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Wikipedia has separate articles on "Brownian motion" and "Wiener process" (http://en.wikipedia.org/wiki/Brownian_motion and http://en.wikipedia.org/wiki/Wiener_process ). I am not an expert, but that strikes me as dubious. That is not the subject of my question, however, and that may be a better issue to discuss at Wikipedia than in MSE.

Both articles say that the Wiener process is "characterized by" (they do not use the words "defined" or "definition") four facts. One of them is essentially the fact that sample paths are almost surely continuous.

In a monograph I will cite below, the continuity of sample paths was not part of the definition. I think it would be better if they gave a mathematician's definition that did not include continuity of sample paths, and then cited the continuity of sample paths as a property that could be proven. Even though people other than mathematicians may be interested in the Wiener process, is is a mathematical idea (according to Wikipedia), and I think it should be defined in mathematical terms.

A good mathematical definition does not include redundant information. For example, when one defines a group, one never states that there is a unique identity element, because the uniqueness can be proven. I don't know if Brown or Wiener used continuity of sample paths as part of their definition, but if they did, I think a modern, leaner definition that omitted unnecessary hypotheses would be better.

I would like to complain about this on the articles' talk pages, but I am not confident enough to be sure this is a valid complaint. Can anyone back me up? Or can anyone justify Wikipedia's "characterization"?

EDIT: This is the definition of Brownian motion from "An Introduction to Stochastic Differential Equations" by L. C. Evans:

(i) $W(0) = 0$ almost surely.

(ii) $W(t)-W(s) \sim N(0,t-s)$ for all $0 \leq s \leq t$

(iii) For all $0<t_1<t_2<\cdots<t_n$, the random variables $W(t_1),W(t_2)-W(t_1),\ldots,W(t_n)-W(t_{n-1})$ are independent.

He then proceeds to prove in a "Theorem" that states "for a.e. $\omega$, the sample path $t \mapsto W(t,\omega)$ is continuous". CORRECTION: He actually proves that the Levy construction of Brownian motion has continuous sample paths. He does not prove continuity follows from (i)-(iii).

Evans is a very good mathematician and a good writer.

If a continuity assumption is necessary, then someone has probably already proven that there exists a stochastic process that satisfies (i)-(iii) above but that is not the same as Brownian motion (as given by several constructions, such as Levy's construction, for example). Has anyone done this?

EDIT: Someone has in fact done this. See the MathOverflow link given in the comment immediately below this answer.

Stefan Smith
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2 Answers2

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There are a number of ways to define Brownian Motion. As you say, many common approaches first construct a suitable collection of processes (often on a product space), then show that in each equivalence class there is a continuous version of the process (generally using Kolmogorov's continuity theorem) and then declare that we are only interested in that element of each equivalence class. That is very different from saying that any process satisfying the other assumptions is necessarily a continuous process. Somewhat paradoxically, in the usual construction on the product space, the event that the process is continuous is not even measurable. In that entirely natural setting, it doesn't even make sense to ask the question of whether or not paths are continuous without doing some kind of a restriction. There is a discussion of this construction and this issue in Durrett, if you are interested.

Since physically (and mathematically) we want to be able to specify that Brownian Motion is continuous pathwise, we have to add the restriction to the continuous element of the equivalence classes into the definition to use the construction on a product space and obtain the process we want.

I think it may be worth looking at the other common, relatively elementary construction as well. Another approach is to construct Brownian Motion directly on the space of continuous functions in the usual topology. But you see in this approach, which I would argue is nicer than the first approach, we implicitly assume that Brownian Motion must be continuous. As a comment, the previous link contains what I generally see referred to as the standard modern mathematical reference on Brownian Motion and the authors do have continuity in the definition.

For what it's worth, I just took a very quick look at Wiener's original paper and it looks (at a very quick glance) like he proves continuity as a theorem.

Chris Janjigian
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  • I thought of a question, hopefully the last one. Analysts often consider functions equivalent if they differ on a set of Lebesgue measure zero. For a "Gaussian process", if a sample path (say, from the set of all Lebesgue-measurable functions from $[0,\infty)$ to $\mathbb{R}$, if this is an appropriate sample space) is selected using a probability distribution that is consistent with requirements (i)-(iii) above, is there a positive probability that it is equivalent to a continuous function.=? – Stefan Smith Nov 03 '13 at 02:46
  • You can do better than that so long as you are extremely careful about what you mean by equivalent. Given such a process $X_t^1$, you can define another process $X_t^2$ on the same space with the property that $P(X_t^1 = X_t^2) = 1$ and so that $X_t^2$ is continuous pathwise. This is different from saying that $X_t^1$ is itself pathwise continuous--in general, that event is not even measurable. – Chris Janjigian Nov 03 '13 at 02:56
  • I said I would have no more questions, but I guess I lied. I don't fully understand your last comment. Suppose I considered sample paths to be "equivalent" if they differ on a set of Lebesgue measure. Then is the "Gaussian process" described by rules (i)-(iii) above (which Evans uses as his definition of "Brownian motion") different from true Brownian motion? – Stefan Smith Nov 03 '13 at 18:14
  • I don't think you're thinking about this right. Even if you take two independent sample paths of standard Brownian Motion drawn from the same distribution, they will almost surely differ on a set of Lebesgue measure zero complement. You are thinking of these as if they were deterministic functions--they aren't. You want to pose these types of questions in terms of the probability space, not pathwise. – Chris Janjigian Nov 03 '13 at 18:36
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The definition may not seem mathematically rigorous, but it is in fact. Perhaps it would be better to say that the Wiener process is the unique process which satisfies these 3 (or 4 depending on how you count) criteria. Of course one has to prove the existence and uniqueness of such a process (the Wiener process article cites Durrett's book, which does). The existence follows from the constructions a Chris mentions in his answer.

The wording of the Wiki article could probably stand to make this more clear, but as stands it is not incorrect.

I will also point out that all the conditions are needed.

D. Kelleher
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  • I am not questioning the rigor of the article or the existence issue, just the necessity of the continuity assumption. L.C. Evans wrote a monograph containing a definition without a continuity assumption (see my edited question) and proceeds to prove continuity as a theorem. Evans knows a lot more than me and is a good writers and I doubt he would omit something necessary from a definition. – Stefan Smith Nov 03 '13 at 01:30
  • Is there a proof that if you don't assume continuity of sample paths, you got something different from what every popular construction of Brownian motion gives you? – Stefan Smith Nov 03 '13 at 01:41
  • That would be in the proof of the uniqueness of Brownian motion. To prove these conditions are sufficient, prove that there is only one probability distribution that satisfies them. To prove that each one is necessary, you come up with a counter example of something else that satisfies the other two but not that one. In the case of continuity, this counterexample is pointed out in Byron's link. – D. Kelleher Nov 03 '13 at 02:40