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I am confused about the notion of $\sigma$-algebras representing information and what information is contained in $\sigma(X)$ for a random variable $X$.

Suppose $(\Omega, \mathcal{F}, \mathbb{P})$ is a probability triple, and $(Y_\gamma : \gamma \in C)$ is a collection of random variables. I am reading that $\sigma(Y_\gamma : \gamma \in C)$ consists of all the events $F \in \mathcal{F}$ such that for all $\omega \in \Omega$ it is possible to determine whether or not $\omega \in F$ given $(Y_\gamma(\omega) : \gamma \in C)$.

I don't even see how this makes sense if we restrict ourselves to a single random variable $X$. Suppose I know $X(\omega)$. I can only be sure $F$ occurred if $F\supseteq X^{-1}(\{X(\omega)\})$ and I can only be sure that $F$ didn't occur if $X(\omega) \notin X(F)$. But we know that $\sigma(X) = \sigma(\{X^{-1}(B) : B \in \mathcal{B}\})$. Where are all the other sets coming from?

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Maybe it will be helpful to look at the case where $X$ has a small finite range, e.g., $X:\Omega\to \{1,2,3,4\}$. In this case, we can partition the points in $\Omega$ into four disjoint subsets, $A:=X^{-1}(1)$, $B:=X^{-1}(2)$, $C:=X^{-1}(3)$, and $D:=X^{-1}(4)$. Given the value of $X$ at some $\omega$, we then know which of $A$, $B$, $C$, or $D$ contains $\omega$, so we can tell whether or not each of the events $A$, $B$, $C$, or $D$ happened, and $\sigma(X)$ must contain at least these four events. But we can also tell whether $B\cup C$ happened, since this event occurs just when $X(\omega)\in\{2,3\}$. So, $B\cup C\in \sigma(X)$ as well, and we also have several more sets; explicitly, $\sigma(X)$ is the Boolean algebra generated by $A$, $B$, $C$, and $D$: $$\sigma(X)=\{\emptyset,A,B,C,D,A\cup B,A\cup C,A\cup D,B\cup C,B\cup D,C\cup D,A\cup B\cup C,A\cup B\cup D,A\cup C\cup D, \Omega\}.$$

In the general case, any element of $\sigma(X)$ will be a union $\cup_{s\in S} X^{-1}(s)$, where $S$ is a measurable subset of the range of $X$.

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Okay, I see how we can tell if anything in $\sigma(X^{-1}(\{s\}) : s \in \Omega)$ happened, but how is it extended $\sigma(X^{-1}(S): S \in \mathcal{B})$? I'm pretty sure singletons don't generate the Borel $\sigma$-algebra. –  nullUser Jan 23 '13 at 21:51
    
If you want to know whether the event $E:=X^{-1}([2,3])$ happened, and you know the value of $X(\omega)$, then $E$ happened iff $2\le X(\omega)\le 3$. So, whether or not $E$ happened can be determined from $X(\omega)$. –  David Moews Jan 23 '13 at 21:56
    
Okay. But then how come I can't determine $X^{-1}(A)$ when $A$ is nonmeasurable? –  nullUser Jan 23 '13 at 22:03
    
You can determine whether or not this event happened (given $X(\omega)$), but you may not be able to find its probability, since $A$ is nonmeasurable. So, it's left out of the $\sigma$-algebra. –  David Moews Jan 23 '13 at 22:12
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A sense in which the $\sigma$-algebra generated by a random variable represents information is given by the Doob-Dynkin Lemma:

Result: Let $(\Omega,\mathcal{F})$ be a measurable space and $f:\Omega\to\mathbb{R}$ be a random variable. Let $g:\Omega\to\mathbb{R}$ be a function. Then $g$ is $\sigma(f)$-measurable if and only if there is a measurable function $h:\mathbb{R}\to\mathbb{R}$ such that $g=h\circ f.$

Here is a sense in which a $\sigma$-algebra may fail to represent information (the example is from Billingsley): Let $\Omega=[0,1]$, $\mathcal{F}$ be the Borel sets and let $\mu$ be Lebesgue measure. Let $\mathcal{C}\subseteq\mathcal{F}$ be the $\sigma$-algebra consisting of countable sets and sets with countable complements. There is no random variable that generates $\mathcal{C}$. Since $\mathcal{C}$ contains all singletons, it should be in some sense perfectly informative. But the conditional expectation $\mathbb{E}_\mathcal{C}$ is equal to a constant function almost surely- and so is every $\mathcal{C}$-measurable function. To see this, note that $\mathcal{C}$ contains only events with Lebesgue measure $1$ or $0$. Let $g:\Omega\to\mathbb{R}$ be $\mathcal{C}$-measurable. Without loss of generality, let $g(\Omega)\subseteq [0,1]$. One of the sets $g^{-1}[0,1/2]$ and $g^{-1}[1/2,1]$ must have measure $1$. Say, it is $g^{-1}[1/2,1]$. Then one of the sets $g^{-1}[1/2,3/4]$ and $g^{-1}[3/4,1]$ must have measure $1$. Continuing this way, we get a decreasing sequence of closed intervals that all have measure $1$ under the distribution of $g$ and their diameter goes to $0$. So there exists a unique point $r$ in the intersection and $g^{-1}\{r\}$ has measure $1$. So $g$ is almost surely equal to the random variable that is constantly $r$. And $\mathcal{C}$ is completely uninformative.

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I was wondering how the lemma is used to explain the sigma algebra generated by a random variable represents information? –  Tim Jan 31 '13 at 2:22
    
@Tim Imagine a decision maker that does not know the true state $\omega\in\Omega$, but only observes a signal $f(\omega)$. In each state $\omega$, she takes some decision $d(\omega)$. Since she does not know the true state, her decision has to depend only on the signal. So it should be of the form $d=g\circ f$ gor some $g$. The Doob-Dynkin lemma establishes that in the measurable framework under the given assumptions, this is exactly the same as $d$ being $\sigma(f)$-measurable. –  Michael Greinecker Jan 31 '13 at 7:48
    
Thanks! ${{{}}}$ –  Tim Jan 31 '13 at 13:23
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