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Suppose I have something like a targeting problem, where I specify an angular dispersion in the up and down direction with two Gaussian distributions, each having a mean of 0 and a std of 0.3 degrees.

As I understand it, the square root of the squares of the random variable pairs from these distributions would form a Rayleigh Distribution, defining the distribution of the dispersion with a single angle. Is this correct?

If so, how can I compute the mean and std of the resulting Rayleigh distribution, using the means and stds of the two original Gaussian distributions?

What if I started with a Rayleigh Distribution? How would I go backwards to break it into the two Gaussians?

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If random variable R ~ Rayleigh($\sigma$), where the Rayleigh distribution has pdf

$ f(x;\sigma) = \frac{x}{\sigma^2} e^{-x^2/(2\sigma^2)}, \quad x \geq 0$

then R is isomorphic to $\sqrt{X^2 + Y^2}$ where X, Y are i.i.d. ~ N(0, $\sigma^2$).

(Note that a Rayleigh variable has a single "scale" parameter. The mean of a Rayleigh variable is always $\sqrt{\pi/2}$ times that parameter.)

If your random variable is Rayleigh distributed with scale $\sigma$, then, in the case that you're working in Cartesian coordinates: you could instead model it as the distance from zero of {x, y} coordinates from two independent Gaussian random variables X and Y with mean 0 and standard deviation $\sigma$.

(The Rayleigh distribution is often used to model "shot dispersion," and a derivation of this isomorphism by transformation through polar coordinates is given by Hogema. This is also explicitly shown here.)

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