Chebyshev's Inequality

Theorem

Let:

• $X$ Random Variable
• ${\sigma }^2\in \mathbb{R}$ Variance of $X$
• $\mu =E(X)$ (by Existence of Lower Order Moments, ${\sigma }^2\in \mathbb{R}$ implies that $E(X)$ exists)

Then, for every $k>0$

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

Or equivalently,

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

Example

If $X$ is a random variable such that $E(X)=3$ and $E(X^2)=13$, use Chebyshev’s inequality to determine a lower bound for the probability $P(-2<X<8)$

We have,

• $\mu =E(X)=3$
• ${\sigma }^2=E(X^2)-[E(X){]}^2=13-3^2=4$
• $\sigma =\sqrt{{\sigma }^2}=\sqrt{4}=2$

Then

$$\begin{array}{rl}P(-2<X<8)& =P(-2-3<X-3<8-3)\\ & =P(-5<X-3<5)\\ & =P(|X-3|<5)\end{array}$$

By Chebyshev’s inequality, the form $P(|X-3|<5)$ has to satisfy $P(|X-\mu |<k\sigma )$. Thus

$$\begin{array}{rl}k\sigma & =5\\ k\cdot 2& =5\\ k& =\frac{5}{2}\end{array}$$

As a result, the lower bound for $P(-2<X<8)$ is:

$$\begin{array}{rl}P(-2<X<8)& =P(|X-3|<5)\\ & =1-\frac{1}{k^2}\\ & =1-\frac{1}{{\left(\frac{5}{2}\right)}^2}\\ & =1-\frac{4}{25}\\ & =\boxed{0.84}\end{array}$$