# Markov's Inequality

## Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $A \in \Sigma$.

Let $f : A \to \overline \R$ be an $A$-measurable function.

Then:

$\ds \map \mu {\set {x \in A: \size {\map f x} \ge t} } \le \frac 1 t \int_A \size f \rd \mu$

for any positive $t \in \R$.

### Corollary

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $X$ be a real-valued random variable on $\struct {\Omega, \Sigma, \Pr}$.

Then:

$\ds \map \Pr {\size X \ge t} \le \frac {\expect {\size X} } t$

for each real $t > 0$.

## Proof

Let $t > 0$ and define:

$B = \set {x \in A: \size {\map f x} \ge t}$

Let $\chi_B$ denote the indicator function of $B$ on $A$.

For any $x \in A$, either $x \in B$ or $x \notin B$.

In the first case:

$t \map {\chi_B} x = t \cdot 1 = t \le \size {\map f x}$

In the second case:

$t \map {\chi_B} x = t \cdot 0 = 0 \le \size {\map f x}$

Hence:

$\forall x \in A: t \map {\chi_B} x \le \size {\map f x}$
$\ds \int_A t \chi_B \rd \mu \le \int_A \size f \rd \mu$
$\ds \int_A t \chi_B \rd \mu = t \int_A \chi_B \rd \mu = t \map \mu B$

Dividing through by $t > 0$:

$\ds \map \mu B \le \frac 1 t \int_A \size f \rd \mu$

$\blacksquare$

## Source of Name

This entry was named for Andrey Andreyevich Markov.