Letting $L = \limsup_n \frac{X_n – \log n}{\log \log n}$, for $\{X_n\}_{n\geq 1}$ iid exponentially distributed random variables with rate $1$, I think I am able to show that $L=1$ (almost everywhere). Assuming I hadn't made some mistake,it boiled down to a direct application of the Borel-Cantelli lemmas via events $E_n = \left\{\frac{X_n – \log n}{\log \log n} > a\right\}$,
$$
\sum P\left(E_n\right) = \sum P \left(\frac{X_n – \log n}{\log \log n} > a\right) = \sum \frac{1}{n (\log n)^a} = \begin{cases} <\infty & \text{when $a > 1$} \\
= \infty & \text{when $a \leq 1$}
\end{cases}
$$
Therefore, by BC
$$
\implies P(E_n, \text{i.o}) \equiv P(\limsup_n E_n) =
\begin{cases}
0 & a > 1 \\
1 & a \leq 1
\end{cases}
$$
So, as $a \to 1$,
$$
1 = P\left( \limsup_n \left\{ \frac{X_n – \log n}{\log \log n} > 1 \right\} \right) \leq P\left( \limsup_n \left\{\frac{X_n – \log n}{\log \log n} \right\} \geq 1 \right) = P(L \geq 1)
$$
Similarly, we can end up showing that $P(L > 1) = 0$, giving us the desired result.
My question is a matter of intuition. I (think) I understand the mathematics, but I just cannot wrap my head around how this is at all possible at any level beyond the symbolic manipulation. If I think about sampling/generating exponential deviates $X_n \stackrel{\text{iid}}{\sim} \text{Exp}(1)$, how is it possible that the ratio $\frac{X_n(\omega) – \log n}{\log \log n}$ could possibly approach 1 (for all $\omega$ except those on measure zero sets). I've tried simulating exponential samples and calculating this ratio, keeping $\omega$ fixed (i.e. the PRNG seed fixed) across $n = 1, …, \texttt{nmax}$. Regardless of how many samples I do, I always end up seeing the behaviour one would expect from, say $\frac{x – \log n}{\log \log n}$ as $n\to\infty$.
Clearly something is either wrong in my calculations above, or something is wrong with my intuition. Any insight would be appreciated!
Best Answer
You ask "how is it possible that the ratio $r_n:=\frac{X_n(\omega) - \log n}{\log \log n}$ could possibly approach 1 (for all $\omega$ except those on measure zero sets)?" but that is not what limsup means. What is true, and follows from your calculation, is that for every fixed $m$, the sequence $$A_m(N):= \max_{n \in [m,N]} \frac{X_n(\omega) - \log n}{\log \log n}$$ approaches 1 as $N \to \infty$ (for all $\omega$ except those on a set of measure zero). In other words, almost surely, there is a subsequence of $r_n$ that tends to 1, and no subsequence can tend to a higher limit.