The proof of Chebyshev's inequality follows from the application of Markov's inequality.
Define . Then is a random variable, and
Applying Markov's inequality to , we see that
![[parent]](http://images.planetmath.org:8080/images/uparrow.png)
![$\displaystyle \mathbb{E}[Y] = \operatorname{Var}[X] = \sigma^2.$](http://images.planetmath.org:8080/cache/objects/3116/l2h/img3.png "$\displaystyle \mathbb{E}[Y] = \operatorname{Var}[X] = \sigma^2.$"){kind=small}
![$\displaystyle \mathbb{P}_{}\left\{\left\vert X - \mu \right\vert \ge t\right\} ... ...\left\{Y \ge t^2\right\} \le \frac{1}{t^2}\mathbb{E}[Y] = \frac{\sigma^2}{t^2}.$](http://images.planetmath.org:8080/cache/objects/3116/l2h/img5.png "$\displaystyle \mathbb{P}_{}\left\{\left\vert X - \mu \right\vert \ge t\right\} ... ...\left\{Y \ge t^2\right\} \le \frac{1}{t^2}\mathbb{E}[Y] = \frac{\sigma^2}{t^2}.$")