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