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Moment-generating Function of Poisson Distribution
Date: 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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