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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.
>
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