1. ## 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) = $\displaystyle e^{mu(s-1)}$

Sorry i have used the word "mu" instead of the symbol, i kept encountering LaTex problems

2. 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.

3. I'm in the same class as whoever posted this, and $\displaystyle \mu$ 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.

4. The expected value of your poisson is $\displaystyle \mu$, 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 $\displaystyle s\mu$, and on the outside, you'll have the e^(mu(s-1)) you're looking for

Originally Posted by chella182
I'm in the same class as whoever posted this, and $\displaystyle \mu$ 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.

5. I know that's how you do it, it's just not coming out right, hence how I'm stuck.

6. 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 $\displaystyle \sum_{x=0}^{+\infty} \frac{(s \lambda)^x}{x!} = e^{s \lambda}$ (that is, recognise the well known Maclaurin series for $\displaystyle e^{t}$).