Assume we have a large number of uniformly randomly chosen points on the surface of a sphere in 3D (x, y, z) space (z being the vertical axis), ignoring for the moment the question of how such a uniform distribution would be accomplished.
My colleague claims that when projected onto the (x, y) plane, the points will be denser near the centre of the resulting circle (these would be the points originally near the poles of the sphere), which sounds dubious to me. Is this claim correct? How would you prove/disprove it?
Additionally, while I don't see why it would, does the answer change at all if you take into account only a single hemisphere (e.g. the northern half of the sphere)?
It's the opposite: The points are denser near the edge of the circle.
There is no (statistical) difference in looking at one hemisphere.
To see geometrically why 1. is true, project a circle (or the top half of a circle) in a plane to the horizontal axis. An element of arc length is "squashed" according to its "steepness". A fixed element of arc length maps to a shorter interval near the left and right edges, so after projection, a horizontal interval of given length contains more points if it lies near the left or right end.
Alternatively, graph $y = \sqrt{1 - x^{2}}$, divide the interval $[-1, 1]$ into $n$ pieces of equal length, and plot the corresponding points on the graph. The intervals near the ends subtend a larger arc on the circle.