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one more fix in formula...
Jukka Kohonen
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Order of orthant probabilities in a prolate multinormal distribution

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).

Jukka Kohonen
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