Re: Efficient Computation involving a Hypergeometric function...
- To: mathgroup at smc.vnet.net
- Subject: [mg72576] Re: Efficient Computation involving a Hypergeometric function...
- From: "sashap" <pavlyk at gmail.com>
- Date: Thu, 11 Jan 2007 02:20:22 -0500 (EST)
- References: <eo259f$ajv$1@smc.vnet.net>
Hi Richard,
As you can see from your own post it looks scrambled, because you did
not convert it to the InputForm first. Doing so may speed up the
response
time :)
Now to your question.The CDF represents the following infinite sum:
1 - p^(1 + x)/Gamma[1 + x]*
Sum[Gamma[2 + x + n + k]/Gamma[2 + n + k]*(1 - p)^(n + k),
{k, 0, Infinity}]
Since n is large (we need it to be much larger than 2+x and 2) we can
apply Erdelyi formula for asymptotic of ratio of Gamma functions:
(k + n)^x*(1+O[1/(k+n)])
See
http://functions.wolfram.com/GammaBetaErf/Gamma/06/02/0007/
Substituting back your CDF becomes a series in LerchPhi:
1 - (1-p)^n*p^(1+x)*LerchPhi[1-p,-x,n]/Gamma[1 + x]+...
Hope it helps.
Oleksandr Pavlyk
Special Functions Developer
Wolfram Research
Richard Palmer wrote:
> I have a discrete distribution:
>
> \!\(f\ \ = \ \ Binomial[x + n, x]\ p\^\(x + 1\)\ \((1 - p)\)\^n\)
>
> Here p and x are fixed, p is real between 0 and 1, x and n are
> integers greater than or equal to zero, and n is the parameter
> (n,0,Infinity).
>
> The CDF is
>
> \!\(1 - \(\((1 -
> p)\)\^\(1 + n\)\ p\^\(1 + x\)\ Ã[2 + n + x]\ \
> Hypergeometric2F1Regularized[1, 2 + n + x, 2 + n, 1 - p]\)\/Ã[1 + x]\)
>
> where n is actually Floor[n].
>
> Does anybody see a simplification or asymtotic for the CDF that
> provides an efficient way to compute this for large n (say n ge
> 1000)? I need to solve CDF==k for various fixed p and x.
>
> Thanks in advance!
> --
> Richard Palmer