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I´m interested in the following problem:

Given a set of vectors $G = \{v_0, \ldots, v_n\}$, what is the probability $P$ that for any vector $u$, there is at least one vector $v_k \in G$ such that there is an acute angle between $u$ and $v_k$ (i.e. $u \cdot v_k > 0$)?

Notes: The vector $u$ has $d$ coordinates with $n \ll d$. Each coordinate of vector $u$ is sampled from a uniform distribution on $\mathbb{R}$ independently.

I'd appreciate it if someone could give me a pointer, and I'd be happy to just prove a loose upper bound on the probability (obviously tighter than $P > 0.5$) as well.

smalldog
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O'ara
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  • What is the distribution of the random vector $u$? The answer depends on this choice. – smalldog Apr 26 '21 at 19:34
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    You cannot tell without knowing the probability distribution on $u$. For example, is $u$ independent of $G$? Is $u$ symmetric around $0$? – eeehm Apr 26 '21 at 19:34
  • This probably goes without saying, but it also depends on $G$. Though the fact that the elements of $G$ are unit vectors isn't actually relevant... – Xander Henderson Apr 26 '21 at 19:37
  • hi @acephalous, The vector u is independent of G and is randomly chosen from the unit hypersphere of dimension d. – O'ara Apr 26 '21 at 19:38
  • @XanderHenderson you are right. I changed the question correspondingly. – O'ara Apr 26 '21 at 19:40
  • Are you using the common terminology for angles as always non-negative, so acute angles are those between $0$ and $\pi/2$? Or are you including those between $-\pi/2$ and $0$? – smalldog Apr 26 '21 at 20:58
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    @acephalous updated the question. should answer your question. – O'ara Apr 26 '21 at 21:29
  • This feels a lot like an XY Problem to me. What are you actually trying to do? What is this set $G$, and why do you care if a vector $u$ forms an acute angle with some element of $G$? Without more information, the answer could be anything from $1/2$ to $1$. – Xander Henderson Apr 27 '21 at 00:20

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The probability depends on $G$, and you cannot improve the bounds $0.5 \leq p \leq 1$ unless further assumption is made on the set. Indeed, taking $G$ to be any single vector $\{v_0\}$ implies $p = 0.5$, while taking $G$ to be $\{v_0, -v_0\}$ for any $v_0$ implies $p=1$.

For the two-dimensional case $d=2$, I'm pretty sure (but still need to prove that) the probability can explicitly be given as $$ p = \min\left\{1, \frac{1}{2}+\frac{1}{2\pi}\max_i \{\theta(v_0, v_i)\}-\frac{1}{2\pi}\min_i \{\theta(v_0, v_i)\}\right\}$$ where $\theta(v_0,v_i) \in (-\pi, \pi]$ is the angle between $v_0$ and each other vector $v_i \in G$. I will try to add a proof shortly. The $d$-dimensional case seems less obvious, I need to think about it.

smalldog
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    what kind of assumption on G would help to improve the bound? Any ideas? – O'ara Apr 26 '21 at 21:27
  • I agree with you. I came up with a similar formula for the d=2 case but I am having a hard time proving it or extending it to higher dimensions. – O'ara Apr 30 '21 at 20:14
  • "I will try to add a proof shortly" was curious if you have made some progress? – Snared Jul 18 '23 at 19:25
  • @Snared Completely forgot about this, apologies. Busy with work atm but would it help you significantly if I provided a proof when I find a moment in the next month or so? – smalldog Jul 20 '23 at 10:16
  • @smalldog I mean yeah in the next month would be nice, no pressure from me though! – Snared Jul 20 '23 at 14:56
  • @Snared Not sure if I'll find the time but if I do, will keep you posted! – smalldog Jul 20 '23 at 17:16