Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 need help to prove the following lower bound for tail probability. I have tried using well-known inequalities like Chebyshev and Paley-Zygmund, but cannot get the required bound.

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and $X$ be a random variable with $\mathbb{P}(X\in [-M,M])=1$ for some positive constant $M$. Show that $\mathbb{P}\left(|X-\mu|\geq \frac{\sigma^2}{4M} \right)\geq \frac{\sigma^2}{8M^2}$, where $\mu= \mathbb{E}(X)$ and $\sigma^2=\text{Var}(X)$.

Thank you.

Additional question: How can I also prove that $\mathbb{P}\left(X\geq \mu+ \frac{\sigma^2}{4M} \right)\geq \frac{\sigma^2}{8M^2}$ ?

share|cite|improve this question
up vote 3 down vote accepted

The function $f(t):= \mathbb P(|X-\mu|\ge t)-\frac t{2M}$ is strictly decreasing on $[0,\infty)$ and we have $f(0)=1$, $f(2M)=-1$. Let $s=\inf\{t\ge0\mid f(t)\le 0\}$. Then $$ \sigma^2 = \int (x-\mu)^2 d\mu =\int_0^\infty 2r\mathbb P(|X-\mu|\ge r) dr\\ \le\int_0^s2r\cdot1\,dr+\int_s^{2M}2r\cdot\frac s{2M}dr\\ =s^2+2Ms-\frac{s^3}{2M}. $$ If $s=0$, then this shows $\sigma^2=0$ and thus the claim we want to show is the trivial statement $\mathbb P(|X-\mu|\ge0)\ge0$. Therefore, we may assume $0< s\le 2M$ and conclude $$ \sigma^2\le s^2+2Ms-\frac{s^3}{2M}< 2Ms+2Ms-0=4Ms.$$ But then $\frac{\sigma^2}{4M}<s$ implies $f\left(\frac{\sigma^2}{4M}\right)>0$, i.e. $$ \mathbb P\left(|X-\mu|\ge \frac{\sigma^2}{4M}\right)>\frac{\sigma^2}{8M^2}.$$ Apparently, the only case where $\le$ canot be replaced with $>$ is the case of an (almost surely) constant variable.

share|cite|improve this answer
Thanks a lot - clever argument! I'm curious... Are there more inequalities of this type? Could you point me towards a reference? – Stefan Wager Nov 2 '13 at 23:07

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.