chebyshev inequality

Chebyshev Inequality

If we have a random variable $X$ with $\mathbb{E}[X]=\mu$ and $\text{Var}(X)={\sigma }^2$, then, for any $a>0$

$$\begin{array}{r}P(|X-\mu |\ge a)\le \frac{{\sigma }^2}{a^2}\end{array}$$

I.e. the probability of $X$ being more than $a$ away from its mean is at most $\frac{{\sigma }^2}{a^2}$:

Proof

Let $Y=(X-\mu {)}^2$, $b=a^2$

$$\begin{array}{rl}P(|X-\mu |\ge a)& =P((X-\mu {)}^2\ge a^2)\\ & =P(Y\ge b)\\ & =P(Y\ge b)\le \frac{\mathbb{E}[Y]}{b}\text{(Markov’s inequality)}\\ & \le \frac{\text{Var}(Y)}{a^2}\\ & \le \frac{{\sigma }^2}{a^2}\end{array}$$

markov’s inequality