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I know that for a continuous local martinagle $M$ we have $\langle M\rangle^\tau = \langle M^\tau\rangle$ for any stopping times. Now if $M,N$ are two local martingale I know that there exists again a process $\langle M,N\rangle$ (continuous, adapted and of bounded variation) such that $MN-\langle M,N\rangle$ is a continuous local martingale. Furthermore I know $\langle M,N\rangle = \frac{1}{4}(\langle M+N\rangle -\langle M-N\rangle)$. Now I want to prove:

$$\langle M,N\rangle^\tau=\langle M^\tau,N^\tau\rangle = \langle M^\tau,N\rangle = \langle M,N^\tau\rangle$$

The first equation is clear, since: $$\langle M,N\rangle^\tau=\frac{1}{4}(\langle M+N\rangle -\langle M-N\rangle)^\tau=\frac{1}{4}(\langle M+N\rangle^\tau -\langle M-N\rangle^\tau)=\frac{1}{4}(\langle M^\tau+N^\tau\rangle -\langle M^\tau-N^\tau\rangle)=\langle M^\tau,N^\tau\rangle $$ Where I used the definition of $\langle M,N\rangle$ and the property $\langle M^\tau\rangle = \langle M\rangle^\tau$. Unfortunately I am not able to prove the second equality. The third will be then the same, I guess. Thank you for your help


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@ hulik : I think you can see this by using the carcterisation often given as the definition of square brackets (the same as for sharp bracket for continuous processes) as the limit in probability of the sum of co-quadratic variations. Then it is simple to see that the 4 equalities coincide. Best regards (look at the theorem definiton in Protter's book for example). – TheBridge Aug 3 '12 at 21:07

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