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Re: Fourier Transform and convolution.


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]


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]


Bob Hanlon

In a message dated 2001/1/9 2:17:00 AM, Adil.Reghai at 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
>compute this convolution we perform 
>a Fourier Transform of the pdf n(0,1) and then square each point. After
>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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