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Let $X = (X_1, \ldots, X_d) \in \mathbb{R}^d$ be a mean-zero Gaussian random vector with identity covariance matrix. Are there upper bounds for $$E \left(\|X\|_{\infty}^k \right)$$ for $k=1, \ldots, 6$ ? It is easy to see for example for $k=1$, the upper bounded is of the order $\sqrt{2\log(2d)}$. In particular for lower order moments (like when $k=6$) can one still obtain upper bounds that are logarithmic in $d$ ?.

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$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$

Let $M:=\|X\|_{\infty}$. Then for $x>0$ \begin{equation} \PP(M>x)\le d\,\PP(|X_1|>x)=2d\, G(x)<2d\, f(x)/x, \end{equation} where $G:=1-F$, $F$ is the standard normal cdf, and $f$ is the standard normal cdf. Also, trivially $\PP(M>x)\le1$. So, \begin{equation} \E M^k=\int_0^\infty kx^{k-1} \PP(M>x)dx\le I_1+2kdI_2, \end{equation}
where $x_d:=\sqrt{2\ln d}$, \begin{equation} I_1:=\int_0^{x_d} kx^{k-1}dx=x_d^k=(2\ln d)^{k/2}, \end{equation} \begin{equation} I_2:=\int_{x_d}^\infty x^{k-2}f(x)\,dx \sim x_d^{k-3}f(x_d)\ll(\ln d)^{(k-3)/2}/d=o((\ln d)^{k/2}/d), \end{equation} where the asymptotic equivalence $\sim$ follows by l'Hospital's rule. So, \begin{equation} \E M^k\le(2\ln d)^{k/2}(1+o(1)); \end{equation} the convergence everywhere here is as $d\to\infty$.


This asymptotic bound is in fact the best possible one. Indeed, take any $c\in(0,2)$. Then for $y=y_{d,c}:=\sqrt{c\ln d}$ we have \begin{multline} \PP(M>y)=1-(1-G(y))^d\ge1-\exp\{-d\,G(y)\} \\ =1-\exp\Big\{-d\,\exp\Big(-\frac{y^2}{2+o(1)}\Big)\Big\} =1-\exp\big\{-d^{1-c/2+o(1)}\big\}\to1 \end{multline} and hence $\PP(M>x)\to1$ uniformly over all $x\in[0,y]$. So, \begin{multline} \E M^k=\int_0^\infty kx^{k-1} \PP(M>x)dx\ge \int_0^y kx^{k-1}\,dx\,(1-o(1)) \\ =y^k(1-o(1))=(c\ln d)^{k/2}(1+o(1)). \end{multline} Taking now $c$ to be arbitrarily close to $2$, we conclude that \begin{equation} \E M^k=(2\ln d)^{k/2}(1+o(1)). \end{equation}

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  • $\begingroup$ I have added a consideration showing that the previously given upper bound is the best possible. $\endgroup$ – Iosif Pinelis Apr 13 '18 at 16:00
  • $\begingroup$ Thanks a lot Iosif for the proof. Previously I was trying to use Gumbel distribution approximation of max of Gaussians to get its moments. This is more elegant. $\endgroup$ – Kcafe Apr 13 '18 at 22:53

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