Convergence in probability of parameters of normal distribution implies convergence in distribution

Question

Suppose $(\mu_n, \sigma_n^2) \stackrel{p}{\rightarrow} (\mu, \sigma^2)$. Does this imply that we have $\mathcal{N}(\mu_n, \sigma_n^2) \rightarrow \mathcal{N}(\mu, \sigma^2)$?

I.e., Does convergence in probability of the parameters of the normal distribution imply convergence to a normal distribution? If not, are there some additional conditions that need to be imposed?

Answer 1

Yes. Use characteristic functions and note that weak convergence is preserved by continuous functions.