📂Lemmas

Proof of Markov's Inequality

Theorem¹

For a random variable $X$, define a function $u(X) \ge 0$. If $E \left( u(X) \right)$ exists, then for $c > 0$ $$ P(u(X) \ge c) \le {E \left( u(X) \right) \over c} $$

Explanation

This is a lemma used in countless proofs, and Chebyshev’s inequality is a more convenient version of it.

Seeing that the condition requires the existence of the first moment, one might regard it as an all-too-easy and obvious condition. Well, that is true to some extent, but if you are at least an undergraduate, you should at least know the fact that such existence is not entirely obvious.

Proof

Strategy: Split the range of integration into two parts based on $c$ and use only order relations to reduce it to a simpler form. This proof is for continuous probability distributions, but the same method can be used to prove it for discrete probability distributions as well.

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Define the set $A := \left\{ x : u(x) \ge c \right\}$ and the probability density function $f$ of the random variable $X$.

Since $\mathbb{R} = A \cup A^c$, $$ E(u(X)) = \int _{-\infty} ^{\infty} u(x)f(x)dx = \int _{A} u(x)f(x)dx + \int _{A^c} u(x)f(x)dx $$ If $u(x)f(x) \ge 0$, then $\displaystyle \int _{A^c} u(x)f(x)dx \ge 0$, so $$ E(u(X)) \ge \int _{A} u(x)f(x)dx $$ Since $u(x) \ge c$, $$ E(u(X)) \ge c \int _{A} f(x)dx $$ Since $\displaystyle \int _{A} f(x)dx = P(X \in A) = P(u(X) \ge c)$, $$ E(u(X)) \ge c P(u(X) \ge c) $$ Dividing both sides by $c$ gives $$ {E(u(X)) \over c} \ge P(u(X) \ge c) $$

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