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Dec 23, 2018 at 8:28 vote accept Emerald
Dec 23, 2018 at 8:28 vote accept Emerald
Dec 23, 2018 at 8:28
Dec 21, 2018 at 20:39 answer added Dirk Werner timeline score: 2
Dec 4, 2018 at 6:41 comment added Emerald @Dirk, I mean, if $ω ↦K(s,ω) $is measurable (and integrable) for each $ ω$, need $ω↦∫^t_0 K(s,ω)ds$ be measurable? I'm looking for a counterexample.
Dec 3, 2018 at 22:34 comment added Dirk Werner Thanks for the clarification, but ... may I suggest to use precise wording in a question? For the record, in stochastic analysis a process $(\xi_s)_s$ is called measurable if $(s,\omega)\mapsto \xi_s(\omega)$ is measurable for the product $\sigma$-algebra. Incidentally, $\xi_s(\omega)$ is a number, not a process. Returning to what you wanted to ask, it seems to me now that the question is: If $s\mapsto K(s,\omega)$ is measurable (and integrable) for each $\omega$ and $\omega\mapsto K(s,\omega)$ is measurable for each $s$, need $\omega\mapsto \int_0^t K(s,\omega)\,ds$ be measurable?
Dec 3, 2018 at 14:27 comment added Emerald @Dirk We don't know if the process will be joint measurable. So everything is not obvious.
Dec 2, 2018 at 22:40 comment added Dirk Werner I think your question doesn't really have to do with filtrations and stochastic processes. You are given a mapping $K: (s,\omega)\mapsto \xi_s(\omega)$ on $[0,t]\times \Omega$, and you seem to be assuming ("measurable random process") that $K$ is measurable for the product $\sigma$-algebra of $[0,t]\times \Omega$. But then it is embodied in the proof of Fubini's theorem that $\omega\mapsto \int_0^t K(s.\omega)\,ds$ is measurable. Maybe I'm missing something?
Dec 1, 2018 at 18:04 history edited Emerald CC BY-SA 4.0
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S Dec 1, 2018 at 17:32 history suggested Amir Sagiv CC BY-SA 4.0
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Dec 1, 2018 at 17:31
Dec 1, 2018 at 17:07 history asked Emerald CC BY-SA 4.0