\endgroup$
    – Marcus M
    Jun 2 '17 at 15:48
  • $\begingroup$ @MarcusM If you don't mind, I have one more follow-up question. Do you happen to know if there are any additional assumptions we could make about $(X_n)$ to get the result, besides making $(X_n)$ $\mathcal{F}_n$-measurable? $\endgroup$
    – aduh
    Jun 2 '17 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.