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?
Best Answer
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.