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Let $\vec u\in\mathbb{R}^d$ be a random unit vector, with uniform distribution on the surface of the unit sphere. For a fixed unit vector $\vec v$, what is the following probability?

$$ \Pr\left[ \left| \langle \vec u, \vec v \rangle \right| \geq \xi \right] $$

My guess is that it would be something like $1-\xi$, since (presumably, I'm not sure about that either) $\langle \vec u, \vec v \rangle$ also has a uniform distribution for fixed unit vector $\vec v$ and uniformly-distributed-on-surface-of-unit-sphere $\vec u$, but I'm not aware of distributions on the surface of spheres at all, so if someone could enlighten me, please do.

Jay
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1 Answers1

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For unit vectors $u$ and $v$, $u\cdot v$ gives the cosine of the angle between them. Hence $\{u\in\mathbb S^2:|u\cdot v|\ge\xi\}$ comprises two spherical caps, one centred on $v$ and the other on $-v$, each with polar angle $\cos^{-1}\xi$. The area of these two caps combined is $2\cdot2\pi(1-\cos\cos^{-1}\xi)=4\pi(1-\xi)$ (assuming $0\le\xi\le1$), so the probability is simply $1-\xi$.

Parcly Taxel
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  • How are these lemmas that you used generalized to the $d$-dimensional case? In the straightforward way? – Jay May 24 '21 at 22:15
  • Based on https://math.stackexchange.com/questions/1246748/distribution-of-an-angle-between-a-random-and-fixed-unit-length-n-vectors, it seems that what you have written might only hold for 2 dimensions, and in fact in $d$ dimensions it might not be the result that I would expect. What's going on actually? Is what you have written true for $d$ dimensions? – Jay May 24 '21 at 23:05