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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 $\mathbb{P}\left(X\geq \mu+ \frac{\sigma^2}{4M} \right)\geq \frac{\sigma^2}{8M^2}$ ?

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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.

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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
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