a) Let $(X_n : n \ge 1)$ be a sequence of random variables which converges in mean square. Show that $\mathbb{E}([X_n-X_m]^2)\rightarrow 0$ as $m,n\rightarrow \infty$.
b) If $\mathbb{E}(X_n)=\mu$ and $\mbox{var}(X_n)=\sigma^2$ for all $n$, show that the correlation between $X_n$ and $X_m$ converges to 1 as $m,n\rightarrow\infty$.
I am mainly asking about part b). The hint says to use the Cauchy Schwarz inequality: "For random variables $U$ and $V$, $\mathbb{E}(UV)^2 \le\mathbb{E}(U^2)\mathbb{E}(V^2)$" (I don't know if this hint refers to part a or part b though).
I have no idea how to show a correlation between 2 random variables/sequences converges. Is there a formula or theorem I should know? It doesn't mention this in the textbook.
Best Answer
For the sake of simplicity, assume that $\mu=0$ (the generalization is conceptually obvious but algebraically messy). Then, for any positive integers $m$ and $n$, \begin{align*} \big|\mathbb E[X_n X_m]-\sigma^2\big|&=\big|\mathbb E[X_n(X_m-X_n)]+\underbrace{\mathbb E[X_n^2]-\sigma^2}_{=0}\big|=\big|\mathbb E[X_n(X_m-X_n)]\big|\\ &\leq\mathbb E[|X_n(X_m-X_n)|]\underset{\substack{\uparrow\\\text{CS}}}{\leq}\sqrt{\mathbb E[X_n^2\vphantom{)^2}]}\sqrt{\mathbb E[(X_m-X_n)^2]}\\ &=\sigma\sqrt{\mathbb E[(X_m-X_n)^2]} \end{align*} (CS = Cauchy–Schwarz). The term $\mathbb E[(X_m-X_n)^2]$ converges to $0$ as $m,n\to\infty$ because mean-square convergence implies the mean-square Cauchy property. Therefore, $$\lim_{m,n\to\infty}\mathbb E[X_nX_m]=\sigma^2.$$ You can finish from here.