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when toggle format what by license comment
4 hours ago history became hot network question
8 hours ago answer added Iosif Pinelis timeline score: 5
10 hours ago comment added mathworker21 @AlekseiKulikov I think by "max" you mean "min" (and by "$x_{2^k}$" you mean "$X_{2^k}$").
11 hours ago comment added Aleksei Kulikov I believe we can show that almost surely $\limsup \frac{nY_n}{\log \log (n)} \le 10$, by defining $Z_k = \max(X_{2^{k-1}+1}, \ldots,x_{2^{k}})$ (they are now independent), bounding $Y_n$ for $2^{k}+1\le n \le 2^{k+1}$ from above by $Z_k$ and repeating your argument.
12 hours ago comment added mathworker21 The expected value is $\frac{1}{n+1}$, so it can't be that $Y_n = \frac{\log n}{n}$ almost surely (even with probability at least $\frac{1.01}{\log n}$).
12 hours ago history asked Wojowu CC BY-SA 4.0