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Let $X_1, X_2, ...$ be random variables with $\mbox E X^2_j = 1$ for all $j$. Show that $\inf P[|X_n| > \epsilon] > 0$ for some $\epsilon > 0$ if and only if $\inf \mbox E |X_n| > 0$.

I have shown necessity, which is quite easy and doesn't use the hypothesis that $\mbox E X_j ^ 2 = 1$. I'm stuck on sufficiency, and can't seem to figure out how to work $\mbox E X_j ^ 2 = 1$ into anything useful.


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up vote 2 down vote accepted


1) $\Rightarrow$ : $E(|x|) = \int_{|x|\le e}|x| \; p(x) \, dx + \int_{|x|> e}|x| \; p(x) \, dx \ge \epsilon \; P(|x| >\epsilon )$

2) $\Leftarrow$ : Chebyshev's inequality

Added: Say we have $X$ with $E(X^2)=1$ and $E(|X|)\ge a > 0$ Let's apply Chebyshev to $Y=|X|$. $Var(Y) = E(Y^2)-E(Y)^2 \le 1- a^2$ Let's take (say) $\epsilon = a/2$. Then $P(Y<\epsilon) = P(a-Y > a/2)$ corresponds to a "tail", that we can bound strictly below 1, for example with the one-sided Chebyshev inequality.

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I am not seeing how to get Chebyshev to help. It seems I need to bound $E[X_n]$ from above using $P(|X_n| > \epsilon)$. What should I apply Chebyshev to? – guy Jul 1 '11 at 1:28
Perhaps we are speaking of opposites senses of the implication. To show that $E(|X|) >0$ given that $P(|X|>\epsilon)>d$ you use the first one. – leonbloy Jul 1 '11 at 1:32
I have the first result. I can't prove that if $\inf_n E[|X_n|] > 0$ then $\inf P(|X_n| > \epsilon) > 0$ for some $\epsilon > 0$, the Chebyshev hint not withstanding. If I can bound $E[|X_n|]$ from above using $P(|X_n| > \epsilon)$ for some $\epsilon$ then I can prove the contrapositive, but that is as far as I can go. Perhaps I am completely on the wrong track? I can't see how to use Chebyshev to do this. – guy Jul 1 '11 at 1:43
I expanded my hint - still not a complete proof. – leonbloy Jul 1 '11 at 2:07
Okay, I think I see now. So bound $P(Y < \epsilon)$ strictly below $1$ using Cantelli's Inequality which gives a lower bound on $P(Y \ge \epsilon)$, correct? – guy Jul 1 '11 at 2:46

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