# Chebyshev’s Inequality

If we have a random variable $X$ and any number $a$, what is the probability that $X \geq a$? If we know the mean of $X$, Markov’s inequality can give an upper bound on the probability that $X$.  As it turns out, this upper bound is rather loose, and it can be improved if we know the variance of $X$ in addition to its mean. This result is known as Chebyshev’s inequality after the name of the famous mathematician Pafnuty Lvovich Chebyshev. It was  first proved by his student Andrey Markov who provided a proof in 1884 in his PhD thesis (see wikipedia).

How likely is it that the absolute value of a random variable $X$ will be greater than some specific value, say, $a$? The Russian mathematician Andrey Andreyevich Markov proved a simple, yet nice, result which enables us to answer the above question.
Markov’s Inequality states that if $X$ is a random variable with mean $\mu$, and $a>0$ is a positive number, then
$Pr[|X| \geq a] \leq \frac{|\mu|}{a}$