Let $X\sim\mathcal{N}(\boldsymbol{\mu}\_1,\mathrm{\Sigma}\_1)$ and $Y\sim\mathcal{N}(\boldsymbol{\mu}\_2,\mathrm{\Sigma}\_2)$. Then it is know that $\mathbb{P}(X>\boldsymbol{t})\leq\mathbb{P}(Y>\boldsymbol{t})$ implies $\mu\_i\leq \mu^{\prime}\_i$ and $\sigma\_{ii} = \sigma^{\prime}\_{ii}$ (Theorem 10 of Muller 2001, Ann. Inst. Stat. Math. 53(3) 567-575). Is there any similar results for absolute $|X|$ and $|Y|$?



Stochastic ordering of absolute multivariate normal random variables
$\newcommand\si\sigma$The condition $P(|X|>\boldsymbol{t})\le P(|Y|>\boldsymbol{t})$ implies $P(|X\_i|>t)\le P(|Y\_i|>t)$, for each $i$ and all real $t$; this follows by letting $t\_j=0$ for all $j\notin\{i\}$.


Fix any $i$ and, for brevity, let $m:=EY\_i$, $s:=\sqrt{Var\,Y\_i}$, $m\_1:=EX\_i$, and $s\_1:=\sqrt{Var\,X\_i}$, so that $X\_i\sim N(m\_1,s\_1^2)$ and $Y\_i\sim N(m,s^2)$.


Then
$$s\_1^2+m\_1^2=EX\_i^2=\int\_0^\infty 2t\,dt\,P(|X\_i|>t) \\ 
\le\int\_0^\infty 2t\,dt\,P(|Y\_i|>t)=s^2+m^2,\tag{0}$$
so that
$$s\_1^2+m\_1^2
\le s^2+m^2, \tag{1}$$
and, for $t\to\infty$,
$$P(|Y\_i|>t)\ge P(|X\_i|>t) \\ 
=\exp\Big\{-\frac{t^2}{(2+o(1))s\_1^2}\Big\},$$
whence
$$s\_1\le s.\tag{2}$$


Moreover, letting $t\downarrow0$ in $P(|X\_i|>t)\le P(|Y\_i|>t)$, we see that the density at $0$ of the distribution of $X\_i$ is no less that the density at $0$ of the distribution of $Y\_i$, which can rewritten as
$$s\_1^2 e^{m\_1^2/s\_1^2}\le s^2 e^{m^2/s^2}. \tag{3}$$


Thus, the condition $P(|X\_i|>t)\le P(|Y\_i|>t)$ implies (1),
(2), and (3).


One can show that, vice versa, if (2) and (3) hold, then $P(|X\_i|>t)\le P(|Y\_i|>t)$ for all real $t$.




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**Details on the second equality in (0) in response to a comment:** For any random variable $Z$,
$$EZ^2=E\int\_0^{|Z|}2t\,dt
=E\int\_0^\infty 2t\,dt\,1(|Z|>t) \\ 
=\int\_0^\infty 2t\,dt\,E1(|Z|>t)
=\int\_0^\infty 2t\,dt\,P(|Z|>t).$$


