The Kochen-Stone theorem says that if $A_n$ is sequence of events with $\sum_{i=1}^{\infty} P(A_i) = \infty$, then: $$ P(A_n \mbox{ i.o.}) \ge \limsup_{n \to \infty} \frac{(\sum_{i=1}^nP(A_i))^2}{\sum_{i, j= 1}^nP(A_i \cap A_j)} $$
I am interested in cases where the $A_n$ are not mutually independent, but where the inequality is actually an equality. Any hints or references to results or examples of this kind would be much appreciated.
In case it's of interest to others, I can now show that if $A_n$ is any increasing sequence of events then the Kochen-Stone inequality is an equality. This was enough for my application (I am looking at issues of computability and wanted to arrange for equality when the probabilities $P(A_n)$ form a Specker sequence).
To see why the Kochen-Stone inequality is an equality for increasing $A_n$, first note that in A Simple Proof of Two Generalized Borel-Cantelli Lemmas (Springer Lecture Notes in Mathematics vol. 1874), Yan showed that the diagonal terms in the sums in the Kochen-Stone inequality are negligible:
\begin{equation} \label{yan-eq} \limsup_{n \to \infty} \frac{(\sum_{k=1}^n P[A_k])^2}{\sum_{i,k=1}^n P[A_i A_k]} = \limsup_{n \to \infty} \frac{\sum_{1 \le i < k \le n} P[A_i]P[A_k]}{\sum_{1 \le i < k \le n}P[A_iA_k]} \tag{1} \end{equation}
Now assume $A_1 \subseteq A_2 \subseteq A_3 \subseteq \ldots$ and assume, for simplicity, that $P(A_1) > 0$. Put $q_n = P(A_n)$. Then $P(A_n \mbox{ i.o.}) = q$ where $q = \lim_{n \to \infty}q_n$. Noting that $P(A_iA_k) = P(A_i)$ when $i < k$, we can write the fraction on the right-hand side of equation (1) as $w_n = \frac{u_n}{v_n}$ where: \begin{align} u_n &= q_1\sum_{k= 2}^n q_k + q_2\sum_{k=3}^n q_k + \ldots + q_{n-1}q_n\\ v_n &= (n-1)q_1 + (n-2)q_2 + \ldots + q_{n-1}\\ \end{align} and then the Kochen-Stone inequality is $q \ge \limsup_{n \to \infty} w_n$. I claim that, in fact, $w_n \to q$ as $n \to \infty$, so that $q = \lim_{n \to \infty} w_n = \limsup_{n \to \infty} w_n$. To see this, note that:
\begin{align} \sum_{k=i}^nq_k &= q\left(n - i + 1 - \frac{1}{q}\sum_{k=i}^n (q - q_k) \right) \end{align}
Let us write $\sigma_i^j$ for $\sum_{k=i}^j(q - q_k)$. From the above, multiplying by $q_{i-1}$ and summing for $i$ from $2$ to $n$, we have: \begin{equation} u_n = qv_n - \sum_{i=2}^{n} q_{i-1}\sigma_i^n \end{equation}
Define the sequence $r_n$ by: \begin{align} r_n &= \frac{\sum_{i=2}^{n} q_{i-1}\sigma_i^n}{v_n} \tag{2} % % % &= \frac{q_1\left(\sum_{k=2}^{n}(q - q_k)\right) + q_2\left(\sum_{k=3}^{n}(q - q_k)\right) + \ldots + q_{n-1}(q - q_{n-1})} % {q_1 (n-1) + q_2 (n-2) + \ldots + q_n} \end{align} I claim that $r_n \to 0$ as $n \to \infty$ so that $w_n = u_n/v_n = q - r_n \to q$, which is what we want to prove. To see this, given $\epsilon > 0$, put $\epsilon_0 = \frac{q_1}{4}\epsilon$. Then there is an $N$ such that for all $n > N$, $q - q_n < \epsilon_0$. Then for $n > N$, we have: \begin{equation} r_n = \frac{C}{v_n} + \frac{s_n}{v_n} \end{equation} where $C$ and $s_n$ are obtained by grouping the monomials $q_iq_j$ in the numerator of the fraction on the right-hand side of equation (2) so that $C$ comprises the $q_iq_j$ for which $i$ and $j$ are most $N$ and $s_n$ contains all the rest. Observing that $s_n$ contains at most $(n-1)(n - N)$ monomials all of which evaluate to at most $\epsilon_0$ and that $v_n > q_1((n-1) + (n - 2) + \ldots + 1)= q_1\frac{1}{2}(n-1)n$, we have: \begin{equation} r _n \le \frac{C}{v_n} + \frac{1}{q_1}\cdot\frac{(n-1)(n-N)\epsilon_0}{\frac{1}{2}(n-1)n} \to \frac{2}{q_1}\epsilon_0 = \frac{\epsilon}{2}\mbox{ as $n \to \infty$} \end{equation} So there is an $M > N$ such that for all $n > M$, $|r_n - \frac{\epsilon}{2}| < \frac{\epsilon}{2}$, but then $r_n < \epsilon$ and we are done.
