Situation
Let $X_1, X_2, \dots$ be i.i.d. random variables on a shared probability space, with distribution function $F$.
Denote the empirical distribution function (EDF) as $\hat F_n(\cdot) := \frac 1n \sum_{i=1}^n \mathbf 1\{X_i \leq \cdot\}$.
Suppose we have a sequence $x_m \uparrow \infty$ such that $F(x_m)^m \to 1$ as $m \to \infty$.
Additionally, let $m \equiv m_n = o(n)$ with $m_n \to \infty$.
Question
Does it follow that $$\hat F_n(x_m)^m \to 1$$ in any sense (almost surely, in probability, ...) as $n \to \infty$?
Thoughts
It would be nice to have a.s. convergence, but I think the answer is no. Here's my attempt so far:
Suppose that $\hat \Omega$ is a set with $\mathbb P(\hat \Omega) = 1$ and $\|\hat F_n^\omega - F\|_\infty \to 0$ for all $\omega \in \hat \Omega$. (Glivenko-Cantelli tells us this is possible.) Let $\omega \in \hat \Omega$.
Let $\varepsilon > 0$ be arbitrarily small.
There exists $N_\omega^\varepsilon \in \mathbb N$ such that $\| \hat F_n^\omega - F \|_\infty \leq \varepsilon$ for all $n \geq N_\omega^\varepsilon$.
So we have for all $n \geq N_\omega^\varepsilon$ $$ \begin{aligned} 0 \leq 1 - \hat F_n^\omega(x_m)^m &\leq 1 - \big( F(x_m) - \varepsilon \big)^m \\ &= 1 - \sum_{k=0}^m \binom{m}{k} F(x_m)^{m-k} \cdot (-\varepsilon)^k \\ &= 1 - \Big[ F(x_m)^m + \sum_{k=1}^m \binom{m}{k} F(x_m)^{m-k} \cdot (-\varepsilon)^k \Big] \\ &= \underbrace{ 1 - F(x_m)^m}_{\to 0} - \sum_{k=1}^m \binom{m}{k} \cdot F(x_m)^{m-k} \cdot (-\varepsilon)^k \end{aligned} $$ So the issue is the "error term" $$ \text{err} := - \sum_{k=1}^m \binom{m}{k} \cdot F(x_m)^{m-k} \cdot (-\varepsilon)^k, $$ which we need to vanish to achieve the desired limit.
But I don't see how to do that, after all, $$ \begin{aligned} |\text{err}| &\leq \sum_{k=1}^m \binom{m}{k} \left| (-\varepsilon)^k \right| \\ &= \sum_{k=1}^m \binom{m}{k} \varepsilon^k \\ &= (1 + \varepsilon)^m - 1 \end{aligned} $$ As $n$ increases, so does $m \equiv m_n$, driving the right hand side up. On the other hand, higher $n$'s allow us to choose a smaller $\varepsilon$, driving the RHS back toward zero.
It seems that the speed of convergence (which determines the smallest $\varepsilon >0$ for the $n > N$) determines whether convergence is achieved.
More information
In the particular example I'm looking at we have
- $F(x) > 1 - \frac{c}{\log^\delta(x)}$ for large $x$, where $c>0$ and $\delta > 2$; and
- $x_m := \xi \cdot \text e^m$ for $\xi \in (0,1)$.
Maybe the specifics here are helpful.