Let $B, B'$ be two independent Brownian motions. It can be shown that $BB'$ is a martingale. My lecture notes say that because of the martingale property the covariation of $B$ and $B'$ is zero. Why does this hold?
2026-03-25 19:04:54.1774465494
Covaration of independent Brownian Motions is zero
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This is a consequence of the property that for continuous local martingales $B, B'$ we have that $(\langle B, B' \rangle_t)$ is the unique process s.t. $(BB' - \langle B, B' \rangle)_t$ is a continuous local martingale