2
$\begingroup$

Let $M$ be the real square matrix of a typed branching process, such that $M_{ij}$ is the expected value of offspring of type $j$ emanating from type $i$.

We know that if the first eigenvalue if $M$ is smaller than 1, then all types will be extinct, and if it is larger than 1, then with positive probability some types won't get extinct.

Is there some interpretation of the eigenvalue other than that? (its actual value.)

$\endgroup$
2
  • 1
    $\begingroup$ The second "smaller" should be "larger". $\endgroup$ Commented Apr 22, 2013 at 7:22
  • 1
    $\begingroup$ if it is larger than 1, then there are types that won't get extinct... is not accurate: rather, there is a positive probability that some types will not get extinct. $\endgroup$
    – Did
    Commented Apr 22, 2013 at 7:57

1 Answer 1

1
$\begingroup$

You can see this process as a dynamical system or a Markov chain without normalization. If the matrix is irreducible, starting from every initial distribution of number of individuals $w_0$, the process will "converge" (in some suitable sense) to $w_{k}=\alpha \lambda^k v$, for some $\alpha\in\mathbb{R}$, and $(\lambda,v)$ the Perron eigenpair.

Thus, in the stationary limit, the ratios among the number of individuals of different types at each time step $k$ $(w_k)_i/(w_k)_j$ are the ratios of components of the Perron vector $v_i/v_j$, while the number of individuals is multiplied by $\lambda$ at each iteration. So $\\lambda $ is a growth factor for the number of individuals at each iteration.

$\endgroup$
2
  • $\begingroup$ Conditionally on non-extinction. $\endgroup$
    – Did
    Commented Apr 22, 2013 at 7:58
  • $\begingroup$ You are right, sorry for forgetting this part! $\endgroup$ Commented Apr 22, 2013 at 9:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .