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| From | "Sjoerd C. de Vries" <sjoerd.c.devries@gmail.com> |
|---|---|
| Newsgroups | comp.soft-sys.math.mathematica |
| Subject | Re: Expected value of the Geometric distribution |
| Date | 2011-04-29 11:32 +0000 |
| Organization | Steven M. Christensen and Associates, Inc and MathTensor, Inc. |
| Message-ID | <ipe7l9$r0l$1@smc.vnet.net> (permalink) |
| References | <ipbg1f$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...@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
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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
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