# Poisson PGF

• Dec 9th 2009, 09:39 AM
Bibby
Poisson PGF
Im not sure where to start with this. Im thinking i may have to start off by intergrating the poisson distribution over 0 to infinity?

If X ~ P(mu), show that its PGF (Probability Generating Function) is given by:

Gx(s) = $e^{mu(s-1)}$

Sorry i have used the word "mu" instead of the symbol, i kept encountering LaTex problems
• Dec 9th 2009, 12:15 PM
T. Student
Are you starting with the Poisson distribution and replacing
lamda with mu since it can shown using infinite series that the population mean of the Poisson distribution is mu=lamda? I think you can integrate that to get the correct answer, but I haven't had time to try it yet. Let me know what you come up with.
• Dec 9th 2009, 03:03 PM
chella182
I'm in the same class as whoever posted this, and $\mu$ is just the parameter of the distribution... nothing to do with the mean. I also have no idea how to do about proving it, though.
• Dec 9th 2009, 03:59 PM
ampersand
The expected value of your poisson is $\mu$, which is probably why your professor uses it there.

anyways, expand out your p.g.f., rearrange it so that within your infinite sum, you set up a new poisson with the parameter as $s\mu$, and on the outside, you'll have the e^(mu(s-1)) you're looking for

Quote:

Originally Posted by chella182
I'm in the same class as whoever posted this, and $\mu$ is just the parameter of the distribution... nothing to do with the mean. I also have no idea how to do about proving it, though.

• Dec 9th 2009, 04:02 PM
chella182
I know that's how you do it, it's just not coming out right, hence how I'm stuck.
• Dec 9th 2009, 05:37 PM
mr fantastic
Quote:

Originally Posted by chella182
I know that's how you do it, it's just not coming out right, hence how I'm stuck.

The problem is trivial if you realise (and you should at this level) that $\sum_{x=0}^{+\infty} \frac{(s \lambda)^x}{x!} = e^{s \lambda}$ (that is, recognise the well known Maclaurin series for $e^{t}$).