2
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

It is maybe too late, and I am not a specialist of discontinuous processes, thus I will present "hints and directions" more than a reply. Everything I will say should be taken as hypothetical.

First show should know that for non continuous semimartingale there is two different brackets, the angle one and the sharp one (see for example Angle bracket and sharp bracket for discontinuous processes ). No for the sharp one:

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.

I apologize for my poor english and hope it was readable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.