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Markov's inequality

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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 on the probability that a non-negative random variable is greater than or equal to some positive constant. Markov's inequality is tight in the sense that for each chosen positive constant, there exists a random variable such that the inequality is in fact an equality.[1]

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. Markov's inequality can also be used to upper bound the expectation of a non-negative random variable in terms of its distribution function.



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]

When , we can take for to 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 nondecreasing functions


If φ is a nondecreasing nonnegative function, X is a (not necessarily nonnegative) random variable, and φ(a) > 0, then[3]

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

The uniformly randomized Markov's inequality


If X is a nonnegative random variable and a > 0, and U is a uniformly distributed random variable on that is independent of X, then[4]

Since U is almost surely smaller than one, this bound is strictly stronger than Markov's inequality. Remarkably, U cannot be replaced by any constant smaller than one, meaning that deterministic improvements to Markov's inequality cannot exist in general. While Markov's inequality holds with equality for distributions supported on , the above randomized variant holds with equality for any distribution that is bounded on .



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

Hence intuitively , which directly leads to .

Probability-theoretic proof


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


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

Discrete case


We now provide a proof for the special case when is a discrete random variable which only takes on non-negative integer values.

Let be a positive integer. By definition

Dividing by yields the desired result.



Chebyshev's inequality


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

for any a > 0.[3] 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

  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 . Then
    which can be proved similarly.[5]



Assuming no income is negative, Markov's inequality shows that no more than 10% (1/10) of the population can have more than 10 times the average income.[6]

Another simple example is as follows: Andrew makes 4 mistakes on average on his Statistics course tests. The best upper bound on the probability that Andrew will do at least 10 mistakes is 0.4 as Note that Andrew might do exactly 10 mistakes with probability 0.4 and make no mistakes with probability 0.6; the expectation is exactly 4 mistakes.

See also



  1. ^ a b Huber, Mark (2019-11-26). "Halving the Bounds for the Markov, Chebyshev, and Chernoff Inequalities Using Smoothing". The American Mathematical Monthly. 126 (10): 915–927. arXiv:1803.06361. doi:10.1080/00029890.2019.1656484. ISSN 0002-9890.
  2. ^ Stein, E. M.; Shakarchi, R. (2005), Real Analysis, Princeton Lectures in Analysis, vol. 3 (1st ed.), p. 91.
  3. ^ a b Lin, Zhengyan (2010). Probability inequalities. Springer. p. 52.
  4. ^ Ramdas, Aaditya; Manole, Tudor, Randomized and Exchangeable Improvements of Markov's, Chebyshev's and Chernoff's Inequalities, arXiv:2304.02611.
  5. ^ Tu, Stephen (2017-11-04). "Markov's Inequality for Matrices". Retrieved May 27, 2024.
  6. ^ Ross, Kevin. 5.4 Probability inequalitlies | An Introduction to Probability and Simulation.