most recent 30 from http://mathoverflow.net 2013-05-24T14:47:09Z http://mathoverflow.net/feeds/question/45422 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45422/maximal-coordinate-on-a-sphere Fedor Petrov 2010-11-09T11:55:41Z 2010-11-09T12:42:09Z

<p>What is the easiest (preferably without calculations) way to see that the mean value of $\max(x_1,x_2,\dots,x_n)$ on the sphere $\mathbb{S}^{d-1}= \{ (x_1,\dots,x_n):\ x_1^2+\dots+x_n^2=1 \}$ behaves like $\sqrt{\log(n)/n}$, or at least that is is much more then $1/\sqrt{n}$ for large $n$? The same (and less or more a priori equivalent) question concerns the standard Gaussian measure and expectation of $\infty$-norm w.r.t. it.</p> <p>The proofs I know (for example, the one which V. Milman attributes to Figiel) use too many integrals. </p> <p>And by the way, how to put {,} in math here? \ { does not work for me</p>

http://mathoverflow.net/questions/45422/maximal-coordinate-on-a-sphere/45425#45425 Gideon Schechtman 2010-11-09T12:42:09Z 2010-11-09T12:42:09Z

<p>For some positive $c$ bounded away from zero, the probability that a standard gaussian variable is larger than $c\sqrt{\log n}$ is $1/n$. It follows that the probability that at least one variable out of $n$ independent standard gaussians is larger than $c\sqrt{\log n}$ is $1-(1-1/n)^n$ which tends to $1-1/e$. From that one gets that the expectation of the maximum of $n$ standard gaussians is at least (basically) $(1-1/e)c\sqrt{\log n}$. As you said the question about variables on the sphere is equivalent to that.</p>