Let $Χ$ be a random variable for witch $0 < E[X] < \infty$. Prove that: $ \mathbb P[X > 0] \geq \frac{\mathbb E [X]^ 2}{ \mathbb E [X^2]}$
Hint: Use the fact that: $(\mathbb{1}_{\{X>0\}}\ \frac{\mathbb E[X^2]}{ \mathbb E [X]}-X)^ 2 \ge 0.$
I think I am missing something here. I tried to prove this by using the argument that since the hint is greater than $0$ and the hint is a random variable, thn its mean should also be greater than $0$ but I was unable to do anything useful. After that I tried to use the fact that if $Y$ is the r.v. of the hint then:
$\mathbb E[Y]^2 \ge 0 $
$ \mathbb E[\mathbb{1}_{\{X>0\}}\ \frac{\mathbb E[X^2]}{ \mathbb E [X]}-X]^2 \ge 0 $
$ (\frac{\mathbb E[X^2]}{ \mathbb E [X]}\mathbb P[X > 0]-\mathbb E [X])^2 \ge 0$
If I could find a way to get rid of the square this would be correct, but I don't think I can. I think I have to use Jensen's inequality somewhere, but I don't know where. Can anyone suggest a different approach?
You are given the hint that $$\left(\mathbb{1}_{\{X>0\}}\ \frac{\mathbb E[X^2]}{ \mathbb E[X]}-X\right)^2 \ge 0,$$ which implies $$\mathbb{1}_{\{X>0\}}\ \frac{\mathbb E[X^2]^2}{\mathbb E[X]^2} - 2\mathbb{1}_{\{X>0\}}\ \frac{\mathbb E[X^2]}{ \mathbb E[X]}X + X^2 \geq 0.$$ Taking the expectation, we have $$\mathbb{P}[X>0]\ \frac{\mathbb E[X^2]^2}{\mathbb E[X]^2} - 2\ \frac{\mathbb E[X^2]}{ \mathbb E[X]}\mathbb E[\mathbb{1}_{\{X>0\}} X] + \mathbb E[X^2] \geq 0.$$ You can notice that $$\mathbb E[\mathbb{1}_{\{X>0\}} X] \geq \mathbb E[X],$$ and $\mathbb E[X] \geq 0$ from the conditions, so $$\mathbb{P}[X>0]\ \frac{\mathbb E[X^2]^2}{\mathbb E[X]^2} + \mathbb E[X^2] \geq 2\ \frac{\mathbb E[X^2]}{ \mathbb E[X]}\mathbb E[\mathbb{1}_{\{X>0\}} X] \geq 2\ \mathbb E[X^2].$$ Subtracting $\mathbb E[X^2]$ from both sides and then multiplying both sides by $\mathbb E[X]^2 / \mathbb E[X^2]$ gives $$\mathbb P[X > 0] \geq \frac{\mathbb E[X]^2}{\mathbb E[X^2]}.$$