A problem related to stochastic ordering

Question

Let $\boldsymbol{X} = (X_1,X_2)^{\rm T}\sim \mathcal{N}_2(\boldsymbol{\mu}, \mathrm{\Sigma})$, where \begin{eqnarray*} \boldsymbol{\mu} = (\mu_1, \mu_2)^{\rm T}& = &(\sqrt{\xi_1\xi_2/(\xi_1+\xi_2)}, 0)^{\rm T}\\ \mathrm{\Sigma} & = &\begin{pmatrix} 1 & -\rho\\ -\rho & 1\end{pmatrix}\\ \rho & = & \sqrt{\xi_1\xi_3/(\xi_1+\xi_2)(\xi_2+\xi_3)}. \end{eqnarray*} It is given that $\xi_1\leq\xi_2\leq\xi_3$, $\xi_i\geq 0$ and $\sum_{i=1}^3\xi_i = 1$. I have a function \begin{equation*} \pi(\boldsymbol{\mu};\boldsymbol{\xi}) = 1-\mathbb{P}(\boldsymbol{X}\leq \boldsymbol{t}) = 1-\mathbb{P}(X_1\leq t \cap X_2\leq t), \end{equation*} where $t>0$. Under the constraints $\xi_1\leq\xi_2\leq\xi_3$, $\xi_i\geq 0$ and $\sum_{i=1}^3\xi_i = 1$, numerically we are getting the maximum of $\pi(\boldsymbol{\mu};\boldsymbol{\xi})$ where the non-zero mean $\mu_1$ is maximized. This occurs when $\xi_1 = \xi_2 = \xi_3 = 1/3$. Can we use some kind of stochastic ordering arguments to prove the result theoretically?

Answer 1

$\newcommand{\der}{\mathrm{der}}\newcommand{\tdert}{\mathrm{dert}}\newcommand{\erf}{\operatorname{erf}}\newcommand{\eqs}{\overset{\text{sign}}=}\newcommand{\tder}{\widetilde\der}$The answer is no.

Indeed, let $x:=\xi_1$ and $y:=\xi_2$, so that $\xi_3=1-x-y$, $0\le x\le y\le1-x-y$, whence $x\in[0,1/3]$.

Consider further the case $y=x\in(0,1/3]$, so that $$\mu_1=M(x):=\sqrt{x/2},$$ $$\rho=-R(x),\quad R(x):=\sqrt{\frac{1-2x}{2(1-x)}}.$$ Let also $$p_t(x):=1-P(X_1\le t,X_2\le t).$$ Your desired result would imply that \begin{equation*} p_t(x)\le p_t(1/3) \tag{0} \end{equation*} for all $x\in(0,1/3]$ and all real $t>0$.

Note that \begin{equation*} p_t(x)=1-P(X_1\le t,X_2\le t)=P(M(x),-R(x)), \tag{1} \end{equation*} where \begin{equation*} P(m,r):=1-\int_{-\infty}^t du \int_{-\infty}^t dv\, f_{m,r}(u,v) \end{equation*} and $f_{m,r}$ is the density function of the bivariate normal distribution with means $m,0$, variances $1,1$, and correlation $r$.

The key is Plackett's observation (formula (3)) that \begin{equation*} D_r f_{m,r}(u,v)=D_v D_u f_{m,r}(u,v), \end{equation*} where $D_w$ denote the partial derivative with respect to a variable $w$. It follows that \begin{equation*} \begin{aligned} D_r P(m,r)&:=-\int_{-\infty}^t du \int_{-\infty}^t dv \, D_r f_{m,r}(u,v) \\ &=-\int_{-\infty}^t du \int_{-\infty}^t dv \, D_v D_u f_{m,r}(u,v) \\ &=-f_{m,r}(t,t). \end{aligned} \tag{2} \end{equation*} Next, \begin{equation*} \begin{aligned} P(m,r)&=1-\int_{-\infty}^t du \int_{-\infty}^t dv\, f_{0,r}(u-m,v) \\ &=1-\int_{-\infty}^{t-m} dw \int_{-\infty}^t dv\, f_{0,r}(w,v) \end{aligned} \end{equation*} and hence \begin{equation*} \begin{aligned} D_m P(m,r)=\int_{-\infty}^t dv\, f_{0,r}(t-m,v)=\int_{-\infty}^t dv\, f_{m,r}(t,v); \end{aligned} \tag{3} \end{equation*} the latter integral can be easily expressed in terms of the error function $\erf$ and elementary functions.

By (1) and a chain rule of differentiation, \begin{equation*} D_x p_t(x)=D_m P(M(x),-R(x))M'(x)-D_r P(M(x),-R(x))R'(x). \end{equation*}

In particular, \begin{equation} D_x p_{1/10}(x)\big|_{x=1/3}=\erf\left(\frac{1}{60} \left(3 \sqrt{6}-10\right)\right)+1-\frac{3 }{\sqrt{\pi }}\,e^{(30 \sqrt{6}-77)/1800} =-0.739\ldots<0. \end{equation} So, inequality (0) fails to hold for $t=1/10$ and all $x$ in a left neighborhood of $1/3$. $\quad\Box$