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Re: Expected value of the Geometric distribution

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  Re: Expected value of the Geometric distribution Barrie Stokes <Barrie.Stokes@newcastle.edu.au> - 2011-04-29 11:33 +0000

#1978 — Re: Expected value of the Geometric distribution

Hi Tonja

Since GeometricDistribution is a discrete distribution, you need to be calculating sums, not integrals.

PDF[ GeometricDistribution[ p ],k ]
Mean[ GeometricDistribution[ p ] ]//Together
Sum[ (1-p)^k p,{k,0,\[Infinity]} ]
Sum[ (1-p)^k p k,{k,0,\[Infinity]} ]

Cheers

Barrie

>>> On 28/04/2011 at 8:37 pm, in message <201104281037.GAA10803@smc.vnet.net>,
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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