TITLE: Coverage of balls on random points in Euclidean space
QUESTION [6 upvotes]: We have n points randomly distributed in a d-dimensional unit hypercube. We randomly sample k of those points and center a ball with radius r on each of those k points. Does there exist an estimate of the radius r such that r is the smallest radius expected to cover all the points with those k possibly overlapping balls?

REPLY [3 votes]: In an email, Doug Jungreis gave me the following solution to this
question. His solution illustrates the case where $n=10^7$, $k=500$, and
$\epsilon = 1/100$.  It seems to work well in the datasets I'm looking at
(even where $d$ is small). I'm quoting his email directly instead of trying to reformat the notation and risk a mistake.
Doug makes two independence assumptions described by Greg Kuperberg:

The first independence assumption is that if $p$ is a typical point in
  the cube, then whether one ball covers it has little do with whether
  another ball covers it. Obviously this assumption is bad in when $d$ is
  small since points at the end are harder to reach than points in the
  middle. But easily by the time you get to 50 dimensions, almost
  everything is comparably in the middle.
The second independence assumption is that whether one point is
  covered has little to do with whether another point is covered. Again,
  this assumption is bad when $d$ is small, but it's a good assumption by
  the time you get to $d=50$ with the given parameters.

Here is Doug's solution:

Here is an estimate if $d$ is large.  I'll make the same independence
  assumptions as Greg.  If you choose 2 values $x,y$ in the unit interval,
  then their expected distance squared is $\int_0^1 \int_0^1 (x-y)^2 dx\, dy
= 1/6$, and their expected distance^4 is $\int_0^1 \int_0^1 (x-y)^4 dx\, dy
= 1/15$ (if I didn't make a mistake).  So the distance squared has mean
  1/6 and variance $1/15 - (1/6)^2 = 7/180$.  Then the distance squared
  between two points in the $d$-hypercube has mean $d/6$ and variance
  $7d/180$, and assuming $d$ is reasonably large, it is very close to
  normally distributed.  You want a 1/100 chance that any of the 10
  million points misses all 500 spheres, so a $1/10^9$ chance that a
  particular point misses all 500 spheres, so a $(1/10^9)^{1/500} =
.9594$ chance that a particular point misses a particular sphere.  So
  you want a .9594 chance that two particular points have distance^2
  greater than $r^2$.  The value .9594 corresponds to 1.744 standard
  deviations on a normal distribution, so $r^2$ should be roughly 1.744
  standard deviations below the mean, i.e., $r^2 = (d/6) - 1.744
\sqrt{7d/180}$.