Monday 16 February 2015

Lecture 14

In this lecture you have had a taste of the theory of branching processes.The calculations we did today in Theorem 14.2 are similar to the sort of thing you need to do in answering Examples Sheet 3, #10, 14, 15. Question #16 does not look like one about branching processes, but you might think about trying to map into a version of a branching process so that you can apply theorems from this lecture.

If you would like to read something more about branching processes, I recommend to A Short Introduction to Branching Processes by Gesine Reinert (a professor of statistics at Oxford).

Branching processes have interesting applications beyond the obvious ones of population biology, e.g. in rumours and advertising. I asked Gesine if she might recommend something beyond the basics. She has written to me, "One application of branching processes is to assess the shortest path length between two randomly chosen vertices in sparse networks; but that may be a bit much for first-year students."

In talking about branching processes with more than one type of individual I was drawing on  Gesine's slides 26-28, and 30-32. In these you can see the idea of a multi-type branching process. We see that it is possible to prove a theorem similar to our Theorem 14.3, but now the probability of extinction depends on the maximum eigenvalue of the matrix $M=(m_{ij})$, where $m_{ij}$ is the mean number of offspring of type $j$ produced by a parent of type $i$. The branching process becomes extinct with a probability 1 or $<1$ as the eigenvalue is $\leq1$ or $>1$.

Suppose there are two types. The key quantities are the two multi-dimensional p.g.f.s

$F_i(z_1,z_2)=E_i\Bigl[z_1^{X_{i1}}z_2^{X_{i2}}\Bigr]$, $i=1,2$,

where $X_{ij}$ is a r.v. having the distribution of the number of offspring of type $j$ produced by a parent of type $i$.

 Let $q_i$ be the probability of extinction when we start with one individual of type $i$. Then, analogous to  Theorem 14.3, $(q_1,q_2)$ are smallest solutions in $[0,1]^2$ to

$q_1 = F_1(q_1,q_2)$
$q_2 = F_2(q_1,q_2)$.