# Stochastic ordering of absolute multivariate normal random variables

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 $$|Y|$$?

• How is $|X|$ defined? Commented Sep 20, 2021 at 16:55
• @losif |.| represents the absolute element wise. Commented Sep 20, 2021 at 17:00
• @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? Commented Sep 23, 2021 at 2:48

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

• Thank for your reply. Is there no role of means? Commented Sep 20, 2021 at 17:57
• @SatyaPrakash : The means enter inequality (1). Commented Sep 20, 2021 at 18:00
• Expectation of X^2 to integral is not clear. Could you please explain? Commented Sep 21, 2021 at 17:23
• @Pinelis Thank you very much for the detailed explanation. Commented Sep 22, 2021 at 18:06
• @SatyaPrakash : All right. So, are you now satisfied with the answer? Commented Sep 22, 2021 at 18:28