convergence almost surely implies convergence of variance?

Question

Does $X_{n} \to X$ almost surely imply that $Var[X_{n}] \to Var[X]?$

I saw this post

convergence in mean square implies convergence of variance

which states that $X_{n} \to X$ in $L^{2}$ implies that $Var[X_{n}] \to Var[X]$ but does that still hold if the convergence is almost surely instead of $L^{2}$?

Many thanks in advance.

No, because the random variables need not even to be integrable, see e.g. Example 3 here — saz, Mar 03 '19 at 15:32

score 1 · Answer 1 · answered Mar 03 '19 at 15:42

1

Take $U \sim$ Uniform$[0,1]$. Let $$X_n := \sqrt{n} 1_{ \{U \leq \frac{1}{n} \} } - \frac{1}{\sqrt{n}}$$ Then $X_n \to 0$ a.s. But $$\mathbb V [ X_n] = 1 - \frac 1 n \to 1 \neq 0$$

answered Mar 03 '19 at 15:42

Falrach

4,102

convergence almost surely implies convergence of variance?

1 Answers1