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More generally, if $\{X_n\}_{n\in\mathbb{N}}, X$ are real random variables with finite variance such that $X_n\xrightarrow{d}X$, what are some sufficient conditions to assure that $\operatorname{Var}(X_n)\rightarrow \operatorname{Var}(X)$? What if $X$ is known to be Gaussian?

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  • $\begingroup$ wrt the CLT isnt Var(Xn) ~ Var(X) for all n $\endgroup$ – mandata Jun 16 '16 at 3:58
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    $\begingroup$ It's hard to conceive of anything that wouldn't be either equivalent to convergence of the variances or even more complicated than that. The result has nothing to do with the distribution of $X$, Gaussian or otherwise. To see why not, modify $X_n$ into a mixture of its original distribution plus an atom at $n^2$ above its mean, in proportions $n-1$ to $1$. The modified $X_n$ still converge to $X$ in distribution but their variances are at least $n$, which diverges. $\endgroup$ – whuber Jun 16 '16 at 17:30
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    $\begingroup$ If $X_n$ takes on values in a bounded closed interval $I$, then you have convergence of variance by the Portmaneau theorem (and the fact that $f=x$ and $f=x^2$ are bounded on $I$). More generally you need $X_n$ to be uniformly integrable. See here for example: math.stackexchange.com/questions/153293/… $\endgroup$ – Alex R. Jun 16 '16 at 18:34
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Partially answered in comments:

It's hard to conceive of anything that wouldn't be either equivalent to convergence of the variances or even more complicated than that. The result has nothing to do with the distribution of $X$, Gaussian or otherwise. To see why not, modify $X_n$ into a mixture of its original distribution plus an atom at $n_2$ above its mean, in proportions $n−1$ to $1$. The modified $X_n$ still converge to $X$ in distribution but their variances are at least $n$, which diverges. – whuber

If $X_n$ takes on values in a bounded closed interval $I$, then you have convergence of variance by the Portmaneau theorem (and the fact that $f=x$ and $f=x^2$ are bounded on $I$). More generally you need $X_n$ to be uniformly integrable. See here for example: https://math.stackexchange.com/questions/153293/does-convergence-in-distribution-implies-convergence-of-expectation – Alex R.

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