I've been reading through Evans' excellent Introduction to Stochastic Differential Equations (used to be lecture notes available on his website a few years back, and became this book), and got somewhat confused with the definition of $n$-dimensional Wiener process. Here's what he says:

Definition: An $\mathbb{R}^n$-valued stochastic process $\mathbf{W}(\cdot) = (W^1(\cdot), \ldots, W^n(\cdot))$ is an $n$-dimensional Wiener process provided

(i) for each $k = 1, \ldots, n$, $W^k(\cdot)$ is a one-dimensional Wiener process.

(ii) the $\sigma$-algebra $\mathcal{W}^k = \mathcal{U}(W^k(t) | t \geq 0)$ are independent, $k = 1, \ldots, n$.

Just to clarify the above notation, given a probability space $(\Omega, P, \mathcal{U})$ and a real-valued random variable $X$ on $\Omega$, $\mathcal{U}(X) = \{X^{-1}(B) | B \subset \mathbb{R} \text{ Borel} \}$. Also, although it was not stated anywhere, I gather $\mathcal{U}(\{X_t\}_{t \in I})$ is the smallest $\sigma$-algebra containing $\bigcup_{t \in I} \mathcal{U}(X_t).$

In the following results, the author seems to imply that the following "fact" is obvious:

"Fact": If $\mathbf{W}$ is an $n$-dimensional Wiener process, then $\mathbf{W}(t)$ is $N(0, tI)$, i.e., $\mathbf{W}(t)$ has an $n$-variate normal distribution.

Also, given $0 < t_1 < \ldots < t_m$, the random variables $$\mathbf{W}(t_1), \mathbf{W}(t_2) - \mathbf{W}(t_1), \ldots, \mathbf{W}(t_m) -\mathbf{W}(t_{m-1})$$ are independent (as $n$-dimensional random variables).

Note that this holds for one-dimensional Wiener processes by definition.

I thought about it for a while, but it's not clear to me why this "fact" holds. For instance, I see that $W^1(t_1), W^2(t_1), W^1(t_2) - W^1(t_1), W^2(t_2) - W^2(t_1)$ are all pairwise independent, but this answer convinced me that pairwise independence does not imply mutual independence even for normal random variables (nor does it imply that they have a joint normal distribution, obviously).

As further evidence for the "fact", this document gives two alternative definitions of multidimensional Wiener process (or Brownian motion) - Definitions 2.2.2 and 2.2.3 on pages 17 and 18 - which happen to be equivalent if the "fact" holds.

Could someone please clarify if and why the "fact" holds?

  • $\begingroup$ May I ask why did you give an example mathoverflow.net/questions/118880/… of something so specific, unrelated to this multidimensional Wiener process? $\endgroup$ – Creator Oct 17 '18 at 0:52
  • $\begingroup$ @Creator The point here is how to get from $n$ independent one-dimensional Wiener processes to $n$-dimensional Wiener process such that $\mathbb{W}(t)$ follows a multivariate normal distribution. When analyzing the individual components, one might get the impression that "everything is independent" and so the "fact" is obvious (at least the second part). However, pairwise independence (which is what you get from (i) and (ii)) does not imply joint normal distribution, and one counterexample is the one in the link. $\endgroup$ – Pedro M. Oct 17 '18 at 1:26
  • $\begingroup$ As I understand we are postulating independence, so why asking it is independent or not? $\endgroup$ – Creator Oct 17 '18 at 1:28
  • $\begingroup$ @Creator The random variables $W^1(t), \ldots, W^k(t)$ are independent, that is clear. However, I do not see how that implies that they are jointly normally distributed. I also do not see how that implies that $W^1(t_1), W^1(t_2) - W^1(t_1), W^2(t_1), W^2(t_2) - W^2(t_1)$ are independent (but I do see that they are pairwise independent, which is strictly weaker). $\endgroup$ – Pedro M. Oct 17 '18 at 1:36
  • $\begingroup$ I am not sure about your questions. Distribution has nothing to do with independence. We are postulating Gaussian distributions. Each sample is independent over space and time both. So I really do not see the question. The specific example you have given is that if you have a constraint as in the example, than it may not be independent. $\endgroup$ – Creator Oct 17 '18 at 1:44

Initially, I underestimated how strong the independence of $\mathcal{W}^k$ was; many thanks to @saz and @Creator for pointing me in the right direction.

In case anyone else wonders about this, the fact can be proven easily as follows:

First, because the sum of independent normal variables is normal, it follows that any linear combination $\sum_{k=1}^n a_k W^k(t)$ is normal, which shows that $\mathbf{W}(t)$ is $N(0,tI)$.

Second, note that if $0 = t_0 < t_1 < \ldots < t_m$ and $\mathbf{x}_1, \ldots, \mathbf{x}_m \in \mathbb{R}^n$, $$P\left(\bigcap_{j = 1}^m\left\{\mathbf{W}(t_j) - \mathbf{W}(t_{j-1}\right) \leq \mathbf{x}_j\}\right) = P\left(\bigcap_{j = 1}^m\bigcap_{k = 1}^n\left\{W^k(t_j) - W^k(t_{j-1}\right) \leq x_j^k\}\right)\\ =\prod_{k=1}^n P\left(\bigcap_{j = 1}^m\left\{W^k(t_j) - W^k(t_{j-1}\right) \leq x_j^k\}\right) \text{[by the independence of the }\mathcal{W}^k] \\ =\prod_{k=1}^n\prod_{j=1}^m P\left(\left\{W^k(t_j) - W^k(t_{j-1}\right) \leq x_j^k\}\right) \text{[because each component is a Wiener process] } \\ =\prod_{j=1}^m P\left(\left\{\mathbf{W}(t_j) - \mathbf{W}(t_{j-1}\right) \leq \mathbf{x}_j\}\right) \text{[again by the independence of the }\mathcal{W}^k],$$

which shows that the random vectors $\{\mathbf{W}(t_j) -\mathbf{W}(t_{j-1})\}_{j=1}^m$ are independent.


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