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Moment-generating Function of Poisson Distribution

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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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Associated Topics:
College Statistics

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