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Chebyshev's inequality

Chebyshev's inequality is a probabilistic inequality. It provides an upper bound to the probability that the absolute deviation of a random variable from its mean will exceed a given threshold.


The following is a formal statement.

Proposition Let X be a random variable having finite mean mu and finite variance sigma^2. Let [eq1] (i.e., k is a strictly positive real number). Then, the following inequality, called Chebyshev's inequality, holds:[eq2]

The proof is a straightforward application of Markov's inequality.


Since [eq3] is a positive random variable, we can apply Markov's inequality to it:[eq4]By setting $c=k^{2}$, we obtain[eq5]But [eq6] if and only if [eq7], so we can write[eq8]Furthermore, by the very definition of variance, [eq9]. Therefore,[eq10]


Suppose we extract an individual at random from a population whose members have an average income of $40,000, with a standard deviation of $20,000. What is the probability of extracting an individual whose income is either less than $10,000 or greater than $70,000? In the absence of more information about the distribution of income, we cannot compute this probability exactly. However, we can use Chebyshev's inequality to compute an upper bound to it. If X denotes income, then X is less than $10,000 or greater than $70,000 if and only if[eq11]where $mu =40,000$ and $k=30,000$. The probability that this happens is:[eq12]Therefore, the probability of extracting an individual outside the income range $10,000-$70,000 is less than $4/9$.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let X be a random variable such that[eq13]Find a lower bound to its variance.


The lower bound can be derived thanks to Chebyshev's inequality:[eq14]Thus, the lower bound is[eq15]

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