In Limit theorems for stochastic processes, by Jacod and Shiryaev we have the existence of a predictable quadratic covariation process stated as the following theorem

$\mathbf{Theorem}$ To each pair $(M,N)$ of locally square integrable martingales one associates a predictable process $\langle M,N\rangle \in \mathcal{V}$, unique up to an evanescent set, such that $MN - \langle M,N\rangle$ is a local martingale. Moreover, $$\langle M,N\rangle =\frac{1}{4}(\langle M + N,M + N\rangle - \langle M - N,M - N\rangle )$$ and if $M,N\in\mathcal{H}^2$ then $\langle M,N\rangle\in\mathcal{A}$ and $MN -\langle M,N\rangle \in\mathcal{M}$. Furthermore $\langle M, M\rangle$ is non-decreasing, and it admits a continuous version if and only if $M$ is quasi-left-continuous.

where $\mathcal{V}$ is the space of processes with finite variation, $\mathcal{A}$ the space of processes with integrable variation, $\mathcal{H}^2$ the space of square integrable martingales and $\mathcal{M}$ the space of uniformly integrable martingales.

In this paper by Protter and Shimbo they define on page 269 in (4) the processes $Z_t$ and mention that this need not be continuous. Hence $Z$ is a general local martingale. On the same page after $(7)$ they write: "Since $M$ is continuous there is no issue about the existence of $d\langle Z,M\rangle_s$ ."

Why is the existence of the predictable quadratic covariation process still valid for a pair $(M,N)$ of local martingales, one continuous and the other not?

I asked this question also on MSE but no one has answered it yet (I also started a bounty without success.). So I hope this is the right place to ask this question.


Your Answer

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

Browse other questions tagged or ask your own question.