Re: Expected value of the Geometric distribution
- To: mathgroup at smc.vnet.net
- Subject: [mg118480] Re: Expected value of the Geometric distribution
- From: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>
- Date: Fri, 29 Apr 2011 07:33:05 -0400 (EDT)
- References: <ipbg1f$ahf$1@smc.vnet.net>
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... at 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