# How far can a particle travel from its origin if it exhibits self-avoiding Brownian motion in two-dimensions?

Let's say I have a point-like Brownian particle undergoing two-dimensional diffusion on an infinite plane with the caveat is that the particle can never return to a coordinate that it previously visited (these coordinates become fully reflecting to the particle). As such, the particle may 'trap' itself if its trajectory generates a closed loop.

Assuming the Brownian particle has some diffusion coefficient, $D$, is there a known mean and variance for the maximum distance the particle will travel from its origin before trapping itself?

For the discretized / random-walk version of this question - Is there a known result for the mean maximum distance from origin (and associated variance) for a random walker on an infinite two-dimensional lattice of some degree $m$?

Edit - To try to better clarify what I mean by the particle being 'trapped', I would define this as the state where the furthest point from the origin in the particle's trajectory is fixed due to the particle being unable to escape from a closed curve. I don't mean to imply that the point-like Brownian circle is "frozen in place", akin to a random walker where all edges from its position connect to previously occupied vertices. I fail to see how this situation is possible without some sort of discretization.

In response to Nate Eldredge's comment, there are certainly some serious issues with defining how the Brownian particle acts at the reflecting boundary its trajectory generates. Loosely I would like to say that the particle reflects at a random angle, and I would appreciate any help with making this better defined. Regarding the second point, that "scaling and the Blumenthal 0-1 law would suggest that the trapping happens immediately", does this also hold true if we assign the Brownian particle some drift parameter $\beta$?

In any case, the discrete or lattice version of this question (mentioned above) is very interesting to me.

• You should ask the question for the lattice version only--- the Brownian version can't trap itself. – Ron Maimon Jul 29 '11 at 2:18
• @Ron Maimon, can you help me understand your point? If the Brownian particle can interact with a reflecting boundary, why wouldn't it be able to generate a closed loop and trap itself inside by reflecting there? – Rob Grey Jul 29 '11 at 2:27
• I'm not sure that your question is meaningful as stated. First, it's typically hard to define what it means to "reflect" off a boundary which is rough, and Brownian sample paths are about as rough as they come. So I think your process may not be well-defined. Second, if it does have a positive probability of "trapping itself" (whatever that means), scaling and the Blumenthal 0-1 law would suggest that the trapping happens immediately, almost surely. – Nate Eldredge Jul 29 '11 at 2:42
• You can find the answer to the trapping questions numerically by simulating an infinite non-trapping walk, and looking at the probability cost of enforcing no-trapping at each time-step (I did this recreationally once, for the getting the asymptotic number of self-avoiding walks of length n-- it gives the wrong answer). Maybe you can get a different universality class for random paths which are allowed to collide, but not cross. It is possible that this condition makes sense, even if the original question is not the right one. – Ron Maimon Jul 30 '11 at 5:10
• However you are defining the process, it can't be done in a scale invariant way. If W(t) is a BM then so is $a^{-1}W(a^2t)$. So, if T was the first time that it 'traps itself' then T has the same distribution as $a^2T$. This is only possible for $T=0$ or $T=\infty$ almost surely. – George Lowther Jul 31 '11 at 16:40

As per Ron Maimon's suggestion, I did a little simulation for the lattice version, counting the number of paths that get trapped at or before $n$ steps. For example, here is a trapped path of 45 steps: Because there is a positive probability at any point of forming a shape like the letter 'G' (if there is sufficient room), the probability of trapping goes to 1 as $n \rightarrow \infty$. For $n=100$, the probability is already over $\frac{3}{4}$. 