3
$\begingroup$

I would like to know more about the two-dimensional processes derived from Brownian motion by the following stochastic differential equation (in the Ito sense)

$$dX_t = f(X_t) dt + \mathcal{R}(f(X_t)) dB_t$$

where

  • $X_t$ is the two-dimensional stochastic process
  • $f$ is a smooth vector field
  • $\mathcal{R}$ is the linear map which rotates a vector through a quarter turn
  • $B_t$ is a standard one-dimensional Brownian motion

So $X_t$ follows the field lines of $f$ apart from a lateral "shake" whose intensity is proportional to the strength of the field $f$ at $X_t$.

In particular I am curious about what is known when $f$ is conservative (i.e. $f = \nabla g$ for some real-valued field $g$). In this case, can the trajectory of $X_t$ self intersect?

References to further properties of this process would also be appreciated.

$\endgroup$

1 Answer 1

3
$\begingroup$

In general, it can self-interesect, as can be seen in the example of the simple rotation $$ f(x,y) = (-y,x)\;,\qquad \mathcal{R} f(x,y) = (x,y)\;. $$ In this case, the deterministic solution turns around the origin at unit speed and there is no reason to believe that there won't be any self-intersection for $t > 2\pi$.

In the case where $f = \nabla g$ for some globally defined smooth function $g$ you cannot have self-intersections except in the trivial case when you start on a critical point for $g$, simply because $g$ is strictly increasing along solutions. (Assuming that your stochastic differential is interpreted in the Stratonovich sense.)

$\endgroup$
3
  • $\begingroup$ I'm more interested in the Ito sense. Does that still imply $g$ strictly increases? It seems unlikely to me. For example it seems that the process should be able to "cross over" from one side of a ridge to another. But even if $g$ is not strictly increasing, can we can get non-self intersection from other considerations? $\endgroup$
    – Tom Ellis
    Feb 10, 2014 at 18:10
  • $\begingroup$ (I edited my question to specify that I'm more interested in Ito, and corrected "irrotational" to "conservative") $\endgroup$
    – Tom Ellis
    Feb 10, 2014 at 18:18
  • 2
    $\begingroup$ If the equation is interpreted in the Itô sense, then I would think that paths can intersect in general. $\endgroup$ Feb 10, 2014 at 18:51

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.