Markov's inequality

Statement

Let $X$ be a random variable with values in $[0, \infty)$ and let $a > 0$. Then $$P(X \ge a) \le \frac{E[X]}{a}.$$

Remarks

Proof: $a \, 1_{\{X \ge a\}} \le X$ pointwise. Take expectations: $a \, P(X \ge a) \le E[X]$. More generally, for each nondecreasing $g \colon [0, \infty) \to [0, \infty)$ with $g(a) > 0$, $P(X \ge a) \le E[g(X)]/g(a)$. Applied to $(X - E[X])^2$, the inequality gives Chebyshev's inequality. Applied to $e^{tX}$, it gives the Chernoff bound.

History

The inequality is named after Andrey Markov. It appears earlier in the work of Pafnuty Chebyshev, the teacher of Markov.

Explanations

L1 human

A nonnegative quantity with a small average cannot often be large. If the mean income is $30000$, then at most one person in five can have an income of $150000$ or more.