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 yaxis: $$ f_x(x)= \frac{1}{2\pi}\int _{ \infty} ^\infty rK_0(r)dy $$ By a change of variable $y=\sqrt{r^2  x^2}$ I get $$ f_x(x)= \frac{1}{\pi}\int _{x} ^\infty rK_0(r)\frac{dy}{dr}dr= \frac{1}{\pi}\int _{x} ^\infty \frac{ r}{\sqrt{r^2x^2}} K_0(r) dr $$ Remarkably this definite integral evaluates numerically to a simple doublesided 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.
1 Answer
$\newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$
Here is a rather elementary derivation of the doublesided exponential distribution for the real part (say $R$) of the product of two standard independent complex random variables (r.v.'s). We have $R=UV$, where $U$ and $V$ are iid r.v.'s such that $U=XY$, where $X$ and $Y$ are iid standard normal (realvalued) 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(1it\sin2\thh)/2}
=\frac1{2\pi}\,\int_0^{2\pi}\frac{d\thh}{1it\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 doublesided exponential distribution, as desired.

$\begingroup$ That nails it !. Though I remain curious about Carlo Beenakker's Wolfram Alpha solution comment. Was that obtained through AI and can it be deconstructed to something like Losif Pinelis's working. $\endgroup$ Apr 2, 2018 at 9:08

$\begingroup$ As an afterthought the CF of the result reminded me of the product of two Gaussians. So the two complex Gaussians could be written $\endgroup$ Apr 3, 2018 at 14:34

$\begingroup$ As an afterthought the CF of the result reminded me of the product of two Gaussians. So the two complex Gaussians could be written z1=x1+iy1, z2=x2+iy2. and the real part of z1*z2 = x1*x2 y1*y2 which is the difference of two real Gaussian products and in isolation would have the 1/(1+t^2) CF. However the real and imaginary parts of z1*z2 are not then independent, so there may not be symmetry as Arg(z1*z2) varies. $\endgroup$ Apr 3, 2018 at 14:43