Is convergence in distribution transitive?

Question

I'm trying to prove that a sequence of statistics $s_n$ is asymptotically normal.

So far I've managed to show two things concerning $s_n$ and a second statistic $s_n^*$:

1. $\lim_{n \to \infty} |s_n - s^*_n| = 0$.

2. $s^*_n$ converges in distribution to $N(0,\sigma_s)$

My question is: what conditions are necessary (if any) to conclude that since $s_n \to s_n^*$ and $s_n^*$ converges in distribution to a centred normal distribution, that $s_n$ is also asymptotically normal?

The statistics in question are the corrected sample standard deviation $s_n$ and the uncorrected sample standard deviation $s_n^*$ taken from a sample $X_1, X_2, \dots, X_n$. Here, the underlying sample $X_i$ is iid but non-normal.

Answer 1

If by 1) you mean that $s_n-s_n^{*} \to 0$ almost surely (or just in probability) then it is true that $s_n$ converges in distribution to $N(0,\sigma_s)$. This can be proved using characteristic functions: $E(e^{its_n}-e^{its_n^{*}})=E(e^{its_n}(1-e^{it(s_n^{*}-s_n)}) \to 0$ by DCT and convergence in distribution follows from this. Actually, $s_n-s_n^{*} \to 0$ in distribution implies $s_n-s_n^{*} \to 0$ in probability so the conclusion holds even in this case.