2
$\begingroup$

A few years ago, when I was working on first-passage percolation problems, I thought about the following problem. Recently it came back to my mind.

Consider, for some $\delta=n^{-1}>0$, the grid $\delta\mathbb{Z} \times \delta\mathbb{Z}$ and associate with each edge $e$ an independent random weight $w(e)$. The edge weights may be assumed to be positive and drawn from a continuous distribution. Then, with probability one, there exists a unique path $p^\delta$ from, say, $(0,0)$ to $(1,0)$ that minimises the sum of edge weights along the path. The distribution of $p^\delta$ defines a measure $\mu^\delta$ on the space of non-intersecting paths in $\delta\mathbb{Z} \times \delta\mathbb{Z}$ from $(0,0)$ to $(0,1)$.

Question: Is anything known about $\lim_{\delta\to 0}\mu^\delta$?

Intuitively, if the limit exists, it could be interpreted as a measure on the space of continuous curves in $\mathbb{Z}^2$ from $(0,0)$ to $(1,0)$. I am interested in any references, partial/conjectural results, or simulations.

$\endgroup$
2
  • $\begingroup$ If that is last passage percolation instead, such a limit is related to directed polymers in random media and with KPZ and stochastic heat equations. Maybe a google scholar search would be helpful here. $\endgroup$ Mar 19, 2013 at 17:51
  • $\begingroup$ @Leonid Petrov: Thank you for your comment. I did have a look at the literature, but the problem I'm interested it doesn't seem to have received a lot of attention in the past. $\endgroup$
    – Eckhard
    Mar 19, 2013 at 18:29

1 Answer 1

1
$\begingroup$

Due to what's written in the text of Benjamini, I would expect it to converge to a straight line, (unless, perhaps, if your law has too heavy log-tails, I'm not sure): see http://www.wisdom.weizmann.ac.il/~itai/randomplanar2.pdf, section 2: «Large balls converge after rescaling to a convex centrally symmetric shape...»

(If this shape in strictly convex, what seems to me more or less natural to expect, then the geodesic for the limit metric is unique, and is the straight line.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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