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$\{A_i, i \in \mathbb{N} \}$ are defined to be independent, if $P(\cap_{k=1}^{n} A_{i_k}) = \prod_{k=1}^{n} P(A_{i_k}) $ for any finite subset of $\{A_i, i \in \mathbb{N} \}$.

  1. We know $P(\cup_{i=1}^{\infty} A_i) = \sum_{i=1}^{\infty} P(A_i) $ iff $\{A_i , i \in \mathbb{N}\}$ are disjoint, which is independent of the probability measure and purely depends on the relation between the sets. I was wondering if it is possible to similarly characterize/interpret $\{A_i , i \in \mathbb{N}\}$ being independent purely from relation between sets, and make it independent of the probability measure as much as possible if completely is impossible?
  2. Is the definition of $\{A_i, i \in \mathbb{N} \}$ being independent equivalent to $P(\cap_{i=1}^{\infty} A_{i}) = \prod_{i=1}^{\infty} P(A_{i}) $. What is the purpose of considering any finite subset instead?
  3. Is generalization of independence from probability space to general measure space meaningful?

    The only interpretations of independence I know are: measure can be exchanged with product/intersection on independent sets, and intuitively, independent events occur independently of each other. Are there other interpretation, especially in the general measure space setting?

Thanks and regards!

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Part 1. $P(\cup_{i=1}^{\infty} A_i)= \sum_{i=1}^{\infty} P(A_i)$ does not necessarily implies $\{A_i , i \in \mathbb{N}\}$ are disjoint, not even in the two events case, because the intersection can be nonempty, but has zero probability. –  GWu Apr 13 '11 at 18:24
    
Your iff statement in (1) is false. For a simple counterexample, consider abutting closed intervals on $[0,1]$ and Lebesgue measure. More "exotic" examples can also be constructed. –  cardinal Apr 13 '11 at 18:25
    
@cardinal: we are talking about the same thing. One can fix the measure and find "exotic" sets, or one can also fixed any non-disjoint set and simply define a probability measure, with the intersection having zero measure. –  GWu Apr 13 '11 at 18:28
    
@GWu, there was just a severe (several minute) delay in my comment getting posted due to my internet connection. Your first comment wasn't there when I initially submitted mine. :) –  cardinal Apr 13 '11 at 18:31
    
@GWu: Thanks! Then is it correct that measure and union/summation can be exchanged iff the intersection of any sub-collection of the class of sets has zero measure? –  Tim Apr 13 '11 at 19:14
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2 Answers

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Not sure your point 2 was addressed, so let me state that defining independence as suggested would lead to a trivial notion, quite different from independence as one wants it.

To wit, any collection of sets $(A_i)_{i\ge1}$, finite or infinite, could be made part of a larger collection $(A_i)_{i\ge0}$ such that the condition stated in 2 holds: simply add $A_0=\emptyset$. One would be led to say that a sequence is independent while one of its subsequences is not.

So, the problem has nothing to do with infinite sequences: to define the independence of $A$, $B$ and $C$ by the only condition that $P(A\cap B\cap C)=P(A)P(B)P(C)$ (thus forgetting the supplementary conditions that $P(A\cap B)=P(A)P(B)$, $P(B\cap C)=P(B)P(C)$ and $P(A\cap C)=P(A)P(C)$) already leads to a notion too weak to model any kind of independence, since the supplementary conditions I just wrote can fail.

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Thanks for clarification! –  Tim Apr 19 '11 at 9:01
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For the first question - you can easily find any counterexamples. Independence is strongly connected with a probability measure. Each two sets $A,B$ can be made independent (non-independent) by choosing an appropriate measure $P$ if $A\cap B \neq \emptyset$. If $A\cap B = \emptyset$ then $A,B$ are always non-independent.

For the second question - it seems to me that definitions are the same. Just for the classic definition you don't use infinite product which is a messy stuff.

Edited: for non-intersecting $A,B$ one can define $P:P(A) = P(B) = 0$ but in my opinion these sets are not independent in the usual sense since their measure is zero. Indeed, if the r.v. takes a value in $A$ it never can take a value in $B$ - this is a non-trivial information which makes occurrence of $A$ and $B$ dependent.

About interpretation - I was also asking the same question some time ago on the other site. I will say that probability theory is measure theory + independence + conditioning (as for me, $P(\Omega)=1$ is just a rescaling though very important). Independence in fact comes from the Cartesian product of sets which then becomes independent for the product space.

Imagine e.g. rectangle - then for any $x$ the interval for $y$ is the same. But on the circle for any $x$ the interval for $y$ changes, so $y$ "depends" on $x$ and circle cannot be presented as a Cartesian product in $x,y$. so, independence comes from this facts. Conditioning comes from the experiments and it's more non-trivial property then independence (in the sense that it appears only in the probability theory).

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Thanks! (1) Is generalization of independence from probability space to general measure space meaningful? (2)The only interpretations of independence I know are: measure can be exchanged with product/intersection on independent subsets, and intuitively independent events occur independently of each other. Are there other interpretation, especially in the general measure space setting? –  Tim Apr 13 '11 at 17:01
    
@Gortaur, counterexample to your statement on nonindependence: Suppose $A \neq \emptyset$ but $\mathbb{P}(A) = 0$. –  cardinal Apr 13 '11 at 18:21
    
@cardinal. Then what? –  Ilya Apr 13 '11 at 18:28
    
@Gortaur: Did you get it? –  cardinal Apr 13 '11 at 18:34
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@Gortaur, here's an example. Consider the space $([0,1], \mathcal{B}, \mathcal{L})$ and take $A = \mathbb{Q} \cap [0,1]$ and $B = [0,1] \setminus A$. Then $A \cap B = \emptyset$, but $A$ is independent of $B$. This "feature" of the definition of independence that you appear to consider pathological actually has quite a deep consequence in the form of the Kolmogorov 0-1 law. –  cardinal Apr 13 '11 at 22:51
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