# Kochen-Stone Borel-Cantelli

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## Theorem

Let $A_n$ be a sequence of events with $\displaystyle \sum \map \Pr {A_n} = \infty$ and:

$\displaystyle \liminf_{k \mathop \to \infty} \frac {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} < \infty$

Then there is a positive probability that $A_n$ occur infinitely often.

## Proof

Fix $\ell < k$.

Let $\displaystyle X = \sum_{n \mathop = \ell}^k 1_{A_n}$. It follows that:

$\displaystyle \expect X = \sum_{n \mathop = \ell}^k \Pr(A_n)$

and:

$\displaystyle \expect {X^2} = \sum_{\ell \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m}$

Using Paley-Zygmund inequality for $\theta = 0$, we obtain:

 $\displaystyle \map \Pr {\bigcup_{n \mathop = \ell}^k A_n}$ $=$ $\displaystyle \map \Pr {X > 0}$ $\displaystyle$ $\ge$ $\displaystyle \frac {\displaystyle \paren {\sum_{n \mathop = \ell}^k \map \Pr {A_n} }^2} {\displaystyle \sum_{\ell \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} }$ $\displaystyle$ $\ge$ $\displaystyle \frac {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \map \Pr {A_n} }^2} {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} - \sum_{1 \mathop \le m, n \mathop < \ell} \map \Pr {A_n \cap A_m} }$

Now, it holds that:

$\displaystyle \map \Pr {A_n \text { occurs i.o.} } = \lim_{\ell \mathop \to \infty} \lim_{k \mathop \to \infty} \map \Pr {\bigcup_{n \mathop = \ell}^k A_n}$

We have:

$\displaystyle \frac {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \map \Pr {A_n} }^2} {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} - \sum_{1 \mathop \le m, n \mathop < \ell} \map \Pr {A_n \cap A_m} } \ge \frac {\displaystyle \left(\sum_{n \mathop = 1}^k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \Pr (A_n)\right)^2} {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} }$

Fix $l \in \N_1$.

Since $\displaystyle \lim_{k \mathop \to \infty}\sum_{n \mathop = 1}^k \map \Pr {A_n} = \infty$, by assumption, if $k$ is sufficiently large:

$\displaystyle \paren {\sum_{n \mathop = 1}^ k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \map \Pr {A_n} }^2 \approx \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2$

So if $k$ is sufficiently large:

$\displaystyle \frac {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \map \Pr {A_n} }^2} {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } \approx \frac {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } = \frac 1 {\frac {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} }$

Suppose:

$\displaystyle \liminf_{k \mathop \to \infty} \frac {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} = c < \infty$

This means that no matter how large $k$ is, there is always some $k' \ge k$, such that:

$\displaystyle \frac {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k'} \map \Pr {A_n \cap A_m} } {\displaystyle \paren {\sum_{n \mathop = 1}^{k'} \map \Pr {A_n} }^2} \approx c$

So for infinitely many $k$s:

$\displaystyle \map \Pr {\bigcup_{n \mathop = \ell}^k A_n} \ge \frac 1 {\frac {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\paren {\displaystyle \sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} } \approx \frac 1 c$

Since $\displaystyle \map \Pr {\bigcup_{n \mathop = \ell}^k A_n}$ is decreasing in $k$, we have, approximately:

$\displaystyle \lim_{k \mathop \to \infty} \map \Pr {\bigcup_{n \mathop = \ell}^k A_n} \ge \frac 1 c$

Since $l$ was arbitrary:

$\displaystyle \lim_{l \mathop \to \infty} \lim_{k \mathop \to \infty} \map \Pr {\bigcup_{n \mathop = \ell}^k A_n} \ge \frac 1 c > 0$

$\blacksquare$