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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:
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 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.
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