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Re: Expected value of the Geometric distribution

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The geometric distribution is discrete (of course), so...

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

(1 - p)/p

Bobby

On Thu, 28 Apr 2011 05:37:38 -0500, Tonja Krueger <tonja.krueger@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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-- 
DrMajorBob@yahoo.com

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Re: Expected value of the Geometric distribution DrMajorBob <btreat1@austin.rr.com> - 2011-04-29 11:33 +0000

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