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Re: Expected value of the Geometric distribution

  • To: mathgroup at smc.vnet.net
  • Subject: [mg118459] Re: Expected value of the Geometric distribution
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Fri, 29 Apr 2011 07:29:18 -0400 (EDT)

Tonja Krueger 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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> 

There are two issues here.

(1) This is a discrete distribution. So you will want a sum rather than 
an integral.

(2) To obtain an expectation one uses a definite rather than indefinite 
"integral" (discrete sum in this case, but that is a type of integral in 
mathematics).

Sum[(1-p)^k*p*k, {k,0,Infinity}] // InputForm
Out[128]//InputForm= (1 - p)/p

Daniel Lichtblau
Wolfram Research


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