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

**References**:**Binomial ratio expectation***From:*Ismo Horppu <ishorppu@NOSPAMitu.st.jyu.fi>