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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 dot products $e_i \cdot u$?

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).

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 $k=2$).

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 $p_i = \mathbb{P} (\forall j=1,\ldots,k: \; X_i e_{ij} > 0)$ 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 dot products $e_i \cdot u$?

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).

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 dot products $e_i \cdot u$?

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).

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 $k \ge 3$ 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 $e_i = (\pm 1, \ldots, \pm 1)/\sqrt{k}$ be the unit vector pointing to the "center" of the $i$th orthant. Let $p_i = \mathbb{P} (\forall i: \; X_i e_i > 0)$ 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 dot products $e_i \cdot u$?

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).

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 $p_i = \mathbb{P} (\forall j=1,\ldots,k: \; X_i e_{ij} > 0)$ 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 dot products $e_i \cdot u$?

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).