05 - Chebyshev's Inequality

Chebyshev's Inequality

Let $Y$ be a random variable with finite mean and variance then

For any $ϵ>0$

$$\mathbb{P}(|Y-\mathbb{E}[Y]|\ge ϵ)\le \frac{\text{var}(Y)}{{ϵ}^2}$$

Proof

$$\begin{array}{rl}\mathbb{P}(|Y-\mathbb{E}[Y]|\ge ϵ)& =\mathbb{P}([Y-\mathbb{E}[Y]{]}^2\ge {ϵ}^2)\\ & \le \frac{\mathbb{E}([Y-\mathbb{E}[Y]{]}^2)}{{ϵ}^2}\\ & =\frac{\text{var}(Y)}{{ϵ}^2}\end{array}$$