Editing Borel–Kolmogorov paradox (section)

== A great circle puzzle ==
Suppose that a [[random variable]] has a [[Uniform distribution (continuous)|uniform distribution]] on a [[unit sphere]]. What is its [[conditional distribution]] on a [[great circle]]? Because of the symmetry of the sphere, one might expect that the distribution is uniform and independent of the choice of coordinates. However, two analyses give contradictory results.  First, note that choosing a point uniformly on the sphere is equivalent to choosing the [[longitude]] <math>\lambda</math> uniformly from <math>[-\pi,\pi]</math> and choosing the [[latitude]] <math>\varphi</math> from <math display="inline">[-\frac{\pi}{2},\frac{\pi}{2}]</math> with density <math display="inline">\frac{1}{2} \cos \varphi</math>.<ref name=Jaynes>{{harvnb|Jaynes|2003|pages=1514–1517}}</ref> Then we can look at two different great circles:
# If the coordinates are chosen so that the great circle is an [[equator]] (latitude <math>\varphi = 0</math>), the conditional density for a longitude <math>\lambda</math> defined on the interval <math>[-\pi,\pi]</math> is <math display="block"> f(\lambda\mid\varphi=0) = \frac{1}{2\pi}.</math>
# If the great circle is a [[line of longitude]] with <math>\lambda = 0</math>, the conditional density for  <math>\varphi</math> on the interval <math display="inline">[-\frac{\pi}{2},\frac{\pi}{2}]</math> is <math display="block">f(\varphi\mid\lambda=0) = \frac{1}{2} \cos \varphi.</math>

One distribution is uniform on the circle, the other is not. Yet both seem to be referring to the same great circle in different coordinate systems.

{{Quote
 | Many quite futile arguments have raged — between otherwise competent probabilists — over which of these results is 'correct'.
 | [[E.T. Jaynes]]<ref name=Jaynes/>
}}