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Let $X_t$ be a continuous semimartingale. I want to prove that $[X-[X]]_t = [X]_t$. I understand that $[[X]]_t = 0$. In my script, it says that $$[X-a[X]]_t = [X]_t-2a[X,[X]]_t+a^2[X]_t = [X]_t-2a[X,[X]]_t$$ for all $a>0$.

Then it is argued that $[X,[X]] >0 $ but I don't understand why this is true. Can you please help me with understanding this?

Believing this, I understand how to deduce $[X-[X]]_t = [X]_t$.

Rooibos
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  • Since $[X]_t$ is of bounded variation and $X$ is continuous you can use this answer with $M=X$ and $V=[X]$ (only the continuity of $M$ was used) to show that $[X,[X]]_t=0,.$ This implies what you want to prove. – Kurt G. Jun 04 '23 at 19:26
  • Note that quadratic variation is an increasing function a.s. as a function of $t$. Now, recall that increasing functions have bounded variation. Lastly, the covariation of a continuous process with a process of bounded variation is always zero. – Oscar Jun 16 '23 at 15:58

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