I want to check that if $X: \Omega \to \mathbb{C}$ is a random variable, then the inequality $| \mathbb{E} X| \le \mathbb{E} |X|$ also holds like in the real case.
We can write $$X = \Re X + i \cdot \Im X$$
And $$|X| = \sqrt{(\Re X)^2 + (\Im X)^2}, \ \ \ \mathbb{E}|X|= \int_{\Omega} \sqrt{(\Re X)^2 + (\Im X)^2}$$
$$|\mathbb{E}X| = | \int_{\Omega} (\Re X + i \cdot \Im X) \text{d}P| = | \int_{\Omega} \Re X \text{d}P + i \int_{\Omega} \Im X \text{d}P| = \sqrt{ ( \int_{\Omega} \Re X \text{d}P)^2 + ( \int_{\Omega} \Im X \text{d}P ) ^2} $$
What can I use now to finish the argument? Could you give me a hint?
Best Answer
If $EX = 0$, then the inequality is immediate. So suppose $EX \neq 0$. There exists a complex number $c$, with $|c| = 1$, such that $c\,\Bbb EX = |\Bbb E X|$. Then $|\Bbb E X| = \Bbb E(cX)$. Use the fact that $|\Bbb EX|$ is real to justify the equation $|\Bbb E X| = \Bbb E(\Re(cX))]$. Use the inequality $\Re z \le |z|$, $z \in \Bbb C$, to deduce $\Bbb E[\Re(cX)] \le \Bbb E|X|$.