concentration of maximum of gaussians

Question

Let $X=(X_1,\ldots,X_n)$, where $X_i \sim N(0,1)$ are iid.

I'm looking for a result (and a proof outline) on the concentration of the max abs value of these Gaussians, $\|X\|_\infty$. That is, some result of the form $P(\bigl | \|X\|_\infty -\sqrt{2\log (2n)}\bigr |>t)<o(t)$, where $o(t)$ is any reasonable function that goes to $0$ as $t$ gets large.

I know these results: $E \|X\|_\infty \leq \sqrt{2 \log (2n)}$, $P(\|X\|_\infty \geq \sqrt{2 \log (2n)}+t)\leq 2\exp(-t^2 /2)$, which seems to be the "right tail" of the result I'm looking for.

Answer 1

So you just need to show that $P(\|X\|_\infty \leq \sqrt{2 \log (2n)}-\epsilon)$ is small. This is easy because $$ P\left(\|X\|_\infty \leq a_n\right) = P\left(|X_1|\leq a_n\right)^n = (2\Phi(a_n) - 1)^n\le \left(1-2\frac{a_n}{\sqrt{2\pi}(a_n^2 + 1)}e^{-a_n^2/2}\right)^n. $$ Plugging $a_n = \sqrt{2\log (2n)-\delta}$,
$$ P\left(\|X\|_\infty \leq a_n\right) \le \left(1-\frac{\sqrt{2\log (2n)-\delta}}{\sqrt{2\pi}n(2\log (2n)-\delta + 1)}e^{\delta/2}\right)^n\\ \le \left(1-\frac{1}{\sqrt{2\pi}n(\sqrt{2\log (2n)-\delta} + 1)}e^{\delta/2}\right)^n\\ \le \left(1-\frac{e^{\delta/2}}{\sqrt{2\pi}n(\sqrt{2\log (2n)} + 1)}\right)^n\\ \le \exp\left\{-\frac{e^{\delta/2}}{\sqrt{2\pi}(\sqrt{2\log (2n)}+ 1)}\right\}, $$ So you can take $\delta = K\log\log n$ with some $K$ large enough in order to make this small. You might be disappointed by the fact that this goes to infinity. This is not so bad as in fact the corresponding $\epsilon$ is of order $$ \frac{c\log \log n}{\sqrt{\log n}} $$ and does go to zero. If you want to have it fixed, you will get even some exponentially small estimates for probability.

Answer 2

I have the inequality you are looking for ! It gives concentration inequality reflecting the behaviour of extremes convergence (by this I mean, if $M_n$ is the maximum of your Gaussian variables, then $a_n(M_n-b_n)$ goes in distribution toward a Gumbel). Thus, the decay must be look like the Gumbel tails.

You can look at my article : http://perso.math.univ-toulouse.fr/ktanguy/files/2012/04/Article-3-brouillon.pdf

which give concentration inequality for extrema of stationary Gaussian processes wich look like

$\mathbb{P}( a_n(M_n-b_n)>t)\leq ce^{-ct}, t>0, n\geq 1$ with $a_n=\sqrt{\log n}$ and $b_n=\sqrt{\log n}+o(\sqrt{\log n})$ (the renomelazing constant of extremes theory) same inequality for the other side also holds

Tell me if it helps. (The absolute value must not be an issue, you can also look to an article of Gideon Schechtmann about random Dvoretsky's theorem for $l_\infty$ balls (see here)).

Answer 3

For the sake of completeness, I will prove here the inequality \begin{equation} P(\|X\|_\infty \geq \sqrt{2 \log (2n)}+t)\leq \frac{1}{2}\exp(-t^2 /2) \quad (t>0), \end{equation} which is slightly stronger than the one quoted by Jason in his post. We have for any $a > 0$: \begin{equation} P(\|X\|_\infty < a) = (2 \Phi(a)-1)^n=(1-2Q(a))^n, \end{equation} where $Q(x)=1-\Phi(x)$.By using Bernoulli's inequality we then get \begin{equation} P(\|X\|_\infty < a) \geq 1-2nQ(a). \end{equation} Now we use the elementary inequality for the tail of the normal distribution: \begin{equation} Q(a) \leq \frac{1}{2} e^{-\frac{a^2}{2}}, \end{equation} so that \begin{equation} P(\|X\|_\infty < a) \geq 1-n e^{-\frac{a^2}{2}}. \end{equation} By putting now $a=\sqrt{2 \log(2n)} + t$, we get \begin{equation} P(\|X\|_\infty < \sqrt{2 \log(2n)} + t) \geq 1-n \exp \left(-\frac{(\sqrt{2 \log(2n)} + t)^2}{2} \right). \end{equation} Since \begin{equation} \exp \left(-\frac{(\sqrt{2 \log(2n)} + t)^2}{2} \right) \leq \exp \left( - \frac{2 \log(2n) + t^2}{2} \right), \end{equation} we get \begin{equation} P(\|X\|_\infty < \sqrt{2 \log(2n)} + t) \geq 1 - \frac{1}{2} e^{-\frac{t^2}{2}}, \end{equation} or \begin{equation} P(\|X\|_\infty \geq \sqrt{2 \log(2n)} + t) \leq \frac{1}{2} e^{-\frac{t^2}{2}}. \end{equation}