I need some help from an Inverse Fourier Transform Expert
- To: mathgroup at smc.vnet.net
- Subject: [mg34607] I need some help from an Inverse Fourier Transform Expert
- From: Richard Palmer <dickp at bellatlantic.net>
- Date: Wed, 29 May 2002 02:46:42 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
--
I want to get a PDF for the sum of n terms of a truncated normal
distribution.
I begin with a Normal Distribution
Normal=\!\(\(1\/\(\[Sigma] \@\( 2\ \[Pi]\)\)\)
Exp[\(-\(\((x -
\[Mu])\)\^2\/\(2\ \[Sigma]\^2\)\)\)]\)
I can write a truncated below at zero version of the distribution in the
regular way as
Truncated=\!\(\(\[ExponentialE]\^\(-\(\((x - \[Mu])\)\^2\/\(2\
\[Sigma]\^2\)\)\)\ \@\(2\
\/\[Pi]\)\)\/\(\[Sigma]\ \((1 + Erf[\[Mu]\/\(\@2\
\[Sigma]\)])\)\)\)
The cf representing the expectation of the sum of n terms is
Cf = \!\(\((\(\[ExponentialE]\^\(1\/2\ \[ImaginaryI]\ t\ \((2\ \[Mu] +
\
\[ImaginaryI]\ t\ \[Sigma]\^2)\)\)\ \((1 + Erf[\(\[Mu] + \[ImaginaryI]\ t\
\
\[Sigma]\^2\)\/\(\@2\ \[Sigma]\)])\)\)\/\(1 + Erf[\[Mu]\/\(\@2\
\
\[Sigma]\)]\))\)\^n\)
InverseFourierTransform[Cf,t,x,FourierParameters->{0,0}] does not evaluate
in Mathematica 4.1 on a MAC. Can someone suggest a transformation or
alternate approach to getting a formula for the PDF I need.
Thanks in advance.
Richard Palmer