# Lower bound on probability of pairwise intersection of events

Let $$A_1,\cdots, A_n$$ be $$n$$ events of some probability space, such that for all $$k\in [n]$$ we have $$\text{Pr}[A_k]=\epsilon$$ for some $$\epsilon\in (0,1)$$.

I'm interested in the following quantity:

$$f(n)=\max_{i\not=j}\text{Pr}[A_i\cap A_j]$$

My question is: what is the minimum of $$f(n)$$ expressed in terms of $$n$$ and $$\epsilon$$ over all possible probability space and events?

In particular, do we have $$f(n)\rightarrow \epsilon$$ as $$n\rightarrow \infty$$?

A naive application of inclusive exclusive principle yields $$f(n)\ge \epsilon^2$$ for large enough $$n$$. Can we improve this?

I'm sure this problem must appears somewhere in the literature, but I can't find a keyword to search.

This is not a complete answer, but it shows an explicit example, where $$f(n) = \epsilon^2$$ for all $$n\geq 2$$ and thus in general there cannot be a better bound than $$\epsilon^2$$.
We will consider the probability space $$[0,1]$$ and $$\mathbb{P}=Uniform(0,1)$$ as our probability measure. We consider $$\epsilon = \frac{1}{m}$$, where $$m\in \mathbb{N} \setminus \{1\}$$. We define $$A_n = \bigcup_{k=0}^{m^{n-1}-1}[\frac{k}{m^{n-1}},\frac{k}{m^{n-1}}+\frac{1}{m^n})$$ such that $$A_n$$ is a union of $$m^{n-1}$$ disjoint intervals of length $$\frac{1}{m^n}$$ and thus $$\mathbb{P}(A_n)=\frac{m^{n-1}}{m^n} = \frac{1}{m}$$. For example $$A_1 = [0,\frac1m)$$, $$A_2=[0,\frac{1}{m^2})\cup[\frac{1}{m},\frac1m + \frac{1}{m^2}) \cup \dots \cup [\frac{m-1}{m}, \frac{m-1}{m}+\frac{1}{m^2})$$ and so on. If $$i, then for each of the $$m^{i-1}$$ intervals in $$A_i$$, this interval contains exactly $$m^{j-i-1}$$ intervals of $$A_j$$ and since each intersection of intervals will have length $$m^{-j}$$, the total length of the intersection will be $$m^{i-1}m^{j-i-1}m^{-j}=m^{-2}$$. And therefore $$\mathbb{P}(A_i \cap A_j) = \frac{1}{m^2} \quad \text{for all } i\neq j$$