# Intensity and compensator for a jump process

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, $$X_t = \sum\limits _{i = 1} ^\infty I\{ \tau _i where $$\{\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.

• 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 $A_t = \int _0 ^t \alpha _s ds$? – Sinusx Feb 14 at 14:08
• 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. – S.Surace Feb 14 at 14:34