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Re: Expected value of the Geometric distribution

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  Re: Expected value of the Geometric distribution Bob Hanlon <hanlonr@cox.net> - 2011-04-29 11:32 +0000

#1988 — Re: Expected value of the Geometric distribution

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@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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