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
be a random variable having finite mean
and finite variance
.
Let
(i.e.,
is a strictly positive real number). Then, the following inequality, called
**Chebyshev's inequality**,
holds:

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

Proof

Since is a positive random variable, we can apply Markov's inequality to it:By setting , we obtainBut if and only if , so we can writeFurthermore, by the very definition of variance, . Therefore,

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 denotes income, then is less than $10,000 or greater than $70,000 if and only ifwhere and . The probability that this happens is:Therefore, the probability of extracting an individual outside the income range $10,000-$70,000 is less than .

Below you can find some exercises with explained solutions.

Let be a random variable such thatFind a lower bound to its variance.

Solution

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

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