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

$|B^i_t-B_t^{i+1}|\leq 1,\quad \forall_{1\leq i\leq n-1}, \forall_{t\geq 0}.$

(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 $\infty$).

Is such process known? What is its behaviour?

My conjecture is that:

  1. 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).
  2. The process of fluctionations around $Z_t$, i.e. $\hat{X}^i_t := X^i_t - Z_t$ is well concentrated (e.g. $\sup_{i} \hat{X}^i_t \sim \sqrt{\log{n}}$).

(I am much less sure of 2 then 1).

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 $[n,n+1]$.

The further questions would be:

  1. 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.

  2. 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 $Z^{i+1}_t - Z^{i}_T \leq 2$.

  • $\begingroup$ A small variation of the problem is to add one more constraint: $|B^1_t -B^n_t|\leq 1$. This should not change the result much but adds additional symmetry. $\endgroup$ Jan 10, 2012 at 18:05

3 Answers 3


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.

  • $\begingroup$ I am not sure if I understand well. Is the $n-1$-dimensional BM (in the second part): $(B_2 -B_1, B_3 - B_2,\ldots, B_n - B_{n-1})$? If, yes, it is not a standard BM, i.e. its components are not independent. $\endgroup$ Jan 10, 2012 at 8:54
  • $\begingroup$ It is rather the motion with coordinates $B_j-\frac 1n\sum_j B_j$, which is the orthogonal projection of the full motion to the plane $\sum x_j=0$. Now, the orthogonal projection of a standard BM is a standard BM again, just in lower dimension. The real catch is the shape of the restriction domain P, which is here a parallelepiped vs. the shape of the desired concentration domain C. You basically want to say that a standard BM conditioned on staying in P stays in C most of the time. Such natural problem must have been studied. I just don't know the literature. $\endgroup$
    – fedja
    Jan 10, 2012 at 11:33
  • $\begingroup$ Thx, now I agree. $\endgroup$ Jan 10, 2012 at 15:42

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.

To estimate fluctuations around $Z_t$, you need to estimate the leading eigenfunction for the Laplacian on the domain given by the constraints $|B^i-B^{i+1}|<1$ -- a parallelogram. In the case $n=3$ this is Piotr's figure (with the additional constraint $|B^1-B^3|<1$).

  • $\begingroup$ Thanks for the answer. You are right. One way of calculating the eigenfunction would be to use the formula in my previous post (as explained in P.P.S.). But, as for a moment we do not know how to handle the formula, may be there are methods of proving some concentration properties of the eigenfunction without finding it explicitly. $\endgroup$ Jan 24, 2012 at 10:10

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

$h_t(x,y) := \sum_{g\in G} (-1)^{r(g)}p_t(x,g(x)),$

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).

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.

So, the questions are:

  1. Is there a way of presenting the above sum in by a closed formula.

  2. What is asymptotic behaviour of $h_t(x,y)/\int h_t(x,y)$?


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.S. Following Omer's comment, it should be not hard to prove that

$l(y):=\lim_{t\rightarrow +\infty} h_t(0,y)/h_t(0,0)$

exists and $l$ is the leading eigenvalue of the Laplacian in the considered domain.


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.