I'm looking at a variation of Chebyshev and having a hard time convincing myself why the following is correct:
Given a non-negative, integer valued RV $X$
$$\mathbb{P}(X > 0) \geq 1 – \frac{Var(X)}{\mathbb{E}^2(X)}$$
The proof I'm reading goes:
$$\mathbb{P}(X=0)\overset{1}{\leq} \mathbb{P}(|X-\mathbb{E}[X]|\geq \mathbb{E}[X])\overset{2}{\leq} \frac{Var(X)}{\mathbb{E}^2(X)}$$
Step 2 is simply the application of Chebyshev, but could someone elucidate why step 1 is correct? Seems trivial but can't wrap my head around it at the moment…
Best Answer
Hint
I guess $\mu=\mathbb E[X]$. Since $X\geq 0$ a.s., $$X=0\iff X\leq 0\iff X-\mathbb E[X]\leq -\mathbb E[X]\implies |X-E[X]|\geq \mathbb E[X].$$