Holder's Inequality

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and $X, Y: \Omega \rightarrow \mathbb{R}$ be random variables.

Holder's Inequality. Let $p, q \in \mathbb{R}^+$ , such that $1 / p + 1 / q = 1$ . Then
$\mathbb{E}\left[ |X| \cdot |Y| \right] \le \mathbb{E}^{1 / p} \left[ |X|^p \right] \cdot \mathbb{E}^{1 / q} \left[ |Y|^q \right]$

Remark 1. If $|X|^p$ and $|Y|^q$ are integrable, so is $|X| \cdot |Y|$ .

Proof. The claim is trivial, if either $\mathbb{E}[ |X|^p ] = \infty$ or $\mathbb{E}[ |Y|^q ] = \infty$ .

Therefore, we consider just the case when both $|X|^p$ and $|Y|^q$ are integrable.
By concavity of the function $f(t) = \ln t$ :

\frac{1}{p} \ln a^p + \frac{1}{q} \ln b^q \le \ln \left( \frac{a^p}{p} + \frac{b^q}{q} \right), \qquad \forall a, b > 0.

We obtain the Young's Inequality:

a b \le \frac{a^p}{p} + \frac{b^q}{q}, \qquad \forall a, b > 0.

Let $s = \mathbb{E}^{1 / p} \left[ |X|^p \right]$ and $t = \mathbb{E}^{1 / q} \left[ |Y|^q \right]$ . Then for each $\omega \in \Omega$ , it holds that

\frac{|X(\omega)|}{s} \cdot \frac{|Y(\omega)|}{t} \le \frac{|X(\omega)|^p}{p \cdot s^p} + \frac{|Y(\omega)|^p}{q \cdot t^q}

Define a new random variable

Z = \frac{|X|^p}{p \cdot s^p} + \frac{|Y|^p}{q \cdot t^q}.

By monotonicity and linearity of integration,

\mathbb{E}\left[ \frac{|X|}{s} \cdot \frac{|Y|}{t} \right] \le \mathbb{E}\left[ \frac{|X|^p}{p \cdot s^p} \right] + \mathbb{E}\left[ \frac{|Y|^p}{q \cdot t^q} \right] = 1.

$\square$

Reference

[1] P. Billingsley, Probability and measure Anniversary Edition, 3rd ed. New York: Wiley, 1995.