# Why does jump process has to be Cadlag and not the other way around

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.

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).
• 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? – Kenneth Chen Jun 23 '16 at 18:49
• 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 ? – MJ73550 Jun 24 '16 at 9:12
• MJ73550 why a caglad process is $t_{+}$ measurable please? That is indeed weird if that's the case. – Kenneth Chen Jul 4 '16 at 14:08
• 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. – MJ73550 Jul 4 '16 at 19:19