I have samples of a 2D multivariate Gaussian as complex numbers, if I rotate the samples by the squared magnitude of itself, are the samples still Gaussian distributed?
$$ x\sim \mathcal{N}\in\mathbb{C} $$
$$ x\mathrm{e}^{j|x|^2} \quad {}_{\sim}^{?} \quad \mathcal{N}\in\mathbb{C} $$
On
If you rotate around the origin, yes. Consider the level sets of $e^{-(x^2+y^2)}$, so $(x,y)\in\mathbb{R}^2$ values where $e^{-(x^2+y^2)}=c \in \mathbb{R}$. Rearranging, you get $(x^2+y^2)=-\ln(c)=\ln\left(\frac{1}{c}\right)$, or
$$x^2+y^2=\left(\sqrt{\ln\left(\frac{1}{c}\right)} \right)^2$$
Which are concentric circles with radius $r=\sqrt{\ln\left(\frac{1}{c}\right)}$.
Meaning that rotating a random normal variable around the origin, the resulting transformed variable has the same distance from the origin, $r = \sqrt{\ln\left(\frac{1}{c}\right)}$, so it's rotationally symmetric.
Multiplying by $\frac{1}{(2\pi)^{n/2}}$ also won't change this fact.
Let $\phi(z) = ze^{j|z|^2}$, $X$ be the 2D multivariate Gaussian as complex numbers, and $Y = \phi(X)$. We are interested in the distribution of $Y$. First we note that $\phi : \mathbb{C} \to \mathbb{C}$ is a bijective map and its inverse is given by $\phi^{-1}(z) = ze^{-j|z|^2}$. The probability density $f_Y$ of $Y$ can be calculated using $$f_Y(y) = f_X(\phi^{-1}(y)) \left| \det (D\phi^{-1})(y) \right|$$ where $D\phi^{-1}$ is the Jacobian of $\phi^{-1}$ (where we view it as a mapping from $\mathbb{R}^2$ to itself). Given that $f_X$ is symmetric we have the equality $f_X(\phi^{-1}(y)) = f_X(y)$. The determinant above can be shown to be always 1. We thus have that $$f_Y(y) = f_X(y)$$ and so $Y$ is indeed normally distributed, as it has the same density as $X$.