Markov's inequality

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Markov's inequality gives an upper bound for the probability that X lies within the set indicated in red.
Markov's inequality gives an upper bound for the probability that X lies within the set indicated in red.

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).

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.

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[edit] Statement

In the language of measure theory, Markov's inequality states that if (X,Σ,μ) is a measure space, f is a measurable extended real-valued function, and t > 0, then

\mu(\{x\in X|\,|f(x)|\geq t\}) \leq {1\over t}\int_X |f|\,d\mu.

For the special case where the space has measure 1 (i.e., it is a probability space), it can be restated as follows: if X is any random variable and a > 0, then

\textrm{Pr}(|X| \geq a) \leq \frac{\textrm{E}(|X|)}{a}.

[edit] Proofs

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.

[edit] Special case: probability theory

For any event E, let IE be the indicator random variable of E, that is, IE = 1 if E occurs and = 0 otherwise. Thus I(|X| ≥ a) = 1 if the event |X| ≥ a occurs, and I(|X| ≥ a) = 0 if |X| < a. Then, given a>0,

aI_{(|X| \geq a)} \leq |X|.\,

which is clear if we consider the two possible values of I(|X| ≥ a). Either |X| < a and thus I(|X| ≥ a) = 0, or I(|X| ≥ a) = 1 and by the condition of I(|X| ≥ a), the inequality must be true.

Therefore

\operatorname{E}(aI_{(|X| \geq a)}) \leq \operatorname{E}(|X|).\,

Now observe that the left side of this inequality is the same as

a\operatorname{E}(I_{(|X| \geq a)})=a\Pr(|X| \geq a).\,

Thus we have

a\Pr(|X| \geq a) \leq \operatorname{E}(|X|)\,

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

[edit] General case: measure theory

For any measurable set A, let 1A be its indicator function, that is, 1A(x) = 1 if x ∈ A, and 0 otherwise. If At is defined as At = {x ∈ X| |f(x)| ≥ t}, then

0\leq t\,1_{A_t}\leq |f|1_{A_t}\leq |f|.

Therefore

\int_X t\,1_{A_t}\,d\mu\leq\int_{A_t}|f|\,d\mu\leq\int_X |f|\,d\mu.

Now, note that the left side of this inequality is the same as

t\int_X 1_{A_t}\,d\mu=t\mu(A_t).

Thus we have

t\mu(\{x\in X|\,|f(x)|\geq t\}) \leq \int_X|f|\,d\mu,

and since t > 0, both sides can be divided by t, obtaining

\mu(\{x\in X|\,|f(x)|\geq t\}) \leq {1\over t}\int_X|f|\,d\mu.

Q.E.D.

[edit] Examples