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 ****************************************************************** * DELEMAR Olivier | Room : 527 * * ICP/INPG | Phone : 76-57-48-27 * * 46 Av. Felix VIALLET | Fax. : 76-57-47-10 * * 38031 GRENOBLE Cedex - FRANCE | e-mail : delemar@icp.grenet.fr * ******************************************************************