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:
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, f a measurable extended real-valued function, and ε > 0, then
This measure-theoretic definition is sometimes referred to as Chebyshev's inequality.
Extended version for monotonically increasing functions
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 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 .
Proof in the language of probability theory
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
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.
In the language of measure theory
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 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):
- The "monotonic" result can be demonstrated by:
- The result that, for a nonnegative random variable X, the quantile function of X satisfies:
- the proof using
- 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.
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. (September 2010)