2
$\begingroup$

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.

Thanks.

1 Answers 1

2

Hint:

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.

  • 0
    Yes. The idea is that knowing that $Y$ has mean $a$ (or more) and finite variance makes impossible that all its probability mass is concentrated below (say) $a/2$.2011-07-01