You may also like

problem icon

Ball Bearings

If a is the radius of the axle, b the radius of each ball-bearing, and c the radius of the hub, why does the number of ball bearings n determine the ratio c/a? Find a formula for c/a in terms of n.

problem icon

Overarch 2

Bricks are 20cm long and 10cm high. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? How big an overhang is it possible to make like this?

problem icon

Cushion Ball

The shortest path between any two points on a snooker table is the straight line between them but what if the ball must bounce off one wall, or 2 walls, or 3 walls?

Population Ecology Using Probability

Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Branching Processes

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.

Probability Generating Functions

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:

  1.  All probabilities need to be positive  $ p_k \geq 0 $
  2. 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: $$ G_X(s)=p_0+p_1 s+p_2 s^2+...=E(s^X)$$


Random Sums Formula

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 jth 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 nth 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.