Independence of the components of a multidimensional Brownian motion Let $B = (B^1, \dots, B^n)$ be an $n$-dimensional ($n \in \{1, 2, \dots\}$) Brownian motion (i.e. $B = (B_t)_{t \geq 0} \in \Omega \rightarrow (\mathbb{R}^n)^{[0,\infty)}$ has continuous paths, $B_0 = 0$ almost surely, $B$ has independent increments, and, for every $s, t \geq 0$ with $s < t$, $B_t - B_s \sim N(0, (t - s) E_n)$, where $E_n$ is the identity matrix in $\mathbb{R}_{n, n}$) adapted to the filtration $\mathfrak{F}$ (which is not necessarily the natural filtration). Is it the case that the components $B^k$ are independent?

EDIT
I have written all my thoughts on the topic in my answer below.
 A: The following answer is based on the proof of theorem 7.8 ("Multidimensional Brownian Motion") in Jochen Wengenroth's textbook "Wahrscheinlichkeitstheorie" (de Gruyter, 2008), p. 136.
We will show, w.l.g., that $B^1 = (B^1_t)_{t \geq 0}$ and $C := (B^2_t, \dots, B^n_t)_{t \geq 0}$ ($B^1$ is a one-dimensional stochastic process and $C$ is an $n - 1$-dimensional stochastic process) are independent. If $\mathfrak{b}$ is a $\pi$-system that generates $\sigma(B^1_t :\mid t \geq 0)$, and if $\mathfrak{c}$ is a $\pi$-system that generates $\sigma(C_t :\mid t \geq 0)$, it suffices to show that $\mathfrak{b}$ and $\mathfrak{c}$ are independent. Denote by $\mathcal{P}_0$ the collection of all non-empty, finite subsets of $[0, \infty)$. Then $\bigcup\{\sigma(B^1_J) = \sigma(B^1_{j_0}, \dots, B^1_{j_m}) :\mid J = \{j_0, \dots, j_m\} \in \mathcal{P}_0\}$ is a $\pi$-system that generates $\sigma(B^1)$, and $\bigcup\{\sigma(C_J) = \sigma(C_{j_0}, \dots, C_{j_m}) :\mid J = \{j_0, \dots, j_m\} \in \mathcal{P}_0\}$ is a $\pi$-system that generates $\sigma(C)$.
Let then $J, J' \in \mathcal{P}_0$. We will show that $\sigma(B^1_J)$ and $\sigma(C_{J'})$ are independent, or, equivalently, that $B^1_J$ and $[C_{J'}]$ are independent, where the notation $[v_0, \dots, v_k]$ denotes the vector obtained by concatenating the components of the individual vectors $v_0, \dots, v_k$. Define $I := J \cup J'$. Then it suffices to show that $B^1_I$ and $[C_I]$ are independent.
Write $I = \{i_0 < \dots < i_m\}$ and denote by $\beta$ the $(m+1)n$ vector $[B_{i_0}, B_{i_1} - B_{i_0}, \dots, B_{i_m} - B_{i_{m - 1}}]$. $\beta$ is multivariate normal, since, due to the fact that $B$ has independent increments, any linear combination of $\beta$'s components is a one-dimensional normal random variable. Since $\beta$'s components are uncorrelated, they are independent. Hence $\alpha := (B^1_{i_0}, B^1_{i_1} - B^1_{i_0}, \dots, B^1_{i_m} - B^1_{i_{m - 1}})$ and $\gamma := [C_{i_0}, C_{i_1} - C_{i_0}, \dots, C_{i_m} - C_{i_{m - 1}}]$ are independent. Since $B^1_I$ is a linear transformation of $\alpha$ and $[C_I]$ is a linear transformation of $\gamma$, $B^1_I$ and $[C_I]$ are independent, Q.E.D.
