Convergence in distribution of a product of R.V.s? If we have a sequence of variables $X_n$ that converges in distribution to $X$, and a sequence $Y_n$ that converges in distribution to $Y$, then does $X_nY_n$ converge in distribution to $XY$. Assume $X_n$ and $Y_n$ are independent. 
Weirdly, I can’t find a theorem on this in my textbook on asymptotics. 
 A: Yes, $X_n\cdot Y_n$ converges in distribution to XY.
Method 1 :
$X_n$ $\rightarrow$ $X$
$\implies$ $\lim_{n\rightarrow ∞}$ ⁡$F(X_n)$ = $F(X)$
Here, $F(X_n)$ and $F(X)$ are the cumulative distribution functions of $X_n$ and $X$, respectively.
$Y_n$ $\rightarrow$ $Y$
$\implies$ $\lim_{n\rightarrow ∞}$ ⁡$F(Y_n)$ = $F(Y)$
Again, $F(Y_n)$ and $F(Y)$ are the cumulative distribution functions of $Y_n$ and $Y$, respectively.
$\lim_{n\rightarrow ∞}$ $F(X_n\cdot Y_n)$  = $\lim_{n\rightarrow ∞}$ $F(X_n)\cdot \lim_{n\rightarrow ∞}$ $F(Y_n)$  [Due to independence]
$\implies$ $\lim_{n\rightarrow ∞}$ $F(X_n\cdot Y_n)$  = $F(X)\cdot F(Y)$
$\implies$ $X_n\cdot Y_n$ $\rightarrow$ $XY$
Method 2 :
Check out the link provided here.
https://en.wikipedia.org/wiki/Proofs_of_convergence_of_random_variables
We know, convergence in probability implies convergence in distribution but the opposite is not true. Hence, if convergence of two sequences in probability implies joint convergence in probability, it will surely converge into distribution assuming the sequence of random variables are independent.
