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 :

  1. W(0) = 0.
  2. 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.
  3. 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.

  • $\begingroup$ Define the $\alpha$-Holder constant of a function $f$ on a set $S$ by $C_\alpha(f,S)=\sup_{x,x'\in S}|f(x)-f(x')|/|x-x'|^\alpha$. Now fix $\alpha<1/2$. I think you can prove the result that you want by looking at Brownian motion restricted to $S_n$, the set of multiples of $2^{-n}$ in $[0,1]$. Define $X_n(\omega)=C_\alpha(B_t(\omega)\vert_{S_n}$. Then you can use Borel-Cantelli to show that $X_n(\omega)$ is almost surely bounded. $\endgroup$ – Anthony Quas Feb 6 '12 at 23:52
  • $\begingroup$ Nice ! This is cleanly and succinctly stated. I regard this as a valid answer but I'm wondering why books that contains the line of thought you describe, take continuity of brownian motion as an axiom... Is there a catch ? I can't see any... $\endgroup$ – Samuel Vidal Feb 6 '12 at 23:58
  • $\begingroup$ @AnthonyQuas I do not think you can prove the continuity of Brownian path with probability one without assuming $X(t)$ is continuous as a function at $t=0$. $\endgroup$ – Henry.L Apr 17 '17 at 15:47

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.

| cite | improve this answer | |
  • $\begingroup$ Many thanks. Indeed this address the difficulty at hand. $\endgroup$ – Samuel Vidal Feb 6 '12 at 23:25

The continuity (or more precisely the existence of an almost sure continuous modification) is a direct consequence of your condition 3 by Kolmogorov's continuity criterion(or more generally the Theorem of Kolmogorov-Chentsov). More details can be found e.g in the book of Revuz and Yor. (and, by the way, for standard Brownian motion $\sigma^2$ should be 1).

Concerning your uniform convergence idea you might have a look on Schilder's Theorem (explained e.g. In the book of Dembo and Zeitouni on large deviations).

| cite | improve this answer | |
  • $\begingroup$ Many thanks. I'll look at the reference you give very carefully. $\endgroup$ – Samuel Vidal Feb 6 '12 at 23:27

Searching the web I've found some documents and among them I've stumbled across this one.

Brownian Motion by Peter Mörters (University of Bath).

It supposes as always that the brownian motion is continuous... But looking at Theorem 1.1 page 2 (attributed to Wiener 1923) saying that "Brownian motion exists". The steps of the proof are the following:

  1. Construct a discreet version of the brownian motion on a diadic grid (I mean, by successive refinement, dividing every time interval in two equal parts at each iteration.
  2. Consider the sequence of piecewise linear function obtained at each stage by linear interpolation
  3. Prove almost sure uniform convergence

This seems totally satisfactory to me. Piecewise linear functions are continuous, uniform limit of continuous functions are continuous...

Anyone has an idea why on earth one has to suppose first that brownian motion is continuous ?

Am I missing something ?

[Edit : Ok, I now understand this is not a proof of continuity. I now understand that this is a proof of existence (non-contradiction), hence the name of the theorem. I doesn't prove that any stochastic process verifying the three axioms is indeed continuous. But still, such proof must exists.]

| cite | improve this answer | |
  • 3
    $\begingroup$ You do have to suppose it's continuous. If $B_t(\omega)$ satisfies the axioms of Brownian motion, then defining $\tilde B_t(\omega)$ to be $B_t(\omega)$ if $t\ne U$ and 0 if $t=U$ where $U$ is an independent uniform random variable gives another process that satisfies the (other) axioms of Brownian motion. $\endgroup$ – Anthony Quas Feb 6 '12 at 23:55
  • $\begingroup$ Are you sure ? I think that with your construction, axiom 3 is violated : the increment between say 1/2 and 1/3 although normal of zero mean, doesn't have the prescribed standard deviation. $\endgroup$ – Samuel Vidal Feb 7 '12 at 0:04
  • 2
    $\begingroup$ Yes. The point is that you screw up the BM at a random point. This means that for any fixed $t_1,t_2,..,t_4$ you're not changing anything with probability 1, so all of the conditions still hold almost surely. $\endgroup$ – Anthony Quas Feb 7 '12 at 0:20
  • 1
    $\begingroup$ In more technical language, $\tilde B_t(\omega)$ is called a version of Brownian motion. It satisfies all of your axioms (1), (2) and (3). For any version of Brownian motion, there is another version in which all sample paths are continuous. The argument I gave earlier shows that with probability 1, all sample paths are uniformly continuous when restricted to the dyadic rationals. They therefore have a unique continuous extension to $[0,1]$. Nevertheless there are families of random variables satisfying the axioms that are not continuous (such as $\tilde B_t$). $\endgroup$ – Anthony Quas Feb 7 '12 at 1:18
  • 1
    $\begingroup$ By the way, in case it would help you to see it written in a book as well as on MathOverflow, the same example is mentioned leading up to Example 1.3 of www.stat.berkeley.edu/~peres/bmbook.pdf (which is a draft version of "Brownian Motion" by Mörters and Peres - I highly recommend the book!). $\endgroup$ – James Martin Feb 7 '12 at 9:41

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.