Independance of Sigma Algebras

A book I'm reading gives the following defintion for independance.

Write $J \subset_f I$ if $J$ is a finite subset of $I$. A family $(S_i)_{i\in I}$ of $\sigma$-sub-algebras of $A$ is called independent, if for every $J \subset_f I$ and every choice $A_j \in S_j$ we have $P[\cap_{j\in J} A_j] = \prod_{j\in J} P[A_j]$. A family of sets $(A_i)_{i\in I}$ is called independent, if the $\sigma$-algebras $S_j = \{\emptyset, A_j, A^C_j, \Omega\}$ are independent.

Then they provide the following example:

Let $\Omega = \{1,2,3,4\}$ and consider the two $\sigma$-algebras $A=\{\emptyset,\{1,3\},\{2,4\},\Omega\}$ and $B=\{\emptyset,\{1,2\},\{3,4\},\Omega\}$. $A$ and $B$ are independent.

I don't see how the sigma algebras are independent. In particular, how are they making this assertion without a probability measure on $\Omega$. Do they mean that they are independent for any choice of probability measure? If so, how do I see that?

thanks!

• In general, for all $\sigma$-algebras $E_1,E_2\ne\{\emptyset,X\}$ on $X$ there is a probability $P:\mathcal P(X)\to [0,1]$ such that $E_1$ and $E_2$ are not $P$-independent. In fact, consider $A_i\in E_i\setminus\{\emptyset,X\}$ such that $A_1\setminus A_2\ne\emptyset\ne A_2\setminus A_1$. Two such sets can be found like this: take $B_i\in E_i\setminus\{\emptyset,X\}$; either $A_i=B_i$ is already a good choice, or $B_i\subseteq B_j$ for some $i,j$, in which case $A_i=B_i$ and $B_j=X\setminus B_j$ works. Finally, let $x\in A_1\setminus A_2,\, y\in A_2\setminus A_1$ and set $P(x)=P(y)=1/2$. – Saucy O'Path Jun 2 '18 at 21:57

I guess it should say (maybe it does somewhere) that $P$ is defined in such a way that $P({n})=\frac14$ for $n\in\{1,2,3,4\}$ (or put in other way $P(A)=\frac{\#A}4$).

So you have to prove that for every pair $(E_A,E_B)\in A\times B$ (*), you have $$P(E_A\cap E_B)=P(E_A)\cdot P(E_B).$$

If one of $E_A$ and $E_B$ or both are empty sets, then both sides equal $0$. If $E_A=\Omega$ then both equal $P(E_B)$ and viceversa.

Finally, in other case, you have $\#E_A=\#E_B=2$ and $\#(E_A\cap E_B)=1$, so the left side equals $\frac14$ and the right side equals $\frac12\cdot\frac12=\frac14$.

If you choose a $P$ such that, for instance, $p_1=\frac12$, $p_2=\frac14$ and $p_3=p_4=\frac18$ (where $p_i=P(\{i\})$, $1\le i \le 4$, then it is easy to see that, for instance $P(\{1,3\}\cap\{1,2\})\neq P(\{1,3\})\cdot P(\{1,2\})$.

(*) Going back to the definition, this would actually be the case $J=I$, and here $\#I=2$. Technically there are three other cases which are all fairly trivial: two of them correspond to $\#J=1$ —that is, taking only events in $A$ and taking only events in $B$, which makes the equality evident since it is of the form $P(E)=P(E)$— and the other one is for $J=\emptyset$ which involves an empty intersection and an empty product which conventionally are interpreted as $\Omega$ and $1$, respectively. Of course, you never need to check these trivial cases, but it is interesting to see that they are somehow included in the definition.

My guess would be that the measure is defined as $\mathbb{P}(1)=\frac{1}{4}$ and similarly for $2,3,4$, i.e. a uniform distribution. Then independence is an easy check - e.g.

$$\mathbb{P}(\{1,2\}\bigcap\{1,3\})=\mathbb{P}(\{1\})=\frac{1}{4}=\mathbb{P}(\{1,2\})\mathbb{P}(\{1,3\})$$