Consider a sub-martingale (e.g., a counting process) $N_t$ adapted to a filtration $\mathcal{F}_t$. According to the Doob-Meyer decomposition, we have:

$$N_t = M_t + A_t,$$

where $M_t$ is a martingale and $A_t$ is the compensator. Throughout, we will assume $A_t$ is absolutely continuous and predictable so that the intensity process $\lambda_t$ with $A_t = \int_0^t \lambda_sds$ exists.

Now, consider another filtration $\mathcal{G}_t$. My question is what are the relations between the following:

(a) $\lambda_t$ is adapted to $\mathcal{G}_t$ (i.e., $\lambda_t$ is $\mathcal{G}$-measuable).

(b) $A_t$ is adapted to $\mathcal{G}_t$.

(c) $M_t$ is adapted to $\mathcal{G}_t$.

(d) $N_t$ is adapted to $\mathcal{G}_t$

Are they equivalent? Given (a), can we dirive (b,c,d)?

  • $\begingroup$ I have the following question: we can only say that a process is adapted to a filtration. So do you mean that $\mathcal F = (\mathcal F_t)_{t\geq 0}$ is a filtration? The same goes for $\mathcal G$. $\endgroup$
    – Mushu Nrek
    Nov 5, 2023 at 9:30
  • $\begingroup$ Yes, I mean $\mathcal{F}, \mathcal{G}$ are both filtrations. I have edited the problem above. $\endgroup$ Nov 6, 2023 at 1:26
  • $\begingroup$ Concerning the second filtration $\mathcal G$: do we know that $\mathcal G\subseteq \mathcal F$? $\endgroup$
    – Mushu Nrek
    Nov 7, 2023 at 7:21
  • $\begingroup$ Yes, one may assume $\mathcal{G}\subseteq \mathcal{F}$. $\endgroup$ Nov 7, 2023 at 8:06
  • $\begingroup$ In this case, you might first want to prove the following: d) implies all others. You have already seen that (a) and (b) are equivalent. I am, however not to sure about the other equivalences... Consider again the unit Poisson process with compensator $A_t = t$. Then, clearly, $A_t$ is adapted to the trivial filtration given by $\mathcal G_t = \{\emptyset, \Omega\}$ for all $t\geq 0$. The process itself, however, will never be measurable with respect to the trivial filtration, because it is not deterministic. Perhaps you can find an counter-example to the implication from c) to d). $\endgroup$
    – Mushu Nrek
    Nov 7, 2023 at 9:07

1 Answer 1


After some searching, I find a partial answer to my question: yes, in some sense, (a),(b),(d) are equivalent. Whether (c) is equivalent to (a,b,d) remains unknown. I post my results below.

(a) $\Leftrightarrow$ (b). This is prove that the Lebesgue integral (as pointed out by M.N.) of a measurable function is measurable, and that the derivative of a measurable function is measurable.

The first one can be proved with the fact that "conditional expectation and integral can be exchanged", while the second one can be proved with the property that "The point limit of measurable function (if exists) is measurable".

Specifically, we have:

$$E(A(t)|\mathcal{G})=E(\int_0^t \lambda(s)ds|\mathcal{G})=\int_0^t E(\lambda(s)|\mathcal{G})ds=\int_0^t \lambda(s)ds = A(t),$$

therefore, $\Rightarrow$ is true.

We also have:

$$\lambda(t)=\lim_{n\to \infty} n(A(t-\frac{1}{n})-A(t)),$$

therefore, $\Leftarrow$ is true.

(a) $\Leftrightarrow$ (d). This requires the property:


i.e., the intensity process describe the (average) changing rate of $N_t$. Therefore, $\lambda_t$ is measurable iff $dN_t$ (or $\frac{dN_t}{dt}$) is measurable.

To prove the property, note that:


because $M_t$ is a martingale.

The LHS has $E(M_t|\mathcal{F}_{t-})=E(N_t-A_t|\mathcal{F}_{t-})=E(N_t|\mathcal{F}_{t-})-E(A_t|\mathcal{F}_{t-}) = E(N_t|\mathcal{F}_{t-}) - A_{t}$ because $A_t$ is a predictable process (i.e., measurable to $\mathcal{F}_{t-}$).

The RHS has $M_{t-}=N_{t-}-A_{t-}=E(N_{t-}|\mathcal{F}_{t-})$ because $N_{t-}$ is $\mathcal{F}_{t-}$-measurable by definition.

Therefore, combining the LHS and RHS, we have:


  • 1
    $\begingroup$ In your first equivalence, you should be very careful: even though $A$ is absolutely continuous, it doesn't mean that it is the Riemann integral of $\lambda$. You only know that it is the Lebesgue integral! So instead, you want to use the fact that you can exchange the integral and the conditional expectation! For the converse, I believe you are missing a minus sign in the limit no? $\endgroup$
    – Mushu Nrek
    Nov 7, 2023 at 7:07
  • $\begingroup$ For the second equivalence, I am not able to follow: with what you write, I find that $$E(N_t - N_{t-} \vert \mathcal F_{t-}) = A_t - A_{t-},$$ not $dA_t$. (Similarly, I would not understand by $dN_t = N_t - N_{t-}$, but that might be just a different notation.) Take for example the Poisson process with unit rate $\lambda_t \equiv 1$. Then $A_t = t$ and $A_t - A_{t-} = 0$ and not $\lambda_t dt = dt$ (what I think you wanted to put in the last line, because it fits better with your introductory claim). $\endgroup$
    – Mushu Nrek
    Nov 7, 2023 at 7:14
  • $\begingroup$ Thanks for your interest in this question! Regarding the first equivalence, you are right about the Lebesgue integral (I have modified the proof as suggested). What do you mean by "missing a minus sign"? Regarding the second equivalence, I got the idea and notations from pg. 52, Statistical Models based on Counting Process (Andersen et al. 1993). $\endgroup$ Nov 7, 2023 at 8:13
  • 1
    $\begingroup$ In the paper On Cox Processes and Credit Risky Securities D. Lando goes to the bottom of this. In short: ${\cal G}_t$ is a smaller filtration to which the intensity process is adapted. $N_t$ is a counting process with that intensity. If you boil it down to the Poisson process where $\lambda_t$ is deterministic, hence adapted to the trivial filtration ${\cal G}_t=\{\emptyset,\Omega\}$ it becomes intuitively clear that the natural filtration of $N_t$ must be much larger. Sure: there is additional randomness coming from the event if there is a jump at $t$ or not. $\endgroup$
    – Kurt G.
    Nov 7, 2023 at 9:06
  • 1
    $\begingroup$ If $A_t$ is absolutely continuous, it is differentiable almost everywhere. The same comes from the fact that the compensator is increasing. @MingzhouLiu $\endgroup$
    – Mushu Nrek
    Nov 8, 2023 at 22:41

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