2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ There certainly should be some extra assumptions. There is no chance that the (Haar) square function of a bounded function on the interval $[0,1]$ is always constant. $\endgroup$
    – fedja
    Aug 25, 2016 at 3:40

1 Answer 1

1
$\begingroup$

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 $\mathbb EV_n^2=\mathbb EM_n^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.