Convergence in distribution of truncated Gaussian variables

Question

Let $X$ be a random variable that is distributed normally with mean $\mu=0$ and variance $\sigma=1$.

Suppose that $X_n$ is a random variable for any positive integer $n$ with truncated normal distribution with $\mu=0$, $\sigma=1$ and $a=-n$ and $b=n$.

I expect that the sequence $(X_n)_{n \geq 1}$ of bounded random variables converges in distribution to the unbounded random variable $X$.

1. Is is true?

2. Do you know a reference with a proof of this convergence?

Answer 1

It is enough to show that the MGFs of $X_n$ converge point-wise to the MGF of $X$. We know that $$ M_{X_n}(t)=e^{\mu t + 1/2\sigma^2t^2}\left(\frac{\Phi(\beta_n-\sigma t)-\Phi(\alpha_n-\sigma t)}{\Phi(\beta_n)-\Phi(\alpha_n)}\right) $$ where $\beta_n:=\frac{n-\mu}\sigma$, $\alpha_n:=\frac{-n-\mu}\sigma$ and $n\in\mathbb N$. So, by the continuity of the CDF $\Phi$, $$\lim\limits_{n\to\infty}M_{X_n}(t)=M_{X}(t).$$

Answer 2

It's true and you don't need anything fancy to show it. Apply this common definition and just work with the CDF in the link you gave.