If $E(X_n^2)<\infty$, then for a Martingale $E(X_n^2)<M$ iff $\sum_{n=1}^\infty E[(X_n-X_{n-1})^2]<\infty$

Question

Let $\{X_n\}_{n\geq0}$ be a martingale with $E(X_n^2)<\infty$ for all $n$. How to prove that:

$E(X_n^2)<M$ for all $n$, if and only if $\sum_{n=1}^\infty E[(X_n-X_{n-1})^2]<\infty$.

The hunch is to use the optional stopping time theorem, but it's hard to see how to apply it to the case. A tip is very much appreciated.

Answer 1

It is in fact related to the quadratic variation process of $X_n$. Note that $$ X_n^2 - \sum_{k=0}^n E[(X_k-X_{k-1})^2\;|\mathcal{F}_{k-1}],\quad n\geq0, $$ is a $\{\mathcal{F}_n\}$-martingale.

Answer 2

By the martingale property you can show $$E[(X_n - X_{n-1})^2] = E[X_n^2] - E[X_{n-1}^2].$$