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This is the definition of "measurability":

Let $\mathit X$ be a random variable defined on $(\Omega,\mathcal F,\mathbb P)$. Let $\mathcal G$ be a $\sigma$-algebra of subsets of $\Omega$. If $\sigma(\mathit X) \subseteq \mathcal G$, we say that $\mathit X$ is $\mathcal G$-measurable.

I can understand this definition, but what does it mean that "a random variable $\mathit X$ is $\mathcal G$-measurable if and only if the information in $\mathcal G$ is sufficient to determine the value of $\mathit X$"?

My questions are:

  1. What is the "information" in $\mathcal G$? As far as I know, $\mathcal G$ is just a set of subsets of $\Omega$. Thus $\mathcal G$ contains lots of "events". Then, when we say "information", what exactly do we mean？ I can't imagine other information except for the elements of $\mathcal G$.
  2. How can the "information" in $\mathcal G$ determine the value of $\mathit X$? Given $\mathcal G$, what we know is just "the elements of $\mathcal G$". How could this "infromation" help us determine the value of $\mathit X$?

Example:

This is an example of rolling a die with $\Omega = \{1, 2, 3, 4, 5, 6\}$:

$$\mathit X_1(\omega)=\omega$$

$$\mathit X_2(\omega)=\begin{cases} 1, & \omega\in \{1,3,5\} \\-1, & \omega\in \{2,4,6\}\end{cases}$$

$\mathit X_1$ and $\mathit X_2$ are both random variables. $\mathit X_1$ gives the exact outcome of the roll, and $\mathit X_2$ is a binary variable whose value depends on whether the roll is odd or even.

Let $\mathcal G = \{\emptyset, \Omega, \{1,3,5\}, \{2,4,6\}\}$, then $\mathit X_2$ is measurable w.r.t $\mathcal G$ but $\mathit X_1$ is not measurable w.r.t $\mathcal G$. I know this because I can check that $\sigma(X_2)=\mathcal G$ according to the defination of $\sigma(X_2)$.

Here are the questions:

  1. What's the information in $\{\emptyset, \Omega, \{1,3,5\}, \{2,4,6\}\}$? What else can we know except for those four elements in $\{\emptyset, \Omega, \{1,3,5\}, \{2,4,6\}\}$?

  2. If this is the only 'information' that we can obtain from $\mathcal G$ (i.e., there are four elements in $\mathcal G$, and those elements are $\emptyset$, $\Omega$, $\{1,3,5\}$ and $\{2,4,6\}$), how can we determine what the value of $\mathit X_2$ will be?

Gödel
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  • What is your definition of $\sigma$? – Fibonacci Cube K May 23 '20 at 06:13
  • This intuition about 'the information in G' is clarified by thinking about $G$ as being $G = \sigma( Y_1, \ldots, Y_m)$, where the $Y_i$ are measurements that someone is making about some experiment for which $X$ is another possible measurement. If $X \in \sigma(Y_1, \ldots, Y_m)$, you can express the value of $X$ in terms of the values of $Y_i$ (literally there is some measurable function $f$ so that $X = f(Y_1, \ldots, Y_n$) ) -- hence you can express $X$ in terms of the information in the $Y_i $ which is to say, the information in $G$. – Elle Najt May 23 '20 at 06:17
  • @DEATH_CUBE_K Hi, $\sigma$-algebra is the subset of $\Omega$ that includes $\Omega$ itself, is closed under complement, and is closed under countable unions. I can understand this concept. – Gödel May 23 '20 at 06:18
  • One common place for this 'information' intuition to come up is if you have a stochastic process of some kind -- say a random walk -- and the information at time $n$ refers to the sigma algebra $F_n$ generated by the first $n$ positions in the random walk. If some random variable is in $F_n$, then you can calculate it based on seeing the first $n$ steps of the walk -- you have enough information -- but if some random variable is not in $F_n$, you cannot calculate it at that time, because you do not have enough information. – Elle Najt May 23 '20 at 06:20
  • That doesn't explain what $\sigma(X)$ means. What is the definition? – Fibonacci Cube K May 23 '20 at 06:20
  • $\sigma(X_1)$ is the sigma algebra generated by all events of the form ${X_1 \leq a}$ for $a \in \mathbb{R}$. – Elle Najt May 23 '20 at 06:20
  • Got it, thanks! – Fibonacci Cube K May 23 '20 at 06:21
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    Saying $\sigma(X) \subseteq G$ is the same as saying that $X$ is $G$ measurable, because it is asserting that all events ${X \leq a }$ are in $G$. – Elle Najt May 23 '20 at 06:21
  • @DEATH_CUBE_K It's $\sigma$-algebra generated by random variable $\mathit X_1$. – Gödel May 23 '20 at 06:22
  • @LorenzoNajt Thanks! It seems like the Doob–Dynkin lemma. I know there is some measurable function f so that $\mathit X=f(\mathit Y)$ if $\mathit X$ is measurable w.r.t $\sigma(\mathit Y)$. But I still can't understand what information is in $\mathcal G$ except for knowing the elements of it. – Gödel May 23 '20 at 06:43
  • @LorenzoNajt and, I read somewhere that "if we know $\mathcal G$, then we can know the value of $\mathit X$". But what we know about $\mathcal G$ when we say we know it? – Gödel May 23 '20 at 06:45
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    "I still can't understand what information is in G except for knowing the elements of it. " -- if you want, you can think about the 'information' in $G$ as observations about which events in $G$ occured in our experiment: We get some example $\omega$ from our probability distribution, and then the information in G is the information of, for each $A \in G$, whether $\omega \in A$ (that is, whether $A$ occurred in the universe/outcome $\omega$). – Elle Najt May 23 '20 at 06:49
  • @LorenzoNajt It's really intuitive to think about information in $\mathcal G$ like this. Thanks for your explanation. – Gödel May 23 '20 at 07:01
  • You're welcome. :-) – Elle Najt May 23 '20 at 07:08
  • See also this question and the first part of this answer – saz May 23 '20 at 07:10
  • @LorenzoNajt That means, if we know for every event in $\mathcal G$ whether it occurred or not, then we can know $\mathit X(\omega)$ even though we still don't know $\sigma$. Right? – Gödel May 23 '20 at 07:12
  • @saz Thanks for your reference. Actually I searched for this topic and have read these two Q&As. It's very helpful, but I'm still not sure about how to think about information. – Gödel May 23 '20 at 07:18

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