Markov's inequality

From Wikipedia, the free encyclopedia
Jump to navigation Jump to search
Markov's inequality gives an upper bound for the measure of the set (indicated in red) where exceeds a given level . The bound combines the level with the average value of .

In probability theory, Markov's inequality gives an upper bound for the probability that a non-negative function of a random variable is greater than or equal to some positive constant. It is named after the Russian mathematician Andrey Markov, although it appeared earlier in the work of Pafnuty Chebyshev (Markov's teacher), and many sources, especially in analysis, refer to it as Chebyshev's inequality (sometimes, calling it the first Chebyshev inequality, while referring to Chebyshev's inequality as the second Chebyshev inequality) or Bienaymé's inequality.

Markov's inequality (and other similar inequalities) relate probabilities to expectations, and provide (frequently loose but still useful) bounds for the cumulative distribution function of a random variable.


If X is a nonnegative random variable and a > 0, then the probability that X is at least a is at most the expectation of X divided by a:[1]

Let (where ); then we can rewrite the previous inequality as

In the language of measure theory, Markov's inequality states that if (X, Σ, μ) is a measure space, is a measurable extended real-valued function, and ε > 0, then

This measure-theoretic definition is sometimes referred to as Chebyshev's inequality.[2]

Extended version for monotonically increasing functions[edit]

If φ is a monotonically increasing nonnegative function for the nonnegative reals, X is a random variable, a ≥ 0, and φ(a) > 0, then

An immediate corollary, using higher moments of X supported on values larger than 0, is


We separate the case in which the measure space is a probability space from the more general case because the probability case is more accessible for the general reader.


where is larger than or equal to 0 as r.v. is non-negative and is larger than because the conditional expectation only takes into account of values larger than which r.v. can take.

Hence intuitively , which directly leads to .

Probability-theoretic proof[edit]

Method 1: From the definition of expectation:

However, X is a non-negative random variable thus,

From this we can derive,

From here, dividing through by allows us to see that

Method 2: For any event , let be the indicator random variable of , that is, if occurs and otherwise.

Using this notation, we have if the event occurs, and if . Then, given ,

which is clear if we consider the two possible values of . If , then , and so . Otherwise, we have , for which and so .

Since is a monotonically increasing function, taking expectation of both sides of an inequality cannot reverse it. Therefore,

Now, using linearity of expectations, the left side of this inequality is the same as

Thus we have

and since a > 0, we can divide both sides by a.

Measure-theoretic proof[edit]

We may assume that the function is non-negative, since only its absolute value enters in the equation. Now, consider the real-valued function s on X given by

Then . By the definition of the Lebesgue integral

and since , both sides can be divided by , obtaining


Chebyshev's inequality[edit]

Chebyshev's inequality uses the variance to bound the probability that a random variable deviates far from the mean. Specifically,

for any a > 0. Here Var(X) is the variance of X, defined as:

Chebyshev's inequality follows from Markov's inequality by considering the random variable

and the constant for which Markov's inequality reads

This argument can be summarized (where "MI" indicates use of Markov's inequality):

Other corollaries[edit]

  1. The "monotonic" result can be demonstrated by:
  2. The result that, for a nonnegative random variable X, the quantile function of X satisfies:
    the proof using
  3. Let be a self-adjoint matrix-valued random variable and a > 0. Then
    can be shown in a similar manner.


Assuming no income is negative, Markov's inequality shows that no more than 1/5 of the population can have more than 5 times the average income.

See also[edit]


  1. ^ "Markov and Chebyshev Inequalities". Retrieved 4 February 2016.
  2. ^ Stein, E. M.; Shakarchi, R. (2005), Real Analysis, Princeton Lectures in Analysis, vol. 3 (1st ed.), p. 91.

External links[edit]