MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Imagine I have an $(p_1, ..., p_N) \in P$ points, on a two-dimensional plane, patterned in a rectangular or hexagonal lattice arrangement in a circle of radius $R_c$, with a spacing between the points of $r_s$.

Let $C_P$ be the centroid of the $P$ points. If I randomly select some subset of $k$ points from $P$, and I compute the centroid of these $k$ points, $C_k$, what is the probability that the distance between $C_p$ and $C_k$ is $\leq D$?

Update: I have specified that the $P$ points in the circle are in a rectangular or hexagonal lattice arrangement (whichever is easiest to analyze).

share|cite|improve this question
The optimal circle packings in a disk are complicated and not known in general, and I doubt you really care about that level of complexity. The exact probability for small $D$ will depend on the exact optimal configuration. If you instead let the $P$ points be IID uniformly random, then you can say something. You can also compare the centroid of $k$ points with the center of the disk. One tool for this is the Central Limit Theorem, although you may want to use an effective version such as the Berry-Esseen Theorem or a large deviations result to get a bound for a particular $k.$ – Douglas Zare Aug 27 '12 at 13:28
@Douglas Zare I have updated the question to specify that the $P$ points should be patterned in a rectangular or hexagonal lattice arrangement (whichever is easiest to analyze). Also, I do care about circle packing, it's really interesting, but it was inappropriate for me to confound this question with a very difficult geometry (et. al.?) problem. – CKura Aug 27 '12 at 14:44
up vote 1 down vote accepted

I might not be understanding the question, but the centroid is just the mean of the sample, so for mildly large samples from a mildly large set it will be (bivariate) normally distributed, and all the statistics are easy to compute.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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