Challenge Level

Branching processes, or tree graphs, model the growth and eventual size of a population. If we know the probabilities of the number of offpsring produced at each generation, then we can determine the probability of ultimate extinction, or the eventual population size.

Consider a variable *X,* where $P(X=0)=p_0, P(X=1)=p_1, ...$

This is an integer valued variable with its mass function as a sequence. We set two conditions:

- All probabilities need to be positive $ p_k \geq 0 $
- Only one event can and must occur, so $p_0+p_1+...=\displaystyle\sum\limits_{k=0}^{\infty} p_k =1$

The *probability generating function* *G*, is an ordinary function in terms of *s:* $$G_X(s)=p_0+p_1 s+p_2 s^2+...$$ **Question:** What is the value of *G(s)* when $s=0$? And when $s=1$?

**Example:** Consider a random variable *Y* with the geometric distribution with parameter *p*.

Then $P(Y=k)=p(1-p)^{k-1}=pq^{k-1}$ for $k=0,1,...$.

So *Y* has PGF given by: $$\begin{align*} G_Y(s) & = \displaystyle \sum_{k=1}^{\infty} p q^{k-1} s^k \\ &= ps \displaystyle \sum_{k=0}^{\infty} (qs)^k \\ &= \frac {ps}{1-qs} \end{align*}$$

We can relate the PGF to the mean, or *expectation*. Recall that: $$E(X)=\bar x = \displaystyle \sum_{all x}^{ } xP(X=x)$$We can extend this definition to not just a variable, but to a function of a variable: $$E(g(X))=\bar{g}(x) = \displaystyle \sum_{all x}^{ } g(x) P(X=x)$$This definition reminds us of our PGF polynomial, with the important result: $$ G_X(s)=p_0+p_1
s+p_2 s^2+...=E(s^X)$$

Consider a population of meerkats, where each individual has a random number of offspring in the next generation. Using this information, we can determine the total expected number of offspring in future generations.

First let $N, X_1, X_2, ...$ be independent variables, with $X_1, X_2, ...$ all having the same probability generating function *G*. Think of these *X* as the individual meerkats in our population. This also means that our PGF is given by $G(s)=p_0+p_1s+p_2s^2+...$, where $p_0=P(\text{no offspring}), p_1=P(\text{one offspring}) , ...$

We are interested in finding the PGF of the sum $X_1+X_2+...+X_N$ $$\begin{align*} G_T(s) & = E[s^T] \\ &= \displaystyle \sum_{n=o}^{\infty} E\Big [s^T|N=n\Big ] P(N=n) \\ & = \displaystyle \sum_{n=o}^{\infty} G(s)^n P(N=n) \\ & = E[G(s)^n] \\ &= G_N \Big( G(s) \Big) \end{align*} $$**Example:** Elephants (in most cases) only have
one offspring at a time, with probability *p*, say. We can model the number of offspring using the Bernoulli distribution with parameter *p.*

Generation *n+1* consists of the offspring of generation *n*.

Let $Z_{n+1}= \displaystyle \sum_{j=1}^{Z_n} X_j$ , where $X_j$ is the number of offspring of the *j*th individual in generation *n.*

In the first generation: $G_{Z_1} (s)=G_X(s)=(1-p)+ps$

In the second generation: $G_{Z_2} (s)=G_{Z_1} \bigg(G_X (s) \bigg)=(1-p)+p\big((1-p)+ps\big)=(1-p^2)+p^2 s$

Continuing, we see that at the *n*th generation: $G_{Z_n} (s)=(1-p^n)+p^n s$

Now click here to find out about branching processes and how we can use probability to determine the likelihood of a population becoming extinct.