MathGroup Archive 2012

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

Search the Archive

Re: How to simplify hypergeometrics

  • To: mathgroup at smc.vnet.net
  • Subject: [mg128411] Re: How to simplify hypergeometrics
  • From: "Dr. Wolfgang Hintze" <weh at snafu.de>
  • Date: Tue, 16 Oct 2012 20:13:11 -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> <k5g11t$3oi$1@smc.vnet.net>

On 15 Okt., 05:49, Roland Franzius <roland.franz... at uos.de> wrote:
> 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] Pochhamme=
r[
> >     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

Thank you for your hints.
Unfortunately your sum is different from the one I'm looking for.
I still need to consider your interesting remark on the
OrderDistribution.

Regards,
Wolfgang



  • Prev by Date: Nonlinear curve fit as a Button function fails quietly
  • Next by Date: "Discovering" a grid in an image?
  • Previous by thread: Re: How to simplify hypergeometrics
  • Next by thread: D under Sum