Is it true that if $E[X]>0$ then $P(X>0)\ge E[X^2]/(E[X])^2$

Question

If $E[X]>0$ then does $P(X>0)\ge \frac{E[X^2]}{(E[X])^2}$?

Background: I've taken an introductory course in probability.

Answer 1

No. Let $X=1$ with probability .6 and $X=-1$ with probability .4.

Answer 2

No, it doesn't. It's quite close to something that is true though. If you suppose that $X \ge 0$, then we do in fact have $$P(X > 0) \ge \frac{(E(X))^2}{E(X^2)}.$$ This is called the "second moment method", and it follows straightforwardly from Cauchy-Schwarz. (The RHS is the reciprocal of what you've got.)

However, this doesn't hold for general $X$ with $E(X) > 0$. As a counter-example, take $X_\epsilon \sim N(\epsilon,1)$. Then $E(X_\epsilon) = \epsilon$ and $E(X_\epsilon^2) = 1 + \epsilon^2$. Consider $0 < \epsilon \ll 1$. Then $E(|X_\epsilon|) \simeq E(|X_0|) = 1$ (you can work this out by hand without much difficulty, I leave that as an exercise). Then the RHS is approximately $1$, but the LHS is approximately $1/2$.