Let $X_n$, $n\geq 0$, be a martingale and let $\xi_n = X_n – X_{n-1}$ for $n \geq 1$. If $E(X_0^2)$ and $\sum_{m=1}^{\infty}E(\xi_m^2) < \infty$, then $X_n \rightarrow X_\infty$ a.s. and in $L^2$.
My attempt: I used $L^p$ convergence theorem to proved this question.
This theorem is stated as If $X_n$ is martingale with $\sup E(|X_n|^p) < \infty$ where $p>1$ then $X_n \rightarrow X$ a.s. and in $L^p$.
The only thing I need to show $\sup E(|X_n|^2) < \infty$ . For. this I proved that If $X_n$ and $Y_n$ are martingale with $E(X_n^2)<\infty$ and $E(Y_n^2)<\infty$, then
\begin{equation*}
E(X_nY_n) – E(X_0Y_0) = \sum_{m=1}^n (X_m – X_{m-1})(Y_m – Y_{m-1})
\end{equation*}
My question is how do I show that $E(X_n^2)<\infty$?
Best Answer
We may use a telescoping sum to get: $$X_n = X_0 + \sum_{k=1}^n (X_k - X_{k-1})$$ Squaring both sides, we get: $$X_n^2 = X_0^2 + \sum_{k=1}^n (X_k - X_{k-1})^2 + \sum_{i \neq j}(X_i -X_{i-1})(X_j - X_{j-1}) + 2X_0 \sum_{k=1}^n (X_k - X_{k-1})$$ By the orthogonality of non-overlapping $L^2$-martingale increments $$\mathbb{E}\left[\left(X_i -X_{i-1}\right)\left(X_j - X_{j-1}\right)\right] = 0 \qquad i \neq j$$ It follows that $$\mathbb{E}\left(X_n^2\right) = \mathbb{E}\left(X_0^2\right) + \sum_{k=1}^n \mathbb{E}(\xi_k^2) \leq \mathbb{E}\left(X_0^2\right) + \sum_{k=1}^\infty \mathbb{E}\left(\xi_k^2\right) < \infty$$ The right-hand side is a uniform bound that does not depend on $n$, so that $\sup_n \mathbb{E}(X_n^2) < \infty$. By the martingale convergence theorem, the result follows.