I'm trying to do the following exercise from the mentioned book.
As for $(M^{(1)}, M^{(2)})$ that's obvious from Levy's characterization. However, I'm now sure how to deal with $(M^{(1)},M^{(3)})$. I guess I have to calculate the square bracket, but then, why is the three-dimesional process not Brownian motion? Is there a problem with adaptability?
$(M^{(1)},M^{(2)},M^{(3)})$ is not a Brownian motion because we lose independence of the three components. More precisely, $M^{(1)}_1 \times M^{(2)}_1 \times M^{(3)}_1 = \big|W^{(1)}_1 \times W^{(2)}_1 \times W^{(3)}_1 \big|$, so one you know the sign of $M^{(1)}_1$ and $M^{(2)}_1$, you can deduce the sign of $M^{(3)}_1$.
Yet, since the random variable $\epsilon := \mathrm{sign}(W^{(1)}_1) \mathrm{sign}(W^{(2)}_1) \mathrm{sign}(W^{(3)}_1)$ takes values in $\{-1,1\}$ and is independent of $(W_1,W_3)$ [to see this, compute the law of $\epsilon$ given $(W_1,W_3)$], the process $(M^{(1)},M^{(3)}) = (W^{(1)},\xi W^{(3)}) $ is still a Brownian motion.
