Re:Fourier Transform PDF Characteristic Function
- To: mathgroup at smc.vnet.net
- Subject: [mg14994] Re:[mg14955] Fourier Transform PDF Characteristic Function
- From: "Tomas Garza" <tgarza at mail.internet.com.mx>
- Date: Wed, 2 Dec 1998 03:59:14 -0500
- Sender: owner-wri-mathgroup at wolfram.com
Yves: I am pasting a copy of a session in my machine (PC, Windows 95,
Mathematica 3.0), where the InverseFourierTransform gives the correct
result. Perhaps you should contact Wolfram support.
In[8]:=
<<Calculus`FourierTransform`
In[14]:=
ft=FourierTransform[\!\(E\^\(-\(x\^2\/2\)\)\/\ at \(2\ \[Pi]\)\),x,t]
Out[14]=
\!\(E\^\(-\(t\^2\/2\)\)\)
In[15]:=
InverseFourierTransform[ft,t,x]
Out[15]=
\!\(E\^\(-\(x\^2\/2\)\)\/\ at \(2\ \[Pi]\)\)
Good luck,
Tomas Garza
Mexico City
Yves Gavreau wrote:
>I'm trying to find the PDF from a Characteristic Function using
>Mathematica. I'm not a mathematician just a curious guy. Here is the
>problem (this is a paste from the notebook).
----------------------------------------------------------------------------
--------------------------------------------------------
>Load the needed packages
<<Statistics`NormalDistribution`
<<Calculus`FourierTransform`
>This is the CharacteristicFunction of a NormalDistribution In[3]:=
>CF=CharacteristicFunction[NormalDistribution[\[Mu],\[Sigma]],t] Out[3]=
\!\(E\^\(I\ t\ \[Mu] - \(t\^2\ \[Sigma]\^2\)\/2\)\) Also this is suppose
>to be the FourierTransform of a NormalDistribution PDF In[5]:=
ft=FourierTransform[PDF[NormalDistribution[\[Mu],\[Sigma]],x],x,t]
Out[5]=
\!\(E\^\(I\ t\ \[Mu] - \(t\^2\ \[Sigma]\^2\)\/2\)\) Now with the
InverseFourierTransform we are suppose to get back the
NormalDistribution PDF
In[6]:=
InverseFourierTransform[ft,t,x]
Out[6]=
\!\(InverseFourierTransform[E\^\(I\ t\ \[Mu] - \(t\^2\
\[Sigma]\^2\)\/2\), t,
x]\)
>As you can see it doesn't give back the NormalDistribution PDF