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RFurball September 7th, 2007 05:57 AM

pdf of Rayleigh squared
Suppose I have a voltage waveform that has a Rayleigh distibution
f(x)=[x/σ^2]*e^(-x^2/2*σ^2) where x>0

I want to find the distribution y=x^2

My solution is as follows:
1) I start with fx(x)=[x/σ^2]*e^(-x^2/2*σ^2) where x>0

2) Going through my probabiliy text (Papoulis) I find that the distribution of y=x^2 is fy(y)=[1/(2y^0.5)]*[fx(y^0.5)+fx(-y^0.5)].

3)Since my domain is limited to x>0. I eliminate the fx(-y^0.5).

4) I then substitute in fx(y^0.5) and get:
fy(y) = [1/(2y^0.5)]*[[(y^0.5)/σ^2]*e^(-(y^0.5)^2/2*σ^2)
which reduces to
fy(y) = [1/(2*σ^2)]*e^(-y/(2*σ^2))

The above distribution is not what I expected. Also the integral of fy
(y) is greater than 1. What did I do wrong?

RFurball September 10th, 2007 05:02 AM

Result is fine
After going over the numbers and consulting some experts my calculations were correct.

The answer is not intuitive and that caused confusion. I even set up a numerical example which showed the results to be correct.

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