1
$\begingroup$

Let $S^{d-1}=\{(x_1,\cdots,x_{d})\in {\mathbb R}^{d}|{x_1}^2+\cdots+{x_d}^2=1\}$, and let us call the intersection of any $d-1$ dimensional subset which passes through the origin and $S^{d-1}$ 'a great circle'. Also, let us call one of two components gotten by cutting $S^{d-1}$ by a great circle '$d-1$ dimensional semi-spherical surface'. Also, suppose that a point set on $S^{d-1}$ is semi-spherical when it is included in a $d-1$ dimensional semi-spherical surface.

Then, here is my question.

Question : Letting $d\ge 2$, find the probability $f(d,n)$ that $n$ points which are randomly selected from $S^{d-1}$ are semi-spherical.

Remark : Suppose that $n$ points are selected randomly with equable probability.

Example : We can easily get $f(2,n)=n\cdot 2^{1-n}.$

Motivation : I've known the $d=2$ case, but I'm facing difficulty for $d$ in general. Can anyone help?

$\endgroup$

2 Answers 2

2
$\begingroup$

This is answered in painstaking detail here.

$\endgroup$
1
  • $\begingroup$ Wow! Thank you very much for great info. $\endgroup$
    – mathlove
    Oct 23, 2013 at 17:09
0
$\begingroup$

Is this not the probability that if you toss a coin n times then you get either n heads or n tails? Paul

$\endgroup$
2
  • 1
    $\begingroup$ Nope, it is not. $\endgroup$
    – Igor Rivin
    Oct 23, 2013 at 16:59
  • 2
    $\begingroup$ I think the question is looking for the probability that the points are in some hemisphere (possibly at a funny angle), not necessarily one of two fixed hemispheres corresponding to a single great circle. $\endgroup$
    – Henry Cohn
    Oct 23, 2013 at 17:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.