Convergence in distribution and convergence of variance Let $X_n \overset d \equiv N(0,\sigma^2_n)$ and $X_n \overset d \rightarrow X $. Further suppose that $\sigma^2_n \overset p \rightarrow \sigma ^2>0$. Then does is follow that $X \overset d \equiv N(0,\sigma^2)$? 
If not, can you give a counterexample? What if $\sigma^2_n\rightarrow \sigma^2$ in L1 or a.s.?
Thanks and regards.
EDIT: Here $\sigma^2_n$ is a random variable and $\sigma ^2$ is a constant. They are dependent to $X_n$... Is this assumption crucial?
 A: I've adapted the proof from the classical result (which is true) in the case where your $\sigma_n$ are non random. We use the equivalence : convergence in distribution iff convergence of characteristic functions. We know from your assumption that : 
$$ \mathbb{E}(e^{itX_n}) \rightarrow \mathbb{E}(e^{itX}).$$
Let's compute it in another way : conditionnaly to $\sigma_n$, you have : 
$$  \mathbb{E}(e^{itX_n}) = \mathbb{E}(\mathbb{E}(e^{itX_n}|\sigma_n)) = \mathbb{E}(e^{-\frac{\sigma_n^2t}{2}})$$ 
But then, since you have convergence in probability you can always extract a subsequence from your $\sigma_n$ converging almost surely (we will still note it $\sigma_n$) to $\sigma$, which is according to your hypothesis non random (this will be crucial). So by dominated convergence theorem : 
$$\mathbb{E}(e^{itX_n}) \rightarrow  \mathbb{E}(e^{-\frac{\sigma^2t}{2}}) = e^{-\frac{\sigma^2t}{2}} $$
which is the caracteristic function of a gaussian random variable with mean 0 and variance $\sigma$.  
A: I solved it....
$$(X_n,\sigma_n^2)\rightarrow ^d(X,\sigma^2)$$
Hence, $$\frac{1}{\sigma_n}X_n\rightarrow ^d\frac{1}{\sigma}X$$ and $$\frac{1}{\sigma_n}X_n\rightarrow ^dN(0,1)$$
