I'm reading through the following theorem:

Let $X=\{X_t,t\in T\}$ be a stochastic process. Then $\sigma (X)=\sigma ( \cup_{t\in T} \sigma (X_t))$

From my basic knowledge of measure theory, I know $\sigma (S)$ as the smallest $\sigma$-algebra containing the system of subsets $S$. In probability theory we're defining $\sigma (X) $ as the $\sigma$-algebra generated by the random variable $X$ (Definition: Let $X:(\Omega , $ $\mathcal{A}$ $ )\rightarrow (S,$ $\mathscr{S} $ $ )$ then we define $\sigma(X)=\{[X\in B], B\in $ $\mathscr{S} $ $\}$ )

That's where the confusion comes in:

a) What is the connection between those two (if there is any) ?

b) In the theorem, is the equality then "The $\sigma$-algebra generated by $X$ equals to the smallest $\sigma$-algebra containing the union of $\sigma$-algebras generated by the $X_t$ ?

Thank you.

  • 1
    $\begingroup$ How do you define $\sigma(X)$ for a stochastic process $X$? (Just asking, because the definition I know is exactly the one stated in the theorem.) $\endgroup$
    – saz
    Commented Oct 11, 2014 at 9:55
  • $\begingroup$ I added the definition in the original post. $\endgroup$
    – Dahn
    Commented Oct 11, 2014 at 10:05
  • $\begingroup$ Well, you added the definition of a sigma-algebra generated by a random variable, but a stochastic process is a family of random variables. Of course, we can consider the process as a random variable $X: T \times \Omega \to \mathbb{R}$, but then we end up with a different $\sigma$-algebra than $\sigma(\bigcup_t \sigma(X_t))$. $\endgroup$
    – saz
    Commented Oct 11, 2014 at 10:53
  • $\begingroup$ We've defined them as the smallest sigma-algebra containing something we called "measurable cylinders". I could write down the definition, but is that relevant? Say this is the definition and not a theorem - then my question is that I don't understand the definition properly. $\endgroup$
    – Dahn
    Commented Oct 11, 2014 at 15:09
  • $\begingroup$ Yes, my confusion arose from the fact that you use $\sigma(X)$ for two different things and I wasn't sure which one you are talking about. I'll write an answer... $\endgroup$
    – saz
    Commented Oct 11, 2014 at 15:13

1 Answer 1

  1. By definition, $\sigma(X)$ denotes the smallest $\sigma$-algebra such that $X$ is measurable, i.e. $$\sigma(X) = \{X^{-1}(B); B \in \mathscr{S}\}. \tag{1}$$ Define $$S = \{C; \exists B \in \mathscr{S}: C = f^{-1}(B)\}.$$ Note that $S$ is a $\sigma$-algebra as a pre-image $\sigma$-algebra; therefore, $$\sigma(X) \stackrel{(1)}{=} S = \sigma(S).$$
  2. Yes, this is a correct. Another one: "The $\sigma$-algebra generated by $X$ equals the smallest $\sigma$-algebra such that $X_t$ is measurable for any $t \in T$."

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