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:


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.

Many thanks in advance!


1 Answer 1


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

  • 3
    $\begingroup$ +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. $\endgroup$
    – whuber
    Nov 1, 2013 at 13:00
  • $\begingroup$ 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]. $\endgroup$
    – Joel
    Aug 23, 2017 at 3:30

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