I'm proving the Second Moment Method. I'm aware of the stronger solution. But I'm trying to figure out a detail in the "weak" version. We need to prove that $$P(X=0) \leq \frac{Var(X)}{(E[X])^{2}} = \frac{E[X]^{2}}{(E[X])^{2}} - 1.$$
The first step is of course to use the Chebychev's inequality, which gives us
$$P(|X - E[X]| \geq E[X]) \leq \frac{Var(X)}{(E[X])^{2}} = \frac{E[X]^{2}}{(E[X])^{2}} - 1.$$ My question is now, how do we get that $P(X = 0) \leq P(|X - E[X]| \geq E[X])$? Can someone give an intuitive explanation on how we get here. If feel like it makes sense, but I'm just not satisfied with "feeling" good about it just being true. Thanks in advanced.
I think you at least need an assumption that $E[X] \ge 0$.
If $X=0$ then $|X-E[X]| =|E[X]| = E[X]$, so $\{X = 0\} \subseteq \{|X-E[X]| \ge E[X]\}$.