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In all books and references that I have been exposed to, the jump processes have been defined to be Cadlag(right continuous with left limits). But no one has explained why this is the preferable case, why can't it be Caglad?

I suspect it has something to do with filtration, but I don't know the exact reasoning.

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I don't know if this is enough. But here is my understanding.

Let's imagine a simple process like a Poisson process. It is naturally cadlag, because at the time you jump, you jump. Just before, you have not jumped. Mathematically, if the first jump occurs at $t$, $\forall s<t, N_s=0$ and $N_t=1$. It means that the jump occuring at time $t$ is $t$-measurable (even if it is not predictible).

So a cadlag process means that at the time of the jump, you see the process jumping.

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    $\begingroup$ Interesting answer. It may also be related to the set-up of the filtration and the integral definition etc. $\endgroup$
    – Gordon
    Commented Jun 23, 2016 at 15:11
  • $\begingroup$ That makes sense, but I guess a left continuous could also be $t-measurable$. The portfolio position, for example, has usually been assumed to be a predictable process. In that sense, of course it's hard to imagine what a Cadlag position will be, it's the trader's own decision to change position so there's should be no surprise. If jumps are "surprises" then I agree it's naturally Cadlag, but is there a mathematical reasoning for it? $\endgroup$
    – Kenneth Chen
    Commented Jun 23, 2016 at 18:49
  • $\begingroup$ speaking of the poisson process caglad would mean that jumps at time $t$ is $t_+$-measurable for the process. Do you agree that it would be weird that the jump time $\tau$ to be not a stopping time of the filtration of the process ? $\endgroup$
    – M. Jeunesse
    Commented Jun 24, 2016 at 9:12
  • $\begingroup$ MJ73550 why a caglad process is $t_{+}$ measurable please? That is indeed weird if that's the case. $\endgroup$
    – Kenneth Chen
    Commented Jul 4, 2016 at 14:08
  • $\begingroup$ I did not say that the process is $t_+$ measurable. I said that knowing jump has occurred before t I.e $\tau\leq t $ would be $t_+$ measurable due to the caglad behavior (draw it to convince you). That's why we like cadlag processes. $\endgroup$
    – M. Jeunesse
    Commented Jul 4, 2016 at 19:19
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Perhaps not the answer you are expecting but quoting "An introduction to the theory of point processes: Volume I, Elementary theory and methods. Springer, 2002.", you do not always take càdlàg processes. It really depends on what you want to model.

  1. It makes sense to have non-previsible jumps: this is this idea of càdlàg, since it is continuous from the right;

  2. but in the case of point proceses where you have an underlying intensity, one would want the intensity to be continuous from the left! Because you want the conditional intensity to be defined by its history, not by the point itself.

Perhaps some insightful keywords would be: you want the intensity to be "predictable", but processes with jumps to be "adapted".

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