4.6 Chebychev’s inequality

A first step to understanding why expectation and variance matter is given by Chebychev’s inequality. Let $R$ be any random variable. Suppose $\mathrm{E}(R)=m$ and $\mathrm{Var}(R)={\sigma }^2$. Let $c>0$ be any constant: we will find a bound on the probability

$$\mathrm{P}(|R-m|>c\sigma )$$

that $R$ is more than $c$ standard deviations away from its expected value.
Let $A$ be the event that $|R-m|>c\sigma$, and let $I_A$ be the indicator of $A$. Recall that

$$\mathrm{E}(I_A)=1\times \mathrm{P}(I_A=1)+0\times \mathrm{P}(I_A=0)=\mathrm{P}(A).$$

Also define the function $g(r)={(r-m)}^2/{(c\sigma )}^2$, and notice that

$g(r)$

$\ge$

$0$

for all $r$, and

$g(r)$

$\ge$

$1$

whenever $|R-m|>c\sigma$, i.e. whenever $A$ occurs.

So if $A$ does not occur, then $I_A=0\le g(R)$.
And if $A$ does occur, then $I_A=1\le g(R)$.
So $I_A\le g(R)$ and it follows that

	$\mathrm{P}(A)$	$=$	$\mathrm{E}(I_A)$
		$\le$	$\mathrm{E}[g(R)]$
		$=$	$\mathrm{E}[{(r-m)}^2/{(c\sigma )}^2]$
		$=$	${\displaystyle \frac{1}{c^2{\sigma }^2}}\mathrm{E}[{(R-m)}^2]$
		$=$	${\displaystyle \frac{1}{c^2}}.$

We see that

$$\mathrm{P}(|R-m|>c\sigma )\le \frac{1}{c^2}.$$