Convergence in distribution of conditional expectations

Question

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.)

Answer 1

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, ...).

Answer 2

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 $P[Y_n/a_n = 0] =1-2^{-n} \rightarrow 1$.