Stochastic ordering of absolute multivariate normal random variables

Question

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 $|Y|$?

Answer 1

$\newcommand\si\sigma$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 $Z$, $$EZ^2=E\int_0^{|Z|}2t\,dt =E\int_0^\infty 2t\,dt\,1(|Z|>t) \\ =\int_0^\infty 2t\,dt\,E1(|Z|>t) =\int_0^\infty 2t\,dt\,P(|Z|>t).$$