Chebyshev's 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, 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