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Re: Expected value of the Geometric distribution

  • To: mathgroup at smc.vnet.net
  • Subject: [mg118497] Re: Expected value of the Geometric distribution
  • From: Peter Breitfeld <phbrf at t-online.de>
  • Date: Fri, 29 Apr 2011 07:36:12 -0400 (EDT)
  • References: <ipbg1f$ahf$1@smc.vnet.net>

"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
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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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