Let $X_i,i=1,2,...,$ be a sequence of i.i.d. random variables and $u_n$ be a sequence of positive real numbers. I was motivated by the fact that if
$$P(\max_{1\leq i\leq n}|X_i|\geq u_n)\to 0 \quad \text{as}\quad n\to\infty. $$
then
$$P(\max_{1\leq i\leq n}|X_i|\geq u_n) \sim nP(|X_1|\geq u_n)$$
This statement is a direct result of Chung's textbook: A course in Probability Theory; Third Edition; Page 67; Exercise 17; which says
If $\forall n:\{E_j^{(n)},1\leq j\leq n\}$ are independent events, and $$P\left(\bigcup_{j=1}^n E_j^{(n)}\right)\to 0\quad\text{and}\quad n\to\infty.$$ Then $$P\left(\bigcup_{j=1}^n E_j^{(n)}\right) \sim \sum_{j=1}^n P\left(E_j^{(n)}\right)$$ Then I made a conjecture based on this.
If
$$P(\max_{1\leq i<j\leq n}|X_iX_j|\geq u_n)\to 0 \quad \text{as}\quad n\to\infty. $$ then $$P(\max_{1\leq i<j\leq n}|X_iX_j|\geq u_n) \sim \frac{n^2}{2}P(|X_1X_2|\geq u_n)$$ The problem is now that the events $\{|X_iX_j|\geq u_n\}$, $1\leq i<j\leq n$ are not mutually independent anymore, since there are some overlaps in the indices, and this makes the conjecture hard to prove or even disprove.
Any comment and discussions are appreciated for the proof or disproof of this conjecture.
I decided to trace through the intuition in my above comments, and it seems to work. Here is an explicit counter-example:
Consider $\{X_i\}$ i.i.d. with a continuous CDF that satisfies: $$ Q(u) = P[X_i>u] = \frac{1}{\log(e-1+u)} \quad , \forall u \geq 1 $$ In particular, $X_i \geq 1$ with probability 1.
Define $u_n$ so that $Q(\sqrt{u_n}) = 1/n^2$ for all $n \in \{1, 2, 3,...\}$. We have: \begin{align*} P\left[\max_{i,j \in \{1, ..., n\}} X_iX_j > u_n\right] &\leq 1-P[\cap_{i=1}^n \{X_i \leq \sqrt{u_n}\}]\\ &=1-(1-Q(\sqrt{u_n}))^n\\ &= 1- (1-\frac{1}{n^2})^n \rightarrow 0 \end{align*} On the other hand: $$P[X_1X_2 > u_n] \geq P[X_1> u_n] = Q(u_n) $$ Thus, for all $n \in \{1, 2, 3, ...\}$ we have \begin{align*} n^2P[X_1X_2>u_n] &\geq n^2Q(u_n) \\ &=\frac{n^2}{\log(e-1+u_n)} \\ &= \frac{(1/2)n^2}{\log(\sqrt{e-1+u_n})}\\ &\overset{(a)}{\geq} \frac{(1/2)n^2}{\log(e-1 + \sqrt{u_n})}\\ &= (1/2)n^2Q(\sqrt{u_n}) \\ &= 1/2 \end{align*} where the inequality (a) follows because $\sqrt{e-1+u_n} \leq \sqrt{e-1} + \sqrt{u_n} \leq e-1 + \sqrt{u_n}$ (since $u_n\geq 0$). So we see that $n^2P[X_1X_2>u_n]$ does not even converge to zero!