# Probability that the convex hull of random points contains sphere's center

What is the probability that the convex hull of n+2 random points on n-dimensional sphere contains sphere's center?

-
On the surface of the sphere? (And a 1-dimensional sphere = circle right?) – kennytm Aug 2 '10 at 14:01
@KennyTM: The circle is the two dimensional sphere, I think. Otherwise the answer would be 0. – Jens Aug 2 '10 at 14:09
@Jens fixed, thanks (for 1-dimensional sphere should be 3 points, of course) – Grigory M Aug 2 '10 at 14:17
@KennyTM Yes (to both questions) – Grigory M Aug 2 '10 at 14:18
@Noldorin: What makes you think this question is not suitable here? (Unrelated comment: this result (Wendel's theorem) is also proved (similar to the answers below) in The art of mathematics: coffee time in Memphis by Béla Bollobás, p. 315-317 (currently inaccessible).) – ShreevatsaR Aug 2 '10 at 19:02

This is one of those old chestnuts that come up again and again. To be precise, the probability that the convex hull of $n+2$ points in $S^n$ (the unit sphere in $\mathbb{R}^{n+1}$) contains the origin is $2^{-n-1}$.

There's a brief argument at Wolfram's mathworld which I don't find entirely convincing but which certainly can be patched to form a convincing argument. In brief, show that for random points $P_1,\ldots,P_{n+2}$ on the sphere, then with probability one, exactly one choice of signs will put the centre in the convex hull of $\pm P_1,\pm P_2,\ldots,\pm P_{n+1}$ and $P_{n+2}$.

Added (3/8/2010) Thanks to Grigory for his comment. Changing the notation slightly, one can show that under some fairy weak hypotheses, if we choose $m+1$ points randomly and indepedently in $\mathbb{R}^m$ the probability their convex hull contains the origin is $2^{-m}$.

Take a probability distribution on $\mathbb{R}^m$ and choose a sequence of points (which we identify with vectors) independently from that distribution. Our first condition on this distribution is that $m$ vectors $v_1,\ldots,v_m$ chosen independently from it are linearly independent with probability one. This can fail if say some point occurs with nonzero probability or the distribution lies in a hyperplane through the origin. Assume this condition.

Now a sequence $v_0,v_1,\ldots,v_m$ of random points chosen according to our distribution are linearly independent: there are reals $a_i$ not all zero with $\sum_i a_i v_i=0$. By our condition, with probability one, the sequence $(a_0,\ldots,a_m)$ is unique up to constant multiple, and moreover all the $a_i$ are nonzero. So we may assume $a_0=1$ and $a_1,\ldots,a_m$ are nonzero and uniquely determined. Then the convex hull of the $v_i$ contains the origin if an only if all the $a_i$ are positive.

Now we introduce another condition: that the distribution is centrally symmetric; in detail the probability that a random vector $v$ lies in a set $A$ equals the probability that $-v$ lies in $A$. A condition like this is clearly necessary; it stops the distribution being supported on a small region far from the origin. This condition shows that all the $2^m$ possibilities of signs for $a_1,\ldots,a_m$ are equiprobable; since changing the sign of some $v_i$ changes the sign of $a_i$.

To conclude, if our probability distribution on $\mathbb{R}^m$ satisfies these two condition, the probability that the convex hull of $m+1$ indepdendently chosen points contains the origin is $2^{-m}$. These conditions are satisfied by the uniform distribution on a sphere with centre at the origin, but also by many others.

-
To make it more convincing one may add some words about barycentric coordinates: a point O lies inside a convex hull of $P_1,...,P_{n+2}$ iff all of its barycentric coordinates (w.r.t. these points) are positive (and switching sign of $P_i$ is almost the same thing as switching sign of i-th barycentric coordinate of O). – Grigory M Aug 2 '10 at 16:10
thanks for detailed answer – Grigory M Aug 3 '10 at 17:19

This problem is discussed in J. G. Wendel; A Problem in Geometric Probability, Mathematica Scandinavica 11 (1962) 109-111. Wendel showed, that the probability of $N$ random points lying on the surface of the unit sphere in dimension $n$ all lie on one hemisphere is

$2^{-N+1}\sum_{k=0}^{n-1} {{N-1}\choose k}$

I've found this here.

-
interesting generalization, thanks – Grigory M Aug 2 '10 at 17:23
@Jens: I am reading J. G. Wendel, A Problem in Geometric Probability mscand.dk/article.php?id=1647. I understand all the other parts, including the part leading up to the equation, except the equation $2^N p_{n,N} = \sum_s \text{Pr}{A_s}$. Could any of you explicate this step? Thank you. – Hans Nov 17 '13 at 20:07