The question:Let N points be scattered at random on the surface
of the unit sphere in n-sphere.The problem of the title is to
evaluate
hemisphere.
, the probability that all the points lie on some
The first solution:
A Problem In Geometric Probability
J.G.Wendel
Let N points be scattered at random on the surface of the unit sphere
in n-sphere.The problem of the title is to evaluate
, the probability
that all the points lie on some hemisphere. I shall show that
（1）
I first heard of the problem from L.J.Savage, who had been
challenged by R.E. Machol
showed that
to evaluate
. Savage
, and more generally that
(2)
Then I was able to obtain the relation
(3)
and D.A. Darling proved that
became
,which on setting N=n+2
（4）
Equation (3) and (4) suggested the attractive “duality relation”
(5)
which was found to hold generally. The results (2), (3) and (5) then led to
conjecture (1).Since (5) is a corollary to (1) it seems superfluous to give a
separate proof; instead I proceed now to the proof of (1),and in s slightly
more general setting.
Let
be random vectors in
whose joint distribution
is invariant under all reflections through the origin and is such that with
the probability one all subsets of size n are linearly independent; for
example, the
may be uniformly and independently distributed over the
surface of the unit sphere. The probability
probability that all
is now interpreted as the
lie in a half-sphere, i.e. that for some vector y the
inner products
are all positive. I shall show that
the recurrence relation
satisfies
(6)
Since the right member of (1) also satisfies (6),together with the evident
boundary conditions
,
if
, this will
complete the proof of (1).
PROOF OF (6). It is sufficient to evaluate the corresponding
conditional probability when the
are non-zero and lie on fixed lines
through the origin. Suppose that y is perpendicular to none of these lines.
Then the sequence
is a random point in the set
of all ordered N-tuples consisting of plus and minus signs. A
specified s is said to occur if there is a y such that
the event that s occurs, and let
be the indicator of
, where
be
. By definition
. Since any s can be
changed into any other by reflecting appropriate
follows that all
. Let
through the origin it
are equally likely. Hence
(7)
say, with
being the number of different s that occur.
Ostensibly Q is a random variable, but in fact a simple argument now
shows that Q is a constant not depending on the direction of the fixed
lined, providing of course that they are linearly independent in sets of n,
Let
be the hyperplane perpendicular to
. Then Q is just the
number of components (maximal connected subsets) complementary to
all the
for which
in
, because each component consists of all the vectors y
have a fixed value.
In order to count the components, consider the effect of deleting one
hyperplane , say
.There remain N-1 hyperplanes, with
complementary set composed of
components. These components
are of two kinds: (i) those which meet
, and (ii) those not meeting
. In an obvious notation we have
. When
is restored it cuts each component of type (i) into two and does not
disturb the others. Therefore
(8)
I claim now that
. In fact, the sets
hyperplanes in the (n-1)-dimensional space
linearly independent in sets of n-1. Therefore .
are
, and their normals are
has
components in
, and it
is easy to see that these are just the intersections of the original type (i)
components with
, establishing the claim. Substituting into (7) and
recalling that
we obtain (6). This completes the proof.
The argument given above is essentially the same as the presented by
Schlafli [ 1,pp.209-212] ,but is included here for the sake of completeness.
I am obliged to H. S. M. Coxeter for the reference. It may also be
remarked that the form of the result (1) shows that
equals the
probability that in tossing an honest coin repeatedly the n’th
“head” occurs on or after the N’th toss. But it does not seem possible to
find an isomorphism between coin-tossing and the given problem that
would make the result immediate.
Reference
1. Ludwig Schlafli, Gesammelte mathematische Abhandlungen I ,
Basel , 1950.
That’s over.
The second solution:
Given a random point with a non-singular positional distribution, we can neglect as having
zero probability the situation where that point lies on a subsurface of positive codimension
since the subsurface has measure zero with respect to the positional distribution. For
example, a random point uniformly distributed on a sphere is almost never on the equator.
Consider n points {p_k} on the unit sphere in R^n, S^{n-1}. For each point, p_k, the other
n-1 points select an n-1 dimensional hyperplane, h_k, that contains the origin and divides
S^{n-1} into two hemispheres; one which contains p_k and one which doesn't. The n
hyperplanes {h_k} divide S^{n-1} into 2^n pieces, each of which is the intersection of one
hemisphere for each p_k. The piece formed by the intersection of the hemispheres that
contain their p_k is the spherical n-hedron with {p_k} as its vertices. Being the spherical
analog of the convex hull, we will call this the spherical hull of {p_k}.
Now consider the sphere, S^{n-2}, as the equator of S^{n-1}. We can generate uniformly
distributed points on the equator by generating uniformly distributed points on the sphere
and mapping them to the equator using the map P: S^{n-1} -> S^{n-2} defined by
(x_1,x_2,...x_{n-1})
P(x_1,x_2,...x_n) = ---------------------|(x_1,x_2,...x_{n-1})|
[1]
For each n, there is such a P. We will write P^m for the composition of m such P which
would map S^{n-1} to S^{n-1-m}.
Notice that the convex hull of {P^m(p_k)} in R^{n-m} contains the origin precisely when the
spherical hull of {p_k} touches the codimension n-m subspace where x_k = 0 for 1 <= k <=
n-m.
Thus, the probability that the convex hull of n uniformly distributed random points on
S^{n-1-m} contains the origin is the same as that of the spherical hull of n uniformly
distributed random points on S^{n-1} touching the codimension n-m subspace where x_k
= 0 for 1 <= k <= n-m.
Going back to the hyperplanes {h_k} that divide S^{n-1} into 2^n pieces, using the result
from http://www.whim.org/nebula/math/spherediv.html, we get that the codimension n-m
subspace mentioned above is divided into
m-1
--2 > C(n-1,j)
--j=0
[2]
pieces by these same hyperplanes.
This means that this subspace almost always touches that many of the 2^n pieces into
which S^{n-1} is divided. This means that the piece which is the spherical hull of our n
points has that many chances out of 2^n of touching that subspace.
Therefore, the probability of the convex hull of n points uniformly distributed on the unit
sphere in R^{n-m} containing the origin is
m-1
1-n --2 > C(n-1,j)
--j=0
[3]
Changing m to n-m, we get the following
Theorem
------The probability of the convex hull of n points uniformly distributed on the unit sphere in
R^m containing the origin is
n-1-m
1-n --2
> C(n-1,j)
--j=0
[4a]
which, since C(n-1,j) = C(n-1,n-1-j), is the same as
n-1
1-n --2
> C(n-1,j)
--j=m
[4b]
Taking the complement of [4], we get the following
Corollary
--------The probability of n points uniformly distributed on the unit sphere in R^m all residing in
one hemisphere is
m-1
1-n --2
> C(n-1,j)
---
[5]
j=0
which is what is cited in Robin Chapman's article.