# Azema martingale and quadratic covariation

Given a filtered space $(\Omega,\mathcal F,\mathbb F,\mathbb P)$ supporting a Brownian Motion $B$, where the filtration $\mathcal F$ is the augmented Brownian filtration, the Azema's martingale is defined by $M_t=\mathbb E(B_t|\mathcal G_t)$, where $\mathcal G_t=\sigma(sign(B_s):s\leq t)$, completed over all $\mathbb P$ null sets. It can be shown that the filtration generated by $M$ is exactly $\mathbb G$, so that $M$ is a martingale under its own filtration. It can also be shown that $M$ is not an $\mathbb F$-semimartingale. My question is, is the quadratic covariation between $M$ and $B$ well-defined? That is, given a sequence of partition $\pi_n$ with $|\pi_n|$ goes to zero, does $\sum_{\pi_n\cap[0,t]}(M_{t_i}-M_{t_{i-1}})(B_{t_i}-B_{t_{i-1}})$ converges in any sense?

Any advice and help are greatly appreciated!

In any interval where $B$ does not vanish, M is smooth (I think that in such an interval, $dM=f(M) dt$ for some deterministic smooth $f$). Thus the only contribution comes from the zero's.
Thus $M$ is the sum of a (F-predictible) jump process and a smooth one. For the smooth one, it's easy to show that it does not contribute to the limit. For the jump process, I think there is an orthogonality result between jump process and continuous martingale (but precisely, it may be orthogonality w.r.t. the angle bracket...). Thus I think the limit is $0$. An idea to prove or disprove it manually may be so distinguish between the left-isolated zeros (which are fewer) and the others (for which $M$ does not really jump). There is maybe some book or article by Jacod to answer you.