# property of the bracket process and stochastic integral

suppose $M$ is a continuous local martingale and $H$ is a predictable process such that $E[\int_0^{\tau_n} H_s^2d\langle M\rangle ]<\infty$, where $\tau_n$ is a sequence of stopping times converging to $\infty$ P-a.s. Then I know that $H\cdot M$, the stochastic integral, is a local martingale. Furthermore we have the following property $\langle H\cdot M, K\rangle = \int H d\langle M,K\rangle$ for every continuous local martingale $K$.

$$\langle H\cdot M\rangle = \langle H\cdot M, H\cdot M\rangle = \int Hd\langle M,H\cdot M\rangle$$

My question is, can I also use this inequality in the second argument, i.e

$$\int Hd\langle M,H\cdot M\rangle= \int H^2d\langle M,M\rangle = \int H^2d\langle M\rangle$$

If so, why is it true, and why is there not a double integral?

hulik

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Yes you can, as it is just using associativity of the integral/change of variables. There is no double integral involved –  George Lowther Jun 12 '12 at 14:02
@ George Lowther: Thanks for your comment. What do you mean by change of variables? I thought, generally $\langle \cdot,\cdot\rangle$ is not symmetric. So why can I write $\langle M,H\cdot M\rangle =\langle H\cdot M,M\rangle$ ? –  user20869 Jun 12 '12 at 15:12
Check again.  –  Did Jul 15 '12 at 9:38