The max of n independent Gaussians concentrates around sqrt(2log(n)). The max central limit theorem allows to quantify this. en.wikipedia.org/wiki/Fisher…

Nov 26, 2020 · 6:00 AM UTC

Replying to @gabrielpeyre
Some intuition: this rate is the inverse of the Gaussian pdf kernel.
Exactly :) This can be seen in the Fisher–Tippett–Gnedenko theorem (central limit theorem for the max) en.wikipedia.org/wiki/Fisher…
Replying to @gabrielpeyre
I am confused isn't \sqrt{2\log(n)} ~o(\sqrt{\log(n)})
Maybe you think about O(...) instead of o(...) ?
What does the big O mean in this case?
It is not a big O but a little o (negligible terms).
Replying to @gabrielpeyre
if the red lines are max_{i ≤n} |X_i|, why are they much higher and lower than the blue lines (which I assume are the samples)? I'm confused.
The blue shows a single realization corresponding to one of the many red curves (and I guess it is hard to see which one it corresponds to). Drawing many red curves might give the impression they are higher maybe?
Replying to @gabrielpeyre
For idd exponential, the max grows ~log(n) for large n. Actually, it's the sum of n terms of the harmonic series, first shown by Alfréd Rényi. en.m.wikipedia.org/wiki/Orde… As a curiosity, it's also proportional to the time for n random walkers to find a target in 1D
Replying to @gabrielpeyre
Just had this in a homework assignment last week. Cute proof, interesting result!