From: horst.kraemer@t-online.de (Horst Kraemer) Subject: Re: Poisson, how do I love thee? Date: Fri, 17 Mar 2000 06:32:02 GMT Newsgroups: sci.math Summary: [missing] On Thu, 16 Mar 2000 15:16:46 -0500, ian wrote: > > If I have three Poisson processes going on--say transmission of messages > over some radio frequency--with rates a, b, and c, from different > sources, is that the same as one Poisson process with rate a+b+c from a > single source? My heart tells me yes, but I'm wondering what the > newsgroup has to say on this highly controversial subject.... Your heart is correct and this subject isn't controversial in mathematics. The superposition of two independent Poisson processes A and B with rates a and b is a Poisson process with rate a+b. The probability that k events from any of the processes are occurring in an interval of size T is P(k) = Pr { (0 events from A and k events from B) or (1 event from A and k-1 events from B) or (2 events from A and k-2 events from B) ..... or (k events from A and 0 events from B) } As the processes are assumed to be independent Pr { (x events from A) and (y events from B) } = Pr { x events from A } * Pr { y events from B } and as the events combined by 'or' are mutually exclusive we can write P(k) = Pr {0 events from A} * Pr { k events from B) + Pr {1 event from A} * Pr { k-1 events from B) + Pr {2 events from A} * Pr { k-2 events from B) ... + Pr {k events from A} * Pr { 0 events from B) This yields k (aT)^i * (bT)^(k-i) P(k) = Sum ------------------- * exp(-aT)*exp(-bT) i=0 i! * (k-i)! 1 k k! = --- Sum -------- (aT)^i * (bT)^(k-i) * exp(-aT)*exp(-bT) k! i=0 i!(k-i)! ( by the binomial theorem (x+y)^k = ..... ) 1 = --- * (aT + bT)^k * exp(-aT)*exp(-bT) k! ((a+b)T)^k = ---------- * exp(-(a+b)T) k! q.e.d. This extends to the sum of 3 or more independent Poisson processes, of course. Regards Horst ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: fourier transform question Date: 10 Oct 2000 13:56:28 -0400 Newsgroups: sci.math Summary: [missing] In article <39E33D34.A7337078@email.sps.mot.com>, Peter Brooker wrote: :I want to take the Fourier transform of the function defined by: : :h(x) = sum_n (delta(x-n) ) : :The sum goes from n=-inf to n=+inf. n is an integer : :The above function is referred to as the comb(x) function. The :answer is suposed to be : :g(f) = comb(f) : :I cannot get this. What I get is the following. : :g(f) = int_x [ sum_n ( delta(x-n)) * exp(-i *2*pi*fx) * dx] : : = int_x [ sum_n ( delta(x-n)) * exp(-i*2*pi*fn) * dx ] : : = sum_n ( exp(-i*2*pi*fn) * int_x [ delta(x-n) dx] ) : : = sum_n ( exp(-i*2*pi*fn) ) : :I do not recognize this as the comb function. Am I doing something :wrong? : :Any help would be greatly appreciated. " :thanks-Peter brooker This is a form of Poisson's summation formula: if u is a suitable function (infinitely differentiable, with derivatives vanishing faster than negative powers near infinity) and if U is its Fourier transform, with variable scaled as you indicated (by 2*pi), then sum_n u(n) = sum_n U(n) and that's the same as your formula, if you "integrate against u". For proof: I find the one in D.H. Griffel: Applied Functional Analysis John Wiley & Sons, 1985, ISBN 0-479-27196-5 or 0-85312-226-1 quite simple (you re-scale the variable afterwards). Bad news: The book seems to be out of print. Cheers, ZVK(Slavek).