Talk:Join (sigma algebra)

Doubts
1. What does it mean, $$ \{A_i \cap B_j \; | \; A_i\in\mathcal{A}, B_j\in\mathcal{B} \} ?$$ Maybe just $$ \{A \cap B \; | \; A\in\mathcal{A}, B\in\mathcal{B} \},$$ the set of intersections? No, this is not a sigma-algebra (it is closed under countable intersections only). Or maybe $$ \{ \cup_{i,j} A_i \cap B_j \; | \; A_i\in\mathcal{A}, B_j\in\mathcal{B} \},$$ the set of countable unions of intersections?

2. What does it mean, "The above defines the join on sub-sigma algebras of a single common sigma algebra, as it is not generally possible to define the join of two unrelated sigma algebras"? The set of all subsets (of the given set) is evidently a sigma-algebra, and evidently contains all other sigma-algebras. How could they be "unrelated"?

Boris Tsirelson (talk) 17:38, 28 July 2012 (UTC)

3. "The two definitions are equivalent: The unions of elements from a partition of a set generate a sigma algebra. Likewise, intersections of elements from a sigma algebra result in a partition of a set." — No. This simple relation between partitions and sigma-algebras holds only when the partition contains at most countable set of atoms. See the last paragraph here.

Ah, it seems, now I understand the strange formula mentioned in Item 1 above. The editor, writing it, had in mind only the case of at most countable set of atoms, that is, only the discrete case.

Boris Tsirelson (talk) 17:49, 28 July 2012 (UTC)


 * Hi Boris, I slapped this article together very quickly, and clearly, as you point out, I failed to proof-read it, and left many errors in it. I plan to come back to it after a brief detour. The join is essentially a way of talking about the conditional probability, using a more measure-theoretic notation.  I've detoured because it suddenly shot into my head that a join looks somewhat like a 2-cocycle, viz. a cartesian product with an extra cocycle condition on it; or maybe one could say its kind-of-like a limit of some certain diagram (category theory).  Not sure, might be barking up the wrong tree.  I'll try to come back to this topic in the next few weeks or month and clean it up and finish it. linas (talk) 17:02, 13 August 2012 (UTC)


 * I guess, this should be $$ \sigma ( \{A \cap B \; | \; A\in\mathcal{A}, B\in\mathcal{B} \} ) $$, as this is a sigma-algebra and it appears to have a chance to be equal to $$ \sigma ( \mathcal{A} \cup \mathcal{B} ) $$. -- 178.25.47.228 (talk) 17:47, 15 October 2012 (UTC)
 * Sure, $$ \sigma ( \{A \cap B \; | \; A\in\mathcal{A}, B\in\mathcal{B} \} ) = \sigma ( \mathcal{A} \cup \mathcal{B} ) .$$
 * Proof (sketch): on one hand, $$ \mathcal{A} \cup \mathcal{B} \subset \{A \cap B \; | \; A\in\mathcal{A}, B\in\mathcal{B} \} ;$$ and on the other hand, $$ \{A \cap B \; | \; A\in\mathcal{A}, B\in\mathcal{B} \} \subset \sigma ( \mathcal{A} \cup \mathcal{B} ) .$$ --Boris Tsirelson (talk) 18:46, 15 October 2012 (UTC)

Redirected
As stated on this page is not usefull. This notion is entirely explained in sigma algebra's page anyway. Arthur MILCHIOR (talk) 14:11, 30 May 2018 (UTC)