Re: Fourier Transform and convolution.

*To*: mathgroup at smc.vnet.net*Subject*: [mg26616] Re: [mg26583] Fourier Transform and convolution.*From*: BobHanlon at aol.com*Date*: Sat, 13 Jan 2001 22:36:01 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Needs["Statistics`NormalDistribution`"]; dist = NormalDistribution[mu, sigma]; p = PDF[dist, x] 1/(E^((x - mu)^2/(2*sigma^2))*(Sqrt[2*Pi]*sigma)) f = FourierTransform[p, x, w, FourierParameters -> {1, 1}]//PowerExpand E^(I*mu*w - (sigma^2*w^2)/2) f == CharacteristicFunction[dist, w] True InverseFourierTransform[f^2, w, x, FourierParameters -> {1, 1}]//PowerExpand 1/(E^((x - 2*mu)^2/(4*sigma^2))*(2*Sqrt[Pi]*sigma)) This is N(2*m, Sqrt[2] * sigma) % == PDF[NormalDistribution[2*mu, Sqrt[2]*sigma], x] True Bob Hanlon In a message dated 2001/1/9 2:17:00 AM, Adil.Reghai at dresdnerkb.com writes: >We wanted to perform the distribution of a sum of two independent variables >X and Y which are both N(0,1). >The way We wanted to do it is to use the convolution property. In order >to >compute this convolution we perform >a Fourier Transform of the pdf n(0,1) and then square each point. After >that >we performed an inverse Fourier transform. >Instead of obtaining a N(0,sqrt(2)) we obtain a distribution which is >shifted in the form of a U. >