# Does the difference between a sequence of random variables and their conditional expectations converge in $L^1$?

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.

• 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 $|X_n - E(X_n | \mathcal{F_n})| = 1$. Jun 1 '17 at 16:12

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

• Thanks, that's helpful. I always (stupidly) forget to check simple iid examples like this.
• Even $X_n=\xi_{n+1}$ works like a charm.
• @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.
• @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?