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.)

Your thoughtful comments are much appreciated!

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