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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> - 2011-04-28 10:37 +0000
    Re: Expected value of the Geometric distribution Gary Wardall <gwardall@gmail.com> - 2011-04-29 11:30 +0000
    Re: Expected value of the Geometric distribution Stefan <wutchamacallit27@gmail.com> - 2011-04-29 11:32 +0000
    Re: Expected value of the Geometric distribution "Sjoerd C. de Vries" <sjoerd.c.devries@gmail.com> - 2011-04-29 11:32 +0000
    Re: Expected value of the Geometric distribution Peter Breitfeld <phbrf@t-online.de> - 2011-04-29 11:36 +0000

#1962 — Expected value of the Geometric distribution

Hi all,
I want to calculate expected value of diverse distributions like the Geometric distribution (for example).
As I understand this, the expected value is the integral of the density function *x.
But when I try to calculate this:
Integrate[(1-p)^k*p*k,k]
I get this as the answer:
((1 - p)^k p (-1 + k Log[1 - p]))/Log[1 - p]^2
Instead of: (1-p)/p.
I would be so grateful if someone could explain to me what I'm doing wrong.
Tonja
___________________________________________________________
Empfehlen Sie WEB.DE DSL Ihren Freunden und Bekannten und wir  
belohnen Sie mit bis zu 50,- Euro! https://freundschaftswerbung.web.de

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#1971

On Apr 28, 5:37 am, "Tonja Krueger" <tonja.krue...@web.de> wrote:
> Hi all,
> I want to calculate expected value of diverse distributions like the Geometric distribution (for example).
> As I understand this, the expected value is the integral of the density function *x.
> But when I try to calculate this:
> Integrate[(1-p)^k*p*k,k]
> I get this as the answer:
> ((1 - p)^k p (-1 + k Log[1 - p]))/Log[1 - p]^2
> Instead of: (1-p)/p.
> I would be so grateful if someone could explain to me what I'm doing wrong.
> Tonja
> ___________________________________________________________
> Empfehlen Sie WEB.DE DSL Ihren Freunden und Bekannten und wir  
> belohnen Sie mit bis zu 50,- Euro!https://freundschaftswerbung.web.de

Tonia,

I think the Geometric Distribution is discrete, in which case you
should use the Sum[  command.

Note:

Sum[(1 - p)^k*p*k, {k, 1, Infinity}]

Yields:

(1 - p)/p

Good Luck

Gary Wardall

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#1983

On Apr 28, 6:37 am, "Tonja Krueger" <tonja.krue...@web.de> wrote:
> Hi all,
> I want to calculate expected value of diverse distributions like the Geometric distribution (for example).
> As I understand this, the expected value is the integral of the density function *x.
> But when I try to calculate this:
> Integrate[(1-p)^k*p*k,k]
> I get this as the answer:
> ((1 - p)^k p (-1 + k Log[1 - p]))/Log[1 - p]^2
> Instead of: (1-p)/p.
> I would be so grateful if someone could explain to me what I'm doing wrong.
> Tonja
> ___________________________________________________________
> Empfehlen Sie WEB.DE DSL Ihren Freunden und Bekannten und wir  
> belohnen Sie mit bis zu 50,- Euro!https://freundschaftswerbung.web.de

Tonja,
  Two things to be considered. First you are on the right track
regarding the definition of expected value of a random variable. Note
that the integral though should not be an indefinite integral, but one
over whatever domain the variable takes its values from. In this case,
0 to Infinity. Your second mistake though, was to use an *integral* to
compute the expected value of a *discrete* random variable. The
geometric distribution is discrete and so any expected values should
be computed using sums, in this case from 0 to Infinity. The line
you're looking for is
Sum[(1 - p)^k*p*k, {k, 0, Infinity}]
= (1-p)/p
Hope this helps.
-Stefan S

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#1990

Hi Tonja,

The problem is that this is a *discrete* probability distribution.So,
instead of integrating you need to sum the terms:

In[15]:= Sum[(1 - p)^k*p*k, {k, 1, \[Infinity]}]

Out[15]= (1 - p)/p

As of Mathematica 8 Mathematica knows a lot of distribution stuff, so you coul
also say:

In[14]:= Expectation[x, x \[Distributed] GeometricDistribution[p]]

Out[14]= (1 - p)/p

Cheers -- Sjoerd


StackOverflow for fast answers to Mathematica questions
http://stackoverflow.com/questions/tagged/mathematica


 On Apr 28, 12:37 pm, "Tonja Krueger" <tonja.krue...@web.de> wrote:
> Hi all,
> I want to calculate expected value of diverse distributions like the Geometric distribution (for example).
> As I understand this, the expected value is the integral of the density function *x.
> But when I try to calculate this:
> Integrate[(1-p)^k*p*k,k]
> I get this as the answer:
> ((1 - p)^k p (-1 + k Log[1 - p]))/Log[1 - p]^2
> Instead of: (1-p)/p.
> I would be so grateful if someone could explain to me what I'm doing wrong.
> Tonja
> ___________________________________________________________
> Empfehlen Sie WEB.DE DSL Ihren Freunden und Bekannten und wir  
> belohnen Sie mit bis zu 50,- Euro!https://freundschaftswerbung.web.de

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#2003

"Tonja Krueger" wrote:

> Hi all,
> I want to calculate expected value of diverse distributions like the
> Geometric distribution (for example).  
> As I understand this, the expected value is the integral of the
> density function *x. 
> But when I try to calculate this:
> Integrate[(1-p)^k*p*k,k]
> I get this as the answer:
> ((1 - p)^k p (-1 + k Log[1 - p]))/Log[1 - p]^2
> Instead of: (1-p)/p.
> I would be so grateful if someone could explain to me what I'm doing wrong.
> Tonja
> ___________________________________________________________
> Empfehlen Sie WEB.DE DSL Ihren Freunden und Bekannten und wir  
> belohnen Sie mit bis zu 50,- Euro! https://freundschaftswerbung.web.de
>

The Geometric Distribution is discrete, so you must use Sum instead of
Integrate:

Sum[(1-p)^k p k, {k,1,Infinity}]

Out= (1-p)/p

This is also the result of
Mean[GeometricDistribution[p]]
as well as of
ExpectedValue[#&,GeometricDistribution[p]]
-- 
_________________________________________________________________
Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de

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