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

DefinitionGiven 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.

independencewhich 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 19:08factorizationbecome 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