I am reading a paper which uses the following and I am struggling to show it. We let $M_n$ be a martingale with bounded increments, wrt the natural filtration $\mathcal{F}_n$. Suppose $M_0=0$. Let $Y_n=M_n-M_{n-1}$. Let $V_n^2=\sum\limits_{j=1}^n \mathbb{E}[Y_j^2 \mid \mathcal{F}_{j-1}]$ and $s_n^2=\mathbb{E}[M_n^2]$. I can't seem to show that $\frac{V_n^2}{s_n^2}\to 1$ in probability as $n\to\infty$. Could anyone please help?

## 1 Answer

So here is an explicit counter-example. Let $M_0=0$ and let $Z$ be 1 or 2 with equal probability. Now let $Y_n=\pm Z$, so that the process is a simple symmetric random walk with step size $Z$ (which is unchanged throughout). Now $V_n^2$ is either $4(n-1)+\frac 52$ if $Z=2$ (since $Z$ is $\mathcal F_k$-measurable for $k\ge 1$) or $n-1+\frac 52$ if $Z=1$. On the other hand, $\mathbb EM_n^2=\frac 52n$.

Notice that it *is* true that $\mathbb EV_n^2=\mathbb EM_n^2$.