Berry-Esseen bound for martingale sequence with varying and dependent variances

Question

Let $(X_{1},\ldots,X_{k},\ldots)$ be a martingale difference sequence, i.e. $$ E[X_{k}|\mathcal{F}_{k-1}] = 0 $$ where $\mathcal{F}_{k-1}$ is the $\sigma$-algebra filtration at $k-1$.

Let $\sigma_{k}^2 = E[X_{k}^2|\mathcal{F}_{k-1}]$. Here note that $\sigma_{k}^2$ is a random variable measurable w.r.t. $\mathcal{F}_{k-1}$. If we know that $\sigma_{k}^2 \in [\underline{\sigma}^2, \overline{\sigma}^2]$, where both $\underline{\sigma}$ and $\overline{\sigma}$ are positive constants, then do we have, as $\beta\to 0^+$, $$ \frac{\sum_{k=1}^\infty (1-\beta)^k X_{k}}{\sqrt{\sum_{k=1}^\infty (1-\beta)^{2k} \sigma_{k}^2}} \rightarrow N(0,1) $$ in distribution, and what will be the correct order of Berry-Esseen bound?

Answer 1

In the setting you describe, there's generally no CLT. Let me describe a counterexample showing that $\sum_{i=1}^{k-1} X_i / \sqrt{\sum_{i=1}^{k-1} \sigma_i}$ doesn't tend to normal. I'm pretty sure the same holds for the discounted averages as well.

The example is a random walk which is lazy when positive. More precisely, conditioned on $\mathcal{F_{k-1}}$, if $\sum_{i=1}^{k-1} X_i <0$ then $X_k$ is uniformly distributed in $\{-1,1\}$ and if $\sum_{i=1}^{k-1} X_i \ge 0$ then $X_k$ is uniformly distributed in $\{-1,0,1\}$. The ratio between $P( \sum_{i=1}^{k-1} X_i <0)$ and $P(\sum_{i=1}^{k-1} X_i >0)$ tends to the ratio between the speed of the random walk when negative and when positive, which is 2:3, so the probability that the sum is positive tends to $\frac35$. I expect the same limit also for the discounted average.