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Characteristic Function of Pareto distribution
*To*: mathgroup at smc.vnet.net
*Subject*: [mg104541] Characteristic Function of Pareto distribution
*From*: fd <fdimer at gmail.com>
*Date*: Tue, 3 Nov 2009 02:56:27 -0500 (EST)
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?
Thanks in advance.
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