Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

As the title (hopefully) suggests, I've been trying to prove (or disprove) the existence of a limiting distribution for a certain projection in a statistical physics model. I'll give the details of the model below, but if you are reading this, I'm particularly interested in proofs that seem 'quantifiable' (e.g. maybe implicitly has some estimate of distance to the limiting distribution). I think it is unlikely that anyone reading this has familiarity with the below model, so reasonable-sounding references would also be great.

The model is the infinite 'East Model', discussed in e.g. 'The Asymmetric One-Dimensional Constrained Ising Model: Rigorous Results'. For a short description, the process is on (0,1) labellings of the positive integers, and at time 0, all labels are 0 except for 1, which is labeled 1 (and for convenience we write that 0 is also labeled 0). Each positive integer k has its own Poisson-1 clock, and when the clock goes off, the label does nothing (if the label of k-1 is 0) or flips with probability 0.5 (if the label of k-1 is 1).

At each time, there is a highest non-zero entry. I am interested in the vector consisting of the labellings of the n entries below the highest non-zero entry. This vector isn't by itself a Markov chain, and if it has a limiting distribution, it isn't quite a Bernoulli process (which is the limit for the East process itself). Any insight into existence of a limit, convergence, and how it might be affected by starting in some different initial state with finitely many 1's would be welcome.

share|cite|improve this question
Nice problem. You write that "This vector isn't by itself a Markov chain": why? – Did May 6 '10 at 0:25
You can lose the top 1 and move one step back getting a new bit on the bottom whose value is very unlikely to be determined by the current state alone even in the sense of Markov chains. On the other hand, most of the time you go forward at linear speed thus making any attempts to play with the bottom bits quite inefficient for controlling the distribution of the top ones. I have put this question on my list of things to think of but, alas, my free time right now is very limited :-(. – fedja May 6 '10 at 22:52

1 Answer 1

you might get more insight with pictures: the whole process (at time when something happened at the east-end) whole process and the last $50$ non-zero sites

enter image description here

share|cite|improve this answer
Hi Alekk, thanks for the reply. I've never known how to interpret 'geometric' pictures to conclude anything about limiting distributions - any comments on this? I have run a number of simulations looking at the long-time distribution of various linear functionals to see if they have limits, since the computer can draw a bunch of empirical CDFs on top of each other, and the answer seems to be yes for the random functionals my computer has turned up... – passing by May 4 '10 at 19:31
Do you have any example of a similarly defined process where there is no convergence ? Is it reasonable to expect a nice description of the limiting distribution (Boltzman distribution of a concrete energy function for example?) – Alekk May 4 '10 at 20:53
I haven't been able to think of anything that is similar where there isn't convergence. I have a few silly examples of pieces of infinite Markov chains where there isn't convergence, but they are very made-up and lack convergence for some straightforward projections. Essentially, these sorts of chains sometimes don't converge if labellings close to 0 are both persistent and have long-term and long-range influence; neither of those are true here. As for the limiting distribution, I expect it to be worse than the Derridas' process, which is already hard to find (though solved now). – passing by May 4 '10 at 22:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.