Order of orthant probabilities in a prolate multinormal distribution

Question

This is inspired by the negative answer to the conjecture in Which orthant probabilities are the largest? (For a multivariate normal distribution).

Suppose $X$ has the $k$-dimensional multivariate normal distribution $N(0, \Sigma)$, where $k \ge 2$ and $\Sigma$ has two distinct eigenvalues: the larger is $\lambda > 1$ and single, and the smaller is $1$ and $(k-1)$-fold. A practical interpretation is that we take the $k$-dimensional standard normal distribution, stretch it $\sqrt{\lambda}$-fold in one direction, and rotate arbitrarily. So the distribution is "prolate in one direction". Let $u$ be the eigenvector associated to the larger eigenvalue.

Number the $2^k$ orthants $i=1,\ldots,2^k$ in some convenient order, and let $e_i = (e_{i1},\ldots,e_{ik}) = (\pm 1, \ldots, \pm 1)/\sqrt{k}$ be the unit vector pointing to the "center" of the $i$th orthant. Let $p_i = \mathbb{P} (\forall j=1,\ldots,k: \; X_j e_{ij} > 0)$ be the probability that $X$ is in the $i$th orthant.

Question. Are the orthant probabilities $p_i$ in the same numerical order as the squared dot products $(e_i \cdot u)^2$?

Intuition. The dot products measure how elongated the distribution is towards that orthant.

Empirical support. I have created $>10\;000$ random instances, with dimensions uniformly random between $3$ and $7$, the stretching factor uniformly random between $1.01$ to $10$, and random rotation. To guard against numerical inaccuracy, I searched for cases where some two orthant probabilities would be in the wrong order and separated by more than $0.003$. No such cases were found.

Note. The case $k=2$ is easy, since we have closed-form expressions for the quadrant probability. Whenever the correlation between $X_1$ and $X_2$ is positive, the positive-positive quadrant has $> 1/4$ probability.

Note. The case $k=3$ might be easy using some known closed-form expressions, and that would be already interesting (but a positive answer here would not solve the general case).

Note. In Which orthant probabilities are the largest? (For a multivariate normal distribution) the distribution was assumed to have two distinct eigenvalues, but the larger eigenvalue could be multiple. A corresponding conjecture turned out to be false already in dimension $4$ when each eigenvalue was double (the distribution was "stretched uniformly in two directions" before rotation).

Edit. The first version of the question asked about dot products instead of their squares. That version would not make much sense (the answer would be negative already in $k=2$).

Answer 1

$\newcommand\la\lambda\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand\R{\mathbb R}\newcommand{\Si}{\Sigma}$The answer is negative, even for $k=3$.

A counterexample is as follows: $$\lambda=2,\quad u=\frac{(4,-3,-3)}{\sqrt{34}},$$ $$e_1:=(1,1,1)/\sqrt3,\quad e_2:=(-1,1,1)/\sqrt3.$$ Then $e_1\cdot u>e_2\cdot u$, whereas $p_1=0.100\ldots<0.183\ldots=p_2$.

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The OP has changed the question, replacing the comparisons of the values of $e_i\cdot u$ by the comparisons of the values of $(e_i\cdot u)^2$.

Then the answer becomes positive, at least for $k=3$.

Indeed, if $X_1,X_2,X_3$ are jointly Gaussian zero-mean random variables with correlation coefficients $\rho_{ij}=Corr(X_i,X_j)$, then, by the 3rd display from the bottom of p. 355 of [Plackett's paper], \begin{equation*} P(X_1>0,X_2>0,X_3>0) \\ =\frac1{4\pi}\,(2\pi-\cos^{-1}\rho_{12}-\cos^{-1}\rho_{23}-\cos^{-1}\rho_{31}). \end{equation*}

If $a=(a_1,a_2,a_3)$ is an eigenvector of the covariance matrix $\Si$ of the random vector $X=(X_1,X_2,X_3)$ corresponding to the eigenvalue $t:=\la>1$, with the other two eigenvalues each equal $1$, then \begin{equation*} \Si=\frac1{a_1^2+a_2^2+a_3^2} \left( \begin{array}{ccc} a_1^2 t+a_2^2+a_3^2 & a_1 a_2 (t-1) & a_1 a_3 (t-1) \\ a_1 a_2 (t-1) & a_2^2 t+a_1^2+a_3^2 & a_2 a_3 (t-1) \\ a_1 a_3 (t-1) & a_2 a_3 (t-1) & a_3^2 t+a_1^2+a_2^2 \\ \end{array} \right). \end{equation*}

It follows that for any $\ep=(\ep_1,\ep_2,\ep_3)\in\{-1,1\}^3$ \begin{equation*} \begin{aligned} p_{a,\ep}:=4\pi P(\ep_1 X_1>0,\ep_2 X_2>0,\ep_3 X_3>0)-2\pi \\ =-\cos ^{-1}R_{12}-\cos ^{-1}R_{23}-\cos ^{-1}R_{31}, \end{aligned} \end{equation*} where \begin{equation*} R_{ij}:=\frac{(t-1)a_i a_j \ep_i \ep_j } {\sqrt{\big((t-1)a_i^2+a_1^2+a_2^2+a_3^2\big) \big((t-1)a_j^2+a_1^2+a_2^2+a_3^2\big) }}. \end{equation*}

We want to show that for any $a=(a_1,a_2,a_3)\in\R^2\setminus\{0\}$ and any $\de=(\de_1,\de_2,\de_3)$ and $\ep=(\ep_1,\ep_2,\ep_3)$ in $\{-1,1\}^3$ such that \begin{equation*} (a\cdot\de)^2\le(a\cdot\ep)^2 \tag{1}\label{1} \end{equation*} we have \begin{equation*} p_{a,\de}\le p_{a,\ep}. \tag{2}\label{2} \end{equation*}

Note that $p_{a,\ep}$ is even in $a$ and in $\ep$ ($p_{-a,\ep}=p_{a,\ep}=p_{a,-\ep}$), and $(a\cdot\ep)^2$ is even in $a$ and in $\ep$ as well. Also, $p_{a,\ep}$ is permutation invariant: $p_{a,\ep}$ remains invariant if the indices of $a$ and $\ep$ are permuted by the same permutation. So, without loss of generality (wlog) \begin{equation*} \de=(-1,1,1)\quad\text{and}\quad\ep=(1,1,1), \end{equation*} and then conditions \eqref{1} and \eqref{2} can be rewritten as \begin{equation*} a_1(a_2+a_3)\ge0\tag{1a}\label{1a} \end{equation*} and \begin{equation*} \sin^{-1}S_{12}+\sin^{-1}S_{13}\ge0, \tag{2a}\label{2a} \end{equation*} where \begin{equation*} S_{ij}:=S_{ij}(a):=\frac{(t-1)a_i a_j } {\sqrt{\big((t-1)a_i^2+a_1^2+a_2^2+a_3^2\big) \big((t-1)a_j^2+a_1^2+a_2^2+a_3^2\big) }}. \end{equation*} Also, $a_1(a_2+a_3)$ and the $R_{ij}$'s are even in $a$, and the $S_{ij}$'s are homogeneous in $a$: $S_{ij}(ka)=S_{ij}(a)$ for any real $k\ne0$. Also, by continuity, wlog the inequality in \eqref{1a} is strict. So, wlog $a_1=1$ and $a_2+a_3>0$.

So, wlog $a_2=a$ and $a_3=b$ for some real $a$ and $b$ such that $a>0$ and $a+b>0$.

If $b\ge0$, then $S_{12}\ge0$ and $S_{13}\ge0$, so that \eqref{2a} is obvious.

If $b<0$, then $-a<b<0$ and hence $b^2<a^2$. Then ($S_{12}>0$ and) \begin{equation*} r(t):=\frac{S_{13}^2}{S_{12}^2}=\frac{b^2 \left(a^2 t+b^2+1\right)}{a^2 \left(a^2+b^2 t+1\right)}, \end{equation*} which is increasing from $r(1+)=\frac{b^2}{a^2}<1$ to $r(\infty-)=1$ as $t$ is increasing from $1$ to $\infty$, so that $|S_{13}|<S_{12}$ and \eqref{2a} follows. $\quad\Box$