Population Ecology using Probability

7316 /www/nrich/html/content/id/7316/ 1 2011-11-28T15:23:21 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><h3>Branching Processes</h3> 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. <h3 style="text-align: center;"><mdo:image alt="" src="tree-with-no-leaves-md.png" style="width: 153px; height: 119px;"></mdo:image></h3> <h3>Probability Generating Functions</h3> 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: <ol> <li> All probabilities need to be positive $ p_k \geq 0 $</li> <li>Only one event can and must occur, so $p_0+p_1+...=\displaystyle\sum\limits_{k=0}^{\infty} p_k =1$</li> </ol> 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: $$\begin{align*} G_Y(s) & = \displaystyle \sum_{k=1}^{\infty} p q^{k-1} s^k \\ &= ps \displaystyle \sum_{k=0}^{\infty} (qs)^k \\ &= \frac {ps}{1-qs} \end{align*}$$ <h3>Expectation</h3> 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: $$ G_X(s)=p_0+p_1 s+p_2 s^2+...=E(s^X)$$ <h3>Random Sums Formula</h3> 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. <mdo:image alt="" src="meerkats-1.jpg" style="width: 285px; height: 190px;"></mdo:image> 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 <a href="http://nrich.maths.org/7254?part=index">here</a> to find out about branching processes and how we can use probability to determine the likelihood of a population becoming extinct.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Note that $G_X(0)=p_0$ and $G_X(1)=1$. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> possible question Question: A simple example of using tree diagrams is in dominant and recessive gene inheritance. Consider the allele pair of genes, R and B, which only affect the women in a family. <mdo:image alt="" src="gene%20transfer%201.jpg" style="width: 286px; height: 173px;"></mdo:image> Let the probabilities of this gene being passed down be: <table align="center" border="0" cellpadding="1" cellspacing="1" style="width: 312px; height: 150px"> <tbody> <tr> <td style="text-align: center;"> </td> <td style="text-align: center;">Daughter has R gene</td> <td style="text-align: center;">Daughter has B gene</td> </tr> <tr> <td style="text-align: center;">Mother has R gene</td> <td style="text-align: center;">2/3</td> <td style="text-align: center;">1/3</td> </tr> <tr> <td style="text-align: center;">Mother has B gene</td> <td style="text-align: center;">1/3</td> <td style="text-align: center;">2/3</td> </tr> </tbody> </table> What is the probability that woman A has two daughters with the B gene, and one with the R gene? What is the probability that a woman with four daughters, has three with the B gene, and one with the R gene? Don't forget to include the likelihood of a woman having a female child. </mdoxml> 2 5 0 0 0 0 1 Population Ecology using Probability An advanced mathematical exploration supporting our series of articles on population dynamics for advanced students. Applications biology Collections Population Dynamics Pre-Calculus and Calculus Calculus generally Using, Applying and Reasoning about Mathematics Mathematical modelling