Calculating expectations of sum of martingales

Question

Let $\left(X_{n}\right)_{n \geq 0}$ is a martingale with $X_{0}=0$. Assume

$$ \sum_{n=1}^{\infty} E\left(\left(X_{n}-X_{n-1}\right)^{2}\right)<\infty . $$

Show that $X=\lim _{n \rightarrow \infty} X_{n}$ exists $P$-a.s.

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I know how to proceed and have notes (from professor) with answers. However, when I calculate the sum I get

\begin{align*} \sum_{n=1}^\infty E((X_n-X_{n-1})^2)=\sum_{n=1}^\infty E(X_n^2+X_{n-1}^2-2X_{n-1}X_n)=\sum_{n=1}^\infty E(X_n^2)+E(X_{n-1}^2)-2E(X_{n-1}X_n) \end{align*}

Now lets consider the expectation $E(X_{n-1}X_n)$:

\begin{align*} E(X_{n-1}X_n)=E(E(X_{n-1}X_n|\mathcal{F}_{n-1}))=E(X_{n-1}E(X_n|\mathcal{F}_{n-1}))=E(X_{n-1}^2) \end{align*}

Thus we now have

\begin{align*} \sum_{n=1}^\infty E(X_n^2)+E(X_{n-1}^2)-2E(X_{n-1}X_n)&= \sum_{n=1}^\infty E(X_n^2)+E(X_{n-1}^2)-2E(X_{n-1}^2)&\\=\sum_{n=1}^\infty E(X_n^2)-E(X_{n-1}^2) \end{align*}

but my notes are saying that the last sum should be $$=\sum_{n=1}^\infty E(X_n^2)-E(X_{n-1})$$ which I do not understand why that would be.

Answer 1

Your computation is correct. It shows that $(EX_n^{2})$ is bounded. Any $L^{2}$ bounded martinagle, more generally any uniformly integrable martingale converges almost surely.

[The $n-$th partial sum of $\sum [a_n-a_{n+1}]$ is $a_n-a_0$. In our case it is $EX_n^{2}$ It is given that the series converges, so the partial sums form a bounded sequence. Hence, $(EX_n^{2})$ is bounded].

Answer 2

I think you want to apply your calculation to $\sum_{n=1}^k$ instead of $\sum_{n=1}^\infty$. This would show you that $$\sum_{n=1}^k E((X_n - X_{n-1})^2) = \sum_{n=1}^k E(X_n^2) - E(X_{n-1}^2) = E(X_k^2) - E(X_0^2)$$ because the second sum telescopes. But the sum on the left is less than $\sum_{n=1}^\infty E((X_n - X_{n-1})^2)$ for all $k$, so we conclude that $E(X_k^2)$ is bounded.