Rayleigh pdf ratio and parameter estimation

Simo · #1 $Old$ July 4th, 2009, 02:54 AM

Dear all,
I've a problem regarding Rayleigh pdf.
In particular, I know the ratio $X$ between two part of the probability density function, at the right and at the left of a given threshold point $th$ , so

$\int_{th}^{\infty}f(x,\sigma)dx=X\int_{0}^{th}f(x,\sigma)dx$

and from the former I need to compute the value of the parameter $\sigma$ of the Rayleigh distribution $f(x,\sigma)$ .

I've made the first steps, reducing the former in

$\int_{0}^{th}f(x,\sigma)dx=\frac{1}{1+X}$

and solving the definite integral, obtaining

$\frac{\frac{1}{\sigma^{2}}\left( e^{\frac{-th^{2}}{2\sigma^{2}}}-1\right)}{1-\frac{1}{\sigma^{2}}}=\frac{1}{1+X}$

However, is now possible to solve the last equation in $\sigma$ ? otherwise, is there any other method to compute (or, at least, estimate) the value of $\sigma$ from the first equation?

Thank you in advance for any suggestion!
Simo

Moo · #2 $Old$ July 4th, 2009, 10:09 AM

Hello,

The cumulative density function of a Rayleigh distribution is (according to the wikipedia) :

$\int_0^t f(x,\sigma) ~dx=1-\exp\left(\frac{-t^2}{2\sigma^2}\right)$

So you actually have :

$1-\exp\left(\frac{-th^2}{2\sigma^2}\right)=\frac{1}{1+X}$

But it looks like you don't have the same definition of the Rayleigh distribution than the wikipedia...

matheagle · #3 $Old$ July 4th, 2009, 04:12 PM

That's why I wanted more info.
A lot of these densities can be written several ways.
These posters need to write down their densities instead of having me guess how their parameters are written.

Simo · #4 $Old$ July 6th, 2009, 01:06 AM

Moo,
you are absolutely right!
...and this solution is also more "tractable"

Thank you very much!