About convergence in distribution

Question

Suppose that a sequence of real valued random variables $(X_n)_{n\geq 1}$ converges in distribution to a random variable $X$. Does the couple $(X_n, X)_{n\geq 1}$ converge in distribution to $(X, X)$?

Answer 1

No. Suppose $P(X=1) = P(X=-1) = 1/2$ and that $X_n = - X$. Then all the $X_n$ and $X$ have the same distribution, and $X_n$ converges in distribution to $X$. But the distribution of $(X_n,X)$ is supported on $\{(1,-1),(-1,1)\}$ and that of $(X,X)$ is supported on $\{(1,1), (-1,-1)\}$. Let $f(x,y)=\sin(x,y)$. It is continuous and bounded, but $Ef(X_n,X)$ does not converge to $E f(X,X)$. So the distribution of $(X_n,X)$ does not converge to that of $(X,X)$.

Or, with more moving parts, let $X_n = (-1)^nX$, and slightly more complicated calculations.

Answer 2

If $X$ is independent of each $X_n$ and $F_X$ (the CDF of $X$) is continuous, then for any $z,x\in \mathbb{R}$, $$ F_{(X_n,X)}(z, x)=F_{X_n}(z)F(x)\to F_X(z)F_X(x). $$ However, $F_{(X,X)}(z,x)=F_X(z\wedge x)$.

Answer 3

No, otherwise the convergence in distribution would imply the convergence in probability, by using the function $(x,y)\mapsto x-y$.