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

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?

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.