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Let $X$ and $Y$ be independent continuous semimartingales on a probability space. I know that we should have $[X, Y] = 0$.

I am able to prove that if $M$ and $N$ are independent continuous local martingales, that $[M, N] = 0$, this is easy by using the sum definition. I want to say "if $M$ and $N$ are the local martingale parts (in the Doob-Meyer decomposition) of $X$ and $Y$ respectively, they are independent and hence $[X, Y] = [M, N] = 0$". Is this true, and if so how do I prove it?

George C
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  • Using the same approach as in my answer here you should be able to show that the covariations of the bounded variation parts of $X$ and $Y$ with anything else are zero. Then $[X,Y]=[M,N]$ (even without independence). – Kurt G. Jun 08 '23 at 12:22
  • @KurtG. That is useful for getting rid of the cross-terms, but how do I conclude that $[M, N] = 0$ if I'm unsure as to whether $M$ and $N$ are independent? – George C Jun 08 '23 at 14:01
  • Without independence you cannot get $[M,N]=0,.$ – Kurt G. Jun 08 '23 at 16:28
  • @KurtG. How can I deduce $M$ and $N$ are independent from the fact that $X$ and $Y$ are? – George C Jun 08 '23 at 16:47
  • $X$ remains a continuous semimartingale with respect to its natural filtration $\mathcal F^X_t:=\sigma{X_s: 0\le s\le t}$. Let $X= M^X + V^X$ be its Doob-Meyer decomposition. Do the same for $Y$, obtaining $Y=M^Y+V^Y$. Clearly $M^X$ and $M^Y$ are independent. You should be able to finish from here, using the fact that $X$ is a semimartingale with respect to $\mathcal F^{X,Y}_t:=\sigma{X_s,Y_s: 0\le s\le t}$, with the same D-M decomposition (because of independence), and likewise for $Y$. – John Dawkins Jun 09 '23 at 18:42
  • That clears it up, thanks! @JohnDawkins – George C Jun 10 '23 at 16:05

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