Prove $H_n\sim N(\xi_n, \sigma_n^2) \stackrel{d}\to H\sim N(\xi, \sigma^2)$ iff $\xi_n \to \xi$ and $\sigma_n^2 \to \sigma^2$

Question

Prove that if $H_n$ is the normal distribution with mean $\xi_n$ and variance $\sigma_n^2$, then $H_n$ tends to normal distribution $H$ with mean $\xi$ and variance $\sigma^2$ if and only if $\xi_n \to \xi$ and $\sigma_n^2 \to \sigma^2$.

Proof :

if $\xi \to \xi$ and $\sigma_n^2 \to \sigma^2$, then $H_n\sim N(\xi_n, \sigma_n^2) \to H_n \sim N(\xi, \sigma)$

Let $Z_n\sim N(0,1)$ and $Z\sim N(0,1)$; $Z_n \to Z$, and apply Slutsky's theorem,

$\sigma_n^2 + \xi_n Z_n \to \sigma^2 + \xi Z$ qed.

Does it seem right?

and How do I prove the converse?

if $H_n\sim N(\xi_n, \sigma_n^2) \to H_n\sim N(\xi, \sigma)$, then $\xi \to \xi$ and $\sigma_n^2 \to \sigma^2$ .

Answer 1

Your proof is OK for one implication. The best way to prove the reverse implication is to use characteristic functions. Since $\left|Ee^{itH_n}\right|^{2} \to \left|Ee^{itH}\right|^{2}$ we get $e^{-t^{2}\sigma_n^{2}} \to e^{-t^{2}\sigma^{2}}$ for all $t$. Take logarithm to see that $\sigma_n^{2} \to \sigma^{2}$. Now use the fact that $e^{it\xi_n}e^{-t^{2}\sigma_n^{2}/2}= e^{it\xi_n}e^{-t^{2}\sigma_n^{2}/2} $ to show that $\xi_n \to \xi$.