# Chebyshev's Inequality

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This article is complete as far as it goes, but it could do with expansion.In particular: According to 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.), there are more of these than just this one. Refactoring and renaming as appropriate can be done when we document them.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding this information.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Expand}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## 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, we see that:

- $\map f x \le \size {x - \mu}^2 = \paren {x - \mu}^2$

for all $x$.

This means that:

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

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 | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 {k^2}\) |

$\blacksquare$

## Source of Name

This entry was named for Pafnuty Lvovich Chebyshev.

## Sources

- 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**Chebyshev's inequalities**