# $X_n$ is bounded in probability and $Y_n$ converges to 0 in probability then $X_nY_n$ congerges to probablity with 0

I want to show : $$X_n$$ is bounded in probability and $$Y_n \rightarrow 0$$ in probability then $$X_nY_n \rightarrow 0$$ in probablity. I know the following definitions that is

Definition 2.17 : We say that $$X_n$$ is bounded in probability if $$X_n = O_P (1)$$, i.e. if for every $$\epsilon$$ > 0, there exist $$M$$ and $$N$$ such that $$P(|Xn| < M) > 1 − \epsilon$$ for $$n > N$$.

So I want to show that for every
$$\epsilon$$ > 0 and $$\epsilon '$$ > 0 there exist $$M>0$$ and $$n_o$$ such that

$$P(|X_n| 1 -\epsilon/2$$

$$P(|X_n| <\epsilon /M) > 1 -\epsilon'/2$$ for every $$n>n_0$$

Then I want to show that

$$P(|X_n| and $$|Y_n| <\epsilon /M) > 1-\epsilon'$$

$$P(|X_nY_n| >\epsilon) \leq P(|X_n| \leq M, |X_nY_n| >\epsilon)+P(|X_n| > M, |X_nY_n| >\epsilon)\leq P(|Y_n| >\frac {\epsilon} M)+[1-P(|X_n|. For $$n >N$$ the second term is less than $$\epsilon$$ and the first term tends to $$0$$ as $$n \to \infty$$.
• @Aryan986 If $|X_nY_n| >\epsilon$ and $|X_n| \leq M$ then $\epsilon <|X_nY_n|\leq M|Y_n|$ which implies $|Y_n| >\frac {\epsilon} M$. Commented Nov 3, 2019 at 6:22
• The first inequality is just by inclusion. $P(A)\leq P(A\cap B) +P(A\cap B^{c})$ because $A \subset (A\cap B)\cup (A\cap B^{c})$ Commented Nov 3, 2019 at 6:24
• Sorry I am still kind of confused as too how that proof's $X_nY_n \rightarrow$ 0 Commented Nov 3, 2019 at 6:28
• You get $P(|X_nY_n| >\epsilon) <2 \epsilon$ for $n$ sufficiently large. This is enough to conclude that $X_nY_n \to 0$ in probability. Commented Nov 3, 2019 at 6:34