when reading some lecture notes of stochastic calculus, I have notice that often the professor seems to use the fact that:
given $B^1$ and $B^2$ as two independent Brownian motion, then we have: $$ \langle B^1, B^2 \rangle_t = 0$$ Is this true? Can anybody explain why?
TBN: angle brackets mean "quadratic variation of the processes".
The process $B = (B^1, B^2)$ is a 2-dimensional Brownian motion, hence a standard Wiener-process (on the natural filtration), so the Levy-theorem states that the covariation matrix is diagonal (we might need that they are continuous). I believe there should be an easier way though.
Edit: Looking back into my lecture, we showed once by direct calculation that the approximation of the quadratic covariation $ \tau_n ( \langle B^1 , B^2 \rangle )_t $ tends to 0 for a Riemann decomposition, which should also yield the claim. I think another option would be to prove that $ B^1 B^2 $ is already a martingale (wrt some natural filtration), as the quadratic covariation is the only process in $ \mathcal{V} $ (the class of continuous processes starting in 0 and with bounded variation) s.t. $ B^1 B^2 - \langle B^1 , B^2 \rangle$ is a local martingale. I'm not sure, but this might also require continuity of $B^1$ and $B^2$ .
I hope this helps