3
$\begingroup$

Let $A\subset\mathbb{R}^n$ be an bounded open convex connected set (possibly with some regularity assumptions).

Now we consider $B_1$ to be a Brownian motion conditioned to stay in $A$ and $B_2$ a Brownian motion reflected on the boundary of $A$.

Is it true that $B_1$ is "better concentrated" than $B_2$?

This vague statement seems to be intuitively appealing (at least for me :)).

I have a few ideas how to formalize "better concentrated" but since I cannot prove (or refute) any of them I prefer to leave it unspecified. I would happy to see any "descent" notion here.

$\endgroup$

1 Answer 1

5
$\begingroup$

Here is one way to formalize "concentrated": for large time, the density of the conditioned process converges to the top eigenfunction of the laplacian with Dirichlet boundary conditions, while the reflected one converges to the uniform density on A. The former puts less mass ``near the boundary'' (take A= square as example). More formally, for a tubular neighborhood of width $\epsilon$ around $\partial A$, the uniform (=reflected) will put mass proportional to $\epsilon$ while the Dirichlet one (=conditioned) with put mass proportional to $\epsilon^2$.

$\endgroup$
1
  • $\begingroup$ Yes. You are right. Still, I was hoping for something more. E.g. that the variance of the conditioned process (or it's marginals) is smaller. Does it follow? Or even further, to get that the stochastic dominance (I am not sure if it is true though). $\endgroup$ Nov 12, 2013 at 11:54

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.