Understanding proof that $P(X=0) \leq \frac{Var(X)}{E(X^2)}$

Solution 1:

Remark that $\mathbb{E}[X] = \mathbb{E}[X1_{X>0}]$ and apply Cauchy-Schwarz inequality: $$\mathbb{E}[X]^2\leq \mathbb{E}[X^2]\mathbb{E}[1_{X>0}] = \mathbb{E}[X^2](1-\mathbb{P}(X=0)). $$ Then conclude $$\mathbb{P}(X=0) \leq 1 - \frac{\mathbb{E}[X]^2}{\mathbb{E}[X^2]} = \frac{\text{Var} X}{\mathbb{E}[X^2]}. $$

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