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I am interested in what circumstances various zero-one laws in probability theory can be relaxed. In particular, independence is a very important factor in such laws.

1) Borel-Cantelli Lemma: Let $A_1, A_2, \cdots$ be a sequence of events. Then $\mathbb{P}(\limsup_{n \rightarrow \infty} A_n) = 0$ if $\displaystyle \sum_{n = 1}^\infty \mathbb{P}(A_n) < \infty$, and $\mathbb{P}(\limsup A_n) = 1$ if $A_1, A_2, \cdots$ are pairwise independent and $\displaystyle \sum_{n = 1}^\infty \mathbb{P}(A_n) = \infty$.

2) Kolmogorov's zero-one law: If $X_1, X_2, \cdots$ are a sequence of random variables, define $H_n = \sigma(X_{n+1}, X_{n+2}, \cdots)$ to be the smallest sigma algebra for which $X_{n+1}, X_{n+2}, \cdots$ are measureable. Then it is clear that $H_n \supset H_{n+1} \supset \cdots$ Let $H_\infty = \bigcap_{n} H_n$. Now suppose that $X_1, X_2, \cdots$ are independent. Then all events $A \in H_\infty$ satisfy $\mathbb{P}(A) = 0$ or $\mathbb{P}(A) = 1$.

I am particularly interested in cases where independence, which is a rather strong assumption and difficult to verify, can be replaced by estimates on various moments of the random variables, their correlation, etc. For example, the original statement of the Borel-Cantelli Lemma assumed that the sequence of events are independent, but this has since been weakned to pairwise independence. Any help would be greatly appreciated.

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As Didier Piau suggests in his answer, pairwise independence is not enough for Kolmogorov's zero-one law. Consider $X_0,X_2,X_4,...$ as independent uniformly random elements of the integers $\mod n$ and choose $X_{2i-1}$ so that $X_{2i-1}=X_{2i}+X_0$. These random variables are pairwise independent. The event that there are infinitely many $i$ so that $X_{2i-1}=X_{2i}$ is in the tail algebra and it has probability $1/n$. A slight modification so that $X_0$ is the checksum for $k$ variables means that it's not enough to assume any $k$-tuple is independent. –  Douglas Zare May 22 '11 at 16:12

2 Answers 2

up vote 5 down vote accepted

Re Borel-Cantelli lemma, if one assumes only the divergence of the series and that $P(A_n\cap A_k)\le cP(A_n)P(A_k)$ for every distinct $n$ and $k$ large enough, one gets that $P(\limsup A_n)\ge1/c$. Proof and situation of the problem by V. V. Petrov here.

Re Kolmogorov's zero-one law, I wonder what kind of weakened conclusion you have in mind.

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Is it possible to replace the constant $c$ with a function $f(n,k)$ and make some non-trivial conclusions? –  Stanley Yao Xiao May 22 '11 at 20:01

The zero-one law is true for extremal Gibbs states.

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@Robert : Hi I think that this claim would deserve further development. What is an extremal Gibbs state ? any references for a proof for this result ? Regards –  The Bridge May 23 '11 at 11:48
Hi The Bridge the book Gibbs Measures and Phase Transition by Hans-Otto Georgii, is a good reference for the details about Israel's answer. –  Leandro May 23 '11 at 21:00
@Leandro: Indeed. However, rather than invoking the 500+ pages of Georgii's book, it should be possible (and probably more helpful to the OP) to provide the corresponding technical statement as a short paragraph here, don't you think? (This remark is The Bridge's first question, formulated differently.) –  Did May 26 '11 at 6:05
Well, here's a try (not as general as possible). We consider Borel probability measures on $\Omega = S^{\cal L}$ where $S$ is a finite set and ${\cal L} = {\mathbb Z}^d$ for some positive integer $d$. For $\omega \in \Omega$ and $\Lambda \subset \cal L$ let $\omega_\Lambda$ be the restriction of $\omega$ to $\Lambda$. (continued in next comment) –  Robert Israel May 26 '11 at 17:37
Consider a system of real-valued continuous functions $H_\Lambda$ on $\Omega$ for each finite $\Lambda \subset \cal L$. A Gibbs state for this system is a probability measure $\mu$ with conditional probabilities $\mu(\omega_\Lambda | \omega_{\Lambda^c}) = \frac{e^{-H(\omega)}} {\sum_{\tau \in S^\Lambda} e^{-H(\tau \times \omega_{\Lambda^c})}}$. These Gibbs states form a compact convex set, and its extreme points are the extremal Gibbs states. –  Robert Israel May 26 '11 at 17:37

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