Stochastic domination of Gaussian random vectors

Question

Let $S$ be the class of all $2$ by $2$ matrices of the form $$\begin{bmatrix} 1 & a \\ a & 1 \end{bmatrix},\, |a|\leq 1.$$ Is there a single matrix $M\in S$ such that for any $N\in S$ and all $x>0$ we have $$\mathbb{P}(||X||_2 \geq x)\geq \mathbb{P}(||Y||_2 \geq x),$$ where $X$ and $Y$ have the Gaussian distribution with covariance matrices $M$ and $N$, respectively? In other words, is it possible to determine which covariance structure gives the fattest tail behavior?

Answer 1

$\renewcommand{\P}{\operatorname{\mathsf P}}\newcommand{\E}{\operatorname{\mathsf E}}$The answer is no. More specifically, let $Y_a$ be a centered Gaussian random vector with covariance matrix $\begin{bmatrix} 1 & a \\ a & 1 \end{bmatrix}$. Then there is some $u_1\in(0,\infty)$ such that \begin{equation} \max_{|a|\le1}\P(\|Y_a\|_2 \ge u) =\begin{cases} \P(\|Y_0\|_2 \ge u) &\text{ if } u\le u_1,\\ \P(\|Y_1\|_2 \ge u) &\text{ if } u\ge u_1. \end{cases} \end{equation}

This is a special case of Theorem 1, which states the following:

For each natural $d$, the equation $\P(\chi_d^2/d>x_d)=\P(\chi_{d+1}^2/(d+1)>x_d)$ in $x\in(0,\infty)$ has a unique root $x_d$, and we have $x_0:=\infty>x_1>x_2>\cdots$; here, $\chi_d^2$ is a random variable (r.v.) with the chi-squared distribution with $d$ degrees of freedom.
For any natural $n$, let $\mathcal Q_n$ be the set of all positive semidefinite quadratic forms $Q_n$ in centered Gaussian r.v.'s such that $\E Q_n=1$. Then \begin{equation} \min_{Q_n\in\mathcal Q_n}\P(Q_n\le x)= \begin{cases} \P(\chi_n^2/n\le x)&\text{ if }x\in[0,x_{n-1}), \\ \P(\chi_d^2/d\le x)&\text{ if }d\in\{n-1,\dots,1\}\ \& \ x\in[x_d,x_{d-1}). \end{cases} \end{equation}