# Convergence in probability of the product of two random sequences.

Let $$X_n\xrightarrow{p}0$$ and $$Z_n\xrightarrow{d}Z$$ with $$\mathbb{E}(|Z|)<+\infty$$ and $$Z$$ is non-degenerate. I want to show that the sequence $$\{X_nZ_n\}$$ converges to $$0$$ in probability using only the definition of convergence in probability. I was thinking about the following: $$\mathbb{P}(|X_nZ_n|>\varepsilon)\leq \mathbb{P}(|X_n|>\sqrt{\varepsilon})+\mathbb{P}(|Z_n|>\sqrt{\varepsilon})$$ and then using Chebyshev's inequality at some point, but then I am not sure how to relate this to the convergence in distribution of $$Z_n$$ to $$Z$$.

## 1 Answer

Let $$\epsilon >0$$. Choose $$N$$ so large that $$P\{|Z| >N\} <\epsilon$$ and $$\pm N$$ are continuity points of $$F_Z$$. Since $$P\{|Zn| >N\} \to P\{|Z| >N\}$$ we get $$P\{|Zn| >N\}<\epsilon$$ for $$n$$ sufficiently large. Now $$P\{X_nZ_n| > \epsilon\} \leq P\{|X_n| >\epsilon /N\} + P\{ |Z_n| >N\}\epsilon /N\}+\epsilon$$ for $$n$$ sufficiently large. Can you take it from here?