1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.)

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.