From: delemar@galet.icp.grenet.fr (DELEMAR)
Newsgroups: sci.math
Subject: Re. summary: pdf's mean value
Date: 19 Apr 1995 11:32:21 GMT

	I posted this question a few days ago :

>	Could anyone help me evaluate this sum :
>
>inf                        
>___                        
>\    d-N     N      d!     
>/   a   (1-a)  ____________
>---            (N-1)!(d-N)!
>d=N			    


	Thanks *A LOT* to Dave Rusin and Terry Moore who gave me the
solution :

p(d) = a^(d-N) . (1-a)^N . (d-1).(d-2)...(d-N+1)/(N-1)! whith d,N
integers and d>=N is a probability density function (it's the duration 
probability of a N-states discrete-time left-right Markov chain). Thus
Sum[p(d)] from N to infinity is 1.

Proof: this is (1-a)^N/(N-1)! Sum[a^(d-N) . (d-1)...(d-N+1)] 
Let k=d-N the sum is : Sum[a^k . (k+1)...(k+N-1)] , 0<=k 
a^k . (k+1)...(k+N-1) is the (N-1)th derivative of a^(k-N+1) and the
previous sum is then the (N-1)th derivative of 
Sum[a^(k-N+1)] = a^(N-1)/(1-a)
Let b=a-1 => db=da  and  a^(N-1)/(1-a) = -(b+1)^(N-1)/b = -1/b + P(b)
where P(b) is an polynomial of degree (N-2) in b. The (N-1)th
derivative of -1/b is -(-1)^(N-1) . (N-1)! / b^N = (N-1)!/(1-a)^N
and Sum[p(d)] from N to infinity is 1.

	The first sum, which is Sum[d.p(d)] is
Sum[a^(d-N).(1-a)^N.N.C(d,N) = N.Sum{a^(d-N).(1-a)^N.[C(d-1,N-1)+C(d-1,N)]}
                             = N.{Sum[p(d)]+Sum[a^(d-N).(1-a)^N.C(d-1,N)]
let N=M-1 :
... = N.{1 + Sum[a^(d-M+1).(1-a)^(M-1).C(d-1,M-1)], d>=M-1}
    = N.{1 + a/(1-a).Sum[p(d)], d>=M} = N/(1-a)

	Easy, isn't it ?
-- 

					DELEMAR Olivier



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