L_infinity norm of two gaussian vector

Question

$X = (x_1,...x_n) \in \mathbb{R}^n, X \sim \mathcal{N}(O, \Sigma_X)$ and $Y = (x_1,...x_n) \in \mathbb{R}^n, Y \sim \mathcal{N}(O, \Sigma_Y)$ are two independent gaussian vectors.

If $\Sigma_Y - \Sigma_X$ is positive semidefinite, then $\forall \alpha \in \mathbb{R}^n$, $\alpha\Sigma_Y\alpha^T \ge \alpha\Sigma_X\alpha^T $. Because $X \alpha^T \sim \mathcal{N}(O, \alpha\Sigma_X\alpha^T)$ and $Y \alpha^T \sim \mathcal{N}(O, \alpha\Sigma_Y\alpha^T)$, we have that $\forall \varepsilon \in \mathbb{R}_+^*$, $ P(|X \alpha^T| \le \varepsilon) \ge P(|Y \alpha^T| \le \varepsilon)$

I am trying to show that it is also true for the infinity norm, ie that $\forall \varepsilon \in \mathbb{R}_+^*$, $ P(||X||_\infty \le \varepsilon) \ge P(||Y||_\infty \le \varepsilon)$, where $ ||X||_\infty = \max_{i = 1..n}|x_i|$. Do you think this is true, and do you have clues for how to show it? Otherwise, is there a counter example?

Answer 1

$\newcommand\R{\mathbb R}$Let $\|\cdot\|$ be any norm on $\R^n$. Take any real $t$. Let $Z$ be a random vector in $\R^n$ such that (i) $Z$ is independent of $X$ and (ii) $Z\sim N(0,\Sigma_Y-\Sigma_X)$. Then $X+Z$ equals $Y$ in distribution.

So, it suffices to show that \begin{equation*} P(\|X\|\le t)\ge P(\|X+Z\|\le t). \tag{1} \end{equation*}

Note that \begin{equation*} P(\|X+Z\|\le t)=Eg(Z), \tag{2} \end{equation*} where \begin{equation*} g(z):=P(\|X+z\|\le t)=\int_{\R^n}dx\, f(x)1(\|x+z\|\le t) \end{equation*} and $f$ is the pdf of $X$. The functions $f$ and $x\mapsto1(\|x+z\|\le t)$ are log concave, and hence the function $x\mapsto f(x)1(\|x+z\|\le t)$ is log concave. So, by the Prékopa–Leindler theorem, $g$ is a log-concave function. Also, the function $g$ is even. So, $g(z)\le g(0)=P(\|X\|\le t)$ for all $z\in\R^n$, and hence (1) follows from (2).