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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?

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  • $\begingroup$ 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. $\endgroup$ – Glen_b Aug 19 '16 at 7:04
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Your check is not quite right: As you set the CDFs to be equal, then really you want to check that the quantiles match up (not the moments).

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).

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