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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

$P(|X_n| <M) > 1 -\epsilon/2$

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

Then I want to show that

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

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1 Answer 1

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$P(|X_nY_n| >\epsilon) \leq P(|X_n| \leq M, |X_nY_n| >\epsilon)+P(|X_n| > M, |X_nY_n| >\epsilon)\leq P(|Y_n| >\frac {\epsilon} M)+[1-P(|X_n|<M)]$. For $n >N$ the second term is less than $\epsilon$ and the first term tends to $0$ as $n \to \infty$.

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  • $\begingroup$ Why is the first inequality true $\endgroup$
    – Aryan986
    Commented Nov 3, 2019 at 5:53
  • $\begingroup$ @Aryan986 If $|X_nY_n| >\epsilon$ and $|X_n| \leq M$ then $\epsilon <|X_nY_n|\leq M|Y_n|$ which implies $|Y_n| >\frac {\epsilon} M$. $\endgroup$ Commented Nov 3, 2019 at 6:22
  • $\begingroup$ The first inequality is just by inclusion. $P(A)\leq P(A\cap B) +P(A\cap B^{c})$ because $A \subset (A\cap B)\cup (A\cap B^{c})$ $\endgroup$ Commented Nov 3, 2019 at 6:24
  • $\begingroup$ Sorry I am still kind of confused as too how that proof's $X_nY_n \rightarrow $ 0 $\endgroup$
    – Aryan986
    Commented Nov 3, 2019 at 6:28
  • $\begingroup$ 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. $\endgroup$ Commented Nov 3, 2019 at 6:34

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