The Chebyshev inequality builds on the [[Markov Inequality]], but makes better boundaries by incorporating the variance of the [[Random Variable]] as well. For the inequality to hold the mean $\mu$ and variance $\sigma^2$ of the r.v. $X$ must be finite. $ \mathbf P\big (\underbrace{\lvert X - \mu \rvert}_{\text{distance from mean}} \ge c \big ) \le \frac{\sigma^2}{c^2} $ **Interpretation:** - The bigger $c$, the smaller the probability that this boundary will be exceeded by the distance to the mean. - The bigger $\sigma$, the bigger the probability that this boundary will be exceeded, because high variance makes more extreme outcomes more likely. **Derivation:** $ \begin{align} \mathbf P(Y \ge a) &\le \frac{\mathbb{E}[Y]}{a} \tag{1}\\[2pt] \mathbf P \big ((X- \mu)^2 \ge c^2) &\le\frac{\mathbb{E}\big[(X-\mu)^2 \big]}{c^2} \tag{2}\\ \mathbf P \big ((X- \mu)^2 \ge c^2)&\le\frac{\sigma^2}{c^2} \tag{3}\\ \mathbf P \big (\lvert X- \mathbb \mu \rvert \ge c)&\le\frac{\sigma^2}{c^2} \tag{4} \end{align} $ where: - (1) Markov inequality. - (2) Let $Y=(X-\mu)^2$. Since this quadratic is always $\ge0$, it is valid to be plugged into Markov inequality. Also we write $a$ as $c^2$ for later convenience. - (3) The numerator $\mathbb E\big[(X-\mu)^2\big]$ is the definition of the variance $\sigma_X^2$. - (4) The following two events are equivalent. $\mathbf P\big((X-\mu)^2\ge c^2\big) \equiv \mathbf P\big(\vert X-\mu \vert \ge c \big)$