Characteristic Function of Pareto distribution
fd
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.
sashap
> 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
>
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.
fd
> On Nov 3, 1:57 am, fd <fdi...@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
Thanks for your answer. Got it now.
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[129]= I (-k t)^\[Alpha] (Gamma[
1 - \[Alpha]] + \[Alpha] Gamma[-\[Alpha]]) Sin[(\[Pi] \[Alpha])/2]
Limit[cc /. k -> 1, \[Alpha] -> 3]
Out[130]= 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[121]:= ca[1, 3]
Out[121]= 1/4 (-2 E^(I t) (-2 + t (-I + t)) +
t^3 (\[Pi] + 2 I CosIntegral[t] - 2 SinIntegral[t]))
Thanks