Given any sequence of random variables $\{ X_n \}$; how do I show that there exists a sequence of real numbers $\{ \alpha_n \}$, such that $\{ \alpha_n X_n \}$ converges in probability to 0 ?
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3$\begingroup$ If any sequence of real numbers is allowed, why not simply pick $\alpha_n=0$ for all $n$? $\endgroup$– Christoph HanckCommented Jun 25, 2016 at 11:26
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$\begingroup$ The problem is that $ \left( \alpha_{n} X_{n}\right) $ is an element, not a sum. The only way that any scalar element $ \left( X_n \right) $ can "converge" to zero is if it is multiplied by zero. Is there a sum in this? (I'm not sure how to make curly brackets). $\endgroup$– EngrStudentCommented Jun 25, 2016 at 11:31
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$\begingroup$ Answered with a hint since this is a typical textbook question. I think the self-study tag is appropriate here. $\endgroup$– KOECommented Jun 25, 2016 at 16:33
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1 Answer
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As indicated in the comments, you have to state some additional assumptions for this to be interesting. This answer assumes $\alpha_n \neq 0$ for all $n$ and $P(\vert X_n\vert < \infty) = 1$ for all $n$.
Fix $n$ and observe / prove that for every $\epsilon_n >0 $ there exists a constant $M_n = M_n(\epsilon_n)$ such that $P(\vert X_n \vert > M_n) \leq \epsilon_n$. That is,
$$ P(\vert \alpha_n X_n\vert>\vert \alpha_n\vert M_n)\leq \epsilon_n. $$
Can you pick $\epsilon_n$ and $\alpha_n$ so as to make $P(\vert \alpha_nX_n\vert > \delta) \to 0$ for any given $\delta>0$?
