马尔可夫不等式 Markov's inequality

If X is any nonnegative integrable random variable and a > 0, then

\mathbb{P}(X \geq a) \leq \frac{\mathbb{E}(X)}{a}.
\mu(\{x\in X:|f(x)|\geq \varepsilon \}) \leq {1\over \varepsilon}\int_X |f|\,d\mu.

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

证明

For any event E, let IE be the indicator random variable of E, that is, IE = 1 if E occurs and IE = 0 otherwise.

Using this notation, we have 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). If X < a, then I(X ≥ a) = 0, and so aI(X ≥ a) = 0 ≤ X. Otherwise, we have X ≥ a, for which I(X ≥ a) = 1 and so aI(X ≥ a) = a ≤ X.

Since \mathbb{E} is a monotonically increasing function, taking expectation of both sides of an inequality cannot reverse it. Therefore

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

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

a\mathbb{E}(I_{(X \geq a)}) = a(1\cdot\mathbb{P}(X \geq a) + 0\cdot\mathbb{P}(X < a)) = a\mathbb{P}(X \geq a).\,

Thus we have

a\mathbb{P}(X \geq a) \leq \mathbb{E}(X)\,

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


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