Take a complex number $z=u+iv$ and call its magnitude $x$.
Consider the real and imaginary parts to be Gaussian distributed:
$$ f_{U}(u)=\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left(\frac{-u^{2}}{2\sigma^{2}}\right) $$ and similarily for $v$.
Then the magnitude of $z$ is distributed as
$$ f(x,\sigma)=\frac{1}{2\pi\sigma^{2}}\int_{-\infty}^{\infty}du\int_{-\infty}^{\infty}dv\,\exp\left(\frac{-u^{2}}{2\sigma^{2}}\right)\exp\left(\frac{-v^{2}}{2\sigma^{2}}\right)\delta\left(x-\sqrt{u^{2}+v^{2}}\right) $$
if we convert to polar co-ordinates then evaluate we have
$$ f(x,\sigma)=\frac{x}{\sigma^{2}}\exp\left(\frac{-x^{2}}{2\sigma^{2}}\right) $$
which is the Rayleigh distribution.
Can anyone explain where the second expression for $f(x,\sigma)$ comes from? How is this the distribution of the length of $z$?
Cheers!
$\endgroup$ 11 Answer
$\begingroup$$P\{Z \leq \alpha\} = P\{U^2+V^2 \leq \alpha^2\}$ is the integral of the joint density of $U$ and $V$ over the disc of radius $\alpha$ centered at the origin. Changing to polar coordinates (as you have done) gives us that $$F_Z(\alpha) = P\{Z \leq \alpha\} = 1 - \exp(-\alpha^2/2\sigma^2)$$ especially after you remember that $(r/\sigma^2)\cdot e^{-r^2/2\sigma^2}$ is a perfect integral whose antiderivative is $-e^{-r^2/2\sigma^2}$). Now differentiate with respect to $\alpha$ to get the density of $Z$.
$\endgroup$ 2
