is allowed, but no or etc is allowed. (Often branching processes are restricted to one type, but you can have multi-type ones.) As for tricks, I know two, and you’re both doing the first one already. That is, find the embedded discrete time Markov chain by ignoring the time it takes to get to the next state, and just look at transition probabilities of going from j to i at the next change. In all cases so far, j has been i-1 or i+1, but it doesn’t have to be.

The other trick I know is probability generating functions, which I bet you do know now I’ve reminded you!

One way in which branching processes can be more general than the reaction networks I’ve seen is that they can have an infinite number of reactions, eg

for every n. (Might be handy for fish which can lay millions of eggs.) This means that the process can ‘explode’, that is, make an infinite number of particles in a finite time. So a ‘non-explosion’ hypothesis is assumed. (I think finite number of reactions implies the hypothesis.) I mention this because of the technical difficulty in the recent blog post The Large-Number Limit for Reaction Networks (Part 3) about passing a limit through a derivative. I bet an exploding process would fail here! But I thought that if you looked at how the ‘non-explosion’ hypothesis is used in branching theory it might help prove the theorem.

]]>It feels good to have this nagging problem settled.

Thanks. By the way, one way to show certain Markov processes have no equilibrium state where the probability of having population 1 vanishes is to copy the argument that if

on some region and is zero on the boundary of the region, then (under certain side conditions). The idea is that the Laplace equation

says that at any point is the average of its values on a small sphere around that point. So, if it’s zero on the boundary of the shape it should be zero everywhere. In the 1-dimensional discrete setting the Laplace equation becomes

Of course, this doesn’t prohibit the solution for but that solution isn’t ‘normalizable’. (That’s part of what I meant by ‘side conditions’.) Any normalizable solution has to be zero.

About self-adjointness, you’re right, but maybe we just need to be using the “right” scalar product in order for the Hamiltonian to become self-adjoint.

Right, there’s a theory of generalized Dirichlet operators on measure spaces, which allows you to adjust the measure and thus the inner product:

• M. Fukushima, *Dirichlet Forms and Markov Processes*, North-Holland, Amsterdam, 1980.

In the book, I wanted to apply this theory to the case of continuous-time Markov process with a finite space of states. If we allow ourselves to use a measure other than counting measure on we get new concepts of ‘self-adjoint’ and ‘infinitesimal stochastic’ operators on and It would be nice, and not too hard, to work out the consequences. But I never got around to it!

There’s also a further generalization of Dirichlet operators that allows them to have a skew-adjoint part as long as it’s dominated by the self-adjoint part:

• Zhi-Ming Ma and Michael Röckner, *Introduction to the Theory of (Non-Symmetric) Dirichlet Forms*, Springer, Berlin, 1992.

Concerning the model with transitions and occurring at equal rates, I had asked:

Your sentiment that eventual extinction will occur with probability 1 sounds right to me; do you have any idea of how to prove it?

I think I know how to do this now. For any population size , consider the next population size that you get as soon as a fission or death event happens. Since the rates are equal, this next population size is either or with equal probability. In this way, we obtain an old friend, namely the ordinary random walk on the integers, modulo the caveat that the walk never leaves the origin once it has arrived there. Due to the recurrence of this random walk, we end up at the origin eventually with probability one, no matter where we started out with. In other words, the above model ends up at extinction with probability one. If you like theatrical analogies, think of this as a mathematical proof showing that doomsday will strike us some day!

John wrote:

But now you are talking about similar things for a situation with infinitely many states… and in some examples, for Hamiltonians that are not self-adjoint.

Countably many states shouldn’t be a problem — people study Laplace operators on countable graphs for example in geometric group theory. About self-adjointness, you’re right, but maybe we just need to be using the “right” scalar product in order for the Hamiltonian to become self-adjoint.

John wrote:

I’d try argue something like this. Suppose that when the population goes to 1 it can never go back to 2, but has a nonzero probability per time of going to 0. Then in any equilibrium the probability of having population 1 must be zero [..]

Beautiful! It feels good to have this nagging problem settled.

]]>Over here, Tobias wrote:

But in order for your statements here about the rate constants to make sense, I assume that you were meaning to write down these two transitions:

Right.

This is still not what I had in mind with “death rate per individual proportional to population size.”

Right. I didn’t see the “per individual” part.

But it’s an interesting model to consider, due to the phenomenon that no new individuals can be born once the whole population has died out. If the reaction rates are and , then the rate equation looks like this:

Since the deficiency is the deficiency zero theorem doesn’t apply here, and so we have to analyze the equilibria by hand.

Right. That seems pretty easy in this case. But if you ever get a harder example with deficiency 1 or more, you might look at the ‘deficiency one theorem’. It’s in here:

• Martin Feinberg, Chemical reaction network structure and the stability of complex isothermal reactors: I. The deficiency zero and deficiency one theorems, *Chemical Engineering Science* **42** (1987), 2229–2268.

and probably also somewhere in these great free online notes:

• Martin Feinberg, *Lectures On Reaction Networks*, 1979.

The deficiency one theorem is more complicated than the deficiency zero theorem, with weaker implications, but still powerful. Despite its name it applies to some cases of deficiency higher than one, too!

As you pointed out, we need in order for an equilibrium to exist, so let’s assume this and put for the sake of simplicity. Then

anyconstant is an equilibrium solution of the rate equation — but only $P=0$ is complex balanced, as you can see by noting that the complex $A+A$ gets produced in the case $P>0$, but never gets destroyed, thereby violating complex balance.

Right.

The master equation looks like this:

I’m not quite sure what to do with this. The Anderson–Craciun–Kurtz theorem won’t tell us much, since we only have a trivial complex balanced equilibrium of the rate equation.

Right, that theorem is useless here.

Your sentiment that eventual extinction will occur with probability 1 sounds right to me; do you have any idea of how to prove it?

I don’t know the tricks for solving this sort of problem, but some people do. There’s a whole industry of showing that random walks of various kinds will almost surely hit some set.

Side question: the Hamiltonians that appear in these master equations vaguely resemble Laplace operators on Cayley graphs, with respect to which the equilibrium distributions are harmonic functions. Is there a connection here?

Continuous-time Markov processes where the Hamiltonian is a Dirichlet operator—both infinitesimal stochastic and self-adjoint—are basically generalizations of the heat equation. For example, when there are finitely many states, a Dirichlet operator is the same as the Laplace operator associated to some *weighted* graph. This is Problem 39 in Section 22.1 of the book Jacob and I are writing.

But now you are talking about similar things for a situation with infinitely many states… and in some examples, for Hamiltonians that are not self-adjoint.

I’m concerned about the possibility of the population dropping below 2, which for us implies guaranteed extinction. I wonder whether this means that we can’t expect our master equations to have any non-trivial equilibrium solutions.

It sounds like in this situation you should be able to prove there aren’t any nontrivial equilibrium solutions, at least if a few reasonable conditions hold. I’d try argue something like this. Suppose that when the population goes to 1 it can never go back to 2, but has a nonzero probability per time of going to 0. Then in any equilibrium the probability of having population 1 must be zero (since otherwise the probability of having population 0 would grow).

But if there’s a nonzero probability per time of the population going from 2 to 1, this implies that in any equilibrium the probability of having population 2 must be zero ((since otherwise the probability of having population 1 would grow).

And so on.

]]>True… I have to admit that I have secretly been thinking in terms of a different model, namely one in which competition is replaced by a “death rate” per individual which is proportional to current population size. This results in a slightly different kind of Hamiltonian with which the population can die out completely.

Right.

(It probably won’t arise from a Petri net, though.)

It does, actually. We just need one species (for ‘amoeba’) and two transitions

with exactly the same rate constant if you want the rate equation to have equilibrium solutions. (Otherwise all solutions of the rate equation will go to infinity (if births outpace deaths) or zero (if deaths outpace births.)

I believe that when the rate constants are equal, the master equation will have the property that with probability 1, the population eventually reaches zero and never bounces back. I haven’t proved this.

Or, if you want something a bit less delicately balanced, you can do something like this:

This shows up in models of fish populations, where the last process represents fishing. For many choices of rate constants the rate equation should have a unique nonzero equilibrium solution… but also zero will be an equilibrium solution. I believe that in these cases, the master equation will have the property that the population eventually reaches zero with probability 1, and never bounces back.

]]>This is still not what I had in mind with “death rate per individual proportional to population size”, since the probability of an individual to die is a constant rather than proportional to population size. But it’s an interesting model to consider, due to the phenomenon that no new individuals can be born once the whole population has died out. If the reaction rates are and , then the rate equation looks like this:

Since the deficiency is the deficiency zero theorem doesn’t apply here, and so we have to analyze the equilibria by hand. As you pointed out, we need in order for an equilibrium to exist, so let’s assume this and put for the sake of simplicity. Then *any* constant is an equilibrium solution of the rate equation — but only $P=0$ is complex balanced, as you can see by noting that the complex $A+A$ gets produced in the case $P>0$, but never gets destroyed, thereby violating complex balance.

The master equation looks like this:

I’m not quite sure what to do with this. The Anderson-Craciun-Kurtz theorem won’t tell us much, since we only have a trivial complex balanced equilibrium of the rate equation. Your sentiment that eventual extinction will occur with probability 1 sounds right to me; do you have any idea of how to prove it? It seems related to the recurrence of a one-dimensional random walk.

Side question: the Hamiltonians that appear in these master equations vaguely resemble Laplace operators on Cayley graphs, with respect to which the equilibrium distributions are harmonic functions. Is there a connection here?

By the way, some friends and I are currently studying mating systems, and we will hopefully be able to put this stuff to good use! I’m concerned about the possibility of the population dropping below 2, which for us implies guaranteed extinction. I wonder whether this means that we can’t expect our master equations to have any non-trivial equilibrium solutions.

]]>In any case, I understand now that infinite absorbing Markov chains can behave quite differently from finite absorbing ones: the latter can’t have non-trivial equilibria, while the former can! Introducing a finite population cap and then letting it tend to infinity will not detect such an equilibrium.

(In order to find it, I guess that one will have to look at the second largest eigenvalue of the transition matrix and see whether it converges to 1 as the population cap tends to infinity.)

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