Re: How to simplify hypergeometrics

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• Subject: [mg128404] Re: How to simplify hypergeometrics
• From: "Dr. Wolfgang Hintze" <weh at snafu.de>
• Date: Sat, 13 Oct 2012 01:04:19 -0400 (EDT)
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• References: <k55o3n\$e7q\$1@smc.vnet.net> <k586hh\$is3\$1@smc.vnet.net>

```On 12 Okt., 06:34, Roland Franzius <roland.franz... at uos.de> wrote:
> 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
>
> --
>
> Roland Franzius

Hello Roland,

1) I extract your idea to use Pochhammer to convert the original
distribution (which IS normalized, contrary to your statement)

p[n_, a_, b_, k_] :=
Binomial[k - 1, a - 1]*(Binomial[n - k, b - a]/Binomial[n, b])

to this form

pp[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])

Comparing the two expressions for concrete values, also checking the
nomalization:

Table[p[10, 3, 4, k], {k, 1, 10}]
Plus @@ %

{0, 0, 1/30, 3/35, 1/7, 4/21, 3/14, 1/5, 2/15, 0}

1

Table[pp[10, 3, 4, k], {k, 1, 10}]
Plus @@ %

{0, 0, 1/30, 3/35, 1/7, 4/21, 3/14, 1/5, 2/15, 0}

1

2a) Ok, so now we calculate again the zeroeth moment of k (which
should give 1, of course).
But, the result does not contain 2F1 anymore but is something
(DifferenceRoot) which I have never seen before, and is not easily
recognized as being = 1

k0 = Sum[pp[n, a, b, k], {k, 1, n}]

(1/(Gamma[a]*Gamma[1 - a + b]*Gamma[1 + n]))*(Gamma[1 + b]*Gamma[1 - b
+ n]*
DifferenceRoot[
Function[{\[FormalY], \[FormalN]}, {(-\[FormalN])*(-\[FormalN] +
a - b +
n)*\[FormalY][\[FormalN]] + (-\[FormalN] - 2*\[FormalN]^2
+
2*\[FormalN]*a - \[FormalN]*b + n + 2*\[FormalN]*n - a*n)*
\[FormalY][
1 + \[FormalN]] + (-1 - \[FormalN] + a)*(-\[FormalN] +
n)*\[FormalY][2 + \[FormalN]] == 0, \[FormalY][1] == 0,
\[FormalY][2] == Gamma[n]/(Gamma[2 - a]*Gamma[a - b +
n])}]][
1 + n])

2b) Trying now to simplify with all information about the parameters
gives ...

FullSimplify[k0, {Element[{n, a, b}, Integers],
Inequality[0, Less, a, LessEqual, b], n >= a}]

FullSimplify::infd:Expression Gamma[1+b] Gamma[1-b+n]
DifferenceRoot[Function[{\[FormalY],\[FormalN]},{-\[FormalN]
Plus[<<4>>] \[FormalY][<<1>>]+Plus[<<7>>] \[FormalY][<<1>>]
+Plus[<<3>>] Plus[<<2>>] \[FormalY][<<1>>]==0,\[FormalY][1]==0,\
[FormalY][2]==Gamma[n]/(Gamma[<<1>>] Gamma[<<1>>])}]][1+n] simplified
to Indeterminate. >>

FullSimplify::infd:Expression (1/(Gamma[a] Gamma[1-a+b]
Gamma[1+n]))Gamma[1+b] Gamma[1-b+n] DifferenceRoot[Function[{\
[FormalY],\[FormalN]},{-\[FormalN] Plus[<<4>>] \[FormalY][<<1>>]
+Plus[<<7>>] \[FormalY][<<1>>]+Plus[<<3>>] Plus[<<2>>] \[FormalY]
[<<1>>]==0,\[FormalY][1]==0,\[FormalY][2]==Gamma[n]/(Gamma[<<1>>]
Gamma[<<1>>])}]][1+n] simplified to Indeterminate. >>

Indeterminate

parameters, gives no evaluation at all

Assuming[{Element[{n, a, b, k}, Integers],
Inequality[0, Less, a, LessEqual, b], n >= a, 1 <= k <= n},
k0 = Sum[pp[n, a, b, k], {k, 1, n}]]

Sum[(b!*Pochhammer[1 - a + k, -1 + a]*
Pochhammer[1 + a - b - k + n, -a + b])/((-1 + a)!*(-a + b)!*
Pochhammer[1 - b + n, b]), {k, 1, n}]

3b) Putting Assuming under the sum

k0 = Sum[Assuming[{Element[{n, a, b, k}, Integers],
Inequality[0, Less, a, LessEqual, b], n >= a, 1 <= k <= n},
pp[n, a, b, k]],
{k, 1, n}]

(1/(Gamma[a]*Gamma[1 - a + b]*Gamma[1 + n]))*(Gamma[1 + b]*Gamma[1 - b
+ n]*
DifferenceRoot[
Function[{\[FormalY], \[FormalN]}, {(-\[FormalN])*(-\[FormalN] +
a - b +
n)*\[FormalY][\[FormalN]] + (-\[FormalN] - 2*\[FormalN]^2
+
2*\[FormalN]*a - \[FormalN]*b + n + 2*\[FormalN]*n - a*n)*
\[FormalY][
1 + \[FormalN]] + (-1 - \[FormalN] + a)*(-\[FormalN] +
n)*\[FormalY][2 + \[FormalN]] == 0, \[FormalY][1] == 0,
\[FormalY][2] == Gamma[n]/(Gamma[2 - a]*Gamma[a - b +
n])}]][
1 + n])

gives us back the expression of attempt 2a).

4) Fixing n in advance results in a funny thing like this one

FullSimplify[
With[{n = 10},
Assuming[{Element[{n, a, b, k}, Integers],
Inequality[0, Less, a, LessEqual, b], n >= a, 1 <= k <= n},
k0 = Sum[pp[n, a, b, k], {k, 1, n}]]]]

(1440*b!*((28*(19 + 8*a - 9*b))/(Gamma[11 - a]*Gamma[2 + a - b]) +
(214 +
3*a^2 + a*(5 - 4*b) + b*(-41 + 3*b))/
(Gamma[8 - a]*
Gamma[6 + a -
b]) + (7/((-8 + a)*(-7 + a)*(-6 + a)*(-5 + a)*Gamma[3 + a -
b]) +
(12600 + a*(-7435 + (2305 - 228*a)*a) - 1385*b -
23*(-5 + a)*a*b + 2*(50 + a)*b^2 - 3*b^3)/Gamma[10 + a -
b])/
Gamma[5 - a]))/
((-a + b)!*Gamma[a]*Pochhammer[11 - b, b])

In summary:
I didn't get the desired result and need more help.

Regards,
Wolfgang

```

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