How does $\arccos$ appear in the expectation of a normal random variable?

Question

I found the following equation in a paper:

$\mathbb{E}_{w \sim N(\mathbf{0},\mathbf{I})} [\mathbb{I}\{w^\intercal x_i \geq 0, w^\intercal x_j \geq 0\}] = \dfrac{\pi - \arccos(x_i^\intercal x_j)}{2\pi}$

Here, $w \in \mathbb{R}^d$ is a d-dim random variable sampled from a normal distribution with $\mu = \mathbf{0}$ and $\Sigma = \mathbf{I}$. $x_i, x_j$ are fixed but unknown vectors in $\mathbb{R}^d$ with unit norm. The function $\mathbb{I}(\cdot)$ is an indicator function denoting if the event happens or not ($\mathbb{I}\{w^\intercal x_i \geq 0, w^\intercal x_j \geq 0\} = 1$ iff $w^\intercal x_i \geq 0 \text{ and } w^\intercal x_j \geq 0$).

Since the paper doesn't have any explanations, I wonder how this equation is derived and how $\arccos(\cdot)$ appears in the calculated expectation.

Any hints or resources are appreciated!

Answer 1

Many thanks to the hint given by Ted Shifrin! It turns that $w^\intercal x_i \geq 0, w^\intercal x_j \geq 0$ only holds when the sampled vector $w$ is in between $-x_j$ and $x_i$, whose angle in between is $\pi - \theta$ (assuming $\theta$ is the angle between $x_i$ and $x_j$).