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

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

1 Answer 1


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.

  • $\begingroup$ Nice! Thanks a lot! $\endgroup$
    – Mad Si
    Apr 9, 2012 at 19:37

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