$X = (x\_1,...x\_n) \in \mathbb{R}^n, X \sim \mathcal{N}(O, \Sigma\_X)$ 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?



L_infinity norm of two gaussian vector
$\newcommand\R{\mathbb R}$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](https://en.wikipedia.org/wiki/Pr%C3%A9kopa%E2%80%93Leindler_inequality#Applications_in_probability_and_statistics), $g$ 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).


