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I have seen quite a few posts but it seems not many answered the question correctly. So I am a bit confused.

My Understanding: Given $(\Omega, \mathcal{A})$ and $(R,B(R))$; let $X$, $Y$ be random variables that $X,Y: \Omega \rightarrow \mathbb R$

$$\sigma(X,Y) = \sigma\left(X^{-1}(B_i)\bigcap Y^{-1}(B_j)\right) = \{ \omega \in \Omega: (X(\omega),Y(\omega)) \in (B_i, B_j)\} \in \mathcal{A} \tag1$$

Or in plain english: $\sigma(X,Y)$ is a sigma algebra that is generated by sets of $\omega$ that satisfy $X(\omega) \in B_i$ and $Y(\omega) \in B_j$ at the same time

Then my gut tells me that because the preimage of the intersection is the intersection of the preimages, which means $$\sigma(X,Y) = \sigma\left(\sigma(X) \bigcap \sigma(Y)\right)$$

I used the following example to test (1) and (2) and it did work $$ X(\omega) = \begin{cases} 1 & \omega \in \{tt, hh\} \\ 0 & \omega \in \{th, ht\} \end{cases} $$

$$ Y(\omega) = \begin{cases} 1 & \omega \in \{tt, hh\} \\ 2 & \omega \in \{th\} \\ 3 & \omega \in \{ht\} \end{cases} $$

$$\Omega = \{tt,hh,th,ht\}$$

Am I correct with (1) and (2)

Snoop
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TJT
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    $\sigma(X,Y) = \sigma(\sigma(X) \bigcup \sigma(Y))$. Also, $X(\omega) =B_i$ does not make sense. – Kavi Rama Murthy Dec 25 '23 at 12:07
  • @geetha290krm I am positive $\sigma(X \bigcup Y) = \sigma(\sigma(X) \bigcup \sigma(Y))$ So $\sigma(X,Y) = \sigma(X \bigcup Y)$? – TJT Dec 25 '23 at 13:33
  • @geetha290krm $\sigma(X,Y) = \sigma({tt,hh}, {th}, {ht})$ and $\sigma(\sigma(X) \bigcup \sigma(Y)) \neq \sigma({tt,hh}, {th}, {ht})$ – TJT Dec 25 '23 at 14:07

2 Answers2

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The intersection of arbitrarily many $\sigma$-algebras is a $\sigma$-algebra, so $\sigma(\sigma(X)\cap \sigma(Y))=\sigma(X)\cap \sigma(Y)$. As pointed out by @geetha290krm in the comments, the definition of a $\sigma$-algebra generated by random variables $X$ and $Y$ is $\sigma(X,Y):=\sigma(\sigma(X)\cup \sigma(Y))$. We can then see that $\sigma(X)\cap \sigma(Y)\subseteq \sigma(X,Y)$. However, the reverse inclusion is not true in general. Consider $X,Y:([0,1],\mathscr{B}([0,1]))\to ([0,1],\mathscr{B}([0,1]))$ where $X(\omega)=0$ and $Y(\omega)=\omega$. Then $$\sigma(X)=\{\emptyset,[0,1]\},\,\sigma(Y)=\mathscr{B}([0,1])$$ So $\sigma(X)\cap \sigma(Y)=\{\emptyset,[0,1]\}$, but $\sigma(X,Y)=\mathscr{B}([0,1])$.

Snoop
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  • Ok, I think I am more confused now. As for one of the example I was given in lecture (illustrated in the question), $\sigma(X,Y) = \sigma({tt,hh}, {th}, {ht})$ and it seems not from union but intersection – TJT Dec 25 '23 at 22:42
  • @TJT $\sigma(X,Y):=\sigma(\sigma(X)\cup \sigma(Y))$. This is well known. It is also true that for real-valued rvs, $\sigma(X,Y)=\sigma({{X\in A,Y \in B};A,B \in \mathscr{B}(\mathbb{R})})$. However, we have seen that $\sigma(X)\cap \sigma(Y)\subseteq \sigma(X,Y)$, while the reverse can be false. – Snoop Dec 25 '23 at 23:01
  • I see. My claim was that $\sigma(X,Y) = \sigma\left(\sigma(X) \bigcap \sigma(Y)\right)$ which should be right as well right since $\sigma(X)\cap \sigma(Y)\subseteq \sigma(X,Y)$ – TJT Dec 25 '23 at 23:12
  • Your claim is false in general; there is an example in the answer. – Snoop Dec 25 '23 at 23:22
  • Thank you so much! I work through the example and see it now. One last thing, I also encountered $\sigma(X \bigcup Y) = \sigma(\sigma(X) \bigcup \sigma(Y))$ before. Does this mean $\sigma(X,Y) = \sigma(X \bigcup Y)$ as well? – TJT Dec 25 '23 at 23:28
  • @TJT the notation $\sigma(X\cup Y)$ is not precise in my opinion; what would $X\cup Y$ mean? I would rely on the definition $\sigma(X,Y):=\sigma(\sigma(X)\cup \sigma(Y))$. – Snoop Dec 25 '23 at 23:32
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    I see. Thank you so much indeed for your help! – TJT Dec 25 '23 at 23:40
  • @Snoop : In your counter-example to JT, why is $\sigma(X)$ different from $\sigma(Y)$. I realize that everything maps to zero in the case of $X$ but $\omega$ can still be anything in [0,1] ? Thanks. – mark leeds Dec 26 '23 at 02:53
  • @markleeds we can see that ${\omega\in [0,1]:Y(\omega)\in B}={\omega\in [0,1]:\omega\in B}=B$ for any Borel set $B$ so $\sigma(Y)={B:B\in \mathscr{B}[0,1]}=\mathscr{B}[0,1]$ – Snoop Dec 26 '23 at 11:17
  • @Snoop: I do get that but can you be bothered to do the same thing for $\sigma(X)$. Thanks and my apologies for being a pest. – mark leeds Dec 26 '23 at 18:01
  • @markleeds no worries, that is a legit question. Have a look here, and let me know if this convinces you. – Snoop Dec 26 '23 at 19:42
  • Thank you. I will have a look. The link is appreciated. – mark leeds Dec 27 '23 at 21:48
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    It does convince me. That link was quite enlightening in all sorts of ways. Thanks a lot. – mark leeds Dec 27 '23 at 22:12
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Let's also see what is right, so we have all pieces of the puzzle: We show

$$\sigma(\mathcal A \cup \mathcal B ) = \sigma(A\cap B, A \in \mathcal A, B \in \mathcal B).$$

(In your case, $\mathcal A = \sigma(X)$ and $\mathcal B = \sigma(Y)$.)

Proof. First, $ A\cap B \in \sigma(\mathcal A \cup \mathcal B )$ for all $ A \in \mathcal A, B \in \mathcal B$ and therefore as $\sigma(A\cap B, A \in \mathcal A, B \in \mathcal B)$ is the smallest $\sigma$-field containing those sets, also $ \sigma(A\cap B, A \in \mathcal A, B \in \mathcal B) \subset \sigma(\mathcal A \cup \mathcal B )$.

On the other hand for $A \in \mathcal A, B \in \mathcal B$, $A = A \cap \Omega \in \sigma(A\cap B, A \in \mathcal A, B \in \mathcal B)$ and $B = \Omega \cap B \in \sigma(A\cap B, A \in \mathcal A, B \in \mathcal B)$. So $\mathcal A\cup \mathcal B \subset \sigma(A\cap B, A \in \mathcal A, B \in \mathcal B)$ and therefore also $\sigma(\mathcal A\cup \mathcal B) \subset \sigma(A\cap B, A \in \mathcal A, B \in \mathcal B)$. QED

Moritz Schauer
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