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Let $(\Omega,\mathscr{F},\mu)$ be a probability space and $\mathscr{G}\subseteq \mathscr{F}$ a $\sigma$-algebra. I have seen it referred to many times that $\mathscr{G}$ is the "information" which is available to us. I think I kinda understand it, but I am not satisfied in my own understanding of it.

Let us say that $A$ is an event and $\mathscr{G} = \{\emptyset, A,A^c,\Omega\}$. Let $\omega$ be a sample point of the experiment. We do not know which of the events in $\mathscr{G}$ contain $\omega$. Now $\Omega$ certaintly contains $\omega$, but we already knew that, so we do not gain any insight. However, either $A$ or $A^c$ will contain $\omega$. If we somehow knew that $A$ contains $\omega$, then that gives us additional insight.

1) Does anyone have a better way of thinking of $\mathscr{G}$ as our "information"?

2) If $\mathscr{G}'\supseteq \mathscr{G}$, then how do we think of $\mathscr{G}'$ as having "more information"? Obviously, it is a larger $\sigma$-algebra, and it has more events, but ignoring set theory, what should one's intuition be for $\mathscr{G}'$?


Follow up question.

3) Let $\xi:\Omega\to\mathbb{R}$ be a $\mathscr{G}$-measurable. I have seen people refer to $\xi$ as "a random variable whose information is known from $\mathscr{G}$", or something along those lines. What is the motivation for this?

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  • $\begingroup$ I don't really understand your first sentence. What is G, just any old sub-$\sigma$-algebra of F? I see that F is the $\sigma$-algebra needed to define the probability space, but what's G doing here? $\endgroup$ – Kodiologist Aug 21 '16 at 6:00
  • $\begingroup$ @Kodiologist When you condition on a $\sigma$-algebra G, G is often referred to as "information that you know". $\endgroup$ – Nicolas Bourbaki Aug 21 '16 at 6:05
  • $\begingroup$ Huh? But you can't condition on G. It's not an event. $\endgroup$ – Kodiologist Aug 21 '16 at 6:08
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    $\begingroup$ @Kodiologist see, e.g., en.wikipedia.org/wiki/… ("see" meaning "see that stuff conditional on sigma-algebras is indeed talked about", not "read and learn" for which the Wikipedia article is probably not the best place). $\endgroup$ – Juho Kokkala Aug 21 '16 at 6:48
  • $\begingroup$ @Kodiologist You should delete your comments. They are not about the nature of my question, but your question about conditioning on $\sigma$-algebras. Which is a separate question in itself. $\endgroup$ – Nicolas Bourbaki Aug 21 '16 at 7:33
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$\mathscr{G}$ is our information in the sense that for all $A \in \mathscr{G}$, we know whether $\omega \in A$.

Let us use the Tickets in a box metaphor, extended to handle $\sigma$-algebras so that the ticket mentions for all $A\in \mathscr{F}$ whether the outcome represented by the ticket belongs to $A$. Now, say that someone else picks the ticket and we don't see it. For any $A \in \mathscr{G}$ we may ask whether the ticket says that the outcome is in $A$ and the person holding the ticket tells us. However, if we ask about some $A \in \mathscr{F} \setminus \mathscr{G}$, we hear "Sorry, you don't know that".

Larger $\sigma$-algebra is more information

This also explains why moving to $\mathscr{G}' \supset \mathscr{G}$ means gaining new information -- now we still get answers to $[X \in A?]$-questions about any$A \in \mathscr{G}$ and additionally to some new questions -- those where $A \in \mathscr{G'} \setminus \mathscr{G}$.

Random variables

So, the tickets also contains the values of random variables. If the random variable $X$ is $\mathscr{G}$-measurable, we get answers to all our questions about its value, such as $[$is $X$ equal to $3]$, since by $\mathscr{G}$-measurability of $X$, $\{\omega \mid X(\omega)=3\}\in\mathscr{G}$. Or, to handle the delicacies of the uncountable case, we may also ask $[$Is $X$ in the set $B]$? (Since for any particular value we think about, the probability of hearing "yes" may be $0$ and that would be boring). So, in this sense we have all information about the realization of the random variable, if our information is $\mathscr{G}$ and the RV is $\mathscr{G}$-measurable.

Caveat: the definition of measurability of random variables restricts the sets $B$ we may ask about. $[$Is $X(\omega) \in B]$ is answered if $B$ is a measurable set in the value space of the random variable (usually Borel $\sigma$-algebra is assumed with $\mathbf{R}$ without mentioning). So, in the uncountable (nondiscrete $X$) case, don't ask whether $X$ is in the Vitali set or the oracle holding the ticket shall be mad.

Reference

I did not cite any reference in the answer but I consulted

  • J. Jacod and P.E. Protter. Probability essentials (2nd edition), Springer, 2004

about the definition of measurability of random variables. (And have learned these things from the same book previously, if I recall correctly).

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