I am trying to prove that
If $X_n \sim N(\mu_n, \sigma_n^2)$ converges to $X$ in probability, then $X$ is also normal, and the convergence is indeed in $L^2$.
by following the argument in https://stats.stackexchange.com/questions/24362/convergence-in-probability-and-l-2-for-normal-random-variables and without using the fact that $X_n \to X$ in distribution and tightness of $X_n$.
I believe it boils down to say $E[X] := \mu < \infty$. If that's the case, I could consider the characteristic function of $X_n – \mu_n\sim N(0,\sigma_n^2)$, and show the limit $X-\mu$ has normal distribution $N(0,\sigma^2)$. Could anyone give me some hint on this?
Best Answer
Using characteristic functions we see that $e^{i\mu_n t} e^{-t^{2}\sigma_n^{2}/2} \to Ee^{itX}$ for all $t$. Take absolute values to see that $\sigma_n $ converges. Then note that convergence of $e^{i\mu_n t}$ for every $t$ implies convergence of $(\mu_n)$. Hence $Ee^{itX} =e^{i\mu t} e^{-t^{2}\sigma^{2}/2}$, so $X$ is normal.