# Proof of Chebyshev's inequality

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$P(|X-\mu|>k\sigma)<\frac{1}{k^2}$

In English: "The probability that the outcome of an experiment with the random variable $X$ will fall more than $k$ standard deviations beyond the mean of $X$ , $\mu$ , is less than $\frac1{k^2}$ ."

Or: "The proportion of the total area under the probability distribution function of $X$ outside of $k$ standard deviations from the mean $\mu$ is at most $\frac1{k^2}$ ."

## Proof

Let $S$ be the sample space for a random variable, $X$ , and let $f_X(x)$ stand for the pdf of $X$ . Let $R_1,R_2,R_3$ partition $S$ , such that for every sample point $x\in S$

$x\in \begin{cases} R_1&x<\mu-k\sigma\\ R_2&|x-\mu|\le k\sigma\\ R_3&x>\mu+k\sigma \end{cases}$

Then

$\sigma^2=\sum_{R_1}(x-\mu)^2f_X(x)+\sum_{R_2}(x-\mu)^2f_X(x)+\sum_{R_3}(x-\mu)^2f_X(x)$

Clearly

$\sigma^2\ge\sum_{R_1}(x-\mu)^2f_X(x)+\sum_{R_3}(x-\mu)^2f_X(x)$

since the term that evaluates to the variance in $R_2$ has been subtracted on the right-hand side.

For any sample point $x\in R_1$

$x<\mu-k\sigma$
$=x-\mu<-k\sigma$
$=(x-\mu)^2>k^2\sigma^2$

Notice that the direction of the inequality changes since squaring causes the right-hand expression to become positive.

And for any sample point $x\in R_3$

$x>\mu+k\sigma$
$=x-\mu>k\sigma$
$=(x-\mu)^2>k^2\sigma^2$

So, for any sample point $x\in R_1$ or $x\in R_3$ , it can be said that $(x-\mu)^2>k^2\sigma^2$ , and so

$\sigma^2>\sum_{R_1}(k\sigma)^2f_X(x)+\sum_{R_3}(k\sigma)^2f_X(x)$

Dividing each side of the inequality by $(k\sigma)^2$ results in

$\frac1{k^2}>\sum_{R_1}f_X(x)+\sum_{R_3}f_X(x)$

Or, in other terms

$\frac1{k^2}>P(|x-\mu|>k\sigma)$

And thus the original claim is proven.

QED