Re: Binomial ratio expectation
- To: mathgroup at smc.vnet.net
- Subject: [mg49936] Re: [mg49905] Binomial ratio expectation
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Fri, 6 Aug 2004 03:10:09 -0400 (EDT)
- Reply-to: hanlonr at cox.net
- Sender: owner-wri-mathgroup at wolfram.com
Needs["Statistics`DiscreteDistributions`"];
dist=BinomialDistribution[n, w];
ev=FullSimplify[ExpectedValue[x/(2+x), dist,x], 0<=w<=1]
(-2*(w - 2)*w*(1 - w)^n - 2*(1 - w)^n +
(n + 2)*w*(n*w + w - 2) + 2)/((n + 1)*(n + 2)*w^2)
Limit[ev , w->0]
0
ev /. w -> 1 // Simplify
n/(n + 2)
ev == FullSimplify[Sum[x/(x+2)*PDF[dist,x], {
x,0,n}], 0<=w<=1]
True
Bob Hanlon
>
> From: Ismo Horppu <ishorppu at NOSPAMitu.st.jyu.fi>
To: mathgroup at smc.vnet.net
> Date: 2004/08/05 Thu AM 09:22:25 EDT
> To: mathgroup at smc.vnet.net
> Subject: [mg49936] [mg49905] Binomial ratio expectation
>
> 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).
>