Complex random variable proof: $|\mathbb EX| \leq \mathbb E |X|$

Question

Let $X = Y + iZ$ be a complex random variable.

How can we prove

$$|\mathbb EX| \leq \mathbb E |X|$$

I understand why this relationship should hold, but I can't prove it in a mathematically rigorous manner.

Answer 1

Put $$ z = \mathbb{E} X, $$ and find a complex number $\alpha$ of unit length such that $$ |z| = z\alpha. $$ This is esentially just the polar decomposition of a complex number. Then $$ |z| = z\alpha = \mathbb{E} (\alpha X) = \text{Re } \mathbb{E} (\alpha X) = \mathbb{E} (\text{Re} (\alpha X)) \leq \mathbb{E} |\alpha X| = \mathbb{E} |X|, $$ where the third sign of equality holds since the left-hand side is a real number, and the inequality holds since $$ \text{Re }\zeta \leq \sqrt{(\text{Re }\zeta)^2} \leq \sqrt{(\text{Re }\zeta)^2+(\text{Im }\zeta)^2} = |\zeta|, $$ for any complex $\zeta$.

Answer 2

By linearity of expectation, $\mathbb{E}[X]=\mathbb{E}[Y+iZ]=\mathbb{E}[Y]+i\mathbb{E}[Z]$. Further, the function $g(y,z)=\sqrt{y^2+z^2}$ is convex. Hence, by Jensen inequality

$$\mathbb{E}[g(y,z)] \ge g(\mathbb{E}[(y,z)])$$

or

$$\mathbb{E}(|X|) \ge g(\mathbb{E}(y,z))=|\mathbb{E}[X]|$$