The definition/notion of independence is always a bit odd in measure theoretic probability theory.

Definition Given a probability space $(\Omega,\mathcal{F},P)$, two sets $A,B\in\mathcal{F}$ are defined to be (piarwise/mutually) independent iff $P(AB)=P(A)P(B)$. For readers who are not familiar with algebraic formulation of $\sigma$-algebra over $\mathbb{Z}_{2}$, this post may be of help. (SE post)

(1)What is the algebraic equivalent of the collection of pairwise independent elements in $\cal{F}$?

Since we can always study the multiplicative structure on $\sigma$-algebra $\mathcal{F}$ defined by $A\cdot B=A\cap B$. Does it mean the probability measure is a (ring) homomorphism on the collection of pairwise independent sets? However, that does not make sense because $$P(A\Delta B)=P((A\setminus B) \cup (B\setminus A))=P(A\setminus B)+P(B\setminus A)=P(AB^{c})+P(A^{c}B)=P(A)P(B^{c})+P(A^{c})P(B)\neq P(A)+P(B)$$ (the last eqality comes from the fact that if $A,B$ are independent so are their generating $\sigma$-algebras) even if they are independent and thus breaks the additive structure of the $\sigma$-algebra. If there is a special term for such a structure formed by independent elements(as set) in the $\sigma$-algebra $\cal{F}$, either in algebra or algebraic geometry, I would like to know.

If not, does that mean there is no means to deal with the concept of independence in algebra?

(2)What is the algebraic equivalent of the collection of mutually independent elements in $\cal{F}$?

Historically it is the notion of mutually independence which is firstly discovered. And mutually independence is actually stronger than pairwise independence (Bernstein example). So I want to know if mutually independent elements correspond to a different algebraic object than pairwise independent.

(3)What is the algebraic equivalent of the sigma field generated by two collections of independent elements in $\cal{F}$?

i.e. Given two subcollection $\mathcal{F}_{1},\mathcal{F}_{2}\subset\cal{F}$, we assume that $\forall C\in \mathcal{F}_{1},D\in \mathcal{F}_{2}$, there exists $P(CD)=P(C)P(D)$. The question is asking the algebraic equivalent of $\sigma( \mathcal{F}_{1}\vee \mathcal{F}_{2})$.

Either answer or reference are welcomed/appreciated.

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    $\begingroup$ The measure couldn't be a ring homomorphism in any reasonable way, because the image is contained in $[0,1]$. The image of a ring homomorphism is a ring. Maybe algebra and measure theory are different fields altogether... $\endgroup$ – R. van Dobben de Bruyn Dec 9 '16 at 19:04
  • $\begingroup$ Perhaps the algebraic structure you are looking for is (pointed) monoid? Monoids are ubiquitous throughout maths, but the way they are used in different fields is often unrelated (even groups vs abelian groups seem to be worlds apart in terms of applications...). $\endgroup$ – R. van Dobben de Bruyn Dec 9 '16 at 19:06
  • $\begingroup$ @R.vanDobbendeBruyn Yes, I wrote that since that is my first thought. But it is kinda weired that probability theory introduced an algebraic structure to describe a notion and later introduced independence which is completely irrelevant/incompatible to the algebraic structure.... And do you mean the independent elements form a monoid? I cannot see how... $\endgroup$ – Henry.L Dec 9 '16 at 19:08
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    $\begingroup$ I've seen an algebraic (in fact, category-theoretic) formulation of pairwise, joint and conditional independence in a talk by Alex Simpson. It arose as an answer to a question here on MO. $\endgroup$ – მამუკა ჯიბლაძე Dec 9 '16 at 21:21
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    $\begingroup$ @მამუკაჯიბლაძე A.Simpson's talk is exactly what I wanted,on the pp.25-26 of his slide he pointed out the equivalence of independent structure "symmetric monoidal structure with jointly monic projections".In fact on p.39 on his later talk(homepages.inf.ed.ac.uk/als/Talks/cambridge14.pdf) he pointed out a local independent structure which is probably more suitable when formalized as a concrete def in measure space category. One interesting thing I noticed is that factorization become part of definition of independence.Thanks for the reference!I was thinking deleting this post earlier... $\endgroup$ – Henry.L Dec 9 '16 at 21:39

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