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 $z$?



1 Answer 1


$P\{Z \leq \alpha\} = P\{U^2+V^2 \leq \alpha^2\}$ 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 $Z$.

  • 1
    $\begingroup$ explicitly what is the joint density function of $u$ and $v$? $\endgroup$
    – apg
    Dec 9, 2013 at 20:50
  • 3
    $\begingroup$ @AlexanderGiles The OP's question implicitly assumes that $U$ and $V$ are independent $N(0,\sigma^2)$ random variables and so the joint density is their product. $\endgroup$ Dec 9, 2013 at 23:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .