Counter-example for the Slutsky Theorem

Question

The Slutsky’s Theorem says that

Let $X_n$ and $Y_n$ random variables such that $X_n\stackrel{d}{\to} X$, with $X$ random variable and $Y_n\stackrel{p}{\to} c$, with $c$ a constant. Then $$X_n+Y_n\stackrel{d}{\to} X+c\quad (1)$$ $$X_n\,Y_n\stackrel{d}{\to} X\,c\quad (2)$$

I am looking for a simple counter-example for the case $(2)$. In particular I need a case in which $(2)$ does not hold because $Y_n\stackrel{p}{\to} Y$ with $Y$ a random variable.

Answer 1

Let $Y_n=Y=X_n$ for every $n$ where $Y$ is not degenerated.

Further let $X\stackrel{d}{=}Y$ and let $X$ and $Y$ be independent.

Then $Y_n\stackrel{p}{\to}Y$ and $X_n\stackrel{d}{\to} X$ and $X_nY_n\stackrel{d}{\to}Y^2$.

But because $Y$ is not degenerated $Y^2$ and $XY$ do not have the same distribution.

Consequently we do not have $X_nY_n\stackrel{d}{\to}XY$.