# Thread: Expected value of a complex rv.

1. ## Expected value of a complex rv.

Let $Z = X + iY$ be complex random variable. Prove that:

$|EZ| \le E|Z|$

2. Originally Posted by albi
Let $Z = X + iY$ be complex random variable. Prove that:

$|EZ| \le E|Z|$
I'd start from the defintion:

$E(Z) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} z \, f(x, y) \,dx \, dy$.

3. First of all we do not know if joint density for the variables exists. However I know (somebody told me) that it can be done in more abstract way using:
1. $X \le Y \Rightarrow EX \le EY$ (for real variables)
2. Cauchy-Schwartz inequality

But I don't know how....

4. Solution

Let $\phi = \arg EZ$

Than: $|EZ| = EZe^{-i\phi} = E(X+iY)(\cos\phi - i \sin \phi) =$
$=E(X\cos\phi + Y \sin \phi) + i E(Y\cos\phi - X \sin \phi)$

Because $|EZ| \in \mathbb{R}$ we have $E(Y\cos\phi - X \sin \phi) = 0$

Hence by Cauchy-Schwartz inequality (and expected value monotonicity)
$
|EZ| = E(X\cos\phi + Y \sin \phi) \le E\left(\sqrt{X^2+Y^2}\sqrt{\cos^2\phi+\sin^2\phi}\ right) = E|Z|$