Let $\xi_{n}=X_{n}-X_{n-1}$ where $EX_{0}^{2}, \sum\limits_{m}E\xi_{m}^{2} < \infty$ then $X_{n} \to X_{\infty}-$a.s. and $L^{2}$

Question

Let $X_{n}, n \in \mathbb N$, be a martingale and $\xi_{n}=X_{n}-X_{n-1}$ where $EX_{0}^{2}, \sum\limits_{m}E\xi_{m}^{2} < \infty$.

Show that $X_{n} \to X_{\infty}-$a.s. and in $L^{2}$

It is clear we need to use the Orthogonality of Martingale Increments to fulfill the $L^{p}$ Convergence Theorem.

Let's define $\tau_{K}:=\inf\{n: X_{n}\geq K\operatorname{or} X_{n}\leq -K\}$ where $K >0$, we know that the stopped process $(X_{n \land \tau_{K}})_{n}$ is a martingale.

This means that $\vert X_{n \land \tau_{K}}\vert\leq K+\sup\limits_{m}\vert \xi_{m}\vert$

It is clear from $\sum\limits_{m}E\xi_{m}^{2}<\infty$ that $\sup\limits_{m} E\xi_{m}^{2}<\infty$. Can I imply from $\sum\limits_{m}E\xi_{m}^{2}<\infty\Rightarrow \sup\limits_{m}\vert \xi_{m}\vert <\infty$? This would be needed to show that $\sup\limits_{n} E[\vert X_{n\land \tau_{K}}\vert ]<\infty$

From there I can say that for any $K>0$: $\sup\limits _{n}E[\vert X_{n \land \tau_{K}}\vert]\leq K+E[\sup\limits_{m}\vert \xi_{m}\vert]< \infty$

We thus let $K\to \infty$ but I am not sure if this helps since then our upper bound is $\infty$

Answer 1

Since $$X_n-X_m = \sum_{j=m+1}^n (X_j-X_{j-1})$$ it follows from the orthogonality of the increments that $$\mathbb{E}((X_n-X_m)^2) = \sum_{j=m+1}^n \mathbb{E}((X_j-X_{j-1})^2).$$ By assumption, the series on the right-hand side is a Cauchy sequence, and so is the left-hand side. By the completeness of $L^2$, this implies that the limit $X_{\infty} = \lim_{n \to \infty} X_n$ exists in $L^2$. Conclude that $(X_n)_{n \in \mathbb{N}}$ is bounded in $L^1$ and apply the martingale convergence theorem to obtain that $X_n \to X_{\infty}$ almost surely.