TITLE: Convergence in distribution of conditional expectations
QUESTION [12 upvotes]: 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.)

REPLY [3 votes]: 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 $P[Y_n/a_n = 0] =1-2^{-n} \rightarrow 1$.