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

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).

• This is true if you assume that $X,Y$ are independent too. (Also it is enough to assume that $X_n$ and $Y_n$ are independent for each $n$.) – zhoraster Apr 23 at 12:05

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.

• I don't understand why $Y_n\overset{d}{\rightarrow}X$ could you expand on why that's the case? – Lorenzo Apr 24 at 13:38
• We have $Y_n\stackrel{d}{\to}Y$ because $Y_n=Y$ for every $n$. But in my answer $X$ and $Y$ have the same distribution so we also have $Y_n\stackrel{d}{\to}X$. Note that convergence in distribution concerns distributions specifically. We can state it as: the distribution of $Y_n$ converges to the distribution of $Y$ (which is also the distribution of $X$). – drhab Apr 24 at 13:46
• But if X and Y have the same distribution, why would (X,X) and (X,Y) not be distributed identically? – Lorenzo Apr 24 at 13:51
• $(X,Y)$ and $(X,X)$ have the same marginal distributions. However not the same distributions, because $X,Y$ are independent and $X,X$ are not. Observe that $(X,Y)$ can take values in $\mathbb R^2$ everywhere, but $(X,X)$ only on the line $\{(x,x)\mid x\in\mathbb R\}$. – drhab Apr 24 at 13:52

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.

• I don't see how your argument leads to the conclusion as you are showing something about a specific function of X and Y. It would be useful if you could expand a bit further. (sorry but I don't have such good knowledge around the topic) – Lorenzo Apr 24 at 13:47
• @Lorenzo I am using some standard facts about characteristic functions. In particular I am using the cat that convergence of distributions is equivalent to convergence of characteristic functions. – Kavi Rama Murthy Apr 24 at 23:05