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If $\sum_n P(A_n) = \infty$ then obviously we can't try to apply inclusion-exclusion directly to evaluate $P\left(\bigcup_n A_n\right)$, without taking limits. But what if $\sum_n P(A_n) < \infty$? Can we define

$$P\left(\bigcup_n A_n\right) = \sum_{n_1 \in {\mathbb N}} P(A_{n_1}) - \sum_{\{n_1,n_2\} \in {\mathbb{N} \choose 2}} P(A_{n_1} \cap A_{n_2}) + \ldots $$

It seems that first all inner series must converge, and then perhaps the outer infinite sum must converge absolutely. At any rate, when is the expression convergent and valid? In particular, what if the $A_n$ are independent?

Michael Hardy
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user2566092
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  • Related. – Did Oct 06 '13 at 19:04
  • @Did Thanks that is indeed related. It says that if the series over all subsets converges absolutely, then the infinite series identity is valid. So I guess my question could potentially be phrased as what general conditions guarantee that the series over all possible subsets converges absolutely, e.g. maybe independence of events is enough. – user2566092 Oct 06 '13 at 20:31
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    Independence + convergence of $\sum P(A_n)$ indeed implies absolute convergence of the series over all subsets. And "independence" can be replaced by negative association. – Did Oct 07 '13 at 06:39

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I'm not sure what happens when the $A_n$ are independent, but if the $A_n$ are not independent then it's possible that the sum of pairwise intersection probabilities diverges even if $\sum_n P(A_n)$ converges. To see this, consider $A_1 \supset A_2 \supset A_3 \cdots$ where each $A_n$ has probability $1/n^2$. Then $\sum_n P(A_n)$ converges but the sum of pairwise intersection probabilities is $\sum_n (n-1)/n^2$ which diverges.

user2566092
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  • So is it possible some particular rearrangement of the series is somehow the "right" one and converges conditionally to the right number? – Michael Hardy Oct 06 '13 at 19:19
  • @MichaelHardy I think that's indeed true, because e.g. we can take the limit of the series $P(\cup_{n=1}^N A_n)$ as $N \to \infty$ and we should get the right answer. This should define a particular rearrangement where we get the right answer. Interestingly, it should be invariant under permutation of the $A_n$, so there should be infinitely many "natural" rearrangements which give the right answer. – user2566092 Oct 06 '13 at 20:33
  • So this should define an equivalence relation on rearrangements. Would it correspond to a finer partition than the relation that says only that two rearrangements are equivalent precisely if they have the same sum? – Michael Hardy Oct 06 '13 at 22:12