# Thread: algebraic manipulation of probability generating functions

1. ## algebraic manipulation of probability generating functions

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

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

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

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

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

Could someone please explain to me how he got this?

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

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

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

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

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

Could someone please explain to me how he got this?
The pgf $\displaystyle \pi (z)$ of a random variable $\displaystyle X$ that takes values in $\displaystyle \{0,\ 1,\ 2,\ ...\}$ is:

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

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

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

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

RonL