Chebyshev's Inequality/Proof 1

Proof
Let $f$ be the function:


 * $\map f x = \begin{cases} k^2 \sigma^2 & : \size {x - \mu} \ge k \sigma \\

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

By construction, we see that:
 * $\map f x \le \size {x - \mu}^2 = \paren {x - \mu}^2$

for all $x$.

This means that:
 * $\expect {\map f X} \le \expect {\paren {X - \mu}^2}$

By definition of variance:


 * $\expect {\paren {X - \mu}^2} = \var X = \sigma^2$

By definition of expectation of discrete random variable, we can show that:

Putting this together, we have:

By dividing both sides by $k^2 \sigma^2$, we get:


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