My question has been asked before at beginner's question about Brownian motion . There was only one answer, which was not accepted. It was probably incorrect, because nothing was said about $\Omega$. So $\Omega$ could contain exactly two points, which would mean that at most two sample paths exist, which is obviously incorrect.

Brownian motion is defined in that question, except for the omission, noted in the answer, that $B_t - B_s$ is independent from $B_r -B_q$ for $0 \leq q < r \leq s < t$. Since Brownian motion is well-known, I will not define it here, but the definition is surprisingly simple, and you can find it at http://en.wikipedia.org/wiki/Brownian_motion . They call it "the Wiener process $W_t$", but it is the same thing. You have to scroll down a lot to get to the definition.

I understand what (the stochastic process) Brownian motion is, but I do not understand what a sample path is. Such a path is often called $X(t,\omega)$, but I have never seen a book or paper that explains exactly what $\omega$ is. Sometimes it is written that $\omega \in\Omega$, but I have never seen $\Omega$ defined. I know how to compute a random walk that simulates a sample path, but that's not what I want.

L.C. Evans wrote an introduction to stochastic differential equations in which he explains Levy's construction of Brownian motion. This defines Brownian motion as a countably infinite sum of "tent functions" weighted by independent, identically distributed normal random variables. I do understand how this construction involves a probability space $\Omega$, consisting of all sequences of real numbers, and I know what probability distribution on $\Omega$ makes sense here, so I understand what "$X(t,\omega)$" might mean if you define Brownian motion this way, and how you can generate sample paths. However, when most people write about Brownian motion, they seem not to use the Levy construction, rather they just assume Brownian motion exists and has all the properties that it's supposed to have, so it's not obvious to me what $\Omega$ is supposed to be, let alone what $\omega$ is and how it is chosen. (Is everyone using Levy's construction but not mentioning it?)

EDIT: MSE is complaining that I am making too many comments, so I'll add to my question. It was suggested that one could take $\Omega$ to be $C(\mathbb{R}^+,\mathbb{R})$, and the sample paths simply elements of $\Omega$. I objected because the fact that sample paths are continuous is not obvious, and most books/papers prove it from the definition of definition of Brownian motion as a stochastic process (of course, they don't bother to define what a sample path is). Then it occurred to me that you could take $\Omega$ to be the set of all functions from $\mathbb{R}$ to $\mathbb{R}$ satisfying some measurability condition, and take the probability distribution on $\Omega$ to be precisely that distribution that ensures that the 4-5 conditions required by the definition of Brownian motion were met. This guess (about the probability distribution on $\Omega$) was confirmed as the correct one by someone who answered this question.


2 Answers 2


A possible choice of space $\Omega$ to define Brownian motion is $\Omega=C(\mathbb R_+,\mathbb R)$, then the Brownian motion $(B_t)_{t\in\mathbb R_+}$ is simply the coordinate process, that is, for every $t$ in $\mathbb R_+$ and $\omega$ in $\Omega$, $B_t(\omega)=\omega(t)$. In this construction, sample paths are the elements $\omega$ of $\Omega$.

But, as is usual in probability, one may prefer not to specify $\Omega$. Then $\Omega$ can be any space large enough for a family $(X_t)_{t\in\mathbb R_+}$ of random variables with the prescribed properties to exist on $\Omega$. Then a sample path is, for some $\omega$ in $\Omega$, the function $X(\omega):\mathbb R_+\to\mathbb R$, $t\mapsto X_t(\omega)$.

Lévy's construction by dichotomy, which you recall, might then correspond to $\Omega=S^\mathbb N$ the product of a countable number of copies of a probability space $(S,\mathcal S,Q)$ being large enough for one standard normal random variable $\xi$ to be defined on it. Then a Brownian motion $(X_t)_{t\in\mathbb R_+}$ on $\Omega$ can be defined as Lévy indicated using the i.i.d. copies of $\xi$ defined on each factor of $\Omega$. Thus, every $\omega$ in $\Omega$ is $\omega=(s_n)_{n\in\mathbb N}$ for some $s_n$ in $S$, and the random variables $X_1$ and $X_{1/2}$, say, are defined by $X_{1}(\omega)=\xi(s_1)$ and $X_{1/2}(\omega)=\frac12\xi(s_1)+\frac12\xi(s_2)$.

Edit: Recall that the $n$th approximation $X^{(n)}$ of the Brownian motion $X$ on $[0,1]$ is piecewise linear on each interval $[(k-1)/2^n,k/2^n]$ with $1\leqslant k\leqslant2^n$. Thus, the $0$th approximation is such that $X^{(0)}_t(\omega)=t\xi(s_1)$ for every $t$ in $[0,1]$, after that, $X^{(n+1)}_t=X^{(n)}_t$ at every $t=k/2^n$ and $2X^{(n+1)}_t=X^{(n)}_{k/2^n}+X^{(n)}_{(k+1)/2^n}+\xi(s_*)/2^{n/2}$ at every $t=(2k+1)/2^{n+1}$, where $*$ is the first index $i$ such that $\xi(s_i)$ is not used yet.

  • $\begingroup$ ?? Lévy's construction needs an i.i.d. sequence of standard normal random variables, no? So I fail to see the generality, in this case. $\endgroup$
    – Did
    Commented Oct 28, 2013 at 16:02
  • $\begingroup$ Yes. $ $ $ $ $ $ $\endgroup$
    – Did
    Commented Oct 28, 2013 at 17:02
  • $\begingroup$ That the limiting process of the linear-by-parts processes corresponding to finer and finer dyadic partitions is indeed in $\Omega$ is a theorem. $\endgroup$
    – Did
    Commented Oct 28, 2013 at 17:58
  • $\begingroup$ The Edit recalls Lévy's construction (even though I feel we begin to be rather far from the question you asked). $\endgroup$
    – Did
    Commented Oct 28, 2013 at 20:21
  • $\begingroup$ Thank you for explaining the linear-by-parts process. I have not seen this construction before. MSE is complaining about too many comments, so I deleted some comments, and added to my question. I think my edit goes back to the idea of my original question. Is the idea I give after "EDIT" in my question reasonable? If so, what kind of measurability, if any, is needed for the functions in $\Omega$? Hopefully this is the last question I will need to ask you. $\endgroup$ Commented Oct 28, 2013 at 20:30

Hi the question is rather long but the answer I think is short, a sample path is simply the whole trajectory of a Brownian motion sampled at one $\omega$.

So if you look at the Brownian motion as a measurable application from $ \Omega \to \mathcal{C}(\mathbb{R}^+,\mathbb{R})$ where $\mathcal{C}(\mathbb{R}^+,\mathbb{R})$ is the Borelian $\sigma$-algebra generated by open sets of the space of continuous functions $\mathcal{C}(\mathbb{R}^+,\mathbb{R})$ equipped with convergence over compact sets (this is the "Wiener measure" point of view on Brownian motion I would say)

When you write $B(\omega,t)$ this is not a trajectory, it is the random variable defined by : $(\Omega,\{t\})\to \mathbb{R}, (\omega,t)\mapsto B(\omega)(t) $ where the trajectory (a continuous function) $B(\omega)$ is evaluated at time $t$.

Hope this make sense to you.

Best regards

  • $\begingroup$ This does make some sense to me, but please note that I am not a probabilist, as should be apparent from my question, so I am having trouble with some things. In my question, I complain that people use $\Omega$ without saying what $\Omega$ means, and you do this in your answer. Maybe all probabilists know what that's supposed to mean, but I don't. You seem to be defining Brownian motion by defining the distribution of sample paths. Maybe this is the best way to do it, but people usually define Brownian motion as a stochastic process, then prove things about "sample paths" without... $\endgroup$ Commented Oct 28, 2013 at 19:30
  • $\begingroup$ ...saying what a "sample path" is. Often I see a sample path called "$X(\omega,t)$", and $\omega$ is presumably an element of $\Omega$, but $\Omega$ is never defined, let alone what the probability distribution on $\Omega$ is. Note that you seem to be assuming sample paths are continuous, whereas it is more common to use the standard definition of Brownian motion to prove sample paths are continuous. Your viewpoint may be better, but I'd like to be able to understand what I've attempted to read. $\endgroup$ Commented Oct 28, 2013 at 19:50
  • 2
    $\begingroup$ @ stefan : I recommend that you read the first chapters of Kartzas and Shreve's book on Brownian motion. They build BM in 3 different ways and by overlaping those methods you could most probably answer any question you have on this kind of topic in my opinion. By the way Did's answer is excellent. Best regards $\endgroup$
    – TheBridge
    Commented Oct 29, 2013 at 10:41

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