Bienaymé-Chebyshev Inequality

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Theorem

Let $X$ be a random variable.

Let $\expect X = \mu$ for some $\mu \in \R$.

Let $\var X = \sigma^2$ for some $\sigma^2 \in \R_{> 0}$.

Then, for all $k > 0$:

$\map \Pr {\size {X - \mu} \ge k \sigma} \le \dfrac 1 {k^2}$

Proof 1

Let $f$ be the function:

$\map f x = \begin{cases} k^2 \sigma^2 & : \size {x - \mu} \ge k \sigma \\ 0 & : \text{otherwise} \end{cases}$

By construction:

$\forall x \in \Dom f: \map f x \le \size {x - \mu}^2 = \paren {x - \mu}^2$

Hence from Expectation Preserves Inequality:

$\expect {\map f X} \le \expect {\paren {X - \mu}^2}$

By definition of variance:

$\expect {\paren {X - \mu}^2} = \var X = \sigma^2$

By definition of expectation of discrete random variable, we can show that:

$\ds \expect {\map f X}$

$=$

$\ds k^2 \sigma^2 \map \Pr {\size {X - \mu} \ge k \sigma} + 0 \cdot \map \Pr {\size {X - \mu} \le k \sigma}$

$\ds $

$=$

$\ds k^2 \sigma^2 \map \Pr {\size {X - \mu} \ge k \sigma}$

Putting this together, we have:

$\ds \expect {\map f X}$

$\le$

$\ds \expect {\paren {X - \mu}^2}$

$\ds \leadsto \ \ $

$\ds k^2 \sigma^2 \map \Pr {\size {X - \mu} \ge k \sigma}$

$\le$

$\ds \sigma^2$

By dividing both sides by $k^2 \sigma^2$, we get:

$\map \Pr {\size {X - \mu} \ge k \sigma} \le \dfrac 1 {k^2}$

$\blacksquare$

Proof 2

Note that as $k > 0$ and $\sigma > 0$, we have $k \sigma > 0$.

We therefore have:

$\ds \map \Pr {\size {X - \mu} \ge k \sigma}$

$=$

$\ds \map \Pr {\paren {X - \mu}^2 \ge \paren {k \sigma}^2}$

$\ds $

$\le$

$\ds \frac {\expect {\paren {X - \mu}^2} } {\paren {k \sigma}^2}$

as $k \sigma > 0$, we can apply Markov's Inequality: Corollary

$\ds $

$=$

$\ds \frac {\sigma^2} {k^2 \sigma^2}$

Definition of Variance

$\ds $

$=$

$\ds \frac 1 {k^2}$

$\blacksquare$

Motivation

While most probability distributions do not come anywhere near the specified limit, the can be useful to identify a strict upper limit which a probability distribution cannot exceed.

Hence, while the is somewhat weak, it does have its uses.

Also presented as

The can also be presented in the form:

$\map \Pr {\size {X - \mu} < k \sigma} \ge 1 - \dfrac 1 {k^2}$

Also known as

The is also known as Chebyshev's Inequality.

However, some sources use this name to mean the Chebyshev's Sum Inequality, which is a completely different result.

Hence, to avoid confusion, the name Chebyshev's Inequality is not used on $\mathsf{Pr} \infty \mathsf{fWiki}$.

Variants on spellings can be seen: Bienaymé-Tchebyshev Inequality, Tchebyshev's Inequality, and so on.

Source of Name

This entry was named for Pafnuty Lvovich Chebyshev and Irénée-Jules Bienaymé.

Historical Note

The result now known as the was first formulated, without proof, by Irénée-Jules Bienaymé in $1853$.

The proof was provided by his friend and colleague Pafnuty Lvovich Chebyshev in $1867$.

Chebyshev's student Andrey Andreyevich Markov provided another proof in his $1884$ Ph.D. thesis.

Sources

1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Tchebyshev's inequality (P.L. Tchebyshev, 1874)
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Tchebyshev's inequality (P.L. Tchebyshev, 1874)
2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Chebyshev's inequalities
2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Chebyshev's inequality