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Let $x_t:\Omega \to \mathbb{R}$ a random variable and $\{x_t\}:\Omega \to \mathbb{R}^\mathbb{Z}$ a stochastic process. Can someone explain to me the notion of the $\sigma$-algebra generated by a stochastic process? I understand that of the $\sigma$-algebra generated by the random variable $x_t$ but I am struggling to extend it to stochastic processes. An example (maybe with discrete $\Omega$) would help too.

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  • $\begingroup$ What is your definition of a stochastic process? $\endgroup$
    – Toni
    Apr 8 '20 at 18:54
  • $\begingroup$ A collection of random variables $\{x_t\}$ defined on $(\Omega, \mathcal{F}, P)$ (and in my definition above indexed by the integers) taking values on some measurable space $(\mathbb{R},\Sigma)$. $\endgroup$
    – econ86
    Apr 8 '20 at 19:01
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Let $(X_t)_{t\in\mathbb{Z}}$ be a (real-valued) stochastic process on some measurable space $(\Omega,\mathcal{F})$, i.e., for all $t\in \mathbb{Z}$ $X_t:\Omega\to\mathbb{R}$ is a random variable.

The sigma algebra generated by the stochastic process $(X_t)_{t\in \mathbb{Z}}$ is the smallest sigma algebra such that $X_t$ is measurable for all $t\in T$, i.e., $$\sigma\left((X_t)_{t\in \mathbb{Z}}\right)=\sigma\left(\bigcup_{t\in\mathbb{Z}}X_t^{-1}(\mathcal{B}(\mathbb{R}))\right)=\left\{A\in\mathcal{F}\mid \exists t\in\mathbb{Z} \;\exists B ∈ \mathcal{B}(\mathbb{R}):\; A=X^{-1}_t(B)\right\},$$ where $\mathcal{B}(\mathbb{R})$ denote the Borel sets over the real line.

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  • $\begingroup$ Thanks. I see the definition, but can you explain further? In particular, how are the sigma-algebras generated by each random variable of the process connected to the sigma-algebra of the process? $\endgroup$
    – econ86
    Apr 9 '20 at 18:55
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    $\begingroup$ The sigma algebra generated by the whole process contains all the sigma algebras generated by the individual random variables, but it possibly contains even more sets because it also contains sets you can get from complements and arbitrary unions of sets from the individual sigma algebras. The following holds $$\sigma\left((X_t)_{t\in \mathbb{Z}}\right)=\sigma\left(\bigcup_{t\in\mathbb{Z}}\sigma(X_t)\right).$$ $\endgroup$
    – Toni
    Apr 9 '20 at 19:24
  • $\begingroup$ Are you missing a $$\sigma$$ sign in the last equality? The class of sets in that last expression is actually the union of individual sigma fields which may itself not be a sigma field. $\endgroup$ Jun 30 at 20:15

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