In probability theory, Chebyshev's inequality (also called the Bienaymé–Chebyshev inequality) guarantees that, for a wide class of probability distributions, no more than a certain fraction of values can be more than a certain distance from the mean. Specifically, no more than 1/k2 of the distribution's values can be k or more standard deviations away from the mean (or equivalently, at least 1 − 1/k2 of the distribution's values are less than k standard deviations away from the mean). The rule is often called Chebyshev's theorem, about the range of standard deviations around the mean, in statistics. The inequality has great utility because it can be applied to any probability distribution in which the mean and variance are defined. For example, it can be used to prove the weak law of large numbers.
Its practical usage is similar to the 68–95–99.7 rule, which applies only to normal distributions. Chebyshev's inequality is more general, stating that a minimum of just 75% of values must lie within two standard deviations of the mean and 88.89% within three standard deviations for a broad range of different probability distributions.
The term Chebyshev's inequality may also refer to Markov's inequality, especially in the context of analysis. They are closely related, and some authors refer to Markov's inequality as "Chebyshev's First Inequality," and the similar one referred to on this page as "Chebyshev's Second Inequality."
History
The theorem is named after Russian mathematician Pafnuty Chebyshev, although it was first formulated by his friend and colleague Irénée-Jules Bienaymé. The theorem was first stated without proof by Bienaymé in 1853 and later proved by Chebyshev in 1867. His student Andrey Markov provided another proof in his 1884 Ph.D. thesis.
Statement
Chebyshev's inequality is usually stated for random variables, but can be generalized to a statement about measure spaces.
Probabilistic statement
Let X (integrable) be a random variable with finite non-zero variance σ2 (and thus finite expected value μ). Then for any real number ,
Only the case is useful. When the right-hand side and the inequality is trivial as all probabilities are ≤ 1.
As an example, using shows that the probability that values lie outside the interval does not exceed . Equivalently, it implies that the probability of values lying within the interval (i.e. its "coverage") is at least .
Because it can be applied to completely arbitrary distributions provided they have a known finite mean and variance, the inequality generally gives a poor bound compared to what might be deduced if more aspects are known about the distribution involved.
Measure-theoretic statement
Let (X, Σ, μ) be a measure space, and let f be an extended real-valued measurable function defined on X. Then for any real number t > 0 and 0 < p < ∞,
More generally, if g is an extended real-valued measurable function, nonnegative and nondecreasing, with then:
The previous statement then follows by defining as if and otherwise.
Example
Suppose we randomly select a journal article from a source with an average of 1000 words per article, with a standard deviation of 200 words. We can then infer that the probability that it has between 600 and 1400 words (i.e. within k = 2 standard deviations of the mean) must be at least 75%, because there is no more than chance to be outside that range, by Chebyshev's inequality. But if we additionally know that the distribution is normal, we can say there is a 75% chance the word count is between 770 and 1230 (which is an even tighter bound).
Sharpness of bounds
As shown in the example above, the theorem typically provides rather loose bounds. However, these bounds cannot in general (remaining true for arbitrary distributions) be improved upon. The bounds are sharp for the following example: for any k ≥ 1,
For this distribution, the mean μ = 0 and the standard deviation σ = , so
Chebyshev's inequality is an equality for precisely those distributions that are a linear transformation of this example.
Proof
Markov's inequality states that for any real-valued random variable Y and any positive number a, we have Pr(|Y| ≥a) ≤ E(|Y|)/a. One way to prove Chebyshev's inequality is to apply Markov's inequality to the random variable with a = (kσ)2:
It can also be proved directly using conditional expectation:
Chebyshev's inequality then follows by dividing by k2σ2.