# 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 $i=2,\ldots,n$

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:

1. 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.
2. 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?

• $1$ and $2$ are equivalent because $A_i$ is independent of $A_j$ is the same as $A_j$ is independent of $A_i$. Further, if you write $A_i$ and $A_j$ in a pairwise matrix form you will notice that the matrix is a symmetric matrix. Moreover, at least for me, $A_i$ is independent of $A_i$ doesn't make sense vis-a-vis the diagonal entries of the matrix are not contributing to the counting process. Commented Nov 2, 2020 at 18:42
• Also, note that pairwise independence is a strong result than $A_1\cup A_2 \cup \cdots\cup A_{i-1}$ and $A_i$ are independent. Commented Nov 2, 2020 at 18:49
• @Kumar If you take three events $A_1,A_2,A_3$ such that any two of them are independent, but they are not jointly independent. Isn't then $\mathcal A_i:=\sigma(A_i)$ a counter-example for "(2.) implies (1.)"? Commented Nov 2, 2020 at 18:49
• I didn't get what you mean by " jointly independent "? If $A_1\cap A_2=\phi$, $A_2\cap A_3=\phi$, and $A_1\cap A_3=\phi$, then $(A_1\cup A_2)\cap A_3=\phi$ Commented Nov 2, 2020 at 18:52
• @Kumar It means that (4.) holds. If each $\mathcal A_i$ is a $\sigma$-algebra, then this reduce to the following: $\mathcal A_1,\ldots,\mathcal A_n$ are (jointly) independent if $\text P[\bigcap_{i=1}^nA_n]=\prod_{i=1}^n\text P[A_i]$ for all $A_1\in\mathcal A_1,\ldots,A_n\in\mathcal A_n$. They are pairwise independent if $\text P[A\cap B]=\text P[A]\text P[B]$ for all $A\in\mathcal A_i,B\in\mathcal A_j$ for all $i\ne j$. Commented Nov 2, 2020 at 18:55

$\mathcal{A}_1,\ldots,\mathcal{A}_n$ 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

• Is it important, that we are considering $\sigma$-algebras? Or would the same argumentation hold, if the $A_i$ would be families of sets? Commented Jul 2, 2015 at 21:46
• the recursion in the last step still works fine even if the family is not a sigma algebra. But this might not give you independence if the family is not a generating class, closed for intersections Commented Jul 2, 2015 at 21:52
• What are the $X_i$'s? Random variables? Are the $A_i$'s elements of the $\sigma$-algebras?
– BCLC
Commented Nov 29, 2015 at 21:04

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 $$i$$
$$P(A_1\cap \dots \cap A_k)=$$ $$P(A_1\cap \dots \cap A_{k-1})P(A_k)=$$ $$P(A_1\cap \dots \cap A_{k-2})P(A_{k-1})P(A_k)=$$ $$\dots$$ $$P(A_1)\cdots P(A_k).$$