4
$\begingroup$

Consider the square grid $$\{-n, -n+1, ..., n-1, n\} \times \{-n, -n+1, ..., n-1, n\} \subset \mathbb{Z}^2$$ As usual, connect $(i,j)$ to $(i',j')$ with an edge if $|i-i'|+|j-j'|=1$.

We randomly specify the environment as follows: At each vertex we designate one of the 4 outgoing directions to have probability 1/2, and the remaining 3 outgoing directions to have probability 1/6. The choice of which outgoing edge gets probability 1/2 is done uniformly and independently at random for each vertex.

Then we consider the random walk starting at $(0,0)$ that moves from each vertex in each direction with the chosen probabilities $1/2$ or $1/6$.

I want to show that for n large, if I fix a random environment and start the random walk at $(0,0)$, then the probability that the random walk hits the top edge of the grid before any of the other edges is very likely to be close to 1/4.

$\endgroup$
  • 2
    $\begingroup$ Is this supported by numerical experiments? $\endgroup$ – Douglas Zare Jan 6 '17 at 9:50
  • 3
    $\begingroup$ This is undoubtedly true but is a well-known open problem, dating back at least to kalikow's paper projecteuclid.org/euclid.aop/1176994306. The walk you describe has the property that every one-dimensional projection is recurrent (by yhe Zerner-Merkl 0-1 law), but I do not thing anything is known rigorously on exit measures. $\endgroup$ – ofer zeitouni Jan 6 '17 at 19:06
  • $\begingroup$ Since it seems the problem is hard, is it at least possible to show that the probability that the random walk hits the top edge of the grid before any other edges is very likely to be at least 1/10^10? $\endgroup$ – Alex Jan 6 '17 at 20:05
  • 1
    $\begingroup$ uniformly in n? seems to me hard. $\endgroup$ – ofer zeitouni Jan 6 '17 at 20:50
  • $\begingroup$ Does the problem become easier if I ask merely if there exists a vertex (i,j) (which may depend on the random environment) such that the probability that the random walk starting at (i,j) hits the top edge of the grid before any of the other edges is very likely to be close to 1/4? $\endgroup$ – Alex Jan 10 '17 at 17:05
3
$\begingroup$

Your random walk can be thought of as a isotropic diffusion with a random iid drift. This is very close to the situation studied here: http://link.springer.com/article/10.1007/s00222-005-0477-5 (ok, they are replacing the discrete model with a continuum one, but this isn't really important and you can find previous work on the discrete model in their references if this makes a difference to you).

Unfortunately, that paper is for dimensions $d>2$ and is perturbative: the drift is required to be smaller than $\delta$ for some very tiny $0<\delta\ll1$. Even in the perturbative regime, the two dimensions is critical in some sense and difficult to analyze. I know some colleagues who are trying to analyze the perturbative regime in $d=2$ and claim to have results "coming soon". This would answer your question if $1/2$ was replaced by $1/4+3\delta$ and the other edges were $1/4-\delta$.

Analyzing the situation outside the perturbative regime seems to be a hugely difficult task and no one knows what to do.

$\endgroup$
  • 2
    $\begingroup$ in $d\geq 3$, the exit measure for the discrete model in the perturbative regime is treated by Bolthausen and myself arxiv.org/abs/math/0607192. We believe the result goes over to d=2 but it will be a while before this is written (maybe we are the colleagues mentioned by user5678 :-). As he points out, outside the perturbative regime the problem is widely open. $\endgroup$ – ofer zeitouni Jan 6 '17 at 19:08
  • $\begingroup$ Yes Ofer, you and Erwin are the colleagues. Happy New Year! $\endgroup$ – Scott Armstrong Jan 7 '17 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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