Re: Characteristic Function of Pareto distribution
- To: mathgroup at smc.vnet.net
- Subject: [mg104608] Re: Characteristic Function of Pareto distribution
- From: sashap <pavlyk at gmail.com>
- Date: Wed, 4 Nov 2009 01:41:53 -0500 (EST)
- References: <hconpj$1kn$1@smc.vnet.net>
On Nov 3, 1:57 am, fd <fdi... at gmail.com> wrote:
> All
>
> I'm computing the Characteristic function of a Pareto distribution
> with Mathematica, but I'm trying to make sense of the answers.
>
> If I define
>
> P=ParetoDistribution[1,3]
>
> In[164]:= PDF[P][x]
>
> Out[164]= 3/x^4
>
> Then, I try
>
> In[165]:= CharacteristicFunction[P, t]
>
> During evaluation of In[165]:= \[Infinity]::indet: Indeterminate
> expression 0 (t^2)^(3/2) ComplexInfinity encountered. >>
>
> Out[165]= Indeterminate
>
> If I try calculating the Fourier transform I get
>
> In[166]:= FourierTransform[PDF[P][x], x, \[Omega], FourierParameters=
-
>
> > {1, 1}]
>
> Out[166]= 1/2 \[Pi] \[Omega]^3 Sign[\[Omega]]
>
> I have some limited understanding of what's going on. In the first
> case the the integration does not include the origin, while with the
> Fourier transform it makes the integration around the origin and
> Mathematica uses the Cauchy integral, thus the result it shows, does
> that make sense?
>
> What seems a problem to me appears if I do not define values for the
> parameters in the Pareto distribution, I get a rather strange answer
>
> P=ParetoDistribution[xm,alpha]
>
> In[170]:= CharacteristicFunction[P, t]
>
> Out[170]=
> alpha (t^2)^(alpha/2) xm^alpha Cos[(alpha \[Pi])/2] Gamma[-alpha] +
> HypergeometricPFQ[{-(alpha/2)}, {1/2,
> 1 - alpha/2}, -(1/4) t^2 xm^2] - (
> I alpha Sqrt[t^2]
> xm HypergeometricPFQ[{1/2 - alpha/2}, {3/2,
> 3/2 - alpha/2}, -(1/4) t^2 xm^2] Sign[t])/(1 - alpha) +
> I (t^2)^(alpha/2) xm^
> alpha Gamma[1 - alpha] Sign[t] Sin[(alpha \[Pi])/2]
>
> The expression I find in reference textbooks don't involve
> Hypergoemetric function of any sort, and is much simpler
>
> alpha (-I xm \[Omega])^alpha Gamma[alpha, I xm \[Omega]]
>
> Can anyone please help me understanding this?
An expression used by Mathematica for characteristic
function of Pareto distribution has removable singularities
for integer values of alpha.
The unexpanded form involves MeijerG function:
alpha /2 1/Sqrt[Pi]
MeijerG[{{}, {1 + alpha/2}}, {{1/2, 0, alpha/2}, {}}, -I (
k t)/2, 1/2]
For alpha == 3 and k ==1:
In[31]:= With[{alpha = 3, k = 1}, alpha/2 1/Sqrt[Pi]
MeijerG[{{}, {1 + alpha/2}}, {{1/2, 0, alpha/2}, {}}, -I (k t)/2,
1/2]] // FunctionExpand // Simplify
Out[31]= 1/4 (4 Cos[t] + 2 I t Cos[t] - 2 t^2 Cos[t] +
2 I t^3 CosIntegral[t] + I t^3 Log[4] + 2 I t^3 Log[-((I t)/2)] -
2 I t^3 Log[t] + 4 I Sin[t] - 2 t Sin[t] - 2 I t^2 Sin[t] -
2 t^3 SinIntegral[t])
Compare with direct integration:
In[32]:= Integrate[
Exp[I t x] PDF[ParetoDistribution[1, 3], x], {x, 1, Infinity},
Assumptions -> Im[t] == 0]
Out[32]= 1/4 (4 E^(I t) + 2 I E^(I t) t - 2 E^(I t) t^2 + \[Pi] t^3 +
2 I t^3 CosIntegral[t] - 2 t^3 SinIntegral[t])
In[33]:= % - %% // FullSimplify[#, t \[Element] Reals] &
Out[33]= 0
Oleksandr Pavlyk
Wolfram Research
>
> Thanks in advance.