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Proposition 3 (ABDL03): If a special semimartingale process $X$ is square integrable with respect to the natural filtration of a standard Brownian motion $W$, then one can write

$X_t - X_0 = \int_0^t \! \mu_u \, \mathrm{d}u + \int_0^t \! \sigma_u \, \mathrm{d}W_u$

where $\mu,\sigma$ are predictable processes.

(Note: I modified the statement of the proposition quite a bit.)

I know that if $X$ is a local martingale, this proposition holds with $\mu \equiv 0$. This is just the martingale representation theorem.

Q: What if $X$ is a finite variation process? Does this proposition hold with $\sigma\equiv 0$? If so, can the sufficient condition of square integrability be relaxed? Is there some standard set of necessary and sufficient conditions for when one can write a finite variation process as $\int_0^t \! \mu_u \, \mathrm{d}u$ with $\mu$ predictable?


Partial Answer: So, if the paths of $X$ are continuously differentiable and bounded on compacts (almost surely?) then by the fundamental theorem of calculus we can write

$X_t - X_0 = \int_0^t \mu_u \, \mathrm{d}u $

where $\mu$ is the continuous, bounded derivative of $X$ (almost surely?) with respect to $t$.

Remaining Q: Have I done this right? Is the existence of a bounded, continuous derivative for the paths of $X$ (a finite variation process) necessary and sufficient for writing it as $\int_0^t \mu_u \, \mathrm{d}u$ with $\mu$ predictable?

UPDATE #2: We might be able to relax this to just differentiable (rather than continuously differentiable) since $X$ already has finite variation.

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Ito integral with respect to the Brownian motion always has a continuous version, hence purely jump processes cannot be written in that form. I also know not much on this subject, but Lévy–Itō decomposition may be helpful. – Sangchul Lee May 3 '12 at 23:05
Poisson process is a finite variation process, but can not be written $X_t - X_0 = \int_0^t \mu_u \mathrm{d} u$, since the integral is continuous. – Sasha May 4 '12 at 1:39
@Sahsa: Right, and so a necessary condition for writing a FV process $X_t - X_0 = \int_0^t \mu_u \, \mathrm{d}u$ (for some predictable $\mu$) is continuity of $X$. I'm looking for necessary and sufficient conditions. Can you check out my update. Do I have this right? (Although this seems like basic calculus, I'm a little thrown off by the stochastic nature of $X,\mu$, hence the "almost surely?"'s) – lowndrul May 4 '12 at 14:58

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