Mean square convergence of the squares

Question

Let $X_n$ be a sequence of random variables converging in mean square to $X$, that is, \begin{equation}\tag{1} \lim_{n\to\infty}\mathbb{E}\Big(\big|X_n-X\big|^2\Big)=0 \end{equation} or, in short-hand notation, $X_n\overset{\mathcal{L}^2}{\longrightarrow}X$

Wish: does $(1)$ imply $X_n^2\overset{\mathcal{L}^2}{\longrightarrow}X^2$?

In the accepted answer to this post it is written that $(1)$ implies $\mathbb{E}(X_n^2)\to\mathbb{E}(X^2)$, but I don't know if it's enough to guarantee the convergence of the squares.

Why is $X^{2}$ in $\mathcal L^{2}$? – geetha290krm Mar 21 '23 at 23:16 — geetha290krm, Mar 21 '23 at 23:16
There’s no reason for it to be. I’m just pretending.. – ric.san Mar 22 '23 at 01:37 — ric.san, Mar 22 '23 at 01:37

score 1 · Accepted Answer · answered Mar 21 '23 at 22:16

1

No, this would require $\lim_{n \rightarrow \infty} \mathbb{E}[|X_n^2-X^2|^2] = 0$, but there is no guarantee that $\mathbb{E}[|X_n^2-X^2|^2]$ is even finite.

answered Mar 21 '23 at 22:16

user6247850

13,426

Oh, ok... but is it true at least that $\mathbb{E}(X_n^2)\to\mathbb{E}(X^2) $? – ric.san Mar 21 '23 at 22:24
@ric.san Yes, that is true, as the post you linked to showed – user6247850 Mar 21 '23 at 23:20

Mean square convergence of the squares

1 Answers1