       Re: How to simplify hypergeometrics

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• Subject: [mg128386] [mg128386] Re: How to simplify hypergeometrics
• From: Roland Franzius <roland.franzius at uos.de>
• Date: Thu, 11 Oct 2012 23:58:47 -0400 (EDT)
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```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:= p[n, a, b, k]
>
> Out= (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:= k0 = Sum[p[n, a, b, k], {k, 1, n}]
>
> Out= ((-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

--

Roland Franzius

```

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