I have a $k$-dimensional multivariate normal distribution $X∼N(0,\Sigma)$ with covariance matrix $\Sigma$. $\Sigma$ has two distinct eigenvalues, say $\lambda_1 > \lambda_2$, with orthogonal eigenspaces $V_1$ and $V_2$. I am interested in the orthant probabilities; given an orthant defined by $\epsilon = (\epsilon_1,\ldots,\epsilon_k)$, with each $\epsilon_i \in \{1,-1\}$, there is an orthant probability $p_\epsilon=Pr(\forall i: \;\epsilon_i X_i \geq 0)$.

There seems to be a literature on finding closed forms for these in special cases, but I do not necessarily need a closed form, but rather I just want to know when $p_{\epsilon} \geq p_{\epsilon'}$. One can write $\epsilon$ as a sum $\epsilon = v_1 + v_2$ with $v_1 \in V_1$ and $v_2 \in V_2$, and $k = ||\epsilon||^2 = ||v_1||^2+||v_2||^2$. My conjecture is that the greater $||v_1||$ is, the greater $p_\epsilon$ is (recall that $\lambda_1 > \lambda_2$). Geometrically, this means that the diagonal vector of the octant defined by $\epsilon$ is closer to the longer axes of the ellipsoids which are the equidensity contour of the distribution.

Is this true? And if so, how does one prove it? I have been stumped (though I know little about probability).

(In the particular case I am interested in, I have a complete graph $K_\ell$, and I am generating a random map from edges of $K_\ell$ to $\mathbb{R}$. The covariance matrix has, as its two eigenspaces, the cut space and the edge space of $K_\ell$.)

I have a $k$-dimensional multivariate normal distribution $X∼N(0,\Sigma)$ with covariance matrix $\Sigma$. $\Sigma$ has two distinct eigenvalues, say $\lambda_1 > \lambda_2$, with orthogonal eigenspaces $V_1$ and $V_2$. I am interested in the orthant probabilities; given an orthant defined by $\epsilon = (\epsilon_1,\ldots,\epsilon_k)$, with each $\epsilon_i \in \{1,-1\}$, there is an orthant probability $p_\epsilon=Pr(\forall i: \;\epsilon_i X_i \geq 0)$.

There seems to be a literature on finding closed forms for these in special cases, but I do not necessarily need a closed form, but rather I just want to know when $p_{\epsilon} \geq p_{\epsilon'}$. One can write $\epsilon$ as a sum $\epsilon = v_1 + v_2$ with $v_1 \in V_1$ and $v_2 \in V_2$, and $k = \lVert\epsilon\rVert^2 = \lVert v_1 \rVert^2+\lVert v_2\rVert^2$. My conjecture is that the greater $\lVert v_1\rVert$ is, the greater $p_\epsilon$ is (recall that $\lambda_1 > \lambda_2$). Geometrically, this means that the diagonal vector of the octant defined by $\epsilon$ is closer to the longer axes of the ellipsoids which are the equidensity contour of the distribution.

Is this true? And if so, how does one prove it? I have been stumped (though I know little about probability).

(In the particular case I am interested in, I have a complete graph $K_\ell$, and I am generating a random map from edges of $K_\ell$ to $\mathbb{R}$. The covariance matrix has, as its two eigenspaces, the cut space and the edge space of $K_\ell$.)