What is the probability of a random normal point falling inside an intersection of two half-spaces? Given two fixed vectors $x_i, x_j \in \mathbb{R}^n$, what is the probability that a random point $y\sim \mathcal{N}(0, \mathbf{Id})$ in $\mathbb{R}^n$ falls inside the intersection of their upper half-spaces i.e. $y \in \{ z \mid z^T x_i > 0 \mbox{ and } z^Tx_j > 0\}$.
I suspect $\frac{\pi - \angle(x_i,x_j)}{2\pi}$ by pure intuition (from a 2D case and the angle of the dual-cone) but any brute-force solution (using say generalized spherical coordinate) seems messy. Is there an easy way to see why this is true? Any pointer is greatly appreciated. Thanks!
 A: Your problem in fact reduces to the 2D case: The spherical normal distribution is invariant under rotations, so we may assume without loss of generality that $x_1 = (1,0,0,\ldots,0)$ and $x_2 = (a_1,a_2,0,\ldots,0)$. Now, your above reasoning applies.
To elaborate a bit further:
First, we may assume that $\|x_1\|_2 = \|x_2\|_2 = 1$ because $z^\top x > 0$ if and only if $z^\top \tilde x > 0$, where $\tilde x = x\,/\,\|x\|$. We also assume $x_1$ and $x_2$ are linearly independent; otherwise, the answer is easy to deduce.
Now, take $U$ to be an orthogonal matrix such that $U x_1 = (1,0,0,\ldots,0)$ and $U x_2 = (a_1, a_2, 0, \ldots, 0)$. (To construct such a $U$, we can take its first row to be $x_1$ and its second row to be the normalization of $x_2 - \langle x_1, x_2\rangle\cdot x_2$; the remaining rows can be chosen arbitrarily to complete an orthonormal basis.) Note that $(Uz)^\top (Ux_k) = z^\top x_k$. Furthermore, $Uz\sim\mathcal N(0, I)$. Hence, it suffices to answer: What is the probability that $\hat z^\top \hat x_k > 0$ for both $k\in\{1,2\}$, where $\hat z = Uz\sim\mathcal N(0, I)$ and $\hat x_k = Ux_k$? In particular, only the first two coordinates of each $\hat x_k$ are non-zero. Hence, we may "forget" about the remaining coordinates and focus only on the first two coordinates of $\hat x_1$ and $\hat x_2$.
Because $\hat x_1$ and $\hat x_2$ both are of unit norm, the probability that $\hat z^\top \hat x_k > 0$ for both $k = 1, 2$ is, as you computed above,
$$\frac 12 - \frac{\arccos(\langle \hat x_1, \hat x_2\rangle)}{2\pi} = \frac 12 - \frac{\arccos(\langle x_1, x_2\rangle)}{2\pi}$$
(where $\arccos$ returns a value in $[0, \pi)$). Note that $\langle \hat x_1, \hat x_2\rangle = \langle Ux_1, U x_2\rangle = \langle x_1, x_2\rangle$ because $U$ is orthogonal. Hence we recover an expression for the probability in terms of $x_1$ and $x_2$.
