User Name Remember Me? Password

 Advanced Statistics Advanced Probability and Statistics Math Forum

 September 7th, 2007, 05:57 AM #1 Member   Joined: Sep 2007 Posts: 30 Thanks: 0 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? September 10th, 2007, 05:02 AM #2 Member   Joined: Sep 2007 Posts: 30 Thanks: 0 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. Tags pdf, rayleigh, squared Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gelatine1 Number Theory 6 October 16th, 2012 11:22 AM Math_Struggles Linear Algebra 1 November 23rd, 2011 07:20 AM bignick79 Algebra 9 June 29th, 2010 12:59 PM Shocker Advanced Statistics 1 February 8th, 2010 04:47 AM cadman12 Advanced Statistics 1 March 27th, 2008 02:53 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      