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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!

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  • $\begingroup$ 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 $P(x)=P(y)=1/2$. $\endgroup$ – Saucy O'Path Jun 2 '18 at 21:57
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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.

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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.

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

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