Markov's inequality

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

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.

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

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.

is larger than or equal to 0 as the random variable

because the conditional expectation only takes into account of values larger than or equal to

, making their average also greater than or equal to

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

, which is clear if we consider the two possible values of

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

is non-negative, since only its absolute value enters in the equation.

, obtaining We now provide a proof for the special case when

is a discrete random variable which only takes on non-negative integer values.

yields the desired result.

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

[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): 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