If $X_n$ and $Y_n$ are independent does $(X_n,Y_n)\overset{d}{\rightarrow}(X,Y)$?

Question

More formally: If $X_n\overset{d}{\rightarrow}X$ and $Y_n\overset{d}{\rightarrow}Y$ and also $X_i$ and $Y_j$ are independent for all i,j; does $(X_n,Y_n)\overset{d}{\rightarrow}(X,Y)$?

I am aware of the Cramer-Wold theorem to prove asymptotic convergence of vector of random variables but I can't quite figure out how to apply it here. (To be honest I don't even know if the statement is true but it feels like it should be).

Answer 1

Assume that it is true and let it be that $X_n=X$ and $Y_n=Y$ for every $n$ where $X$ and $Y$ are iid and non-degenerate random variables.

Evidently for every pair $i,j$ the rv's $X_i$ and $Y_j$ are iid.

Then $(X_n,Y_n)=(X,Y)$ for all $n$ so of course we have: $$(X_n,Y_n)\stackrel{d}{\to}(X,Y)\tag1$$

But we also have $X_n\stackrel{d}{\to}X$ and $Y_n\stackrel{d}{\to}X$ so under the assumption we arrive at the conclusion that: $$(X_n,Y_n)\stackrel{d}{\to}(X,X)\tag2$$

Combining $(1)$ and $(2)$ we find that $(X,Y)$ and $(X,X)$ have equal distributions which cannot be true.

We conclude that the assumption is false.

Answer 2

This just complements the nice answer by drhab. $Ee^{i(tX_n+isY_n)}=Ee^{itX_n} Ee^{isY_n} \to Ee^{itX}Ee^{isY}$ and this implies that $(X_n,Y_n)$ converges in distribution to $(X,Y)$ provided $X$ and $Y$ are independent.