• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $I$ be an index set
  • $\mathbb F=(\mathcal F)_{t\in I}$ be a filtration on $(\Omega,\mathcal A)$
  • $X=(X_t)_{t\in I}$ be a stochastic process on $(\Omega,\mathcal A,\operatorname P)$

If $I=\mathbb{N}_0$, then $X$ is called $\mathbb F$-predictable $:\Leftrightarrow$ $X_0$ is a constant and $$X_n\text{ is }\mathcal F_{n-1}\text{-measurable}\;\;\;\text{ for all }n\in\mathbb N\;.\tag{1}$$ If one thinks about $\mathcal F_n$ as being the information known about $X$ until time $n$, $(1)$ means that at time $n-1$ we already know how $X_n$ will behave.

If $I=[0,\infty)$, then $X$ is called $\mathbb F$-predictable $:\Leftrightarrow$ $X$ is measurable with respect to $$\sigma\left(\left\{ \left\{0\right\}\times F:F\in\mathcal F_0\right\}\cup\bigcup_{0\le s<t}\left\{(s,t]\times F:F\in\mathcal F_s\right\}\right)\;.$$ Let $t>0$. Now, the intuition is much harder to find. There is no single number $s$ immediately before $t$. And clearly, $s<t$ doesn't imply the $\mathcal F_s$-measurability of $X_t$.

So, what is the intuition? Maybe we can show that $X_t$ is measurable with respect to $$\bigcup_{0\le s<t}\mathcal F_s\tag {2}\;.$$ I would be very happy if this would be true and someone could provide a proof.ic

  • $\begingroup$ I'm confused by your notation. By e.g. $\{0\}\times \mathcal F_0$ do you mean $\{(0,F) : F\in\mathcal F_0\}$? $\endgroup$ – Math1000 Jul 6 '15 at 17:10
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    $\begingroup$ @Math1000 I was confused by notation, too. Fixed now. $\endgroup$ – 0xbadf00d Jul 6 '15 at 17:31
  • $\begingroup$ Isn't this the definition of predictable for a continuous-time process? Basically you want $X$ measurable with respect to the $\sigma$ algebra generated by all left continuous adapted processes... which is what you have. Intuition = I know $X_t$ given all $\mathcal{F}_s$ where $s < t$ $\endgroup$ – Jeb Jul 6 '15 at 18:36
  • $\begingroup$ @Jeb No, it's not exactly the definition. However, it seems like one can't expect anything beyond that, since $(2)$ is no $\sigma$-algebra in general. $\endgroup$ – 0xbadf00d Jul 6 '15 at 18:48

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