Re: Distance Between 2 Randomly-chosen Points on a Sphere

From: Nora Baron (norabaron_at_hotmail.com)
Date: 10/09/04 

Date: 9 Oct 2004 12:08:55 -0700

David C. Ullrich <ullrich@math.okstate.edu> wrote in message news:<asjfm0dokt0bgaq7gre7hp8vfm2kpkdq5s@4ax.com>...
> On 8 Oct 2004 12:28:03 -0700, norabaron@hotmail.com (Nora Baron)
> wrote:
>
> >"Eamon Warnock" <ewarnock@gz.cngb.com> wrote in message news:<00c9e4d0a5dcef258d26cda4af481189.61944@mygate.mailgate.org>...
> >> What is the most likely straight-line distance between two randomly
> >> selected points on a sphere?
> >
> > Two randomly chosen points lie on one unique great circle
> >(with probability 1).
>
> Yes.
>
> >You can without loss of generality
> >assume that circle is in the x-y plane and that one of the
> >points has coordinates (1, 0, 0).
>
> Yes.
>
> > The other point has a
> >uniform probability of being anywhere on the circle.
>
> No. Say one point is (1,0,0), and say E is the equator.
> Say p is the actual other point, and p' is "the" point
> on E which is at the same distance from (1,0,0). Say
> d > 0 is small.
>
> The probability that d' lies on an arc on the equator
> of length d which includes the point (1,0,0) is
> proportional to the area of a spherical cap of
> radius d, or roughly d^2. Otoh if you consider an
> arc on the equator of length d which passes through
> the point (0,1,0) the probability that p' lies on
> that arc is the probability that p lies in a
> sort of strip of width d, which is roughly a
> constant times d.
>
> > The
> >distance, however, is not uniformly distributed, and clearly
> >has highest density when the second point is as far as possible
> >from the first. That is, the "most likely" distance is the
> >diameter. The least-likely distance is zero. The expectation
> >of the distance is (2/3) * diameter.
> >

  No - my argument is correct - first, you know that the two
points are on a unique great circle. Then assume that one
of the points is *labelled* as (1, 0, 0). *Conditional on
both of these*, the second point has a uniform distribution
on that great circle. The rest follows.

  You can also prove what I say anyway with no conditioning on the
great circle: the mode of the distribution of the distances is
the diameter. Computing the expectation is straightforward.
 
  Nora B.

> > Nora B.
>
>
> ************************
>
> David C. Ullrich

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