       Fourier Transform PDF Characteristic Function

• To: mathgroup at smc.vnet.net
• Subject: [mg14955] Fourier Transform PDF Characteristic Function
• From: "Yves Gauvreau" <gauy at videotron.ca>
• Date: Fri, 27 Nov 1998 03:49:53 -0500
• Sender: owner-wri-mathgroup at wolfram.com

```Hi,

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).

----------------------------------------------------------------------------
--------------------------------------------------------
<<Statistics`NormalDistribution`
<<Calculus`FourierTransform`
This is the CharacteristicFunction of a NormalDistribution In:=
CF=CharacteristicFunction[NormalDistribution[\[Mu],\[Sigma]],t] Out=
\!\(E\^\(I\ t\ \[Mu] - \(t\^2\ \[Sigma]\^2\)\/2\)\) Also this is suppose
to be the FourierTransform of a NormalDistribution PDF In:=
ft=FourierTransform[PDF[NormalDistribution[\[Mu],\[Sigma]],x],x,t]
Out=
\!\(E\^\(I\ t\ \[Mu] - \(t\^2\ \[Sigma]\^2\)\/2\)\) Now with the
InverseFourierTransform we are suppose to get back the
NormalDistribution PDF
In:=
InverseFourierTransform[ft,t,x]
Out=
\!\(InverseFourierTransform[E\^\(I\ t\ \[Mu] - \(t\^2\
\[Sigma]\^2\)\/2\), t,
x]\)
As you can see it doesn't give back the NormalDistribution PDF even if I
use the constant
In:=
InverseFourierTransform[ft,t,x,FourierOverallConstant->1/ 2 Pi] Out=
\!\(InverseFourierTransform[E\^\(I\ t\ \[Mu] - \(t\^2\
\[Sigma]\^2\)\/2\), t,
x, FourierOverallConstant \[Rule] \[Pi]\/2]\) Or if I manualy try
the integral
In:=
1/2 Pi Integral[ft Exp[I t x],{t,-Infinity,Infinity}] Out=
\!\(\*
RowBox[{\(1\/2\), " ", "\[Pi]", " ",
RowBox[{"Integral", "[",
RowBox[{
\(E\^\(I\ t\ x + I\ t\ \[Mu] - \(t\^2\ \[Sigma]\^2\)\/2\)\), ",",
RowBox[{"{",
RowBox[{"t", ",",
InterpretationBox[\(-\[Infinity]\),
DirectedInfinity[ -1]], ",",
InterpretationBox["\[Infinity]",
DirectedInfinity[ 1]]}], "}"}]}], "]"}]}]\) I even tried
to split the function in Even and Odd parts In:=
\!\(f[x_] := E\^\(I\ x\ \[Mu] - \(x\^2\ \[Sigma]\^2\)\/2\); \n\n
G[x_] := \ 1/2\ \((f[x] + f[\(-x\)])\)\n
H[x_] := \ 1/2\ \((f[x] - f[\(-x\)])\)\) In:=
\!\(\[Integral]\_\(-\[Infinity]\)\%\[Infinity] G[x] Cos[2\ Pi\ t\ x]
\[DifferentialD]x\  -
I\ \(\[Integral]\_\(-\[Infinity]\)\%\[Infinity] H[x] Sin[2\ Pi\ t\
x]
\[DifferentialD]x\)\)
Out=
\!\(\*
RowBox[{
RowBox[{\(-I\), " ",
RowBox[{"If", "[",
RowBox[{
\(Im[2\ \[Pi]\ t - \[Mu]] == 0 && Im[2\ \[Pi]\ t + \[Mu]] == 0
&&
Re[\[Sigma]\^2] > 0\), ",",
\(\(I\ \((
E\^\(-\(\((\(-2\)\ \[Pi]\ t + \[Mu])\)\^2\/\(2\
\[Sigma]\^2\)\)\) -
E\^\(-\(\((2\ \[Pi]\ t + \[Mu])\)\^2\/\(2\
\[Sigma]\^2\)\)\))\)\
\ at \(\[Pi]\/2\)\)\/\ at \[Sigma]\^2\), ",",
RowBox[{
SubsuperscriptBox["\[Integral]",
InterpretationBox[\(-\[Infinity]\),
DirectedInfinity[ -1]],
InterpretationBox["\[Infinity]",
DirectedInfinity[ 1]]],
\(\(1\/2\
\((\(-E\^\(\(-I\)\ x\ \[Mu] - \(x\^2\
\[Sigma]\^2\)\/2\)\) +
E\^\(I\ x\ \[Mu] - \(x\^2\ \[Sigma]\^2\)\/2\))\)\
Sin[2\ \[Pi]\ t\ x]\) \[DifferentialD]x\)}]}], "]"}]}],
"+",
RowBox[{"If", "[",
RowBox[{
\(Im[2\ \[Pi]\ t - \[Mu]] == 0 && Im[2\ \[Pi]\ t + \[Mu]] == 0 &&
Re[\[Sigma]\^2] > 0\), ",",
\(\(\((E\^\(-\(\((\(-2\)\ \[Pi]\ t + \[Mu])\)\^2\/\(2\
\[Sigma]\^2\)\)\) +
E\^\(-\(\((2\ \[Pi]\ t + \[Mu])\)\^2\/\(2\
\[Sigma]\^2\)\)\))\)\
\ at \(\[Pi]\/2\)\)\/\ at \[Sigma]\^2\), ",",
RowBox[{
SubsuperscriptBox["\[Integral]",
InterpretationBox[\(-\[Infinity]\),
DirectedInfinity[ -1]],
InterpretationBox["\[Infinity]",
DirectedInfinity[ 1]]],
\(\(1\/2\
\((E\^\(\(-I\)\ x\ \[Mu] - \(x\^2\ \[Sigma]\^2\)\/2\) +
E\^\(I\ x\ \[Mu] - \(x\^2\ \[Sigma]\^2\)\/2\))\)\
Cos[2\ \[Pi]\ t\ x]\) \[DifferentialD]x\)}]}], "]"}]}]\)
Here I took what seemed to be the real part \!\(fs = \(\((
E\^\(-\(\((\(-2\)\ \[Pi]\ t + \[Mu])\)\^2\/\(2\
\[Sigma]\^2\)\)\) +
E\^\(-\(\((2\ \[Pi]\ t + \[Mu])\)\^2\/\(2\
\[Sigma]\^2\)\)\))\)\
\ at \(\[Pi]\/2\)\)\/\ at \[Sigma]\^2\) Out=
\!\(\(\((E\^\(-\(\((\(-2\)\ \[Pi]\ t + \[Mu])\)\^2\/\(2\
\[Sigma]\^2\)\)\) +
E\^\(-\(\((2\ \[Pi]\ t + \[Mu])\)\^2\/\(2\
\[Sigma]\^2\)\)\))\)\
\ at \(\[Pi]\/2\)\)\/\ at \[Sigma]\^2\) This is a simple verification
In:=
fs/.{\[Mu]->0,\[Sigma]->1,t->0}
Out=
\!\(\ at \(2\ \[Pi]\)\)
In:=
PDF[NormalDistribution[\[Mu],\[Sigma]],t]/.{\[Mu]->0,\[Sigma]->1,t->0}
Out=
\!\(1\/\ at \(2\ \[Pi]\)\)
And as you can see it doesn't come close In:=
\!\(\[Integral]\_\(-\[Infinity]\)\%\[Infinity] fs \[DifferentialD]t\)
Out=
\!\(\*
RowBox[{"If", "[",
RowBox[{
\(Re[\[Mu]\/\[Sigma]\^2] > 0 && Re[\[Sigma]\^2] > 0\), ",", "1",
",",
RowBox[{
SubsuperscriptBox["\[Integral]",
InterpretationBox[\(-\[Infinity]\),
DirectedInfinity[ -1]],
InterpretationBox["\[Infinity]",
DirectedInfinity[ 1]]],
\(\(\(\((E\^\(-\(\((\(-2\)\ \[Pi]\ t + \[Mu])\)\^2\/\(2\
\[Sigma]\^2\)\)\) +
E\^\(-\(\((2\ \[Pi]\ t + \[Mu])\)\^2\/\(2\
\[Sigma]\^2\)\)\))\)\
\ at \(\[Pi]\/2\)\)\/\ at \[Sigma]\^2\)
\[DifferentialD]t\)}]}],
"]"}]\)
I ploted fs and it's not even close but it looks Gaussian In:=
Plot[fs/.{\[Mu]->0,\[Sigma]->1},{t,-3,3},PlotRange->All] My question is
can someone tell me what I'm doing wrong ? The actual characteristic
function I'd like to find the PDF for goes like this
and I can't find it as well.
\!\(Exp[\(c\^\[Alpha]\/Cos[\(\[Pi]\ \[Alpha]\)\/2]\)
\((\(\((t\^2 + \[Lambda]\^2)\)\^\(\[Alpha]\/2\)\)
Cos[\[Alpha]\ ArcTan[t\/\[Lambda]]] -
\[Lambda]\^\[Alpha])\)]\) \[Alpha]\[NotEqual]1
c, \[Lambda] and \[Alpha] are real constant
----------------------------------------------------------------------------
----------------------------------------------------------

Hope someone can help

Thanks,

Yves Gauvreau

```

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