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

Theorem: Let $X$ be a random variable with finite expected value $\mu$ and finite variance $\sigma^2$. Then, for any positive real number $k > 0$, the following inequality holds:

\[\label{eq:cheb-ineq} \mathrm{Pr}\left( |X-\mu| \geq k \sigma \right) \leq \frac{1}{k^2} \; .\]

Proof: Markov’s inequality states that, for any positive real number $a > 0$, the probability that $Y$ is larger than or equal to $a$ is smaller than or equal to the expected value of $Y$, divided by $a$:

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

The variance of a random variable is defined as:

\[\label{eq:var} \sigma^2 = \mathrm{Var}(X) = \mathrm{E}\left[ (X-\mu)^2 \right] \quad \text{where} \quad \mu = \mathrm{E}(X) \; .\]

Let $Y = (X-\mu)^2$ and $a = (k \sigma)^2$. Using Markov’s inequality, we then have:

\[\label{eq:cheb-ineq-qed} \begin{split} \mathrm{Pr}\left( |X-\mu| \geq k \sigma \right) &= \mathrm{Pr}\left( (X-\mu)^2 \geq (k \sigma)^2 \right) \\ &\overset{\eqref{eq:mark-ineq}}{\leq} \frac{\mathrm{E}\left[ (X-\mu)^2 \right]}{(k \sigma)^2} \\ &\overset{\eqref{eq:var}}{=} \frac{\sigma^2}{k^2 \sigma^2} \\ &= \frac{1}{k^2} \; . \end{split}\]
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Metadata: ID: P482 | shortcut: cheb-ineq | author: JoramSoch | date: 2025-01-10, 12:27.