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I'm having trouble understanding the concept of a previsible process in continuous time, so I'm asking this question to get a better idea of what it means for a process to be previsible.

(In what follows, suppose we're working with respect to a fixed filtered probability space.)

We say a process $H$ is previsible if $H$ is $\mathcal{P}$-measurable, where $\mathcal{P}$ is the $\sigma$-algebra generated by events of the form $E\times(s,t]$, where $E\in\mathcal{F}_s$ and $t>s$.

We can easily show that for any previsible process $H$, $H_t$ is $\sigma(\mathcal{F}_s:s<t)$ -measurable. Is every process with this property previsible? If not, is there a simple counterexample?

Thank you.

Ben Derrett
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    The two are equivalent (and the usual word is predictable). See here, for example. – Did Sep 11 '11 at 16:58
  • @didier Thank you for the comment (and the correction). I can't find where he shows equivalence in the reference. Lemma 7, proving an equality of $\sigma$-algebras, seems close, as its definition of $S$ is mine of $P$. – Ben Derrett Sep 11 '11 at 18:07

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Not necessarily. Consider a unit rate Poisson process $N = \{N(t), t\ge 0\}$ with its completed natural filtration $\{\mathcal{F}_t, t\ge 0\}$. Then $\sigma\{\mathcal{F}_s : s<t\} = \mathcal{F}_t$ (because $\mathbf{P}[N(t) \not= N(t-)]=0$ for each fixed $t>0$); therefore $N(t)$ is $\sigma\{\mathcal{F}_s : s<t\}$ -measurable for each $t>0$. But the process $N$ is not previsible. For if it were then the martingale $M(t):=N(t)-t$ would be previsible hence continuous, which it clearly is not.

John Dawkins
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