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The predictable processes are defined as the elements of the predictable sigma algebra, which is generated by the caglad processes. What kinds of extra processes do we get by extending from caglad to predictable processes? Specifically, are there predictable processes that are not left continuous? Apologies if there are lots of obvious examples (it seems like there should be, by analogy with extending from continuous to measurable functions) but it isn't obvious to me from the definition what these processes might be or how to go about constructing them.

mathrat
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    @ mathrat : the following dertiministic process gives an answer : $X_t=1_{\lbrace t \geq 1\rbrace}$. – TheBridge Dec 02 '14 at 21:30
  • @TheBridge: thanks. I see now that any deterministic process is trivially predictable. I suppose I was looking for a more interesting example, but I think I just need to inspect the definitions more carefully. I'll try to post back here myself once I've found a more interesting process (unless someone else beats me to it). – mathrat Dec 03 '14 at 14:39
  • @ mathrat : Any càdlàg process with predictable jumps only would also do the trick. Best regards. – TheBridge Dec 03 '14 at 16:37
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    Another interesting example might be $(\mathrm{sgn}(B_t))_{t\geq 0}$ where $B$ is a one-dimensional standard Brownian motion. This process is neither left- nor right-continuous, though predictable. – user427574 Feb 21 '19 at 11:34

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Answering my own question, but here is what I was looking for. Credit goes entirely to TheBridge for guiding me to this. Let $T$ be a predictable stopping time. Then $X=(X_t)_{t\geq 0}$ defined by $X_t = \mathbb{1}_{\{t \geq T\}}$ for $t\geq 0$ is a predictable process. To prove this, let $T_n$ be an announcing sequence for $T$. Clearly the processes $X^n_t = \mathbb{1}_{\{t > T_n\}}$ are predictable (they are càglàd). Define a stochastic interval as follows \begin{equation*} [T,\infty) = \{(t,\omega) \in \mathbb{R}^+ \times \Omega : T(\omega) \leq t\}. \end{equation*} We can similarly define $(T_n,\infty)$ and observe that $(T_n,\infty) \in \sigma(X^n)$ and therefore these intervals are in $\mathcal{P}$, the predictable $\sigma$-algebra. Furthermore, \begin{equation*} [T,\infty) = \bigcap_n (T_n,\infty) \in \mathcal{P}. \end{equation*} It follows that $\sigma(X) \subset \mathcal{P}$, i.e. $X$ is predictable.

peer
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mathrat
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