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 $|Y|$?

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

1 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 $t$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.