Hi,
I've started studying brownian motion, and gathered some books on the subject but
something looks odd to me : All of the presentations I've seen this far consider the continuity of the brownian motion as an axiom.
Well, if you recast the brownian motion in the wider setting of Levy processes with stable independent increments on non-overlapping intervals then this is a very special properties of the brownian motion. Considering that, there should be a way to prove this almost sure continuity property as a consequence of the other axioms, namely :
- W(0) = 0.
- For all $0 \le t_1 \le t_2 \le t_3 \le t_4$, $W(t_2) – W(t_1)$ and $W(t_4) – W(t_3)$ are independent random variables.
- For all $0 \le t_1 \le t_2$ , $W(t_2) – W(t_1)$ is normally distributed with mean 0 and variance $\sigma^2\,(t_2 – t_1)$.
The third axiom being the one which is special to brownian motion. The normality condition entering it appears to be sufficient to tame any wild excursion from continuity.
What motivates my question is that when doing a numerical simulation on equally spaced time intervals using only these axioms, the continuity property is rather ovious. Especially when one refines the simulation by taking smaller and smaller time intervals.
While the result seems experimentally obvious I can't find anything that doesn't state it as an axiom.
A typical proof could be to look at the probability that the brownian motion on $[0, 1]$ doesn't get too far from the nodes of the simulation in between them.
For instance uniform continuity could be :
The probability of a brownian motion escaping the nodes farther than say $\delta > 0$ is lower than any $\gamma > 0$ provided that the steps of the simulation is smaller than some $\epsilon$ depending on both $\delta$ and $\gamma$.
Does that seems sensible to anyone ? Does anyone know of such result in the literature ?
Any help or reference will be greatly appreciated.
Best Answer
This is a partial answer but shows the kind of subtlety that makes the continuity of Brownian motion non trivial. If you try and take the first three axioms of Brownian motion and try to prove that the process has continuous paths using a central limit theorem argument what you get is that on a probability space $(\Omega,\mathbb{P})$, that $\forall t > 0$
$\mathbb{P}(B_t\ is\ discontinuous\ at\ t) = 0 $
This means that there are null sets $\mathcal{N}_t \subset \Omega$ such that if $\omega \not\in \mathcal{N}_t$ then $B_t(\omega)$ is continuous at time $t$. And here is the delicate part. This does not imply that there exists any single $\omega \in \Omega$ such that $B_t(\omega)$ is continuous for all $t$. I have seen this property called stochastic continuity in some places.
What you usually want is a single null set $\mathcal{N}$ so that for $\omega \not \in \mathcal{N}$ $B_t(\omega)$ is continuous for all $t \ge 0$.
Of course the usual constructions of Brownian motion do take care of this subtlety but some times without mentioning it.