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Let $X$ be a semimartingale and $[X,X]^c$ the continuous part of the quadratic variation of $X$. I need to prove that $[X,[X,X]^c] = 0$.

I thought it might be useful to divide the SM into its continuous and purely discontinuous parts, i.e. $X = X^c + \Delta X$, and prove that both $[X^c,[X,X]^c]$ and $[\Delta X,[X,X]^c]$ are equal to zero.

I have already proved $[X^c,[X,X]^c] = 0$ using the fact that it's possible to write the bracket as the limit of the Riemann sums of the product of the increments of the two processes, but I can't find a way to prove that $[\Delta X,[X,X]^c]$ is zero.

Many thanks

[EDIT]

$\Delta X_t := \sum_{0<s\leq t} \Delta X_s = \sum_{0<s\leq t} (X_s - X_{s-})$

[EDIT: SOLUTION (?)]

Since $\Delta X_t$ counts the number of jumps that $X$ makes in the interval $[0,t]$, it is an increasing process.

Of course $[X,X]^c$ is a continuous process.

So, since the bracket of an increasing process and a continuous process is zero (see Quadratic Variation of Increasing Process?), $[\Delta X, [X,X]^c] = 0$.

maramath
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  • I am used to 'purely discontinuous' being associated to local martingales, and not to semimartingales in general. It may be useful to add definitions, or a reference you are using. – Snoop Feb 22 '23 at 22:37
  • Is there an assumption that $X$ can only jump upwards? – user6247850 Feb 23 '23 at 14:09
  • No, so $\Delta X$ is not necessarily an increasing process...but I thought it might be useful to prove the following: given a general process $X$, if $Y$ is a continuous process of finite variation, then $[X,Y] = 0$ (maybe using the representation of the bracket as the limit of Riemann sums and then bringing out the sup of the increments of $Y$) – maramath Feb 23 '23 at 17:42

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