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- - **pdf of Rayleigh squared**
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pdf of Rayleigh squaredSuppose 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? |

Result is fineAfter 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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