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Let $\mathbf{Z}\sim N(\boldsymbol{\mu},\mathrm{\Sigma})$, where \begin{equation}\label{Eq.Mean} \boldsymbol{\mu}^{\rm T} = \delta[-\sqrt{\frac{x_1x_2}{x_1+x_2}},\frac{-1}2\sqrt{\frac{x_1x_3}{x_1+x_3}},\frac12\sqrt{\frac{x_2x_3}{x_2+x_3}}] \end{equation} and

\begin{equation} \mathrm{\Sigma} = \begin{bmatrix}1 & \frac{x_2x_3}{(1-x_2)(1-x_3)} & \frac{-x_1x_3}{(1-x_1)(1-x_3)}\\ \frac{x_2x_3}{(1-x_2)(1-x_3)} & 1 & \frac{x_1x_2}{(1-x_1)(1-x_2)}\\ \frac{-x_1x_3}{(1-x_1)(1-x_3)} & \frac{x_1x_2}{(1-x_1)(1-x_2)}& 1 \end{bmatrix}, \end{equation} $0\leq x_i\leq 1$, $x_1+x_2+x_3 = 1$ and $\delta>0$. Further suppose for some $a>1$ \begin{equation} \psi_x = 1 - \int_{-a}^{a}\int_{-a}^{a}\int_{-a}^{a}\phi(\boldsymbol{\mu};\mathrm{\Sigma})d\boldsymbol{z}. \end{equation}

When $x_1 = x_2 = x_3 = 1/3$, then \begin{equation} \boldsymbol{\mu}^{\rm T} = \frac{\delta}{2\sqrt{6}}[-2,-1,1] \end{equation} and

\begin{equation} \mathrm{\Sigma} = \begin{bmatrix}1 & \frac{1}{2} & \frac{-1}{2}\\ \frac{1}{2} & 1 & \frac{1}{2}\\ \frac{-1}{2} & \frac{1}{2} & 1 \end{bmatrix}. \end{equation} Now let $y_1=\min(x_1,x_2,x_3)$, $y_2=\text{min}\{(x_1,x_2,x_3)\setminus y_1\}$ and $y_3=\max(x_1,x_2,x_3)$. Let $\psi_y$ be the result of replacing $\boldsymbol{x} = (x_1,x_2,x_3)$ by $\boldsymbol{y} = (y_1, y_2, y_3)$ in the general expressions of $\boldsymbol{\mu}$, $\mathrm{\Sigma}$ and $\psi$.

I am able to show numerically that $\psi_{\boldsymbol{x}=(1/3,1/3,1/3)}>\psi_y$. Can it be proved analytically?

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