TheoremProved

Chebyshev's inequality

Statement

Let $X$ be a random variable with finite mean $\mu = \mathbb{E}[X]$ and finite variance $\sigma^2 = \operatorname{Var}(X)$ . For every real number $\varepsilon > 0$ , $\mathbb{P}(|X - \mu| \ge \varepsilon) \le \frac{\sigma^2}{\varepsilon^2}$ . Equivalently, when $\sigma > 0$ , for every $k > 0$ , $\mathbb{P}(|X - \mu| \ge k\sigma) \le \frac{1}{k^2}$ .

Why is it true?

No matter how irregular or skewed the distribution of $X$ may be, its variance $\sigma^2$ acts as a hard budget on how much probability mass can sit far from the mean $\mu$ : at most a fraction $\frac{1}{k^2}$ of the outcomes can lie $k$ or more standard deviations away.

Proof sketch

Let $A = \{|X - \mu| \ge \varepsilon\}$ . On the event $A$ , we have $(X - \mu)^2 \ge \varepsilon^2$ , so the pointwise inequality $(X - \mu)^2 \ge \varepsilon^2 \mathbf{1}_A$ holds everywhere on the sample space. Taking expectations on both sides and using monotonicity and linearity gives $\sigma^2 = \mathbb{E}[(X - \mu)^2] \ge \mathbb{E}[\varepsilon^2 \mathbf{1}_A] = \varepsilon^2 \mathbb{P}(A) = \varepsilon^2 \mathbb{P}(|X - \mu| \ge \varepsilon)$ . Dividing both sides by $\varepsilon^2 > 0$ yields $\mathbb{P}(|X - \mu| \ge \varepsilon) \le \frac{\sigma^2}{\varepsilon^2}$ .

Topics that use this theorem

Random variables and distributions The law of large numbers and the central limit theorem

Related theorems

Weak law of large numbers Strong law of large numbers Kolmogorov's axioms of probability

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Pafnuty Chebyshev (1867). Des valeurs moyennes
William Feller (1968). An Introduction to Probability Theory and Its Applications, Vol. 1