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MathGroup Archive 2004

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Re: Binomial ratio expectation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49922] Re: [mg49905] Binomial ratio expectation
  • From: DrBob <drbob at bigfoot.com>
  • Date: Fri, 6 Aug 2004 03:09:41 -0400 (EDT)
  • References: <200408051322.JAA05935@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

How about this:

FullSimplify[ExpectedValue[#1/(2 + #1) & , BinomialDistribution[n, w]]]
((1 - w)^n*(-2*(-1 + w)^2 + (1/(1 - w))^n* (2 + (2 + n)*w*(-2 + w +
         n*w))))/((1 + n)* (2 + n)*w^2)

Bobby

On Thu, 5 Aug 2004 09:22:25 -0400 (EDT), Ismo Horppu <ishorppu at NOSPAMitu.st.jyu.fi> wrote:

> I have the following problem, I need to compute
> EXPECTATION[X/(2+X)],
> where X follows Binomial distribution with n trials and success
> probability of w.
>
> I have tried to solve it with Mathematica (version 4.1) as
> Sum[((x)/(2 + x))*Binomial[n, x]*w^x*(1 - w)^(n - x), {x, 0, n}]
>
> I omit here the result which seems to be okay (according to
> simulations) for values 0<w<1. Problem is that result (intermediate
> or full simplified one) is not defined with values 0 or 1 of parameter w.
> However, it is trivial to compute the result by hand on those cases
> (as the X is then a fixed constant, 0 or n).
>
> Does anyone know how to get the full result with Mathematica, or at
> least a warning that the result is partial. I am also interested in
> whether someone knows what kind of summation formula Mathematica uses
> for the sum, some kind of binomial identity formula perhaps? (I am
> unable to find which one, any references would be appreciated).
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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