# Why is $N^\tau ( M - M^\tau )$ a continuous local martingale if $M$ and $N$ are?

Working through my stochastic calculus script, I encountered the following identity, for which no proof is given: $\langle M, N^\tau \rangle = \langle M, N \rangle^\tau$, if $M, N$ are continuous local martingales, null at 0.

I know that (by uniqueness of the bracket) it suffices to show the following:

If both $M,N$ are continuous local martingales and $\tau$ is a stopping time, then $N^\tau ( M - M^\tau )$ is again a continuous local martigale.

But why is this true?

I know how to proof this with the properties of the stochastic integral. But since the above identity is used in our script during the construction of the stochastic integral, I would like to prove it directly. Does anybody know how this can be done?

Trying to prove it, I started like this:

$M,N$ continuous local martingales $\ \Rightarrow \$ there are localizing sequences $\tau_n, \sigma_n$ such that $M^{\tau_n}$ and $N^{\sigma_n}$ are bounded martingales. By the stopping theorem, $N^{\tau_n \wedge \sigma_n \wedge \tau}$ and $M^{\tau_n \wedge \sigma_n \wedge \tau}$ are bounded martingales, too.

Now I would like to proof that $N^{\tau_n \wedge \sigma_n \wedge \tau} (M^{\tau_n \wedge \sigma_n } - M^{\tau_n \wedge \sigma_n \wedge \tau})$ is a martingale, since this would then imply that $N^\tau ( M - M^\tau )$ is a continuous local martigale (with localizing sequence $\tau_n \wedge \sigma_n$).

Thanks a lot for your help! Regards, Si

• The fact that $N$ is a local martingale is irrelevant. Any continuous adapted process will work. Apr 9, 2012 at 15:21

Let $X^{(n)}=N^{\kappa_n}\cdot(M^{\varrho_n}-M^{\kappa_n})$, where $\varrho_n=\tau_n\wedge\sigma_n$ and $\kappa_n=\varrho_n\wedge\tau$ hence $\kappa_n\leqslant\varrho_n$. Then, $\mathrm dX^{(n)}_t=[\kappa_n\leqslant t\leqslant\varrho_n]\cdot N_{\kappa_n}\cdot\mathrm dM_t=[\kappa_n\leqslant t\leqslant\varrho_n]\cdot N^{\kappa_n}_t\cdot\mathrm dM_t^{\tau_n}$ and $M^{\tau_n}$ is a martingale hence $X^{(n)}$ is a martingale.