# Union of random half spaces cover a ray

Let $x, y \in \mathbb{R}^{n}$ be two fixed unit vectors with angle $\alpha \in (\frac{\pi}{2}, \frac{3\pi}{4})$. Define the positive half space associated with a vector $z$ to be $\mathcal{H}(z) = \{h : z^\top h \geq 0\}$.

Choose $m$ unit vectors $\{a_i\}_{i=1}^{m}$ uniform over the set $\mathcal{H}(x) \cap \{h : \|h\|_2 = 1\}$, what is the probability that $y \in \bigcup_{i=1}^m \mathcal{H}(a_i)$?

How does this probability depend on $\alpha, m$ and $n$?

You should look instead at the probability $p_m$ that $$y\notin \bigcup_{i=1}^m \mathcal H(a_i) \;,$$ or, in other words, that $\langle y, a_i \rangle < 0$ for all $a_i$. Since $a_i$ are independent, $$p_m = \left( \mathbf P \{ \langle y, a \rangle < 0 \} \right)^m \;,$$ where $a$ is uniformly distributed on the half-sphere determined by $x$. Finally, $\mathbf P \{ \langle y, a \rangle < 0 \}$ is just the ratio of the volume of the intersection of the half-spheres determined by $x$ and $-y$ and the volume of the unit radius half-sphere in $\mathbb R^n$.
• Thank you for your answer. Is the ratio just $\alpha / \pi$? I wonder whether the ratio depends on the dimensionality n? – Cong Ma Apr 15 '17 at 18:32
• Yes, indeed it's just $\alpha/\pi$ (which can be seen by looking at the plane spanned by $x$ and $y$ and its orthogonal complement). – R W Apr 15 '17 at 18:50