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Re: Expected value of the Geometric distribution

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On Apr 28, 6:37 am, "Tonja Krueger" <tonja.krue...@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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Tonja,
  Two things to be considered. First you are on the right track
regarding the definition of expected value of a random variable. Note
that the integral though should not be an indefinite integral, but one
over whatever domain the variable takes its values from. In this case,
0 to Infinity. Your second mistake though, was to use an *integral* to
compute the expected value of a *discrete* random variable. The
geometric distribution is discrete and so any expected values should
be computed using sums, in this case from 0 to Infinity. The line
you're looking for is
Sum[(1 - p)^k*p*k, {k, 0, Infinity}]
= (1-p)/p
Hope this helps.
-Stefan S

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Thread

Expected value of the Geometric distribution "Tonja Krueger" <tonja.krueger@web.de> - 2011-04-28 10:37 +0000
  Re: Expected value of the Geometric distribution Gary Wardall <gwardall@gmail.com> - 2011-04-29 11:30 +0000
  Re: Expected value of the Geometric distribution Stefan <wutchamacallit27@gmail.com> - 2011-04-29 11:32 +0000
  Re: Expected value of the Geometric distribution "Sjoerd C. de Vries" <sjoerd.c.devries@gmail.com> - 2011-04-29 11:32 +0000
  Re: Expected value of the Geometric distribution Peter Breitfeld <phbrf@t-online.de> - 2011-04-29 11:36 +0000

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