Re: Expected value of the Geometric distribution
- To: mathgroup at smc.vnet.net
- Subject: [mg118482] Re: Expected value of the Geometric distribution
- From: Barrie Stokes <Barrie.Stokes at newcastle.edu.au>
- Date: Fri, 29 Apr 2011 07:33:27 -0400 (EDT)
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 at smc.vnet.net>,
Tonja Krueger <tonja.krueger 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
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