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On protter's book Stochastic Integration and Differential Equations, there is some confusion for me about the decomposition about quadratic variation:

Since the process $[X, X]$ is non-decreasing with right continuous paths, and since $\Delta [X, X]_t = (\Delta X_t)^2 $ for all $t\geq0$ (with the convention that $X_{0-} = 0$), we can decompose $[X, X] $ path-by-path into its continuous part and its pure jump part.

I don't know what is the "pure jump part", and why(and how) we can decompose the quadratic variation pathwise.

Thanks for any comment.

George
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1 Answers1

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So, let $V$ be some right-continuous process of bounded variation (like the quadratic variation process). We can define the jump process of $V$ to be a new process $\Delta V(t) = V(t) - V(t-)$, where $V(t-)$ denotes the left limit $V$ at $t$. Note that $\Delta V(t)$ is different than zero precisely when there's a jump. Now, for each outcome $\omega$ we can consider $V^J :=(\sum \Delta V)(t) = \sum_{s\leq t} \Delta V(s)$. Observe that $V^J$ is absolutely convergent. This way we can decompose $V$ into a continuous part with finite variation and a pure jump part with finite variation as well. $$V = \left( V - \sum \Delta V \right) + \sum \Delta V = V^C + V^J$$

Oscar
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