Prove E(XY) < E|XY|

Question

I know that for vectors $x'y \leq |x'y|$. But how can this be proved for two random variables which are continuous?

$ E(XY) \leq E|XY|$.

Answer 1

Let $(\Omega, \mathcal{F}, P)$ be a probability space and $(\mathcal{S},\mathcal{B})$ be a measurable space where $S \subset \mathbb{C}$. Let $X$ and $Y$ be a measurable function from $\Omega$ to $\mathcal{S}$, then $X$ and $Y$ are measurable functions. Then $XY$ is measurable and by definition \begin{align} E(XY) = \int_{\mathcal{S}} XY dP \end{align}Edit: Notice that $\Omega$ may be a space without an order relation, e.g. $\mathbb{C}$. So in general the "inequality" $E(XY) \leq E(|XY|)$ may not hold as $E(XY)$ may not be able to have an order relation. For example, $E(XY)$ may be a complex number. However, instead by the triangle inequalit y one has \begin{align} |E(XY)| = |\int_{S} XY dP| \leq \int_{S}|XY| dP = E(|XY|)\end{align} Notice we do not have the problem with order relation as $|E(XY)| \in \mathbb{R}$, which is an ordered field.

Answer 2

The quantity $|XY|-XY$ is either $2|XY|$ or $0$. So, it is never negative, and therefore its expectation can never be negative.