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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).

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  • $\begingroup$ 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 $n$.) $\endgroup$ – zhoraster Apr 23 at 12:05
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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.

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  • $\begingroup$ I don't understand why $Y_n\overset{d}{\rightarrow}X$ could you expand on why that's the case? $\endgroup$ – Lorenzo Apr 24 at 13:38
  • $\begingroup$ 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 $X$). $\endgroup$ – drhab Apr 24 at 13:46
  • $\begingroup$ But if X and Y have the same distribution, why would (X,X) and (X,Y) not be distributed identically? $\endgroup$ – Lorenzo Apr 24 at 13:51
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    $\begingroup$ $(X,Y)$ 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 $\{(x,x)\mid x\in\mathbb R\}$. $\endgroup$ – drhab Apr 24 at 13:52
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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.

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  • $\begingroup$ 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) $\endgroup$ – Lorenzo Apr 24 at 13:47
  • $\begingroup$ @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. $\endgroup$ – Kavi Rama Murthy Apr 24 at 23:05

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