Beginner here. I've started reading Bass' book about Stochastic Processes (DOI). I am trying to prove exercise 2.3, that is:
Let $W_t$ be a d-dimensional Brownian Motion starting in $x=0 \in \mathbb R^d$ and let $A$ be a $d\times d$-dimensional and orthogonal matrix (orthogonal: $A^{-1}=A^T$). Then $Y_t=AW_t$ again is a Brownian motion.
First, I figured that $Y_t = (\sum_{i=1}^d a_{1i}W_t^{(i)},\dots, \sum_{i=1}^d a_{i}W_t^{(i)})$ by multiplication so we need to look at each component and determine whether it is one-dimensional Brownian Motion or not.
I then tried to show all the properties of Brownian Motion one after another:
($\mathcal F_t$-measureable): Here, we need that the sum of two random variables is also measureable, meaning that $\{\omega\in\Omega:a_{11}W_t^{(1)}+\dots+a_{1d}W_t^{(d)}\}\in\mathcal F_t$. I don't know why that is.
($Y_t^{(1)}-Y_s^{(1)}$ is normal with $\mu=0,\ \sigma^2=t-s$): We have $$ Y_t^{(1)}-Y_s^{(1)}=a_{11}(W_t^{(1)}-W_s^{(1)}) + \dots + a_{1d}(W_t^{(d)}-W_s^{(d)}) $$ where each $(W_t^{(i)}-W_s^{(i)})$ is normal with $\mu = 0,\ \sigma = t-s.$ What about the sum though?
($Y_t^{(1)}-Y_s^{(1)}$ is independent of $\mathcal F_s$ for all $s<t$): I think this simply follows from the fact that $W_t^{(i)}-W_s^{(i)}$ is independent from $\mathcal F_s$ by definition and so is their sum.
($Y_t$ has continuous paths): I think this also follows directly from the fact that $W_t^{(i)}$ has continuous paths for all $i$.
I'd really appreciate any help!
Edit: As mentioned in the comment below, it is also necessary to show that $(\sum_i a_{ji}W_t^{(i)})_j$ are independent. This statement seems kind of weird to me, since for all $j=1,\dots,d$ there are always the same random variables $W_t^{(j)}$ included. How does this make sense?