Many Brownian motions moving together - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:38:34Z http://mathoverflow.net/feeds/question/85266 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85266/many-brownian-motions-moving-together Many Brownian motions moving together Piotr Miłoś 2012-01-09T16:20:33Z 2012-01-24T10:05:10Z <p>Let $(B^i),\:{{i=1,\ldots,n}}$ be a set of independent Brownian motions. By $(X^i)$ we denote $(B^i)$ conditioned on the event</p> <p>$|B^i_t-B_t^{i+1}|\leq 1,\quad \forall_{1\leq i\leq n-1}, \forall_{t\geq 0}.$</p> <p>(this is a $0$-measure event but one can make the definition correct by conditioning on a finite time horizon and them sending it to $\infty$).</p> <p>Is such process known? What is its behaviour?</p> <p>My <strong>conjecture</strong> is that:</p> <ol> <li>The mass centre of the process, i.e. $Z_t := \frac{1}{n} \sum_{i=1}^{n} X^i_t$ behaves as $n^{-1/2} W_t$ ($W_t$ is a BM again).</li> <li>The process of fluctionations around $Z_t$, i.e. $\hat{X}^i_t := X^i_t - Z_t$ is well concentrated (e.g. $\sup_{i} \hat{X}^i_t \sim \sqrt{\log{n}}$).</li> </ol> <p>(I am much less sure of 2 then 1).</p> <p>These predictions come from considering a very crude version of the model as follows. We let the Brownian motions to move unconstrained for time $[0,1]$ then we calculate their mean $z_1 := n^{-1} \sum_{i=1}^n B^i_1$ and set all process to start from this position, i.e. $B^i_{1+}:= z_1$ . We repeat this procedure on each interval $[n,n+1]$.</p> <p>The further <strong>questions</strong> would be:</p> <ol> <li><p>Can this process be described as a diffusion. A standard way is to perform a $h$-transform but one needs to find a harmonic function first. I tried this but beyond $n=2$ calculations become messy. </p></li> <li><p>Does this process have connections to the random matrices theory? E.g. by defining $Z^i_t:= Z^i_t+i$ one can regard this process as the Dyson Brownian motion with additional conditions $Z^{i+1}_t - Z^{i}_T \leq 2$.</p></li> </ol> http://mathoverflow.net/questions/85266/many-brownian-motions-moving-together/85297#85297 Answer by fedja for Many Brownian motions moving together fedja 2012-01-09T23:05:52Z 2012-01-09T23:05:52Z <p>You can split. I'll do it for $B_1,B_2$. Let's go in small time steps to avoid talking about stochastic differential equations and other stuff I don't really know. Note that the increments $\xi$ and $\eta$ of $B_2-B_1$ and $B_1$ at each step are correlated Gaussians with certain covariance. Now write $\eta=a\xi+\gamma$ where $\gamma$ is a Gaussian orthogonal to and, thereby, independent of $\xi$. Note that then you have free Brownian motion controlled by $\gamma$ that takes care of the overall drift combined with the bounded motion controlled by $((1+a)\xi,a\xi)$, which is conditioned to stay in some domain. The same can be done with any number of $B$'s. Note that the orthogonal projection of $(1,0,\dots,0)$ to the orthogonal complement of the plane $\sum_j x_j=0$ is of length $n^{-1/2}$ confirming your first conjecture. The second conjecture then says (after a linear transformation) that the standard $n-1$-dimensional BM conditioned on staying in a certain parallelepiped stays fairly concentrated. I do not see it immediately but that may be well-known to probabilists. </p> http://mathoverflow.net/questions/85266/many-brownian-motions-moving-together/85406#85406 Answer by Piotr Miłoś for Many Brownian motions moving together Piotr Miłoś 2012-01-11T11:26:24Z 2012-01-24T10:05:10Z <p>In the case of $n=3$ (by the procedure outlined by fedja) the problem boils down to study two dimensional BM in the yellow domain. Let now $G$ be a group generated by the reflections in the blue lines, then the transition density of the BM killed on the hitting boundary, $h_t(x,y)$, is</p> <p>$h_t(x,y) := \sum_{g\in G} (-1)^{r(g)}p_t(x,g(x)),$</p> <p>where $p_t(x,y) = (2\pi t)^{-1} e^{-|x-y|^2/(2t)}$ is the transition density of the BM in $\mathbb{R}^2$ and $r$ is "the rank" of $g$ (to be explained in a moment). </p> <p>I know almost nothing about the group theory but it seems to me that $G$ is what is called "a reflection group" or a special case of a Coxeter group. $r(g)$ as far as I understand is the length of the shortest way in which $g$ can be represented using the generator only. </p> <p>So, the questions are:</p> <ol> <li><p>Is there a way of presenting the above sum in by a closed formula.</p></li> <li><p>What is asymptotic behaviour of $h_t(x,y)/\int h_t(x,y)$?</p></li> </ol> <p><img src="http://www.mimuw.edu.pl/~pmilos/domain.png" alt="Domain"></p> <p>P.S. It is probably not very common to answer own questions but I think it is more then a comment and I do not know how to paste images into comments. </p> <p>P.P.S. Following Omer's comment, it should be not hard to prove that</p> <p>$l(y):=\lim_{t\rightarrow +\infty} h_t(0,y)/h_t(0,0)$ </p> <p>exists and $l$ is the leading eigenvalue of the Laplacian in the considered domain.</p> http://mathoverflow.net/questions/85266/many-brownian-motions-moving-together/86142#86142 Answer by Omer for Many Brownian motions moving together Omer 2012-01-19T21:32:47Z 2012-01-19T21:32:47Z <p>The first claim is correct. As Fedja and Piotr suggest, the key is a change of basis. The centre of mass $Z_t$ has law $W_t/n$ where $W$ is B.M. independent of the differences. Since the condition only affects the differences, the claim follows.</p> <p>To estimate fluctuations around $Z_t$, you need to estimate the leading eigenfunction for the Laplacian on the domain given by the constraints $|B^i-B^{i+1}|&lt;1$ -- a parallelogram. In the case $n=3$ this is Piotr's figure (with the additional constraint $|B^1-B^3|&lt;1$).</p>