
How to simplify hypergeometrics
Posted:
Oct 11, 2012 2:14 AM


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.
Simplifying simply gives
In[71]:= FullSimplify[k0]
Out[71]= ComplexInfinity
but also a qualified Simplify does not help, because in this case Mathematica makes slight cosmetic changes but the result is still far from being recognized as 1:
In[73]:= FullSimplify[k0, {Element[{a, b, n}, Integers], 0 <= a <= b, n >= a}]
Out[73]= (((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]
Any help would be greatly appreciated.
Best regards, Wolfgang

