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Recognize the integrand as a Gaussian scale mixture. If $R$ has the unit Rayleigh distribution and $Z$ is an independent standard normal distribution variable, then the conditional density of $RZ$ given $R=r$ is $(\sqrt{2\pi }r{)}^{-1}{e}^{-x^2/(2r^2)}$. Multiplying by the radius density givesThus use $U_1$ for the independent radius and $U_2,U_3$ for the normal output of the Box-Muller transform:The mixture argument also proves that $g$ integrates to one, by the Tonelli theorem. As a check, $R^2$ is exponential of rate $1/2$, so the characteristic function of $RZ$ is $\mathbb{E}{e}^{-t^2{R}^2/2}=1/(1+t^2)$. This is the Rayleigh-normal scale mixture, with Laplace distribution density $e^{-|x|}/2$.

Squaring a Rayleigh distribution variable produces an exponential distribution. For $w\ge 0$,where the exponential distribution parameter is a rate. Equivalently, $R^2/{\sigma }^2$ has chi-squared distribution with two degrees of freedom (statistics). An amplitude and its square therefore have different probability density functions, even though one determines the other.

This makes the Rayleigh-normal scale mixture transparent: at unit scale, $R^2$ is exponential, allowing a random normal scale to produce a Laplace distribution. Another equivalent parametrization uses a Weibull distribution with shape two and scale $\sqrt{2}\sigma$. Conventions that call this latter scale the Rayleigh scale differ by a factor of $\sqrt{2}$; check the probability density function before comparing numerical parameter values.^[ref]