Does $X_n \stackrel{d}{\to X}$ and $Y_n \to 0$ almost surely imply that $X_n Y_n \to 0$ almost surely?

Question

I'm stuck with the following problem: I have a sequence $X_n$ of random variables, which converge in distribution to some random variable, which is finite almost surely. The other sequence $Y_n$ converges almost surely to $0$. Is it true that $X_n Y_n$ converges to zero almost surely?

Answer 1

No, the product does, in general, not converge to zero almost surely. Below I give an example which shows that $X_n \to 0$ in probability and $Y_n \to 0$ almost surely does not imply $X_n Y_n \to 0$ almost surely.

Let $(X_j)_{j \in \mathbb{N}}$ be a sequence of independent variables such that $$\mathbb{P}(X_j = \sqrt{j}) = \frac{1}{j} \qquad \mathbb{P}(X_j=0) = 1-\frac{1}{j}.$$ It is not difficult to see that $X_j \to 0$ in probability and, hence, in particular $X_j \to 0$ in distribution. If we define

$$Y_j := \frac{1}{\sqrt{j}},$$

then clearly $Y_j \to 0$ almost surely. On the other hand, we have $$\sum_{j \geq 1} \mathbb{P}(X_j = \sqrt{j}) = \sum_{j \geq 1} \frac{1}{j} = \infty,$$ and therefore it follows from the Borel-Cantelli lemma that $$\mathbb{P}(X_j=\sqrt{j} \, \, \text{for infinitely many $j$})=1.$$ Hence, $X_j Y_j = 1$ for infinitely many $j$ with probability $1$, and this means that $X_j Y_j$ does not converge to $0$ almost surely.