MathGroup Archive 2002

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Characteristic function of Caucy distrbution

  • To: mathgroup at smc.vnet.net
  • Subject: [mg35527] Re: Characteristic function of Caucy distrbution
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Thu, 18 Jul 2002 03:06:08 -0400 (EDT)
  • Organization: Universitaet Leipzig
  • References: <ah24j0$r3d$1@smc.vnet.net>
  • Reply-to: kuska at informatik.uni-leipzig.de
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

In[]:=expr = Integrate[
      Exp[I t x] Gamma/(Gamma^2/2 + (x - m)^2), {x, -Infinity,
Infinity}]/Pi;

and

In[]:=expr /. a_.*Integrate[ex_, bound_] :> 
      Integrate @@ {ex /. x -> y + m, bound /. x -> y, 
          Assumptions -> {Im[t] == 0 && Gamma > 0}} // 
  FullSimplify[#, Element[Gamma, Reals]] &

Out[]=(Sqrt[2]*E^(I*m*t - (Gamma*t*Sign[t])/Sqrt[2])*Pi)/Gamma

and t*Sign[t] is Abs[t]

Regards
  Jens

milkcart wrote:
> 
> Hello
> 
> How is Characteristic function of Caucy distrbution  derived properly.
> I wrote this code as
> 
> Simplify[Integrate[E^(I*t*x)/
>      (1 + (x - $B&L(B)^2/$B&R(B^2),
>     {x, -Infinity,
>      Plus[Infinity]}]/
>    (Pi*$B&R(B)]
> 
> .  But this integration return itself.
> 
>  I know that right answer  is Exp[I $B&L(B t -$B&R(B Abs[t]].
> How can  I get right answer.
> 
> Thanks in advance.
> 
> *****************
> milkcart
> milkcart at m17.alpha-net.ne.jp
> *************************


  • Prev by Date: RE: Re: help in generating a gaussian random variable
  • Next by Date: Re: Output of left-hand side of expression
  • Previous by thread: Characteristic function of Caucy distrbution
  • Next by thread: free boundary problem