Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2004
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Binomial ratio expectation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49992] Re: Binomial ratio expectation
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Mon, 9 Aug 2004 04:29:37 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <cetem8$6fs$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <cetem8$6fs$1 at smc.vnet.net>,
 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. 

After loading the statistics stub,

  << Statistics`

the expectation value, simplified assuming 0 < w < 1 is immediate.

  ev = FullSimplify[ExpectedValue[Function[x,x/(2+x)],     
        BinomialDistribution[n, w]], 0 < w < 1];

We need to take a limit (or use Series) as w -> 0.

  Limit[ev, w -> 0]
  0

However, we can just substitute in the value as w -> 1.

  Simplify[ev /. w -> 1]
  n/(n + 2)

> 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).

You can look at the code Mathematica uses to compute the ExpectedValue 
for the BinomialDistribution (see Statistics`DiscreteDistributions`):

BinomialDistribution/: ExpectedValue[f_Function,
       BinomialDistribution[n_, p_], opts___?OptionQ] :=
   Module[{x, sum},
     (
     If[{opts} =!= {}, Message[ExpectedValue::sum]];
     sum
     ) /; (sum = Sum[ Evaluate[f[x] PDF[BinomialDistribution[n, p], x]],
               Evaluate[{x, 0, n}] ];
      FreeQ[sum, Sum])
   ] /; ParameterQ[BinomialDistribution[n, p]]

If you simplify the summand you will see that you are computing

  Sum[x/(x + 2) (1 - w)^(n - x) w^x Binomial[n, x], {x, 0, n}]

One way to compute this sum is to note that it can be generated by 
parametric differentiation and integration from the simpler but more 
general binomial sum:

  FullSimplify[Sum[(1 - a)^(n - x) b^x Binomial[n, x], {x, 0, n}], 
   0 < w < 1]

which evaluates to 

  (b - a + 1)^n

Since 

  Assuming[x > 0, w D[1/w^2 Integrate[b b^x, {b, 0, w}], w]]

evaluates to 

  w^x x/(x + 2)

the sum you need to compute is just

  FullSimplify[w D[1/w^2 Integrate[b (b - a + 1)^n, {b, 0, w}],w] /. 
    a -> w, 0 < w < 1]

and this result agrees with that from using ExpectedValue.

Cheers,
Paul

-- 
Paul Abbott                                   Phone: +61 8 9380 2734
School of Physics, M013                         Fax: +61 8 9380 1014
The University of Western Australia      (CRICOS Provider No 00126G)         
35 Stirling Highway
Crawley WA 6009                      mailto:paul at physics.uwa.edu.au 
AUSTRALIA                            http://physics.uwa.edu.au/~paul


  • Prev by Date: Re: populate a list with random numbers from normal distribution?
  • Next by Date: Re: Re: populate a list with random numbers from normal distribution?
  • Previous by thread: Re: Binomial ratio expectation
  • Next by thread: Simplify or Cancel question