Why does $E(XY)^2 \le E(X^2)E(Y^2)$ fail to hold for complex $X, Y$?

Question

Does the Cauchy-Schwarz inequality hold only for real vectors? Why is that so?

Answer 1

This inequality that you have written is valid for real random variables, since you can not compare complex values with each other. As you know the expectation of a complex random variable is still complex, and hence you cannot state the Cauchy-Schwarz inequality for complex random variables in this way.

Note that according to Hölder's inequality we can say:

if $\frac{1}{p}+\frac{1}{q}=1$ then for any real or complex random variables $X$ and $Y$: $E[|XY|]\leq (E[|X|^p])^\frac{1}{p}(E[|Y|^q])^\frac{1}{q}$

hence if you let $p=q=2$, then you have:

$E[|XY|]^2\leq E[|X|^2]E[|Y|^2]$

which you can use for complex random variables.

Note that in this case, the final values which you have, are real values, hence comparison is valid.

Answer 2

You might be looking for Hölder inequalities. In particular, the Cauchy-Schwarz inequality, if $XY$ is $L^1$ integrable, and $X$ and $Y$ are $L^2$ integrable:

$$ \left| E\left[X\overline{Y}\right]\right| \leq E\left[\left|XY\right|\right] \leq E\left[\left|X\right|^2\right]^{1/2}E\left[\left|Y\right|^2\right]^{1/2}.$$