Moment-generating Function of Poisson DistributionDate: 10/25/1999 at 10:13:27 From: Goksel Ozdilek Subject: Moment-generating Function of Poisson Distribution Dear Dr. Math, I have two questions: 1. The moment-generating function of a Poisson distribution is given by M.G.F. (s,t) = e^(lambda(s-1)t) What does this moment generating function imply? (Is lambda*t the intensity?) 2. mij = 1 + Sk 1 j Pik mkj {mu ij = 1 + SIGMA (k is not equal to j) Pik * mu kj} How can I derive the expectation from this formula? E [Tij] = E [Tj | Xo = i] Tij first passage time from i to j Thanks, Goksel Date: 10/25/1999 at 14:31:20 From: Doctor Anthony Subject: Re: Moment Generating Function of Poisson Distribution 1. The moment generating function is M(t) = Expected value of e^(xt) = SUM[e^(xt)f(x)] and for the Poisson distribution with mean a inf = SUM[e^(xt).a^x.e^(-a)/x!] x=0 inf = e^(-a).SUM[(ae^t)^x/x!] x=0 = e^(-a).e^(ae^t) = e^[a(e^t -1)] The mean of the distribution is the coefficient of t/1! and E(x^2) is the coefficient of t^2/2! in expansion of the MGF as a series. M(t) = 1 + a(e^t -1) + a^2(e^t -1)^2/2! + ... = 1 + a(t + t^2/2! + ...) + a^2(t + t^2/2! + ...)^2/2! + ... = 1 + a(t + t^2/2! + ...) + a^2(t^2 + terms higher than t^2)/2! + ... From this we see that coefficient of t/1! = a (so mean = a) Coefficient of t^2/2! = a + a^2 Therefore E(x^2) = a + a^2 and var(x) = E(x^2) - mean^2 = a + a^2 - a^2 = a Therefore mean and variance are both equal to a. 2. I'm afraid I cannot follow your notation here. However the method is to expand the MGF as a series and then find the coefficient of t. This will give you the mean. The coefficient of t^2/2! will give you E(x^2) and then variance = E(x^2) - mean^2. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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