Bienaymé-Chebyshev Inequality/Proof 2

Theorem

Let $\expect X = \mu$ for some $\mu \in \R$.

Let $\var X = \sigma^2$ for some $\sigma^2 \in \R_{> 0}$.

Then, for all $k > 0$:

$\map \Pr {\size {X - \mu} \ge k \sigma} \le \dfrac 1 {k^2}$

Note that as $k > 0$ and $\sigma > 0$, we have $k \sigma > 0$.

We therefore have:

$\ds \map \Pr {\size {X - \mu} \ge k \sigma}$

$=$

$\ds \map \Pr {\paren {X - \mu}^2 \ge \paren {k \sigma}^2}$

$\ds $

$\le$

$\ds \frac {\expect {\paren {X - \mu}^2} } {\paren {k \sigma}^2}$

as $k \sigma > 0$, we can apply Markov's Inequality: Corollary

$\ds $

$=$

$\ds \frac {\sigma^2} {k^2 \sigma^2}$

Definition of Variance

$\ds $

$=$

$\ds \frac 1 {k^2}$

$\blacksquare$