# Mathematical Statistics Lesson of the Day – Chebyshev’s Inequality

The variance of a random variable $X$ is just an expected value of a function of $X$.  Specifically, $V(X) = E[(X - \mu)^2], \ \text{where} \ \mu = E(X)$.

Let’s substitute $(X - \mu)^2$ into Markov’s inequality and see what happens.  For convenience and without loss of generality, I will replace the constant $c$ with another constant, $b^2$. $\text{Let} \ b^2 = c, \ b > 0. \ \ \text{Then,}$ $P[(X - \mu)^2 \geq b^2] \leq E[(X - \mu)^2] \div b^2$ $P[ (X - \mu) \leq -b \ \ \text{or} \ \ (X - \mu) \geq b] \leq V(X) \div b^2$ $P[|X - \mu| \geq b] \leq V(X) \div b^2$

Now, let’s substitute $b$ with $k \sigma$, where $\sigma$ is the standard deviation of $X$.  (I can make this substitution, because $\sigma$ is just another constant.) $\text{Let} \ k \sigma = b. \ \ \text{Then,}$ $P[|X - \mu| \geq k \sigma] \leq V(X) \div k^2 \sigma^2$ $P[|X - \mu| \geq k \sigma] \leq 1 \div k^2$

This last inequality is known as Chebyshev’s inequality, and it is just a special version of Markov’s inequality.  In a later Statistics Lesson of the Day, I will discuss the motivation and intuition behind it.  (Hint: Read my earlier lesson on the motivation and intuition behind Markov’s inequality.)

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