# 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.

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.

• Thanks! I've tried to prove the first part about tightness for a while now, but I'm not getting anywhere... what properties of convergence in distribution am I supposed to use? The definition I have got is that there is pointwise convergence of the cdf's wherever the limiting distribution is continuous, but that's all I know. – Richard Feb 14 '15 at 9:46
• It's a standard result about weak convergence, in fact. See e.g. Lemma 3.1.6 of these notes for a proof: math.ku.dk/noter/filer/vidsand12.pdf Essentially, the proof works by approximating and indicator function with a continuous (here, even uniformly continuous) and bounded mapping. – Alexander Sokol Feb 15 '15 at 11:30
• I've given this another go, but I believe I'm not getting anywhere -- I can't show that the sequences are bounded, nor arrive at any sort of contradiction assuming boundedness and that there were two limits. It might be just that I just haven't got sufficient mathematical machinery to deal with such a problem yet, but would it be possible for you to type out a fuller solution to this? Though if you're busy and can't do it, it's okay as well. Thanks :) – Richard Feb 15 '15 at 15:47
• show $\lim_{n \to \infty} exp(i \mu _n t)= \phi (t) exp(\sigma^2 t^2 /2)$ and then show above equality with two different limit points of $\mu_n$ (or just consider limsup, liminf) is a contradiction. – izimath Jun 6 '18 at 8:20

A sequence of complex number $$z_n$$ converges to a limit $$z$$ iff $$|z_i| \to |z|$$ and $$\arg(z_i) \to \arg(z)$$. Now we have the following from Levy continuity theorem, $$\lim_{n \to \infty}e^{-\sigma_n^2t^2/2}e^{i\mu_nt} \to r(t)e^{i\theta(t)}, \forall t \in \mathbb{R}$$ $$\iff \lim_{n \to \infty}e^{-\sigma_n^2t^2/2} \to r(t), \lim_{n \to \infty}\mu_nt \to \theta(t), \forall t \in \mathbb{R}$$ Now you can see this implies $$\mu_n,\sigma_n$$ does converge to constant. For example, $$\lim_n \mu_n=\theta(t)/t=const$$ as LHS does not contain $$t$$. Some details here to verify to be completely rigorous.