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I just encountered the following counterexamle that should illustrate that marginal convergence in distribution does not imply joint convergence in distribution:

Let U and V be independent standard normal random variables and define \begin{equation}X_n = U\end{equation} and $$Y_n=(−1)^n \cdot \frac{1}{2} \cdot U+ \sqrt{\frac{3}{4}}\cdot V$$ Then $X_n$ and $Y_n$ are both standard normal for all $n$, and hence trivially converge in law marginally. But $$\text{cov}(X_n, Y_n) = (−1)^n $$ for all $n$, and so the sequence $(X_n, Y_n)$ of random vectors cannot converge in law.

Here is my question (and I feel that the answer is straightforward, I just can't see it): why does the alternating covariance violate convergence of distribution of the vector $(X_n,Y_n)$?

Many thanks for any help, it is much appreciated. And sorry again if the solution is straighforward...

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The covariance is $(-1)^n\frac12$, not $(-1)^n$. – Did Jan 6 '13 at 23:39
I don't understand what is being asked. The random variables $(X_1,Y_1)$, $(X_3,Y_3)$, $(X_5,Y_5), \ldots$ all have one jointly normal distribution with covariance $-\frac{1}{2}$ while $(X_2,Y_2)$, $(X_4,Y_4)$, $(X_6,Y_6), \ldots$ all have a different jointly normal distribution where the covariance is $+\frac{1}{2}$. So in what sense are you thinking that $F_{X_n,Y_n}$ is converging to a fixed $F_{U,V}$? Isn't that something like asking why the sequence $+1, -1, +1, -1, +1, -1, \ldots$ is not a convergent sequence? – Dilip Sarwate Jan 7 '13 at 16:49
@DilipSarwate No theorem states that convergence in distribution requires convergence of the covariance. In fact it does not, see my Edit-edit. – Did Jan 7 '13 at 22:53
up vote 4 down vote accepted

The variance of $X_n+2Y_n$ is $3$ if $n$ is odd and $7$ if $n$ is even hence $$ \mathbb P(X_n+2Y_n\leqslant\sqrt{21})=\begin{cases}\Phi(\sqrt7)\quad\text{if $n$ is odd},\\ \Phi(\sqrt3)\quad\text{if $n$ is even.}\end{cases} $$ Nota: As regards finding a counterexample, I prefer $X_n=U$, $Y_n=(-1)^nU$.

Edit: In the case at hand, the sequences $(X_n)_n$ and $(Y_n)_n$ are uniformly bounded in every $L^p$, for example in $L^4$, hence these families are uniformly square integrable. Assume that the sequence $(X_n,Y_n)_n$ converges in distribution. Then there exists some random variables $(\bar X,\bar Y)$ and some sequences $(\bar X_n)_n$ and $(\bar Y_n)_n$, defined on a common probability space, such that each $(\bar X_n,\bar Y_n)$ is distributed as $(X_n,Y_n)$ and the sequence $(\bar X_n,\bar Y_n)_n$ converges almost surely to $(\bar X,\bar Y)$.
Since the family $(\bar X_n,\bar Y_n)_n$ is uniformly square integrable, $(\bar X_n,\bar Y_n)_n$ converges to $(\bar X,\bar Y)$ in $L^2$, in particular $\mathrm{Cov}(X_n,Y_n)=\mathrm{Cov}(\bar X_n,\bar Y_n)$ converges. This is not so hence $(X_n,Y_n)_n$ does not converge in distribution.

Edit-edit: To see that convergence in distribution can coexist with nonvanishingly oscillating covariances, assume that $\mathbb P(X_n=1/\sqrt{2p_n})=p_n$, $\mathbb P(X_n=-1/\sqrt{2p_n})=p_n$ and $\mathbb P(X_n=0)=1-2p_n$ for some positive sequence $(p_n)_n$ such that $p_n\leqslant1/2$ and $p_n\to0$, and define $Y_n=(-1)^nX_n$. Then $X_n\to0$ in distribution, $Y_n\to0$ in distribution, $\mathrm{Cov}(X_n,Y_n)=(-1)^n$ and $(X_n,Y_n)\to(0,0)$ in distribution. Note that $(X_n)_n$ is not uniformly square integrable since $\mathbb E(X_n^2)=1$ for every $n$ and $X_n\to0$ in distribution.

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thanks, but that does not answer exactly the question about how the alternating covariance is related to the non-convergence. – s_2 Jan 6 '13 at 23:35
The problem is that the convergence in distribution does not imply the convergence of the covariances, except under extra uniform integrability assumptions, which are satisfied in your example but not in general. – Did Jan 6 '13 at 23:42
hi did, many thanks for your excellent and well-explained answer. I really appreciate your support!!! – s_2 Jan 7 '13 at 19:00
Well, thanks. $ $ – Did Jan 7 '13 at 22:52

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