Re: Binomial ratio expectation

*To*: mathgroup at smc.vnet.net*Subject*: [mg49927] Re: Binomial ratio expectation*From*: "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk>*Date*: Fri, 6 Aug 2004 03:09:49 -0400 (EDT)*References*: <cetem8$6fs$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

This does what you want: Load the discrete distributions package (see the documentation in the Help Browser): <<Statistics`DiscreteDistributions` Evaluate the required expectation value and simplify the result: FullSimplify[ExpectedValue[x/(2 + x), BinomialDistribution[n, w], x]] ((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) N.B. I haven't checked the correctness of this result. Steve Luttrell "Ismo Horppu" <ishorppu at NOSPAMitu.st.jyu.fi> wrote in message news:cetem8$6fs$1 at smc.vnet.net... > 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). >