Counter-examples related to Slutsky's Theorem.

Solution 1:

If the pair $(X_n,Y_n)$ converges in distribution to $(X,Y)$, then necessarily $X_n+Y_n$ will converge in distribution to $X+Y$ (by the continuous mapping theorem). So a counterexample would require us to specify $X$ and $Y$ and their joint distribution such that $(X_n,Y_n)$ does not converge in distribution to $(X,Y)$.

So take $X$ to be a nonconstant symmetric random variable, define $X_n:=X$, $Y_n:= X$, and $Y:=-X$. Then trivially $X_n$ converges in distribution to $X$, and $Y_n$ converges in distribution to $Y$ (since $X$ is symmetric). But $X_n+Y_n$ equals $2X$, while $X+Y=0$. Note that, as expected, $(X_n,Y_n)$ does not converge in distribution to $(X,Y)$, which is concentrated on the line $y=-x$; it converges in distribution to $(X,X)$, which is concentrated on the line $y=x$.

Solution 2:

Let $Z_k$ be iid with mean $0$ and variance $1$, and let $X_n = \frac{1}{\sqrt n} \sum_{k=1}^n Z_k$ and $Y_n = -X_n$. By the CLT, both $X_n$ and $Y_n$ converge in distribution to some $X$ which is standard normal. But $X_n + Y_n$ is always zero.