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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.

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up vote 2 down vote accepted

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.)

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Your proof using Lévy's continuity theorem is correct, yes.

So since $X_n \to X$ in $L^2$, $\mathbb{E}X_n=0$ we have

$$\sigma_n^2 = \mathbb{E}((X_n-\mathbb{E}X_n)^2) = \mathbb{E}(X_n^2) \to \mathbb{E}(X^2)=:\sigma^2 \qquad (n \to \infty)$$


$$\mathbb{E}e^{\imath \, \xi \cdot X_n} = \exp \left(-\frac{1}{2} \sigma_n^2 \cdot \xi^2 \right) \to \exp \left(-\frac{1}{2} \sigma^2 \cdot \xi^2 \right)$$

and therefore $X_n \stackrel{w}{\to} X \sim N(0,\sigma^2)$ by Levy's continuity theorem.

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