The definition of an adapted process $X$ is that $X_i$ be $(\mathcal{F_i}, \Sigma)$-measuriable where $\mathcal{F.} = (\mathcal{F_i})_{i \in S}$ is a filtration of the sigma algebra $\mathcal{F}$ (probability space) and $\Sigma$ is part of the measurable space $(S, \Sigma)$. (and there are other required ones, but I will skip those parts.)

And predictable processes are the ones that $X_t$ is measurable with respect to $F_{t-}$.

It seems that adapted left-continuous processes cannot be predictable processes - after all, if they are adapated, $X_t$ must be $F_t$ measurable, while the definition of predictable processes say that $X_t$ must be $F_{t-}$-measurable.

Why are these seemingly nonsenses working?

• ?? Adapted: $X_t$ $F_t$-measurable. Predictable: $X_t$ $F_{t-}$-measurable. Hence predictable $\implies$ adapted but not the other way round. Does this answer your post (which I am not sure to understand)? – Did Apr 6 '13 at 13:39
• @Did Maybe I am confusing concepts. So, being adapted process is usually said informally as "one cannot see into future." Being $\mathcal{F_t-}$-measurable seems to look like "being able to see into future"... Anyway, so being $\mathcal{F_t}$-measurable includes cases where it is $\mathcal{F_s}$-measurable where $s<t$. (to rephrase, the whole thing means that if it is $\mathcal{F_s}$-measurable, then it is $\mathcal{F_t}$-measurable, where $s<t$, right?) – Zeus Apr 6 '13 at 14:05
• Let me suggest to drop (at least for a moment) this talk about "seeing into the future" and the like. Better to stick to the definitions before paraphrasing them. Now: adapted + left-continuous $\implies$ predictable. – Did Apr 6 '13 at 14:15
• @Zeus The title of your question is rather confusing, perhaps you could choose a more appropriate one. – saz Apr 7 '13 at 8:28

If $X_t$ is $\mathcal{F}_t$-measurable, this does not imply that $X_t$ is not $\mathcal{F}_{t-}$-measurable. Actually, left-continuous adapted processes are predictable:

Let $t \geq 0$. The information about the path $[0,t) \ni s \mapsto X(s,\omega)$ already determines the value at time $t$, since $$X(t,\omega) = \lim_{s \uparrow t} X(s,\omega)$$ by the left-continuity of $X$. This means (intuitively) that you can predict $X_t$ by the knowledge of $(X_s)_{s <t}$. Since $X_s$ is $\mathcal{F}_{t-}$-measurable for all $s<t$, this implies the $\mathcal{F}_{t-}$-measurability of $X_t$.

• The "Intuition" step seems like a full proof, quite rigorous. – Did Apr 6 '13 at 16:18
• @Did ... so I removed the "intuition" (and shortened the proof). – saz Apr 6 '13 at 16:27