For a random variable $ X \geq 0 $ and $E[X^2] < \infty $, I'm asked to prove the following: $$ P(X> 0) \geq \frac{(E[X])^2}{E[X^2]}$$

It makes intuitive sense to me that it must be the case, since P(X>0) equals 1 and Var(X), which is the difference between the denominator $E[X^2]$ and the numerator $(E[X])^2$, must be positive. But the question is asking us to prove it by employing Cauchy-Schwarz inequality: $ |\langle u,v \rangle|^2 \leq \langle u,u \rangle \langle v,v \rangle $. It specifically hints us to use E(UV) as a valid inner product defined for U and V on the set of random variables.

But while trying to substitute U and V by X, I've arrived at the following:

$$|E[XX]|^2 = E[XX]^2 \leq E[XX]E[XX] = E[XX]^2$$

Which obviously doesn't get me anywhere. I've spent a lot of time on it, but I can't figure out how to prove it with Cauchy-Schwarz inequality.

  • $\begingroup$ $P(X>0) \neq 1$ if $P(X=0) > 0$, which, given that the constraint on $X$ is just $X \geq 0$, could well be the case. $\endgroup$ – jbowman Apr 15 '19 at 19:10

Provided the expectations exist, Cauchy-Schwarz inequality states

$$E\left[(g(X))^2\right]E \left[(h(X))^2\right]\ge \left(E \left[g(X)h(X)\right]\right)^2$$

Choose $g(X)=I_{X>0}$, the indicator of the event $\{X>0\}$, and $h(X)=X$.

And keep in mind that

$$E(X)=E(X I_{X>0})+E(X I_{X<0})=E(X I_{X>0})$$

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  • 1
    $\begingroup$ Thank you very much $\endgroup$ – Daniel Apr 15 '19 at 20:06

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