Will the variance of some imaginary variable be negative?

Question

From the formula $$\mathrm{Var}[aX]=a^2\mathrm{Var}[X],$$ if we take $a=\sqrt{-1}$ (so all measurements are pure imaginary numbers or zero), does this make the variance of $\sqrt{-1}X$ possibly negative?

Answer 1

It depends on how you define the variance of a complex-valued random variable. For a complex-valued random variable $Z\in\Bbb C$, if you define the variance as $$ \mathsf{Var}Z=\mathsf E|Z-\mathsf EZ|^2 $$ then it is straightforward to show $$ \mathsf{Var}Z=\mathsf{Var}(\Re (Z))+\mathsf{Var}(\Im (Z)). $$ In the special case where $Z=aX$, where $X\in\Bbb R$ is a real-valued random variable and $a\in\Bbb C$ is a complex-valued scalar we obtain $$ \mathsf{Var}Z=\mathsf{Var}(\Re (a)X)+\mathsf{Var}(\Im (a)X)=|a|^2\mathsf{Var}X, $$ which is clearly nonnegative. Choosing $a=i$ then gives $\mathsf{Var}Z=\mathsf{Var}X$.

If, however, you define the variance using the pseudo-variance $$ \mathsf{Var}Z=\mathsf E(Z-\mathsf EZ)^2 $$ then the resulting quantity will be in general complex-valued. In our specific example $$ \mathsf{Var}Z=\mathsf E(aX)^2-(\mathsf E(aX))^2=a^2\mathsf{Var}X. $$ Choosing $a=i$ subsequently gives $\mathsf{Var}Z=-\mathsf{Var}X$.