For $g(n)$ a decreasing function, we have $\limsup Y_n/g(n)\geq 1$ if and only if $Y_n> g(n)(1-\epsilon)$ infinitely often. Based on [an approach suggested by Aleksei Kulikov][1], if $n \in [2^k, 2^{k+1}]$ then this implies $X_m > g(n) (1-\epsilon) \geq g(2^{k+1} ) (1-\epsilon)$ for all $m \leq 2^k $ for infinitely many $k$. This is an event with probability $\approx e^{ - 2^k g(2^{k+1} ) (1-\epsilon)}$ and it can only occur with positive probability for infinitely many $k$ if the sum of these probabilities $\sum_{k=1}^\infty e^{ - 2^k g(2^{k+1} ) (1-\epsilon)}$ is positive (this direction of Borel-Cantelli not requiring independence). 

So $\limsup Y_n/g(n)\geq 1$ with probability $0$ for $g(n) = C \log \log n/n$ as soon as $C>2$.


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Let $A_n$ be the event that $Y_{2^n} \geq C (\log n)/2^n$. Then $\mathbb P(A_n) \approx n^{-C}$. For $n<m$, the intersection $A_n \cap A_m$ occurs when $X_k \geq C ( \log n)/2^n$ for $k\leq 2^n$ and $X_k \geq C (\log m)/2^m$ for $2^m < k \leq 2^n$ which has probability $$\approx e^{ - C (  \log n  + \frac{2^m-2^n}{2^m} \log m )} = e^{ C 2^{n-m} \log m} e^{ - C \log n} e^{-C \log m}. $$

This is $(1+o(1) ) n^{-C} m^{-C}$ as long as $m-n > \log N$, since $e^{ C 2^{ -\log_2 N } \log m } \leq e^{ C 2^{-\log N} \log N} = 1+o(1)$. When $m$ is close to $n$, this probability is still bounded by $e^{- C\log n}$. This gives

$$\sum_{m,n \leq N} \mathbb P(A_n \cap A_m) \leq (1+o(1))\sum_{n,m\leq N}  n^{-C} m^{-C} + \sum_{\substack{ n,m \leq  N \\ |n-m| \leq \log \log N}}e^{ - C \log n} \approx (1+o(1)) ((1-C) N^{1-C})^2 + 2 (1-C) N^{1-C}  \log N = (1+o(1)) ((1-C) N^{1-C})^2  $$ as long as $C<1$.

Now $\sum_{n=1}^N \mathbb P(A_n) \approx  N^{1-C} / (1-C)$.

So the ratio$$  \frac{ (\sum_{n=1}^N \mathbb P(A_n) )^2}{ \sum_{m,n \leq N} \mathbb P(A_n \cap A_m) }$$ converges to $1$. By Kochen-Stone, this implies that $A_n$ occurs infinitely often with probability $1$.

In other words, $\limsup Y_n / (\log \log n/n) \geq 1$ with probability $1$.

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So $\limsup Y_n / (\log \log n/n) \in [1,2]$ with probability $1$. We cannot change the limsup by changing finitely many values of $X_n$ to something nonzero and so by Kolmogorov's zero-one law, there is a fixed deterministic value in $[1,2]$ that the limsup approaches. I am not sure if this is possible to compute.


  [1]: https://mathoverflow.net/questions/484200/asymptotics-for-minimum-of-a-sequence-of-random-variables#comment1261226_484200