Independence of sigma-algebras

Two sigma-algebras are independent if, on a probability space $(\Omega,\mathcal F,\mathbb P)$ , sub-sigma-algebras $\mathcal G_1,\mathcal G_2\subseteq\mathcal F$ satisfy

\mathbb P(A\cap B)\;=\;\mathbb P(A)\,\mathbb P(B)\quad\text{for all }A\in\mathcal G_1,\;B\in\mathcal G_2.

A family $(\mathcal G_i)_{i\in I}$ of sub- $\sigma$ -algebras is independent if for every finite subset $\{i_1,\dots,i_n\}\subseteq I$ and every choice of events $A_k\in \mathcal G_{i_k}$ ,

\mathbb P\!\left(\bigcap_{k=1}^n A_k\right)\;=\;\prod_{k=1}^n \mathbb P(A_k).

This formalizes “independence of information”: events determined by $\mathcal G_1$ do not influence probabilities of events determined by $\mathcal G_2$ . In particular, independence of random variables $X$ and $Y$ can be characterized by independence of the generated $\sigma$ -algebras $\sigma(X)$ and $\sigma(Y)$ .

Examples

In two independent coin flips, let $\mathcal G_1$ be the $\sigma$ -algebra generated by the first flip and $\mathcal G_2$ the $\sigma$ -algebra generated by the second flip. Then $\mathcal G_1$ and $\mathcal G_2$ are independent.
If $\mathcal G_2\subseteq \mathcal G_1$ and $\mathcal G_2$ contains an event $B$ with $\mathbb P(B)\in(0,1)$ , then $\mathcal G_1$ and $\mathcal G_2$ are not independent (since taking $A=B$ violates $\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B)$ ).