Suppose $X_{n}\to X$ in distribution and $Y_{n}\to 0$ in probability, then I want to show that $X_{n}Y_{n}\to 0$ in probability.
My idea is that choose $\delta>0$ and we have $\mathbb {P}(|X_{n}Y_{n}|\geq\epsilon)=\mathbb {P}(|X_{n}Y_{n}|\geq\epsilon,|Y_{n}|\leq \delta)+\mathbb {P}(|X_{n}Y_{n}|\geq\epsilon,|Y_{n}|\geq \delta)$. Then since $|X_{n}Y_{n}|\geq\epsilon$ and $|Y_{n}|\leq \delta$, we get $|X_{n}|\geq \frac {\epsilon}{\delta}$. Then we have $\mathbb {P}(|X_{n}Y_{n}|\geq\epsilon)\leq \mathbb {P}(|X_{n}|\geq \frac {\epsilon}{\delta})+\mathbb {P}(|Y_{n}|\geq\delta)$. When I take the limit, I clearly have $\mathbb {P}(|Y_{n}|\geq\delta)$ goes to $0$, then I left with $\lim_{n\to \infty}\mathbb {P}(|X_{n}Y_{n}|\geq\epsilon)\leq \lim_{n\to \infty}\mathbb {P}(|X_{n}|\geq \frac {\epsilon}{\delta})=1-F_{|X|}\left(\frac {\epsilon}{\delta}\right)$. But I don't know how to argue that $1-F_{|X|}\left(\frac {\epsilon}{\delta}\right)$ goes to $0$. It seems like its value depends on the choice of $\delta$, it needs to be pretty small, like smaller than $\epsilon$.
Best Answer
Yes, $\delta$ should be small enough. So if you let $\delta\rightarrow 0$, the desired result will follow.