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I wanted to prove that if $X_n$ is bounded in probability and $Y_n = o_p(X_n)$, then $Y_n \rightarrow 0$ in probability

I know the following definitions that is $X_n$ is bounded in probability meaning that $P(|X_n|<M)>1-\epsilon$

and I know that $Y_n = o_p(X_n)$ implies that $Y_n/X_n \rightarrow0$ in probability

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$|Y_n| >\epsilon$ implies either $|\frac {Y_n} {X_n}| >\frac {\epsilon} M$ or $|X_n| \geq M$. [You can prove this by contradiction]. Hence $P(|Y_n| >\epsilon) \leq P(|\frac {Y_n} {X_n}| >\frac {\epsilon} M)+P(|X_n| \geq M)$. Can you finish the proof?

Some details: Let $\eta_1$ and $\eta_2 >0$. Choose $\epsilon >0$ such that $\epsilon <\eta_1$ and $\epsilon <\eta_2 /2$ . Note that $|Y_n| >\eta_1$ implies that $|Y_n|>\epsilon$. Now choose $n_0$ such that $P(|\frac {Y_n} {X_n}| >\frac {\epsilon} M) <\eta_2 /2$ for $n \geq n_0$. Now put these together to conclude that $P(|Y_n| >\eta_1) <\eta_2$ whenever $n \geq n_0$. This proves that $Y_n \to 0$ in probability.

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  • $\begingroup$ Would you mind explaining how that contradiction is true also would you mind showing me how to end the proof. I am learning these things from scratch. Thanks you ! @Kabo Murphy $\endgroup$
    – Aryan986
    Commented Nov 5, 2019 at 2:44
  • $\begingroup$ @Aryan986 I have added some details. $\endgroup$ Commented Nov 5, 2019 at 5:18

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