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) 567575). Is there any similar results for absolute $X$ and $Y$?

$\begingroup$ How is $X$ defined? $\endgroup$– Iosif PinelisCommented Sep 20, 2021 at 16:55

$\begingroup$ @losif . represents the absolute element wise. $\endgroup$– Satya PrakashCommented Sep 20, 2021 at 17:00

$\begingroup$ @SatyaPrakash : Do you mean $\left(a,b,c,\ldots)\right = \max\{ a, b, c, \ldots \}$ or $\left(a,b,c,\ldots)\right = a+b+c+\cdots$ or something else? $\endgroup$– Michael HardyCommented Sep 23, 2021 at 2:48
1 Answer
$\newcommand\si\sigma$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 $t$.
Details on the second equality in (0) in response to a comment: For any random variable $Z$, $$EZ^2=E\int_0^{Z}2t\,dt =E\int_0^\infty 2t\,dt\,1(Z>t) \\ =\int_0^\infty 2t\,dt\,E1(Z>t) =\int_0^\infty 2t\,dt\,P(Z>t).$$

$\begingroup$ Thank for your reply. Is there no role of means? $\endgroup$ Commented Sep 20, 2021 at 17:57

$\begingroup$ @SatyaPrakash : The means enter inequality (1). $\endgroup$ Commented Sep 20, 2021 at 18:00

$\begingroup$ Expectation of X^2 to integral is not clear. Could you please explain? $\endgroup$ Commented Sep 21, 2021 at 17:23

$\begingroup$ @Pinelis Thank you very much for the detailed explanation. $\endgroup$ Commented Sep 22, 2021 at 18:06

$\begingroup$ @SatyaPrakash : All right. So, are you now satisfied with the answer? $\endgroup$ Commented Sep 22, 2021 at 18:28