How can I transform a Rayleigh distribution to Standard Normal space?

I need to transform a function of Rayleigh distributed variates $G(X)$ to one in standard normal space $G(U)$.

The transformation is governed by equal exceedance probabilities in both spaces, such that the CDFs of the Standard Normal $\Phi$ and Rayleigh $F$ must be equal:

$\Phi(U)=F(X)$

Where

$\Phi(U) = \frac{1}{2}\Big[1+erf\Big(\frac{U}{\sqrt{2}}\Big)\Big]$

$F(X) = 1-exp\Big[-\frac{X^2}{2\sigma^2}\Big]$

The transformation equation hence reduces to:

$X = \sigma \sqrt{-2\ln{\Big[\frac{1}{2}-\frac{1}{2}erf\Big(\frac{U}{\sqrt{2}}\Big)\Big]}}$

As a check to be sure that this is correct, I had to validate that the origin in the standard normal space would translate to the mean value of $X$, $\mu$, ie:

$U=0, X = \mu$

For Rayleigh distributed variates, the mean $\mu$ and parameter $\sigma$ share the relationship:

$\sigma = \mu / \sqrt{\frac{\pi}{2}}$

However, substituting this into the derived transformation equation doesn't give me the mean:

$U = 0, X = \mu\sqrt{\frac{-4\ln{0.5}}{\pi}} = 0.93944 \mu$

Hence failing the validation check. What gives? Is this an approximation error?

• There's a potential difficulty here you don't yet seem to have touched on -- in practice you don't know the parameters; and if you estimate them from the sample things become a little trickier. Aug 19 '16 at 7:04

So really you want the medians to match up. For a normal distribution (or any symmetric PDF), the mean and the median are the same. But for a Rayleigh PDF, the median is $\sigma\sqrt{2\ln2}$. And this is what you found in your check. So all is good!
Note that in general, quantiles can be "passed through" a monotonic transform, whereas moments cannot (i.e. for $f'[x]>0$, $f[x]_q=f[x_q]$ for any quantile $q$, but $\langle f[x]\rangle\neq f[\langle x\rangle]$ unless $f$ is linear, and never for higher order moments).