Limit of Gaussian random variables is Gaussian?

Consider a sequence $X_n$ of Gaussian random variables with mean $\mu_n$ and variance $\sigma_n^2$, which converges in distribution (to some limiting distribution). Can I then conclude that $\mu_n$ converges to some $\mu$ and $\sigma_n^2$ converges to some $\sigma^2$, with the limiting distribution being $N(\mu, \sigma^2)$? Here, I think I'd allow the degenerate Gaussian distribution with zero variance.

I've tried looking this up elsewhere on this site, but all those I've found assume almost sure/L2 convergence. I was thinking this should be true, and was trying to do this by characteristic functions but without assuming the convergence of the two parameters, I can't seem to conclude anything.

Solution 1:

The following holds: Let $(\xi_n)$ and $(\sigma_n)$ be sequences in $\mathbb{R}$, where $\sigma_n>0$. Let $\mu_n$ denote the Gaussian distribution with mean $\xi_n$ and variance $\sigma_n^2$. Then $(\mu_n)$ converges weakly if and only if $(\xi_n)$ and $(\sigma_n^2)$ both converge, and in the affirmative case, the limiting distribution is a Gaussian distribution with mean $\xi = \lim_n \xi_n$ and variance $\sigma^2 = \lim_n \sigma_n^2$. Here, the case $\sigma=0$ is understood to indicate the Dirac measure in $\xi$.

You are correct in observing that when proving this, the difficult part is to show that $(\xi_n)$ and $(\sigma_n^2)$ when $(\mu_n)$ converges weakly. Here are some hints. Assume that $(\mu_n)$ converges weakly. It then holds that

$$ \lim_{M\to\infty} \sup_{n\ge1} \mu_n([-M,M]^c) = 0, $$ essentially meaning that $(\mu_n)$ is a tight family of measures. Use this and the properties of Gaussian distributions to show that both $(\xi_n)$ and $(\sigma_n^2)$ are bounded sequences. Assume, expecting a contradiction, that the sequences are not convergent. As the sequences are bounded, it must in particular hold that $(\xi_n)$ has two different limit points. Use this to obtain a contradiction. Thus, $(\xi_n)$ is convergent. Use a similar technique to obtain that $(\sigma_n^2)$ is convergent.