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Where $\mathcal{H}$ is a $\sigma$-algebra on $\Omega$, section 9.1 here discusses thinking of a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{H}$ "as the collection of all numerical random variables that are $\mathcal{G}$-measurable".

This answer gives some intuition for $\sigma$-algebras generated by random variables ("restricting the world's probabilism"), which aligns with the lecture notes (the first link). Even so, I don't see where any functions (random variables) are in a set of sets ($\mathcal{G}$).

Can I have some pointers to further reading?

Zhanxiong
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johnsmith
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    A random variable determines a sub sigma algebra (its events are the inverse images of the measurable sets of real numbers). Conversely, a sub sigma algebra determines the space of all measurable functions: the random variables. The correspondence is one-to-one. – whuber Jan 28 '23 at 14:58

1 Answers1

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$\newcommand{\classG}{\mathcal{G}}$

To facilitate discussion, let's first settle the probability space to be $(\Omega, \mathcal{H}, P)$, and let $\mathcal{G}$ be a sub-$\sigma$-algebra of $\mathcal{H}$. The complete sentence that the author wrote is:

Cinlar advocates for thinking of the sub-$\sigma$-algebra $\mathcal{G}$ both as a collection of events (i.e., measurable subsets of $\Omega$), and as the collection of all numerical random variables that are $\classG$-measurable.

The first part of the quote is obvious: by definition $\classG$ (in fact any subclass $\mathcal{A}$ of $\mathcal{H}$) is a collection of events.

Mechanically speaking, the second part of the quote does not make sense, because $\classG$ is a class of sets, instead of random variables (i.e., functions defined on $\Omega$). However, such "thinking" does make sense in view of the following isomorphism $\tau$ between $\classG$ and $\mathscr{I}$, which is defined as the class of indicator functions of events in $\classG$: for every $G \in \classG$, define $\tau(G) = I_G(\omega), \omega \in \Omega$. Now $\mathscr{I} = \{I_G: G \in \classG\}$ is a collection of numerical random variables that are $\classG$-measurable -- intuitively, for each $\omega \in \Omega$, knowing the value of $I_G(\omega)$ is equivalent to knowing if the event $G$ occurs, that is, if $\omega \in G$. In addition, $\tau$ is obviously a bijective between $\classG$ and $\mathscr{I}$. This justifies the second part of the quote.

For more enlightening discussion on sub-$\sigma$-algebra and its heuristic relation to partial information, I recommend you reading the Subfields subsection in Section 4 of Probability and Measure by Patrick Billingsley.

Zhanxiong
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  • To clarify, $\mathscr{I}$ doesn't have all $\mathcal{G}$-measurable numerical random variables, right? – johnsmith Jan 29 '23 at 12:32
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    Right, $\mathscr{I}$ cannot contain all $\mathcal{G}$-measurable functions. But there is a theorem that any $\mathcal{G}$-measurable function is a limit of simple functions (which are finite linear combinations of members in $\mathscr{I}$). So to strictly fit the qualifier all, you may enlarge the definition of $\mathscr{I}$ to the monotone class generated by all simple functions. However, this loses the transparent 1-to-1 correspondence between $\mathcal{G}$ and $\mathscr{I}$. To me, the statement made by the author is largely heuristic, which is probably not mathmatically provable. – Zhanxiong Jan 29 '23 at 13:26
  • @johnsmith The statement is more relevant if $\mathcal{G} = \sigma(X)$ (where $X$ is a given random variable), for which $\mathscr{I}$ can be modified to $\mathscr{I} = {f(X): f \text{ is a Borel function from } \mathbb{R} \to \mathbb{R}}$. In this case $\mathscr{I}$ indeed contains all $\mathcal{G}$-measurable functions since any $\sigma(X)$-measurable function must be a function of $X$. – Zhanxiong Jan 31 '23 at 01:57