Convergence in distribution to normal random random variable

Question

Let $X_n$ be a sequence of random variables on $\mathbb R$ such that $X_n$ converges in distribution to a Gaussian variable $N(0,\sigma^2)$.

Is it true that

$$\mathbb E[e^{-\frac{X_n}{\sqrt n}}]\to 1?$$

Can you help me a bit?

Answer 1

Let $Z\sim\mathcal{N}(0,\sigma^2)$ and $U\sim\mathcal{U}[0,1]$ be independent. Define $X_n$ by

$$ X_n = Z - (\sqrt{n} \log n) \mathbf{1}_{\{U \leq 1/n\}}. $$

Since $X_n \to Z$ $\mathbb{P}$-almost surely, we have $X_n \to Z$ in distribution. However,

$$ \mathbb{E}[e^{-X_n/\sqrt{n}}] = \mathbb{E}[e^{-Z/\sqrt{n}}]\mathbb{E}[n^{\mathbf{1}_{\{U \leq 1/n\}}}] = e^{\sigma^2/2n} \left( 2 - \frac{1}{n} \right) \xrightarrow{n\to\infty} 2. $$