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According to my lecture notes, a stochastic process is a sequence $(X_n)_{n \in \mathbb N}$ of random variables defined on $(\Omega, \mathcal F, P)$ (which is a probability space).

Then follows the definition of a filtration, and then the definition of the canonical filtration as $\mathcal F_n = \sigma(X_1, \dots, X_n)$. I search on this site questions about filtration to have a better understanding of it, and I came accros this question.

Although the answer makes sense, I don't understand how $\sigma(X_1, \dots, X_n)$ can be subset of $\mathcal F$, because for me $(\Omega, \mathcal F)$ is the measurable space over which each $X_i$ is defind, but not the entire sequence of $X_i$. This might be wrong, but then I don't know how to understand the definitions of my lecture.

matyce
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  • What is F here? – vallev Nov 09 '23 at 22:55
  • Well I think that this where I struggle. In the post I linked $\mathcal F$ seems to be $\mathcal P(\Omega)$ with $\Omega = {1, \dots, 6}^n$. However, if I would formally write a sequence of dice throws according to my understanding of the definitions, given that each $X_i$ is a throw, I would set $\Omega = {1, \dots, 6}, \mathcal F = \mathcal P(\Omega)$. But then I would not be able to construct the filtration as defined in the answer... – matyce Nov 10 '23 at 00:13
  • omega is the sample space of ALL outcomes, not the outcomes of a throw – vallev Nov 10 '23 at 00:45

2 Answers2

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Your statement that the entire sequence of $X_i$'s is not defined over the measurable space $(\Omega,\mathcal{F})$ is actually incorrect. I'll try and break this down into steps, our goal is to construct a probability space supporting infinite sequences of random variables. Let's start with something easy and do finite sequences. If each $X_i$ is defined on the probability space $(\Omega_i,\mathcal{F}_i,\mathbb{P}_i)$ then we can define the probability space in which the sequences $(X_n)_{n=1,\ldots,k}$ live as the product space of the individual probability spaces i.e, $$ (\Omega,\mathcal{F},\mathbb{P}) = (\Omega_1 \times \cdots \times \Omega_k,\mathcal{F}_1 \otimes \cdots \otimes \mathcal{F}_k,\mathbb{P}_1 \otimes \cdots \otimes \mathbb{P}_k ). $$ Now if you wish to extend this to infinite sequences (which we do) then things get more tricky. You need to use what's known as Kolmogorov's extension theorem https://en.wikipedia.org/wiki/Kolmogorov_extension_theorem. As I'm not sure at which level of mathematical detail you would like, I'll only give the idea behind this theorem rather than a detailed explanation. Essentially it says in order to define this ''infinite product space'' you need to first construct the finite dimensional distribution of your desired sequence on the state space (which is usually $(\mathbb{R},\mathcal{B}(\mathbb{R}))$) and then show that the defined distribution has ceratin properties which then allows you to apply the theorem to tell you such an ''infinite product space'' exists. This fact is usually ignored in introductory courses on probability as it is very technical.

The main thing for you to take from this answer is that the entire sequence is in fact defined on $(\Omega,\mathcal{F},\mathbb{P})$ as given in your lecture notes so the sigma algebra generated by the first $n$ elements of the sequence is indeed contained in $\mathcal{F}$.

Emmet
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  • Thank you very much for your answer, and sorry for my (very) late reply. In my course, it is written that $\forall t \in \mathcal{T}, \omega \mapsto X_t( \omega)$ is a measurable application from $(\Omega, \mathcal F)$ to the states space, wich lead me to think the entire sequence should be defined on a kproduct of $(\Omega, \mathcal F)$. – matyce Dec 13 '23 at 15:27
  • I've been doing probability and stochastics for a number of years now and I have never heard of a "measurable application". I'm presuming your instructor means measurable function. the statement $\forall t \in \mathcal{T}, \omega \mapsto X_t(\omega)$ is measurable, seems to suggest a definition for an arbitrary indexing set as apposed to the natural numbers which is all thats needed for sequences. If this is the case you need to work even harder to construct the probability space, but again my statement is still true that the entire sequence is defined on $(\Omega,\mathcal{F})$. – Emmet Dec 14 '23 at 11:49
  • If you would like to delve deeper into how these spaces are constructed you should take a look at chapter 2 of "Probability" by Albert Shiryaev. He gives an excellent exposition of how these kinds of probability spaces are constructed. – Emmet Dec 14 '23 at 11:52
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Assuming $\mathcal{F}$ is the sigma-algebra. Since $\mathcal{F}$ contains all possible information about the outcomes in $\Omega$, it certainly contains all the information provided by any subset of the random variables, hence $\sigma(X_1, \ldots, X_n) \subseteq \mathcal{F}$. The indexing $n$ is variable.

vallev
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  • Thank for your answer ! What confuses me is that my notes say that each $X_t$ is defined over $(\Omega, \mathcal F)$... – matyce Dec 13 '23 at 15:29