# A counterexample that marginal convergence in law does not imply joint convergence in law

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 $$X_n = U$$ 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

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.
Well, thanks.  – Did Jan 7 '13 at 22:52