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I'm trying to teach myself probability theory, and an exercise has me stumped. This exercise comes from Alon & Spencer 4.8, in the chapter on the second moment method.

Let $X$ be a random variable taking values in $\mathbb{Z}_{\geq 0}$. Let $E[X^2]$ denote the expectation of its square. Prove that

$\displaystyle Pr[X=0] \leq \frac{Var[X]}{E[X^2]}$

In particular, they prove in the chapter that this is true by replacing $E[X^2]$ with $E[X]^2$ by using a simple application of Chebyshev's inequality (they set $\lambda$ in the theorem to $E[X]/Var[X]$). However, this bound is easily seen to be tighter, and so it seems I need to come up with a shrewder application of the inequality by redefining the random variable or choosing an appropriate constant.

Any hints?

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1 Answer 1


$$P(X=0)=1-P(X \ge 1)$$ $$\frac{\operatorname{Var}(X)}{E(X^2)}=1-\frac{E(X)^2}{E(X^2)}$$ Take conditional expectation and using $E(X \mid X \ge 1)^2 \le E(X^2 \mid X \ge 1)$, calculate both sides and get the result.

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calculations are fairly straightforward do the expansions into infinite sum –  Koushik Dec 21 '12 at 2:36
You can use $\LaTeX$ on this site. I edited your answer to use it. You could double-check that I didn't mess it up. –  Nate Eldredge Dec 21 '12 at 2:51
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