Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2012

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

Search the Archive

Re: How to simplify hypergeometrics

  • To: mathgroup at smc.vnet.net
  • Subject: [mg128420] Re: How to simplify hypergeometrics
  • From: Roland Franzius <roland.franzius at uos.de>
  • Date: Sun, 14 Oct 2012 23:41:01 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • Delivered-to: l-mathgroup@wolfram.com
  • Delivered-to: mathgroup-newout@smc.vnet.net
  • Delivered-to: mathgroup-newsend@smc.vnet.net
  • References: <k55o3n$e7q$1@smc.vnet.net> <k586hh$is3$1@smc.vnet.net>

Am 12.10.2012 06:32, schrieb Roland Franzius:
> Am 11.10.2012 08:14, schrieb Dr. Wolfgang Hintze:
>> Consider the probability function
>>
>> p[n_, a_, b_, k_] :=
>>    Binomial[k - 1, a - 1]*(Binomial[n - k, b - a]/Binomial[n, b]) /; {0
>> <= a <=
>>       b, n >= a}
>>
>> In[68]:= p[n, a, b, k]
>>
>> Out[68]= (Binomial[-1 + k, -1 + a]*Binomial[-k + n, -a + b])/
>> Binomial[n, b]
>>
>> Now let's look for the zeroeth moment k^0 (for the higher ones the
>> situation is similar)
>>
>> In[69]:= k0 = Sum[p[n, a, b, k], {k, 1, n}]
>>
>> Out[69]= ((-Binomial[-1, -a + b])*Binomial[n, -1 + a]*
>>       Hypergeometric2F1[1 - a + b, 1 + n, 2 - a + n, 1] +
>>      Binomial[0, -1 + a]*Binomial[-1 + n, -a + b]*
>>       HypergeometricPFQ[{1, 1, 1 - a + b - n}, {2 - a, 1 - n}, 1])/
>>    Binomial[n, b]
>>
>> This should give 1, but it looks clumsy.
>
> All general expression with HypergeometricPFQ generally will be wrong
> for integer parameters in the numerator parameters.
>
> One has to regularize by extracting and cancelling infinite Gamma
> factors at negative integers but that means to have control of the
> evaluation by understanding the limit formulas.
>
> Contrary to your assumption of a normalized ditribution the direct
> conversion
>
> Binomial[n_,k_]:> Pochhammer[n-k+1,k]/k! leaves you with the expression
>
>    p[n_, a_, b_, k_]:=(b! Pochhammer[1 - a + k, -1 + a] Pochhammer[
>     1 + a - b - k + n, -a + b])/((-1 + a)! (-a + b)! Pochhammer[
>     1 - b + n, b])
>
>
> Simplify[Sum[p[6, a, b, k], {k, 0, n}]] /. {n->6 ,a -> 1, b -> 3}
>
> is not independent of a,b,n
>
>
> Perhaps a normalization constant is missing? In any binomial
> distribution the normalized distribution has to contain n-th powers of
> the parameters


So, at least, I found some material related to your problem.

On the internet:

http://www.math.uah.edu/stat/index.html

Especially

http://www.math.uah.edu/stat/urn/OrderStatistics.html


Mathematica 8 knows about OrderStatistics.


In: Assuming[{{k, n, m, i} \[Element] Integers,
           1 <= m, 1 <= i <= n <= m},
  FunctionExpand[
   Sum[ Binomial[k - 1, i - 1] Binomial[m - k,
       n - i] /. {Binomial[a_, b_] :> a!/b!/(a - b)!}, {k, i,
     m - n + i}]]]

Out: Gamma[1 + m]/(Gamma[1 + m - n] Gamma[1 + n])


In the Mathematica Help  for "OrderDistribution" you will find an 
example, adapted here to the discrete case

In: \[ScriptCapitalD] =
  OrderDistribution[{DiscreteUniformDistribution[{1, n}], a}, b];

In: CDF[\[ScriptCapitalD], x]

Out:
	
\[Piecewise]	
-BetaRegularized[-(1/n)+k/n,b,1+a-b]+
     BetaRegularized[k/n,b,1+a-b]	k>=1&&k-n<0
1-BetaRegularized[1-1/n,b,1+a-b]	k>=1&&k-n==0
-BetaRegularized[-(1/n),b,1+a-b]	k>=1&&k-n<=0
0	True

Perhaps it helps.

-- 

Roland Franzius












  • Prev by Date: Resources on using stylesheets for publishing
  • Next by Date: Re: reverse engineering principal components...
  • Previous by thread: Re: How to simplify hypergeometrics
  • Next by thread: Re: How to simplify hypergeometrics