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I've found this question, with a very good answer, but they don't broach my question.

In Oksendal's Stochastic Differential Equations book, it's written «the stochastic Process is a probability measure (...)» [Page 11]. What does he mean by that?

Based on what I've read on the book and wikipedia, I wonder if it's not the following:

Let $T$ be a parameter set (countable or uncountable), and we have the probability space $(\Omega,\mathcal{F},P)$ with a collection of r.v. $\{X_t:t\in T\}$, where $X_t$ are $(E,\mathcal{E})$-valued r.v..

It can be proved that $X$ can be viewed as $(E^T,\mathcal{E}^T)$-valued r.v., and in that regard we can also view $P\circ X^{-1}$ as a probability measure defined on $(E^T,\mathcal{E}^T)$.

Is this what is meant as a Stochastic Process being a probability measure?

An old man in the sea.
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  • Stochastic process is *not* a probability measure. It does induce a probability measure $P\circ X^{-1}$ on $(E^T, \mathcal E^T)$ (unfortunately, $\mathcal E^T$ is quite poor, especially when $T$ is uncountable). – zhoraster Oct 01 '18 at 19:40
  • @zhoraster thanks for you comment. I wrote the stochastic Process is a probability measure, since it's what Oksendal wrote in page 11 of his Stochastic Differential Equations, 6th edition – An old man in the sea. Oct 01 '18 at 20:42
  • This point of view is very inconvenient. Say, you have two independent processes with the same distribution. Then, by this point of view, they are the same object. – zhoraster Oct 02 '18 at 05:28
  • @zhoraster I've been reading the book, and it seems that this point of view is fundamental to the notion of stochastic integral... Why do you say that it's inconvenient? Could you please elaborate? – An old man in the sea. Oct 03 '18 at 17:34
  • As I wrote, it is inconvenient, because, for example, one has to consider two independent identically distributed processes as the same object. I also don't see how is is 'fundamental' in the definition of stochastic integral. – zhoraster Oct 05 '18 at 11:16

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