Markov's Inequality

Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $f$ be an $A$-measurable function where $A \in \Sigma$.

Then:

$\displaystyle \mu \left({\left\{{x \in A: \vert f \left({x}\right) \vert \ge t}\right\} }\right) \le \frac 1 t \int_A \left \vert f \right \vert \mathrm d \mu$

for any positive $t \in \R$.

Proof

Pick any $t$ and define:

$B = \left\{{x \in A: \left \vert f \left({x}\right) \right \vert \ge t}\right\}$

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:

$t \chi_B \left({x}\right) = t \cdot 1 = t \le \left \vert f \left({x}\right) \right \vert$

In the second case:

$t \chi_B \left({x}\right) = t \cdot 0 = 0 \le \left \vert f \left({x}\right) \right \vert$

Hence:

$\forall x \in A$, $t\chi_B \left({x}\right) \le \left \vert f \left({x}\right) \right \vert$

By the monotonicity of the Lebesgue integral:

$\displaystyle \int_A t \chi_B \mathrm d \mu \le \int_A \vert f \vert \mathrm d \mu$

But by the linearity of the Lebesgue integral:

$\displaystyle \int_A t \chi_B \mathrm d \mu = t \int_A \chi_B \mathrm d \mu = t \mu \left({B}\right)$

Hence, dividing through by $t$, we get:

$\displaystyle \mu \left({B}\right) \le \frac 1 t \int_A \vert f \vert \mathrm d \mu$

$\blacksquare$

Markov's Inequality in Probability

Let $\left({\Omega, \Sigma, \Pr}\right)$ be a probability space.

Markov's inequality asserts that for a random variable $X$:

$\Pr \left({\left \vert {X}\right \vert \ge t}\right) \le \dfrac {\mathrm E \left({\left \vert {X}\right \vert}\right)} t$

for any $t > 0$.

It can then be used to derive the probabilistic form of Chebyshev's inequality.

Source of Name

This entry was named for Andrey Andreyevich Markov.