Measure Theory - Building Intuition

Question

I'm new in this world, so I apologize if this question is missing information.

Let $\Omega = \{\text{all students in the classroom}\}$.

Let $X(\omega) = 0$, if $\omega$ male student, and $X(\omega) = 1$, if $\omega$ is a female student.

Then $\sigma(X)$ consists of 4 subsets. Which ones?

$\emptyset, X^{-1}(0), X^{-1}(1)$, and $\Omega$. The idea is that you start taking inverse image of all measurable sets of the image ${0,1}$. So, inverse image of $\emptyset\subset{0,1}$ is the $\emptyset\subset\Omega$. The inverse image of ${0}$ and ${1}$ are $X^{-1}(0)$ and $X^{-1}(1)$. Finally, the inverse image of ${0,1}$ is $X^{-1}({0,1})=\Omega$. — SphericalTriangle, Mar 21 '18 at 11:45
In other words, it is: $\sigma(X) = { \text{male student, female student}, \emptyset, \Omega}$ ? — gariban, Mar 21 '18 at 11:51
Yes, the smallest $\sigma$-algebra that makes $X$ measurable. — SphericalTriangle, Mar 21 '18 at 11:57
$X^{-1}(\varnothing)=\varnothing$ if that's what mean. Because there is no $\omega\in\Omega$ with $X(\omega)\in\varnothing$. — drhab, Mar 21 '18 at 12:02
$X^{-1}(A)={x\in\Omega:\ X(x)\in A}$. If $A$ is empty, then no $x\in\Omega$ satisfies $X(x)\in A$. Therefore $X^{-1}(\emptyset)=\emptyset$ — SphericalTriangle, Mar 21 '18 at 12:03
@gariban I think you should distinguish the $\emptyset^{\mathbb R}$ and $\emptyset^{\Omega}$. After all, complementation is relative: $(\emptyset^{\mathbb R})^c = \mathbb R$ but $(\emptyset^{\Omega})^c=\Omega$ — BCLC, Apr 08 '18 at 07:12

drhab · Answer 1 · 2018-03-21T11:57:43.910

If $X$ is a random variable then in general: $$\sigma(X)=\{\{X\in B\}\mid B\in\mathcal B\}$$

where $\mathcal B$ denotes the Borel $\sigma$-algebra on $\mathbb R$ and: $$\{X\in B\}:=X^{-1}(B)=\{\omega\in\Omega\mid X(\omega)\in B\}$$

In your case $X$ can only take values in $\{0,1\}$ and consequently the set $\{X\in B\}$ can only take $4$ forms:

$\varnothing$ if $B\cap\{0,1\}=\varnothing$
$\{\text{male students}\}$ if $B\cap\{0,1\}=\{0\}$
$\{\text{female students}\}$ if $B\cap\{0,1\}=\{1\}$
$\Omega=\{\text{all students}\}$ if $B\cap\{0,1\}=\{0,1\}$

So we end up with:$$\sigma(X)=\{\varnothing,\{\text{male students}\},\{\text{female students}\},\Omega\}$$

score 1 · Accepted Answer · answered Apr 08 '18 at 08:08

The $\sigma$-algebra generated by a random variable $X$ in the probability space $(\Omega, \mathscr F, \mathbb P)$ is all the preimages of $X$ for any Borel set. What does this mean?

For any Borel set $B$ (pretend I said 'any subset' for the sake of understanding but then accept that any subset will technically not work) in the real numbers, there exists a subset of the sample space $\Omega$ that is also in $\mathscr F$, i.e., its $\mathbb P$robability can be computed s.t. if the subset is inputted into $X$, $X$ will give an output that is all the values in the Range of $X$ that is in $B$.

This is given by

$$\sigma(X)=\{ X^{-1}(B) | B\in\mathscr B\}$$

where $X^{-1}(B)$ is the said subset for a given $B$. This is given by

$$X^{-1}(B)=\{\omega\in\Omega | X(\omega)\in B\}.$$

Note that

$$X^{-1}(B) \in \mathscr F$$ $$X^{-1}(B) \subseteq \Omega$$ $$B \in \mathscr B$$ $$B \subseteq \mathbb R$$

Here, $X$ is an indicator/characteristic function for females: If you input a female student, $X$ gives you 1 and 0 otherwise (I assume all students are male xor female). This is given by

$$X(\text{(any) female student in the classroom}) = 1$$ $$X(\text{(any) male student in the classroom}) = 0$$

On the other hand, an output of 1 lets you know you've got female, and 0 lets you know you've got male. This is given by

$$X^{-1}(\{1\}) = \{\omega \in \Omega | X(\omega) = 1 \} = \{\omega \in \Omega | \omega \text{ is a female student in the classroom} \} = \{\text{(all) female students in the classroom}\}$$ $$X^{-1}(\{0\}) = \{\omega \in \Omega | X(\omega) = 0 \} = \{\omega \in \Omega | \omega \text{ is a male student in the classroom} \} = \{\text{(all) male students in the classroom}\}$$

However, we need not be limited by singletons. We can use intervals, unions of disjoint intervals, etc. For example,

$$X^{-1}((0.5,1.5] \cup {2} = \{\omega \in \Omega | X(\omega) \in (0.5,1.5] \cup {2} \} = \{\omega \in \Omega | X(\omega) = 1 \} = \{\omega \in \Omega | \omega \text{ is a female student in the classroom} \} = \{\text{female students in the classroom}\}$$

$X$ doesn't care about the numbers in $(0.5,1.5] \cup {2}$ besides $1$ because its range consists only $0$ and $1$, i.e. $Range(X) = \{0,1\}$. Other examples:

$$X^{-1}((\pi,2,000]) = \{\omega \in \Omega | X(\omega) \in (\pi,2,000] \} = \{\omega \in \Omega | X(\omega) \in \emptyset^{\mathbb R} \}$$ $$= \{\omega \in \Omega | \omega \text{ is neither male nor female student in the classroom} \} = \emptyset^{\Omega}$$

$$X^{-1}([0,10]) = \{\omega \in \Omega | X(\omega) \in [0,10] \} = \{\omega \in \Omega | X(\omega) \in \{0,1\} \}$$ $$= \{\omega \in \Omega | \omega \text{ is a male or female student in the classroom} \} = \Omega$$

$$X^{-1}([-1,1)) = \{\omega \in \Omega | X(\omega) \in [-1,1) \} = \{\omega \in \Omega | \omega \text{ is a male student in the classroom} \} = \{\text{male students in the classroom}\}$$

Now, we can use any Borel set $B$. The important thing is whether or not $B$ has any 0's or 1's. Here's what to do:

$$X^{-1}(B) = \{\omega \in \Omega | X(\omega) \in B \} = \{\omega \in \Omega | \omega \text{'s that satisfy the condition} \} = \text{subset of} \ \Omega \ \text{that is in} \ \mathscr F \}$$

In your case I guess $\Omega$ would be finite, and all its subsets are in $\mathscr F$. Anyway, 4 cases

$B$ has $1$ but not $0$

$$X^{-1}(B) = \{\omega \in \Omega | X(\omega) \in B \} = \{\omega \in \Omega | X(\omega) = 1 \} = \{\omega \in \Omega | \omega \text{ is a female student in the classroom} \} = \{\text{female students in the classroom}\}$$

$B$ has neither $1$ nor $0$

$$X^{-1}(B) = \{\omega \in \Omega | X(\omega) \in B \} = \{\omega \in \Omega | X(\omega) \in \emptyset^{\mathbb R} \}$$ $$= \{\omega \in \Omega | \omega \text{ is neither male nor female student in the classroom} \} = \emptyset^{\Omega}$$

$B$ has both $1$ and $0$

$$X^{-1}(B) = \{\omega \in \Omega | X(\omega) \in B \} = \{\omega \in \Omega | X(\omega) \in \{0,1\} \}$$ $$= \{\omega \in \Omega | \omega \text{ is a male or female student in the classroom} \} = \Omega$$

$B$ has $0$ but not $1$

$$X^{-1}(B) = \{\omega \in \Omega | X(\omega) \in B \} = \{\omega \in \Omega | \omega \text{ is a male student in the classroom} \} = \{\text{male students in the classroom}\}$$

Thus, $X^{-1}(B) = $

$\{\text{female students in the classroom}\}$ if $B$ has $1$ but not $0$
$\emptyset^{\Omega}$ if $B$ has neither $1$ nor $0$
$\Omega$ if $B$ has both $1$ and $0$
$\{\text{male students in the classroom}\}$ if $B$ has $0$ but not $1$

Since $B$ cannot have other forms according to whether or not $B$ has $0$ or $1$, the cases are exhaustive. Thus, we know that $X^{-1}(B)$ cannot take any other forms.

Therefore:

$$\sigma(X)=\{ X^{-1}(B) | B\in\mathscr B\} = \{\emptyset^{\Omega},\{\text{male students in the classroom}\},\{\text{female students in the classroom}\},\Omega\}$$

What does $\emptyset^\mathbb{R}$ and $\emptyset^\Omega$ mean? — gariban, Apr 11 '18 at 18:27
@gariban They belong to different sets. See my comment here. $\emptyset^{\mathbb R}$ is in $\mathscr B(\mathbb R)$, so $(\emptyset^{\mathbb R})^c = \mathbb R \ \text{and not} \ \Omega$. — BCLC, Apr 12 '18 at 08:41

Measure Theory - Building Intuition

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