algebraic manipulation of probability generating functions

Quote:

Originally Posted by zealot

I am revising for a statistics test and I have an urgent problem involving probability generating functions. I understand the concept its just the algebraic manipulation that gets me. My exam is only a few days away so if you could help me out that would be GREAT! :)

I'm supposed to determine the probability distribution of X from the pgf below.

Here is the working I have done so far:

for $(0<\alpha<1)$ ,
$\pi(z)=\frac{1-\alpha(1-z)}{1+\alpha(1-z)} = \frac{\frac{1-\alpha}{1+\alpha}+\frac{\alpha}{1+\alpha}z}{1-\frac{\alpha}{1+\alpha}z}$

$= (\frac{1-\alpha}{1+\alpha}+\frac{\alpha}{1+\alpha}z)(1+[\frac{\alpha}{1+\alpha}z] +[\frac{\alpha}{1+\alpha}z]^2+...)$

therefore $P(X=0)=\frac{1-\alpha}{1+\alpha}$

This matches exactly with what the lecturer had done exept I couldn't understand how he leapt to this next part:

and for x=>1, $P(X=x)=(\frac{1-\alpha}{1+\alpha})(\frac{\alpha}{1+\alpha})^x + (\frac{\alpha}{1+\alpha})(\frac{\alpha}{1+\alpha}) ^{x-1}$

$=(\frac{\alpha}{1+\alpha})^x(\frac{1-\alpha}{1+\alpha}+1)=\frac{2\alpha^x}{(1+\alpha)^{ x+1}}$

Could someone please explain to me how he got this?

The pgf $\pi (z)$ of a random variable $X$ that takes values in $\{0,\ 1,\ 2,\ ...\}$ is:

$\pi(z)=\sum_{n=0}^{\infty} P(X=n)z^n$

where $P(X=n)$ is the probability that $X$ takes a value $n \in \{0,\ 1,\ 2,\ ...\}$ .

So if you write the pgf as a power series in $z$ the probability
$P(X=n)$ is equal to the coeficient of $z^n$ in the expansion.

What you lecturer has done is collect together the terms containing the same
powers of $z$ , and the resulting coefficients are the required probabilities.

RonL