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Expected value of the Geometric distribution

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  Expected value of the Geometric distribution "Tonja Krueger" <tonja.krueger@web.de> - 2011-05-03 12:22 +0000
    Re: Expected value of the Geometric distribution Peter Breitfeld <phbrf@t-online.de> - 2011-05-04 23:48 +0000

#2068 — Expected value of the Geometric distribution

Dear everybody,
Thank you all for your kind help. But I'm still stuck trying to find the expected value for a continuous distribution like the Gumbel distribution or GEV, Weibull.
Moment[GumbelDistribution[\[Alpha], \[Beta]], 1]
gives this as result:
\[Alpha] - EulerGamma \[Beta]
But when I try using
Integrate[ E^(-E^(-((x - \[Mu])/\[Beta])) - (x - \[Mu])/\[Beta])/\[Beta]* x, {x, -\[Infinity], \[Infinity]}]
This is what I get:
ConditionalExpression[\[Beta] (EulerGamma + Log[E^(\[Mu]/\[Beta])] - E^-E^((\[Mu]/\[Beta])) Log[E^(-(\[Mu]/\[Beta]))] + Log[E^(\[Mu]/\[Beta])])), Re[\[Beta]] > 0]
I am stumped.
Tonja
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#2106

"Tonja Krueger" wrote:

> Dear everybody,
> Thank you all for your kind help. But I'm still stuck trying to find
> the expected value for a continuous distribution like the Gumbel
> distribution or GEV, Weibull.
> Moment[GumbelDistribution[\[Alpha], \[Beta]], 1]
> gives this as result:
> \[Alpha] - EulerGamma \[Beta]
> But when I try using
> Integrate[ E^(-E^(-((x - \[Mu])/\[Beta])) - (x -
> \[Mu])/\[Beta])/\[Beta]* x, {x, -\[Infinity], \[Infinity]}]
> This is what I get:
> ConditionalExpression[\[Beta] (EulerGamma + Log[E^(\[Mu]/\[Beta])] -
> E^-E^((\[Mu]/\[Beta])) Log[E^(-(\[Mu]/\[Beta]))] +
> Log[E^(\[Mu]/\[Beta])])), Re[\[Beta]] > 0]
> I am stumped.
> Tonja

Because GumpelDistribution ist defined for real mu and positive beta you
should take these Assumptions in Integrate:

Integrate[
 E^(-E^(-((x - mu)/beta)) - (x - mu)/beta)/beta* x,
 {x, -Infinity, Infinity},
 Assumptions -> {mu] \[Element] Reals, beta > 0}]

Out= EulerGamma beta + mu

Alternatively, you can Simplify your ConditionalExpression (I'll name it
int)

simpbeta=Simplify[int,beta>0]
Out

beta (EulerGamma + Log[E^(mu/beta)] - 
   E^-E^((mu/beta)) (Log[E^(-(mu/beta))] + Log[E^(mu/beta)]))

Mathematica assumes all variables to be potentially complex, so Log[E^x]
is not automatically simplified to x because Log is a multivalued
function. You can in this case either use

PowerExpand[simpbeta]

or declare mu to be real

Simplify[int,beta>0 && Element[mu,Reals]]

This will give you the desired result.
-- 
_________________________________________________________________
Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de

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