Markov Inequality

Markov Inequality puts bound on how much probability can be at arbitrarily large values once its mean is specified.

Suppose that X is a random variable such that P(X ≥ 0) = 1. Then for every real number t > 0,

$$P(X \ge t) \le {E(X) \over t}$$

Proof: for convecience, we can assume that X random variable has a discrete distribution. For a discre distribution we have:

$$E(X) = \sum_{x} xf(x) = \sum_{x < t} xf(x) + \sum_{x \ge t} xf(x)$$

Since $$X \ge 0$$, all the terms in the summations are nonnegative, therefore:

$$E(X) \ge \sum_{x \ge t} xf(x) \ge \sum_{x \ge t} tf(x) = tP(X \ge t)$$

And divide both sides by t > 0 to obtain the Markov inequality.

The Markov inequality is primarily of interest for large values of t. In fact, when $$t \le E(X)$$, the inequality is of no interest, since it is known that $$P(X \le t) \le 1$$. However, it is found from the Markov inequality that for every nonnegative random variable X whose mean is 1, the maximum possible value of $$P(X \ge 100)$$ is 0.01. Furthermore, it can be verified that this maximum value is actually attained by every random variable X for which P(X = 0) = 0.99 and P(X = 100) = 0.01.