Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Expected value ▷ Markov's inequality

Theorem: Let $X$ be a non-negative random variable and let $a > 0$ be a positive real number. Then, the probability that $X$ is larger than or equal to $a$ is smaller than or equal to the expected value of $X$, divided by $a$:

\[\label{eq:mark-ineq} \mathrm{Pr}(X \geq a) \leq \frac{\mathrm{E}(X)}{a} \; .\]

Proof: Let us define an indicator variable that is one whenever $X$ is larger than or equal to $a$:

\[\label{eq:I-X-a} \mathrm{I}_{X \geq a} = \left\{ \begin{array}{rl} 0 \; , & \text{if} \; X < a \\ 1 \; , & \text{if} \; X \geq a \; . \end{array} \right.\]

It is clear that the following inequality holds:

\[\label{eq:a-I-X} a \, \mathrm{I}_{X \geq a} \leq X \; .\]

Taking the expectation on both sides, we have:

\[\label{eq:mark-ineq-qed} \begin{split} \mathrm{E}\left( a \, \mathrm{I}_{X \geq a} \right) &\leq \mathrm{E}(X) \\ a \, \mathrm{E}\left( \mathrm{I}_{X \geq a} \right) &\leq \mathrm{E}(X) \\ a \, \left[ 0 \cdot \mathrm{Pr}(\mathrm{I}_{X \geq a}=0) + 1 \cdot \mathrm{Pr}(\mathrm{I}_{X \geq a}=1) \right] &\leq \mathrm{E}(X) \\ a \, \mathrm{Pr}(X \geq a) &\leq \mathrm{E}(X) \\ \mathrm{Pr}(X \geq a) &\leq \frac{\mathrm{E}(X)}{a} \; . \end{split}\]
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Metadata: ID: P481 | shortcut: mark-ineq | author: JoramSoch | date: 2025-01-10, 11:50.