# Why does $X_n \to 0$ in probability and $Y_n \to Z$ in probability imply $X_nY_n \to 0$ in probability when Z is finite?

Why does $X_n \to 0$ in probability and $Y_n \to Z$ in probability imply $X_nY_n \to 0$ in probability when Z is finite?

Using the almost sure convergence along a subsequence I can conclude that if the limit $(X_{n_k}Y_{n_k})_{k \in \mathbb{N}}$ converges to $0$ almost surely and hence in probabilty. But I am unable to show that

$\lim_{n \to \infty} P(\vert X_nY_n\vert \geq \epsilon)=0$. How could I go about showing this? Any hints would be highly appreciated.

• If $Z$ is finite it is bounded and therefore $Y_n$ is also bounded. – Yanko Apr 29 '18 at 10:46

In general if $X,Y,X_n,Y_n$ are random variables (taking values in $\mathbb R$) defined on the same probability space with $X_n\stackrel{p}{\to}X$ and $Y_n\stackrel{p}{\to}Y$ then it follows that: $$X_nY_n\stackrel{p}{\to}XY$$