I never got around to it. But it’s easy to create a number of variants of the rate equation and master equation, depending on whether one is interested in:

• ordered k-tuples of not-necessarily-distinct inputs

• unordered k-tuples of not-necessarily-distinct inputs

• ordered k-tuples of distinct inputs

• unordered k-tuples of distinct inputs

In my notes I’m always considering ordered k-tuples of distinct inputs. These are counted by the falling powers This case turns out to be most easily described using annihilation and creation operators. However, to count unordered k-tuples of distinct inputs, you just divide by . So, you can just divide each term in the Hamiltonian by a suitable factor.

]]>We shouldn’t redefine it’s too beautiful to change. If we have a situation where a transition occurs at a rate proportional to because we need to make some choice among the inputs, we can multiply the Hamiltonian by to account for that. I don’t think duels between rabbits occur twice as fast because there are 2 choices of which rabbit might survive.

]]>Also, if operating with a on includes , the number of ways a rabbit can be chosen from n rabbits, then shouldnâ€™t operating with also include the number of ways an output rabbit can be chosen from the inputs (at least in some cases)?

]]>Arjun wrote:

Why are we considering all the s to appear on the left side of all the s in ?

Because each transition in a stochastic Petri net describes a processes where a bunch of things get used simultaneously to create some other bunch of things.

Surely, there can be situations when the order is not- all inputs, then all outputs. Suppose that 2 rabbits meet and fight and one gets killed. The overconfident victor then challenges any one of the remaining rabbits to a duel and one of them then gets killed. The new victor seeing that he might get killed if he further challenges, stops.

That’s true. I would think of this using a Petri net with a set consisting of several states. For example: ordinary rabbits, and ‘overconfident victor rabbits’, and ‘wisely cautious victor rabbits’. I’d also introduce several transitions in , for example:

ordinary rabbit + ordinary rabbit overconfident victor rabbit

overconfident victor rabbit + ordinary rabbit wisely cautious rabbit

You can try to work out the Hamiltonian for such a process.

Or, you can check to see if the Hamiltonian you propose is an infinitesimal stochastic operator. If it is, it describes some random process. If not, there’s something wrong with it.

]]>Surely, there can be situations when the order is not- all inputs, then all outputs. Suppose that 2 rabbits meet and fight and one gets killed. The overconfident victor then challenges any one of the remaining rabbits to a duel and one of them then gets killed. The new victor seeing that he might get killed if he further challenges, stops. If this constitutes the transition, shouldn’t H be ?

Also, if operating with a on includes n, the number of ways a rabbit can be chosen from n rabbits, then shouldn’t operating with also include the number of ways an output rabbit can be chosen from the k inputs?

]]>Yes, thanks!

]]>Shouldn’t it be

]]>

I wasn’t using Rule 2, I was using the commutation relations and then making a typo, which gave .

I think the typos are mostly gone now, thanks to you. The more important problem here is that there are so many ways to use the rules to do these calculations that it’s hard to find the ‘best’ way.

]]>Thanks. I decided to switch to an argument that seems a bit simpler:

]]>In Puzzle 1, the Hamiltonian for catching rabbits is

so the rate of change of the expected number of rabbits caught is

There are many ways to use our rules to evaluate this. For example, Rule 1 implies that

$

but the commutation relations say

so

and using Rule 1 again we see this equals

Thus we have

It follows that the expected number of rabbits grows linearly: