Chebyshev's inequality

Statement

Let $X$ be a real random variable with finite mean $\mu$ and finite variance $\sigma^2$. Then for each $a > 0$, $$P(|X - \mu| \ge a) \le \frac{\sigma^2}{a^2}.$$ Equivalently, if $\sigma > 0$, then $P(|X - \mu| \ge k\sigma) \le 1/k^2$ for each $k > 0$.

Remarks

Proof: apply Markov's inequality to the nonnegative variable $(X - \mu)^2$ and the level $a^2$. For $k = 2$, at least $3/4$ of the mass is within two standard deviations of the mean, for every distribution with finite variance. The bound is sharp: some distributions reach equality for a given $k \ge 1$. Chebyshev's inequality gives a short proof of the weak law of large numbers for variables with finite variance.

History

The inequality is named after Pafnuty Chebyshev. Irénée-Jules Bienaymé formulated it first.

Explanations

L1 human

The variance controls how often a variable can be far from its mean. For any distribution, at most $1/9$ of the mass lies three or more standard deviations away from the mean.