12
$\begingroup$

This is a companion contrast to the earlier analogous question for unit $n$-cubes, where the answer (provided by several respondents) is $\infty$ .

What is the limit, as $n \to \infty$, of the expected distance between two points chosen uniformly at random within a unit-radius hypersphere in $\mathbb{R}^n$, i.e., in the unit-radius $n$-ball?

Dividing OEIS A093530 by A093531 I see that it appears to be approaching approximately $1.37$ for odd $n$, but I wonder if the limit is actually known, either exactly or to significant precision? I cannot quite extract an answer from the MathWorld article...

This certainly provides a dramatic contrast between the $n$-cube and the $n$-sphere!

$\endgroup$
2
  • 6
    $\begingroup$ I'm not sure there's such a contrast between cubes and spheres. In high dimensions a unit cube is vastly larger than a unit sphere, as measured by diameter or volume, so it's not as fair a comparison as the word "unit" suggests. $\endgroup$
    – Henry Cohn
    Apr 15, 2014 at 0:44
  • 2
    $\begingroup$ If instead we look at the ratio of expected distance to diameter, then for spheres the limit is asymptotically $\frac{1}{\sqrt{2}}$, while for cubes it is (from Nate Eldredge's answer to the linked question) $\frac{1}{\sqrt{6}}$. Is it known whether Spheres maximize this ratio, whether in finite dimensions or asymptotically? $\endgroup$ Apr 15, 2014 at 21:15

1 Answer 1

15
$\begingroup$

Both points will be very close to (let's pretend: on) the surface with prob almost 1. Call the first point the north pole. By concentration of measure for the sphere, a randomly chosen second point is almost guaranteed to be almost on the equator, so the limit should be $\sqrt{2}$.

$\endgroup$
7
  • $\begingroup$ May I ask: Why can you assume that the first point is the north pole? Concentratoin of measure again? $\endgroup$ Apr 15, 2014 at 0:17
  • 1
    $\begingroup$ I condition on the first point, and the sphere looks the same from any point. $\endgroup$ Apr 15, 2014 at 0:18
  • $\begingroup$ Are you reading the question as "two points on the surface of the unit sphere", or ought this be overwhelmingly likely? $\endgroup$
    – usul
    Apr 15, 2014 at 0:57
  • $\begingroup$ Yes, "sphere" (as opposed to "ball"), but even if it's the ball, almost all the volume is near the surface, so it won't matter. $\endgroup$ Apr 15, 2014 at 1:02
  • 1
    $\begingroup$ I should have known, after your first comment (sorry for the fake discussion)... Solution stays the same, though, by also applying the trivial concentration of measure near the surface first. $\endgroup$ Apr 15, 2014 at 1:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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