3 Generating functions 3 Generating functions 3.2 Using pgfs to analyse stochastic processes

3.1 Generating functions and their properties

Definition 3.1.1.

Let $X$ be a nonnegative integer-valued rv with $\mathrm{P}(X=i)=p_{i}$ , for $i=$ 0, 1, 2, 3, …. Then the probability generating function (pgf) of $X$ is

G(z)\ =\ p_{0}+p_{1}z+p_{2}z^{2}+\ldots\ =\ \sum_{i=0}^{\infty}p_{i}\,z^{i}=% \mathrm{E}(z^{X}).

Remark.

$G(z)$ is well defined, i.e. is (absolutely) convergent for any numerical value of $z$ in $|z|\leq 1$ , because $\sum p_{i}$ converges. It may, of course, converge in a larger region than this.

A simple but useful example is when $X$ is a Bernoulli variable with

p_{0}=\mathrm{P}(X=0)=q\mbox{\ \ and\ \ }p_{1}=\mathrm{P}(X=1)=p

with X taking no other values. Then

G(z)\ =\ p_{0}+p_{1}z\ =\ q+pz

which is defined for all $z$ .

Figure 3.1: Link, Caption: none provided

For real $z$ in [0,1], $G(z)$ is increasing from $G(0)=p_{0}$ to $G(1)=\sum p_{i}=1$ provided $X$ has a proper distribution. We shall also consider cases where $X$ is not proper, so that $G(1)<1$ . One example could be the hitting time of $0$ in the gambler’s ruin problem.

Proposition 3.1.2 (Properties of pgfs).

(a)

There is a unique correspondence (1:1) between a pmf $\{p_{i}\}$ and the corresponding pgf $G(z)$ . So if we know that $X$ has a pgf $G(z)$ which may be expanded:

$G(z)=p_{0}+p_{1}\,z+p_{2}\,z^{2}+p_{3}\,z^{3}+\ldots$

then $X$ must have the pmf $\{p_{0}$ , $p_{1}$ , $p_{2}$ , $p_{3}$ , $\ldots\}$ .
(b)

The moments of a (proper) rv $X$ can be derived from $G(z)$ :

$\mu\ =\ \mathrm{E}(X)\ =\ G^{\prime}(1)$

by which we mean $\frac{d}{dz}G(z)$ evaluated at $z=1$ . Also

$\mathrm{E}[X(X-1)]\ =\mathrm{E}(X^{2})-\mathrm{E}(X)\ =\ G^{\prime\prime}(1).$

This gives $\mathrm{E}(X^{2})=G^{\prime\prime}(1)+\mu$ so that $\mathrm{Var}(X)=G^{\prime\prime}(1)+\mu-\mu^{2}$ .
(c)

Distributions of sums of independent rvs can be found.

Let $X$ and $Y$ be independent rvs with pmfs $\{p_{i}\}$ and $\{q_{j}\}$ respectively, and let $S=X+Y$ have pmf $\{r_{k}\}$ . Then $S$ has pgf

$G_{S}(z)=G_{X}(z)G_{Y}(z),$

the product NOT the sum of the pgf’s of $X$ and $Y$ .

Proof.

(a) is a result from mathematical analysis which is not proved here.

For example if $X$ has pgf

G(z)=\frac{1}{2-z}\ =\ \frac{1}{2}+\frac{1}{4}z+\frac{1}{8}z^{2}+\dots

then $X$ has pmf $p_{0}=1/2$ , $p_{1}=1/4$ , $p_{2}=1/8$ , ….

We shall solve some problems by finding an expression for $G(z)$ , and then obtaining the pmf by expanding $G(z)$ as a power series. To do this we shall use some standard expansions. If necessary we use the formula for a Taylor (or Maclaurin) series about $z=0$ .

Proof of (b) - first version. Differentiate $G(z)$ term by term:

\frac{d}{dz}(p_{0}+p_{1}\,z+p_{2}\,z^{2}+p_{3}\,z^{3}+\ldots)=p_{1}+p_{2}\,2z+% p_{3}\,3z^{2}+\ldots

which on setting $z=1$ gives

p_{1}+2\,p_{2}+3\,p_{3}+\ldots\ =\ \sum_{i=0}^{\infty}i\,p_{i}\ =\ \mathrm{E}(% X).

Proof of (b) - second version. Differentiate inside the expectation:

\frac{d}{dz}G(z)\ =\ \frac{d}{dz}\mathrm{E}(z^{X})\ =\ \mathrm{E}\left(\frac{d% }{dz}z^{X}\right)\ =\ \mathrm{E}(Xz^{X-1})

which on setting $z=1$ gives $E(X)$ . This is possible because expectation is linear, i.e. we can let $h\rightarrow 0$ in

\frac{G(z+h)-G(z)}{h}=\frac{\mathrm{E}[(z+h)^{X}]-\mathrm{E}(z^{X})}{h}=% \mathrm{E}\left[\frac{(z+h)^{X}-z^{X}}{h}\right].

Example. Find the mean and variance of a random variable with pgf $G(z)=1/(2-z)$ .

G^{\prime}(z)=1/(2-z)^{2},

and so $\mu=\mathrm{E}(X)=1$ . Further,

G^{\prime\prime}(z)=2/(2-z)^{3},

so $\mathrm{E}[X(X-1)]=2$ and $\mathrm{Var}(X)=2+\mu-\mu^{2}=2$ .

Proof of (c).

G_{S}(z)=\mathrm{E}(z^{S})=\mathrm{E}(z^{X+Y})=\mathrm{E}(z^{X}z^{Y})=\mathrm{% E}(z^{X})\mathrm{E}(z^{Y})=G_{X}(z)G_{Y}(z).

We are using here the fact that $X$ and $Y$ are independent, from which it follows that any function of $X$ is independent of any function of $Y$ . In particular here $z^{X}$ and $z^{Y}$ are independent, and the expectation of the product of independent rvs is the product of their expectations. ∎

Corollary 3.1.3.

Let $X_{1}$ , $X_{2}$ , …, $X_{n}$ be mutually independent rvs each with the same distribution and therefore the same pgf $G(z)$ . Then their sum

S=X_{1}\,+\,X_{2}\,+\,\ldots\,+\,X_{n}

has pgf

G_{S}(z)=G(z)^{n}.

This follows from repeated application of the previous result.

Example. Let $X_{i}$ be a Bernoulli process, so that $G(z)=q+pz$ . Calculate the pgf of $S=\sum_{i=1}^{n}X_{i}$ , and hence its distribution.

By 3.1.3:

G_{S}(z)=G(z)^{n}=(q+pz)^{n}.

Now we can expand this (Binomial expansion) to get

G_{S}(z)=\sum_{i=0}^{n}\left(\begin{array}[]{c}n\\ i\end{array}\right)p^{i}q^{n-i}z^{i}.

Reading off the coefficient of $z^{i}$ , we see that $S$ has a Binomial $(n,p)$ distribution.