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Bernt Øksendal writes in his book Stochastic Differential Equations (page 11) that a stochastic process is a probability measure $P$ on the measurable space $((\mathbb{R}^n)^T, \mathcal{B})$.

The sample space $(\mathbb{R}^n)^T$ is the set of all functions $f: T \rightarrow \mathbb{R}^n$, and the $\sigma$-algebra $\mathcal{B}$ is generated by the set of functions $\{f: f(t_1) \in F_1, \dots, f(t_k) \in F_k \}$, where $F_i \subset \mathbb{R}^n$ are Borel sets.

I understand the point of view that a stochastic process can be thought of as a realisation from $(\mathbb{R}^n)^T$. And I can accept the definition of the $\sigma$-algebra $\mathcal{B}$. But I'm struggling to grasp why a stochastic process is a probability measure $P$ on the space $((\mathbb{R}^n)^T, \mathcal{B})$.

How does $P(f)$, with $f \in \mathcal{B}$, represent a stochastic process? And how does this representation fulfil axioms such as $0 \leq P \leq 1$?

harisf
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  • Do you know Kolomogorov 's Consistency Theorem? (AKA Kolomogorov 's Existence Theorem). – Kavi Rama Murthy Nov 15 '20 at 08:40
  • There's a Kolmogorov theorem on that very page of Øksendal's book called the "Kolmogorov extension theorem". It says that given a family of probability measures $P$ as above, there exists a stochastic process that has its finite-dimensional distribution given by $P$. Ok, so that's what Øksendal means when he writes that "P is a stochastic process"? – harisf Nov 15 '20 at 08:55
  • Yes, that is what he means. – Kavi Rama Murthy Nov 15 '20 at 09:31
  • Bernt Øksendal does not write that. He writes, I quote, "one may also adopt the point of view that a stochastic process is a probability measure $P$ on the measurable space $((\mathbb R^n)^T,\mathcal B)$". The devil is in the details: one may, but one does not have to. Actually, I would not recommend doing this. Say, you have two independent standard Wiener processes in $\mathbb R^n$. They have the same distribution, so by this point of view they should be the same process; however, this is absurd. – zhoraster Nov 15 '20 at 16:57
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    Thank you for the quote. However, my confusion was not about the possibility of adopting this viewpoint. I was concerned with the statement "$P$ is a stochastic process". And from the discussion above it seems that a clearer statement would be "$P$ represents a stochastic process". – harisf Nov 15 '20 at 17:17

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