Bienaymé-Chebyshev Inequality

Theorem
Let $X$ be a random variable.

Let $\mathbb E \left[{X}\right] = \mu$ for some $\mu \in \R$.

Let $\operatorname{var} \left({X}\right) = \sigma^2$ for some $\sigma^2 \in \R_{> 0}$.

Then, for all $k > 0$:


 * $\mathbb P \left({\left\vert X - \mu \right\vert \ge k \sigma}\right) \le \dfrac 1 {k^2}$

Proof
Let $f$ be the function:


 * $f(x) = \begin{cases} k^2\sigma^2 & : |x - \mu| \ge k\sigma \\

0 & : \text{otherwise} \end{cases}$

It can be seen that $f(x) \le (x - \mu)^2$ for all $x$.

This means that $\mathbb E(f(X)) \le \mathbb E((X - \mu)^2)$.

From the definition of variance,


 * $\mathbb E((X - \mu)^2) = \operatorname{var}(X) = \sigma^2$

It can be shown that:


 * $\mathbb E(f(X)) = k^2\sigma^2 \mathbb P(|X - \mu| \ge k\sigma)$

Putting this together, we have:


 * $k^2\sigma^2 \mathbb P(|X - \mu| \ge k\sigma) \le \sigma^2$
 * $\mathbb P(|X - \mu| \ge k\sigma) \le \dfrac 1 {k^2}$