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Suppose $(B_t)_{t\geq0}$ is a Brownian motion and $(A_t)_{t\geq0}$ is a continuous process of bounded variation. I wish to show that $\langle A,B\rangle =0$.

For this, I know that $(B_t-t)_{t\geq0}$ is a martingale and so $\langle B \rangle=t$. Moreover, it is true that $\langle A, B \rangle \leq \langle A \rangle^{1/2}\langle B \rangle^{1/2}$. So if I can show $\langle A \rangle=0$ then we are done, however I am not sure how I would go about this. Any suggestions?

Milly Moo
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    You know more than I: for example that $B_t-t$ is a martingale which is completely new to me despite the fact that I have been working in this field for more than 30 years. The fact $\langle A,B\rangle=0$ can be shown as in this answer. – Kurt G. Dec 16 '22 at 17:14
  • Ah I think I meant $B_t^2-t$ is a martingale. Thank you for the link, this now makes sense to me using the definition – Milly Moo Dec 16 '22 at 17:28
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    That's correct and equivalent to $\langle B\rangle=t,.$ Just like in the link I gave you can show $\langle A\rangle=0$ by taking the limit of the time discretization. – Kurt G. Dec 16 '22 at 17:32

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