Let $(X,\mathcal{B},\mu)$ be some probability space, and let $x_n$ be a sequence of centered, real, Gaussian random variables with variance $\sigma_n^2$. Suppose that $x_n$ converges in $L_2(X,\mathcal{B},\mu)$ to some random variable $x$. I want to show that $x$ also has real, centered, Gaussian distribution. I believe it can be done by using Levy's continuity theorem to imply pointwise convergence of the Fourier transforms, which will imply the convergence of $\sigma_n$ to some $\sigma$ and that the Fourier transform of $x$ is $\exp(-\frac{1}2 \sigma^2t^2)$ which implies that $x$ is a real, centered, Gaussian random variable with variance $\sigma^2$ by the uniqueness of the Fourier transform. First, I want to ask if this proof is correct? Also I want to find a proof that does not use Lévy's continuity theorem.
[Math] Is the $L_2$ limit of centered Gaussian random variables also a centered Gaussian random variable
probability theory
Best Answer
Without Lévy's continuity theorem:
The sequence $(x_n)_n$ converges in $L^2$ hence it is bounded in $L^2$ hence the sequence $(\sigma_n^2)_n$ is bounded. Consider a converging subsequence $\sigma_{\varphi(n)}^2\to\sigma^2$. Then $x_{\varphi(n)}$ converges in distribution to a centered normal random variable with variance $\sigma^2$. Since $x_{\varphi(n)}\to x$ in $L^2$ and convergence in $L^2$ implies convergence in distribution to the same limit, $x$ is a centered normal random variable with variance $\sigma^2$. (Recall that normal distributions with variance zero are Dirac distributions.)