It is often said that the Brownian motion for a particle in a box, thus a finite domain, is described by a uniform probability distribution in the longtime limit. One may easily imagine this maybe intuitively, but is there actually an easy way of showing this? How is it that even though there are boundaries, no bias is introduced in the longtime limit of where the particle may be in the box?
If one assumes the discrete case, so the box is populated with only a finite number of points the Brownian particle can occupy, does the above statement still hold?
Please feel free to give references that you see fit for such questions, I imagine these are all solved problems and that's why most people often state them using the adverb trivially.
Best Answer
I decided to run a simple simulation of a one-dimensional random walk on lattice with closed boundary condition. To implement the boundary conditions, I basically forced to "turn back" the particle every time it hits the boundaries of the segment.
I took every lattice site to be an integer, and the boundaries to be at $[-L,L]$. This is the result for $L=50$ and $10^8$ steps:
So, it looks like in this simple lattice model the distribution is uniform everywhere except from the very boundaries of the segment, i.e. it is uniform in $(-L,L)$.
I don't know if this still holds in a continuous model, also because it becomes more tricky to do the simulation in this way, because the particle can never really hit the boundary, so it is possible that the result depends on how the boundary conditions are implemented exactly.
Relevant literature
This problem is commonly known as reflected brownian motion (RBM). Many articles can be found regarding RBM in open regions (like $[0,\infty)$), but we are more interested in RBM in closed regions.
About this, I found this relevant article:
The Stationary Distribution of Reflected Brownian Motion in a Planar Region - . M. Harrison, H. J. Landau, L. A. Shepp
It gives an explicit -and rather complicated- expression for the stationary probability distribution. The article is quite technical, but from what I understand the stationary distribution is not uniform.