Re: Expected value of the Geometric distribution
- To: mathgroup at smc.vnet.net
- Subject: [mg118477] Re: Expected value of the Geometric distribution
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Fri, 29 Apr 2011 07:32:33 -0400 (EDT)
GeometricDistribution is a discrete distribution
dist = GeometricDistribution[p];
Mean[dist]
-1 + 1/p
Sum[k*PDF[dist, k], {k, 0, Infinity}]
(1 - p)/p
% == %% // Simplify
True
Bob Hanlon
---- 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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