Prove that $\sigma$-algebras $A_1,\ldots,A_n$ are independent if and only if $A_i$ is independent of each $A_1,\ldots,A_{i-1}$, for all $i=2,\ldots,n$

Question

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $k\in\mathbb N$ and $\mathcal F_1,\ldots,\mathcal F_k\subseteq\mathcal A$. Remember the following terminology

1. $(\mathcal F_1,\ldots,\mathcal F_k)$ is called independent if $$\operatorname P\left[\bigcap_{i\in I}A_i\right]=\prod_{i\in I}\operatorname P\left[A_i\right]\tag1$$ for all $I\subseteq\{1,\ldots,k\}$ and $A_i\in\mathcal F_i$ for all $i\in I$.
2. $\mathcal F_1$ and $\mathcal F_2$ are called independent (or $\mathcal F_1$ is called independent of $\mathcal F_2$) if $(\mathcal F_1,\mathcal F_2)$ is independent.

Moreover, remember the following basic results:

3. Let $K\subseteq\{1,\ldots,k\}$ and $(I_k)_{k\in K}$ be a disjoint subdivision of $\{1,\ldots,k\}$. If $(\mathcal F_1,\ldots,\mathcal F_k)$ is independent, then $\left(\bigcup_{i\in I_k}\mathcal F_i\right)_{k\in K}$ is independent.
4. If $\mathcal F_i\cup\{\emptyset\}$ is closed under finite intersections for all $i\in\{1,\ldots,k\}$, then $(\mathcal F_1,\ldots,\mathcal F_k)$ is independent if and only if $(\sigma(\mathcal F_1),\ldots,\sigma(\mathcal F_k))$ is independent.

Question: (a) Are we able to show $(\mathcal F_1,\ldots,\mathcal F_k)$ is independent if and only if $\mathcal F_i$ is independent of $\mathcal F_1\cup\cdots\cup\mathcal F_{i-1}$ for all $i\in\{2,\ldots,k\}$?

(b) Or do we need to replace $\mathcal F_1\cup\cdots\cup\mathcal F_{i-1}$ by $\sigma(\mathcal F_1\cup\cdots\cup\mathcal F_{i-1})$ and/or assume that each $\mathcal F_i$ is a $\sigma$-algebra?

Answer 1

$\mathcal{A}_1,\ldots,\mathcal{A}_n$ are independent if and only if $\mathcal{A}_i$ is independent of each $\mathcal{A}_1,\ldots,\mathcal{A}_{i-1}$, for all $i=2,\ldots,n$

$(\Rightarrow)$ It follows from the definition of independence that $$\Bbb{P}[X_1 \in A_1, \ldots , X_{i-1} \in A_{i-1}, X_i \in A_i] =\\ \Bbb{P}[X_1 \in A_1, \ldots , X_{i-1} \in A_{i-1}, X_i \in A_i, X_{i+1} \in \Bbb{R}\ldots X_n \in \Bbb{R}] = \\ \Bbb{P}[X_1 \in A_1]\ldots \Bbb{P}[ X_{i-1} \in A_{i-1}]\Bbb{P}[ X_i \in A_i]\Bbb{P}[ X_{i+1} \in \Bbb{R}]\ldots\Bbb{P}[ X_n \in \Bbb{R}] = \\ \Bbb{P}[X_1 \in A_1]\ldots \Bbb{P}[ X_{i-1} \in A_{i-1}]\Bbb{P}[ X_i \in A_i] $$

$(\Leftarrow)$ $$\Bbb{P} [X_1 \in A_1, \ldots X_n \in A_n] = \Bbb{P} [X_1 \in A_1, \ldots X_{n-1} \in A_{n-1}] \Bbb{P} [X_n \in A_n] \\ = \cdots = \Bbb{P}[X_1 \in A_1]\ldots \Bbb{P}[ X_{n-1} \in A_{n-1}]\Bbb{P}[ X_n \in A_n]$$

which is the condition of independence

Answer 2

For (a) we have the following simple counterexample. Let $\Omega=\{0,1,2,3\}$, $\mathcal A$ be the family of all subsets of $\Omega$ and $P(A)=|A|/|\Omega|$ for each $A\in\mathcal A$. Let $k=3$. For each integer $i$ from $1$ to $k$ put $A_i=\{0,i\}$ and $\mathcal F_i=\{A_i\}$. Since $$P(A_i\cap A_j)=\frac 14=\left(\frac 12\right)^2=P(A_i)P(A_j)$$ for each distinct integers $i,j$ from $1$ to $k$, we see that $\mathcal F_2$ is independent from $\mathcal F_1$ and $\mathcal F_3$ is independent from $\mathcal F_1\cup \mathcal F_2$. On the other hand, $(\mathcal F_1, \mathcal F_2, \mathcal F_3)$ is not independent, because $$P(A_1\cap A_2\cap A_3)=\frac 14\ne \left(\frac 12\right)^3=P(A_1)P(A_2)P(A_3).$$

Concerning (b) we can prove that $(\mathcal F_1, \mathcal F_2,\dots, \mathcal F_k)$ is independent under weaker assumptions, which are sufficient to justify the following sequence of equalities for any choice $A_i\in\mathcal F_i$ for each $i$
$$P(A_1\cap \dots \cap A_k)=$$ $$P(A_1\cap \dots \cap A_{k-1})P(A_k)=$$ $$P(A_1\cap \dots \cap A_{k-2})P(A_{k-1})P(A_k)=$$ $$\dots$$ $$P(A_1)\cdots P(A_k).$$