       Re: Characteristic Function of Pareto distribution

• To: mathgroup at smc.vnet.net
• Subject: [mg104660] Re: Characteristic Function of Pareto distribution
• From: fd <fdimer at gmail.com>
• Date: Fri, 6 Nov 2009 05:15:57 -0500 (EST)
• References: <hconpj\$1kn\$1@smc.vnet.net> <hcr9ui\$9nb\$1@smc.vnet.net>

```On Nov 4, 6:19 pm, sashap <pav... at gmail.com> wrote:
> 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:= PDF[P][x]
>
> > Out= 3/x^4
>
> > Then, I try
>
> > In:= CharacteristicFunction[P, t]
>
> > During evaluation of In:= \[Infinity]::indet: Indeterminate
> > expression 0 (t^2)^(3/2) ComplexInfinity encountered. >>
>
> > Out= Indeterminate
>
> > If I try calculating the Fourier transform I get
>
> > In:= FourierTransform[PDF[P][x], x, \[Omega],  FourierParameters
>  -
>
> > > {1, 1}]
>
> > Out= 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:= CharacteristicFunction[P, t]
>
> > Out=
> > 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]]
>
>
> 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:= 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= 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 + 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:= Integrate[
>  Exp[I t x] PDF[ParetoDistribution[1, 3], x], {x, 1, Infinity},
>  Assumptions -> Im[t] == 0]
>
> Out= 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:= % - %% // FullSimplify[#, t \[Element] Reals] &
>
> Out= 0
>
> Oleksandr Pavlyk
> Wolfram Research
>
>
>

I did a few other experiments on this matter and I still don't quite
agree with Mathematica not returning a CharactheristicFunction
[ParetoDistribution[1,3],t] that is not usable. It surely returns an
expression for

c2 = CharacteristicFunction[ParetoDistribution[k, \[Alpha]], t]

which for a definite \[Alpha] is the same as the integration from k to
infinity

c1 = Integrate[
Exp[I t x] PDF[ParetoDistribution[k, \[Alpha]], x], {x, k,
Infinity},
Assumptions -> {Im[t] == 0, Im[\[Alpha]] == 0, Re[\[Alpha]] > -1,
Re[k] > 0, Im[k] == 0, \[Alpha] \[Element] Integers}]

cc =
Simplify[c1 - c2,
Assumptions -> {Im[\[Alpha]] == 0, Im[t] == 0, Re[\[Alpha]] > 1,
Re[k] > 0, Im[k] == 0, t < 0}] (**t assumed < 0 to eliminate t-Abs
[t]t**)

Out= I (-k t)^\[Alpha] (Gamma[
1 - \[Alpha]] + \[Alpha] Gamma[-\[Alpha]]) Sin[(\[Pi] \[Alpha])/2]

Limit[cc /. k -> 1, \[Alpha] -> 3]

Out= 0

But it would be more useful if Mathematica returned an usable
expression whether alpha is defined or a symbol

ca[k_, \[Alpha]_] :=
Integrate[
Exp[I t x] PDF[ParetoDistribution[k, \[Alpha]], x], {x, k,
Infinity},
Assumptions -> {Im[t] == 0, Im[\[Alpha]] == 0, Re[\[Alpha]] > -1,
Re[k] > 0, Im[k] == 0, \[Alpha] \[Element] Integers}]

In:= ca[1, 3]

Out= 1/4 (-2 E^(I t) (-2 + t (-I + t)) +
t^3 (\[Pi] + 2 I CosIntegral[t] - 2 SinIntegral[t]))

Thanks

```

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