# Show that for a Geometric distribution, the probability generating function is given by $\frac{ps}{1-qs}$, $q=1-p$

Suppose that $X$ has a geometric distribution with probability mass function $P(X=x) = q^{i-1}p$, $i=1,2,...$ and $q=1-p$

Show that its probability generating function is given by $\pi(s)=\frac{ps}{1-qs}$. Hence show that $E(x)=\frac{1}{p}$ and $Var(X)=\frac{q}{p^2}$

Hi everyone, I am doing this question for exam practice, and I can't seem to get the correct answer. And to be honest, I am just working through it mechanically and don't have a great understanding of the probability generating functions.

Here is what I have:

$$\pi(s)=E(S^X)=\sum^\infty_{i=0}q^{i-1}p\cdot s^i$$ $$= p\sum^\infty_{i=0}q^{i-1}\cdot s^i=p\sum^\infty_{i=0}\frac{q^i}{q}\cdot s^i$$ $$=\frac{p}{q}\sum^\infty_{i=0}(qs)^i$$

Then using the sum of a geometric series formula, I get:

$$=\frac{p}{q}(\frac{1}{1-qs})$$

Now I am stuck. I feel like I am close, but am just missing something. I'll be ok with deriving the expected value and variance once I can get past this part.

As an addition I was wondering if anyone could also give me a bit of an 'idiots' explanation of the probability generating function, as I am struggling to understand it conceptually. $s$ seems to be the dependent variable, but my lecturer hasn't explained what exactly it is.

It's normal you'd arrive at the wrong answer in this case. The problem is that your index is wrong. There are two definitions for the pdf of a geometric distribution. The one you use, where $E(X)=\frac{1}{p}$ is defined from 1 to infinity. At zero it is not defined. So, the generating function needs to take this into account, as well.

$$\pi(s)=E(S^X)=\sum^\infty_{i=1}q^{i-1}ps^i$$ $$= ps\sum^\infty_{i=1}(qs)^{i-1}=ps\sum^\infty_{i=0}(qs)^i$$ $$=\frac{ps}{1-qs}$$

If you use the alternative definition, where $P(Y=y)=q^ip$, then the pdf is defined at zero. In this case the generating function converges to $\frac{p}{1-qs}$.

As for what $s$ represents, as far as I know it represents nothing. Generating functions are derived functions that hold information in their coefficients. They are sometimes left as an infinite sum, sometimes they have a closed form expression. Take a look at the wikipedia article, which give some examples of how they can be used. Here and here.wiki article probability generating functions and wiki article generating functions

• +1 This is a very clear explanation. Concerning pgfs: you are correct that formally $s$ is merely a placeholder to track the probabilities. (Mathematically it generates the maximal ideal in a ring of formal power series.) The beauty of generating functions is that in many cases when we substitute numbers for $s$, the infinite sum exists, creating a function that can be analyzed using methods of Calculus. This often produces further information about the original coefficients that would have been difficult to obtain otherwise. – whuber Nov 1 '13 at 13:00
• Additionally, it's worth noting that there is one case where $s$ is meaningful. If we have a random number of $i$ trials, each with probability of failure $s$, and the number of trials $i$ is geometrically distributed, then $\pi(s)$ is the probability of all trials failing. [but for the most part, the variable $s$ is just a useful auxiliary value rather than having a meaningful interpretation by itself]. – Joel Aug 23 '17 at 3:30