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I'll put the problem and then I'll explain my problem.

Problem: Let ${A_n}$ be events such as

$\operatorname{Cov}(I_{A_i},I_{A_j})=E[I_{A_i}I_{A_j}]-E[I_{A_i}]E[I_{A_j}]\leq 0,\ \forall i\neq j\tag{1}$

If $\sum P(A_i)=\infty$ then $P[\lim \sup A_n]=1$.

Answer: By (1) we have that $\operatorname{Cov}(I_{A^c_i},I_{A^c_j})\leq 0$ too. So $P(\lim \inf A_n^c)=P(\lim \bigcap A_n^c)=\lim P(\bigcap A_n^c)\overset{Q!} {\leq} \lim \prod P(A_n^c)= \lim \prod (1-P(A_n))\leq \lim e^{-\sum P(A_n)}=0$

The detail is that in the inequality marked with a "Q!" I used that $P(\bigcap A_n)\leq \prod P(A_n)$. It is intuitive but i couldn't prove through the problem statement. But it is interesting result. We could use B-C lemma even for correlated events, since they are negatively correlated. What do you guys think about it?

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try showing that the variance is smaller than the mean. Then since the mean in going to $\infty$ the st. dev. is much smaller and by a markov type inequality, the sum must be fairly large. – mike Apr 19 '12 at 22:10
up vote 2 down vote accepted

I think Fristedt & Gray's book on probability proves a version of Borel--Cantelli that assumes only nonpositive correlation rather than independence. In particular, that means pairwise independence is enough.

Later edit: Here's what I find in A Modern Approach to Probability Theory by Bert Fristedt and Lawrence Gray, page 79:

Lemma 5. [Borel-Cantelli] Let $(A_1,A_2,\ldots)$ be a sequence of events in a probability space $(\Omega,\mathcal{F},P)$. Assume that for each $i\ne j$, the events $A_i$ and $A_j$ are either negatively correlated or uncorrelated. Let $A=\lim\sup_{n\to\infty} A_n$. If $\sum_{n=1}^\infty P(A_n)=\infty,$ then $P(A)=1$.

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This is awesome. Thank you! – Rodolfo Apr 19 '12 at 23:54
Fristedt and Gray set an exercise right after this statement: prove the result by using the Kochen-Stone lemma. – Michael Hardy Apr 20 '12 at 19:24

The result you mention as being intuitive does not hold. To see this, one can recall the classical example of three uncorrelated dependent events, which is $A_1=[U=1]$, $A_2=[V=1]$ and $A_3=[UV=1]$, where $U$ and $V$ are i.i.d. $\pm1$ symmetric Bernoulli random variables.

Then the events $A_i$ are pairwise independent, each $A_i$ has probability $\frac12$, and $A_1\cap A_2\cap A_3=[U=V=1]$ hence $\mathrm P(A_1\cap A_2\cap A_3)=\frac14\gt\frac18=\mathrm P(A_1)\mathrm P(A_2)\mathrm P(A_3)$.

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True, my mind mislead me. Do you any idea how to prove that then? Thanks. – Rodolfo Apr 19 '12 at 21:56
This is (a special case of) Kochen-Stone lemma, which is based on Paley-Zygmund inequality. Googling these keywords will unearth several excellent short proofs. – Did Apr 19 '12 at 22:13
Rodolfo: I am lost: did you intend to ask for general versions of BC, or, as explicitely written in your post, for the validity of inequality "Q!"? – Did May 7 '12 at 12:21

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