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.