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Set-up and assumptions. Let $(\mathscr{F}_t, t \geq 0)$ be a right-continuous complete filtration. Let $(X_t, t\geq 0 )$ be a pure jump $\mathbb{R}$-valued process with unit jumps, that is, $$ X_t = \sum\limits _{i = 1} ^\infty I\{ \tau _i <t \}, $$ where $\{\tau _i \}$ is an a.s. increasing sequence of $(\mathscr{F}_t)$-stopping times, a.s. $\lim_{i \to \infty} \tau _i = \infty$. Assume also that $E X _t < \infty$, and that all $\tau _i$ are totally inaccessible.

We know by Doob-Meyer decomposition theorem that there exists a predictable process $(A_t)$ such that $$ X_t - A _t $$ is an $(\mathscr{F}_t)$-martingale.

Assume that we also know that for some uniformly bounded predictable continuous process $(\alpha _t, t \geq 0)$ a.s.

$$ P\big[X_{t + \Delta t} - X _t = 1 \mid \mathscr{F}_t \big] = \alpha _t \Delta t + o(\Delta t), \ \ \ \ \ \ \ (1) $$ $$ \ \ \ \ \ \ \ P\big[X_{t + \Delta t} - X _t = 0 \mid \mathscr{F}_t \big] = 1 - \alpha _t \Delta t + o(\Delta t), \ \ \ \ \ \ \ \ \ \ \ \ $$ $$ P\big[X_{t + \Delta t} - X _t > 1 \mid \mathscr{F}_t \big] = o(\Delta t). \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) $$

Question. Can we prove that $A_t = \int\limits _0 ^t \alpha _s ds$?

Thoughts. Intuitively $\alpha _t$ should be the intensity of jumps for $(X_t)$ and $X _t - \int\limits _0 ^t \alpha _s ds$ should be a martingale. However I have not found any reference confirming this. It is proven in multiple references that the inverse implication is true, that is, if $A_t = \int\limits _0 ^t \alpha _s ds$, then (1) - (2) holds. For example, in Point Processes and Queues. Martingale Dynamics (3.5) in Chapter 2 is very similar to (1)-(2). Another example is Lemma 2.22 in Chapter 2 of Enlargement of Filtration with Finance in View. However I did not find the answer to the posted question and I don't see how to show it myself, and would very welcome suggestions or suitable references.

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Assuming conditions (1)-(2) hold, to prove that the compensator is almost surely absolutely continuous, one needs an extension of the Ethier-Kurtz criterion. This extension is discussed here; see (4) and Ref. [28], which I copy below.

  • Y. Zeng, Compensators of Stopping Times, PhD thesis, Mathematics Department, Cornell University, 2006
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  • $\begingroup$ I came across that paper too, and I see how it is possible to show that the compensator is a.s. absolutely continuous. The question still remains: even if we also know that $(A_t)$ is a.s. absolutely continuous, can we claim that $A_t = \int _0 ^t \alpha _s ds$? $\endgroup$ – Sinusx Feb 14 at 14:08
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    $\begingroup$ I may be missing something, but I believe that once you know that $A_t$ is absolutely continuous, with density $\beta_t$, you may use (1) and (2) to express $E[A_{t+\Delta t}-A_t|\mathscr{F}_t]$ and expand in $\Delta t$. This relates $\alpha$ to the conditional expectation of $\beta$. Thus $\alpha_t=\beta_t$ a.s. iff $\beta$ is $\mathscr{F}_t$-predictable. $\endgroup$ – S.Surace Feb 14 at 14:34

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