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In the lecture about probability theory, we had the following defintion for independence of a finite number of sub-sigma-algebras on the probability space $(\Omega, \mathcal{A},P)$:

The sub-sigma-algebras $\mathcal{B}_1, \dotsc, \mathcal{B}_n \subseteq \mathcal{A}$ are independent if

$$P[B_1 \cap \dotsb \cap B_n] = P[B_1 ] \dotsb P[B_n]$$

for all $B_i \in \mathcal{B_i}, 1 \leq i \leq n$.

I know from an other basic course about probability, that one says that a finite set of events $A_1, \dotsc, A_n$ is independent for all $I \subseteq \{1, \dotsc, n\}$ we have that

$$P[\bigcap_{i \in I} A_i]=\prod_{i \in I} P[A_i].$$

This above definition is for events in the same sigma-algebra, so $A_i \in \mathcal{A}$ for all $i$?

Now I wonder why one would not define sigma-algebras to be independent if for any subset of sigma-algebras the above property (of factorization of the intersection of the events) holds. In particular, the following confuses me: We had a remark that says each subsequence of independent sigma-algebras is independent. Why is this the case? Is it thus obvious that

$$P[B_1 \cap B_2 \cap B_3] = P[B_1] P[B_2]P[B_3]$$

implies

$$P[B_1 \cap B_2] = P[B_1] P[B_2]?$$

Moreover, here I could find a different definition of independence of events based on independence of sigma-algebras. Are both definitions correct (hence equivalent) or am I missing something with the above definition for independence of events? Does one define independence of events with help of independence of sigma-algebras or vice versa?

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    $\begingroup$ To check the independence of sigma-algebras, the "intersection-product" property for $n$ events suffices since one can choose $\Omega$ for some $B_i$s (because $\Omega$ is in every sub-sigma-algebra), say, for every $i$ not in $I$, and then one is left with the "intersection-product" property over $I$. In other words, if $$P[B_1 \cap \dotsb \cap B_n] = P[B_1 ] \dotsb P[B_n]$$ for all $B_i \in \mathcal{B_i}$, $1 \leq i \leq n$, then, for all $I \subseteq \{1, \dotsc, n\}$ and all $B_i \in \mathcal{B_i}$, $i$ in $I$, $$P[\bigcap_{i \in I} B_i]=\prod_{i \in I} P[B_i].$$ $\endgroup$
    – Did
    Oct 30, 2015 at 7:41
  • $\begingroup$ Related? Mutual Independence Definition Clarification, Mutual Independence Definition Clarification $\endgroup$
    – BCLC
    Dec 6, 2016 at 13:29

2 Answers 2

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Events $B_1,B_2,\ldots,B_n$ are mutually independent provided $$ P[\cap_{k=1}^n B_k^*]=\prod_{k=1}^n P[B^*_k] $$ for all possible choices of each $B_k^*$ as either $B_k$ or $B_k^c$. This entail $2^n$ identities that must hold.

Notice that if $B_1,B_2,B_3$ are independent in this sense, then

$\begin{align*}P[B_1\cap B_2] &=P[B_1\cap B_2\cap B_3]+P[B_1\cap B_2\cap B_3^c]\\ &=P[B_1]P[B_2]P[B_3]+P[B_1]P[B_2]P[B_3^c]\\ &=P[B_1]P[B_2]\left\{P[B_3]+ P[B_3^c]\right\}\\ &=P[B_1]P[B_2]\end{align*} $

etc., so that $B_1$ and $B_2$ are independent, yielding the subsequence property you mention.

Also, events $B_1,B_2,\ldots,B_n$ are mutually independent in the above sense if and only if the $\sigma$-algebras $\mathcal B_1,\ldots,\mathcal B_n$ are independent, where $\mathcal B_k=\{\emptyset,\Omega,B_k,B_k^c\}$.

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  • $\begingroup$ Where does the first statement come from? How does is follow from the definition of independence? $\endgroup$
    – user540084
    Oct 4, 2019 at 7:34
  • $\begingroup$ If by "first statement" you mean the first sentence of my answer, it is a special case of the OP's definition of independence of $\sigma$-fields, gotten by choosing those $\sigma$-fields as in the final sentence of my answer. $\endgroup$ Oct 5, 2019 at 14:31
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Note that thet sub $\sigma$-algebras of $\mathscr F$ all contain $\Omega$. Hence

$$P(B_1, ..., B_n) = P(B_1) ... P(B_n )$$

is equivalent to

$$P(B_{i_1}, ..., B_{i_j}) = P(B_{i_1}) ... P(B_{i_j})$$

where $\{i_1,i_2,...,i_j\} \subseteq \{1,2,...,n\}$

by setting $$B_{k}= \Omega \ \forall k \in \{1,2,...,n\} \cap \{i_1,i_2,...,i_j\}^C$$

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