Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question


$$P(X=0)=1-P(X \ge 1)$$ $$\frac{\operatorname{Var}(X)}{E(X^2)}=1-\frac{E(X)^2}{E(X^2)}$$ 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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.