# Convergence in distribution of conditional expectations

I was just reading this question, which is about how the classical central limit theorem can be interpreted as giving a rate of convergence for the law of large numbers for iid random variables. I was wondering whether the same idea can be generalized to martingales.

For example, let $X$ be integrable on $(\Omega, \mathcal{F}, P)$ and assume $\mathcal{F}_n \uparrow \mathcal{F}$. Then, $$E(X \mid \mathcal{F}_n) \to X \ \ \text{a.s.}$$ Is there a sequence $(a_n)_n$ and a non-zero $W$ such that, with $Y_n = E(X \mid \mathcal{F}_n) - X$, we have $$\frac{Y_n}{a_n} \Rightarrow W?$$ (Here, $\Rightarrow$ denotes convergence in distribution.)

• @Michael Your corrections are wrong. The mean is zero by martingale property and the sum is implicit in the fact that each $X_n$ is the sum of all previous differences of the form $X_n-X_{n-1}$ – Bananach May 25 '17 at 13:07
• @Bananach : Thanks, the asymtptoic mean goes to zero I suppose when dividing by $v$ and taking $v \rightarrow \infty$, even if $X_0\neq 0$ or $E[X_0] \neq 0$. Somehow I looked at the formula and thought it would always stay at the same possibly nonzero mean. I deleted my previous comment. The martingale central limit theorem is here en.wikipedia.org/wiki/Martingale_central_limit_theorem – Michael May 25 '17 at 14:23

The answer is generally no, even if the $a_n$ sequence can be chosen based on the particular $X$ and $\mathcal{F}_n$ (Bananach treats the case when $a_n$ must be chosen without knowledge of $X$ and $\mathcal{F}_n$).

Define $X$ uniform over $[-1,1]$. For $n \in \{1, 2, 3, ...\}$ define: $$Z_{n} = X 1_{|X|>2^{-n}}$$

where $1_{\mathcal{A}}$ denotes the indicator function for an event $\mathcal{A}$. If you know $Z_n$ for some $n>1$, then you can infer $Z_1, ..., Z_{n-1}$. Define $\mathcal{F}_n = \sigma(Z_n)$. Then $\mathcal{F}_n \subseteq \mathcal{F}_{n+1}$ for all $n \in \{1, 2, 3, ...\}$, and

$$E[X|\mathcal{F}_n] = E[X|X 1_{|X|>2^{-n}}] = X1_{|X|>2^{-n}}$$

Thus, $E[X|\mathcal{F}_n]\rightarrow X$ almost surely (in fact, surely). Now define $Y_n = E[X|\mathcal{F}_n]-X$. Then $$P[Y_n=0] = P[|X|> 2^{-n}] = 1-2^{-n}$$ and for any sequence $\{a_n\}$ with $a_n\neq 0$ we have $P[Y_n/a_n = 0] =1-2^{-n} \rightarrow 1$.

• A simpler example of course defines $\mathcal{F}_n = \sigma(X)$ and then $E[X|\mathcal{F}_n]=E[X|X]=X$, which means $Y_n=0$ always. Usually central limit theorems are defined in terms of summations of something. It is not clear what motivated you to look at $E[X|\mathcal{F}_n]$. – Michael May 24 '17 at 19:03
• Thanks for the helpful answer. My motivation for looking at conditional expectations would take awhile to explain, but I will eventually write another question about it. – grndl May 25 '17 at 0:31

A single sequence $a_n$ that works for all $X$ and $\mathcal{F}_n$ does not exist. Just take any such inputs and create a new sequence of filtrations by repeating the old filtrations more and more often. More specifically let $$\tilde{\mathcal{F}}_n:=\mathcal{F}_{\sqrt{n}}$$ (Round to nearest integer)

If you had rate $1/2$ convergence before, you will have rate $1/4$ convergence with respect to this new filtration.

Since i.i.d. random variables can be considered as special case of martingales we obviously do have convergence in some situations. To study when this is the case, one way is to have a look at $Z_n :=E[X|F_n]-E[X|F_{n-1}]$ and apply various general forms of the central limit theorem (Lindeberg, ...).

• Thanks! This is helpful. How about I stipulate that the filtration is strictly increasing? Can anything interesting be said then? – grndl May 21 '17 at 15:16
• Doesn't help. The filtration might be increasing but with information independent of $X$ – Bananach May 21 '17 at 15:36
• Could you elaborate some on what you have in mind when you say, "one way is to have a look at $Z_n :=E[X|F_n]-E[X|F_{n-1}]$ and apply various general forms of the central limit theorem"? – grndl May 22 '17 at 13:44