Is the distribution of the real part of product of two independent complex variates exponential?

Question

Trying to find the pdf of the real part x of the product $z_1z_2$ of two uncorrelated complex random Gaussian variates . The pdf of the modulus $r \equiv |z_1z_2|$ is known $ f_r(r)=rK_0(r)$ from Wells, Anderson and Cell 1962: "The Distribution of the Product...", $\theta $ is uniform on $[0,2\pi]$. So the marginal pdf on x is found by integrating on the y-axis: $$ f_x(x)= \frac{1}{2\pi}\int _{- \infty} ^\infty rK_0(r)dy $$ By a change of variable $y=\sqrt{r^2 - x^2}$ I get $$ f_x(x)= \frac{1}{\pi}\int _{|x|} ^\infty rK_0(r)\frac{dy}{dr}dr= \frac{1}{\pi}\int _{|x|} ^\infty \frac{ r}{\sqrt{r^2-x^2}} K_0(r) dr $$ Remarkably this definite integral evaluates numerically to a simple double-sided exponential. To computer precision $f_x(x)=0.5e^{-|x|} \pm 10^{-8}$. Is this a known result, can it be proved or is it trivial? I searched through the Bessel section in Gradshteyn and Ryzhik to no avail.

Answer 1

$\newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$

Here is a rather elementary derivation of the double-sided exponential distribution for the real part (say $R$) of the product of two standard independent complex random variables (r.v.'s). We have $R=U-V$, where $U$ and $V$ are iid r.v.'s such that $U=XY$, where $X$ and $Y$ are iid standard normal (real-valued) r.v.'s. Since the r.v. $V$ is symmetrically distributed, we can write \begin{equation} f=g^2, \end{equation} where $f$ and $g$ are the characteristic functions (c.f.'s) of $R$ and $U$, respectively. Next, using the polar coordinates, for real $t$ we have
\begin{multline*} g(t)=\E e^{itXY}=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{itxy}\frac1{2\pi}\,e^{-(x^2+y^2)/2}dx\,dy \\ =\frac1{2\pi}\,\int_0^{2\pi}d\thh\int_0^\infty r\,dr\, e^{-r^2(1-it\sin2\thh)/2} =\frac1{2\pi}\,\int_0^{2\pi}\frac{d\thh}{1-it\sin2\thh} =\frac1{\sqrt{1+t^2}} \end{multline*} (to evaluate the latter integral, use the standard substitution $u=\tan\thh$, $d\thh=\frac{du}{1+u^2}$, $\sin2\thh=\frac{2u}{1+u^2}$ to reduce it to the integral of a rational function.) So, for all real $t$ \begin{equation} f(t)=\frac1{1+t^2}, \end{equation} which is easily seen to be the expression of the c.f. of the double-sided exponential distribution, as desired.