Stochastic domination of Gaussian random vectors 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?
 A: $\renewcommand{\P}{\operatorname{\mathsf P}}\newcommand{\E}{\operatorname{\mathsf E}}$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}

