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Re: Expected value of the Geometric distribution

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Recall that the geometric distribution is discrete. This means that, when you speak of the integral of the density function (which in this case should be called probability, or mass, function) it should be interpreted as the Riemann-Stieltjes integral, i.e., the sum of the series k (1 - p)^k p. Then you should write
In[1]:= Sum[k (1-p)^k p,{k,1,Infinity}]
Out[1]= (1-p)/p
and get the correct output.
By the way, I guess you are doing this as an exercise, since all these expected values - and much, much more - are available in Mathematica. Take a look at the tutorials on Discrete Distributions and Continuous Distributions.
-Tomas
> Date: Thu, 28 Apr 2011 06:37:38 -0400
> From: tonja.krueger@web.de
> Subject:  Expected value of the Geometric distribution
> To: mathgroup@smc.vnet.net
>
> 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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Re: Expected value of the Geometric distribution Tomas Garza <tgarza10@msn.com> - 2011-04-29 11:31 +0000

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