mean square value of rayleigh distribution

Question

I'm reading a paper which suggests that using random numbers with Rayleigh distribution are necessary to get a good result. from what I can see the only parameter of the Rayleigh distribution is the scale, theta.

However, the parameter they have specified to be used is in terms of the 'mean square', ie.

generate c from a Rayleigh distribution with mean square value of 2S(fn)Af

(numerical simulation of a random sea: a common error and its effect upon wave group statistics, Tucker, Challenor and Carter)

I don't really understand this. How can I choose the mean square value of the distribution? If I google mean square stuff about the mean squared error comes up, but I am not trying to calculate an error since I am not trying to compare how the distribution fits against some data set, I'm trying to generate values.

Can anyone help me out? Is the mean square value the MSE and I just have to do some reverse engineering? or is it something else entirely? Is it literally the square of the mean?

Answer 1

What the paper seems to say is:

Random $a_n$ and $b_n$ are generated from a Gaussian distribution with variance $S(f_n) \Delta f$ .... An alternative is to generate $c_n$ from a Rayleigh distribution with mean square value of $2S(f_n) \Delta f$ and $\phi_n$ from a uniform distribution $(0, 2\pi)$

It is called the mean square value because $\mathbb E(c_n^2) = 2S(f_n) \Delta f$, being $\mathbb E(a_n^2+b_n^2)=\mathbb E(a_n^2) + \mathbb E(b_n^2)$, so the expected value or mean of the square of $c_n$, but not the rather smaller square of the mean

For the Rayleigh distribution, Wikipedia uses a scale parameter called $\sigma$ and MathWorld a scale parameter called $s$. In this context they would both be $\sqrt{S(f_n) \Delta f}$ so without the $2$