Assume I choose three random points on the surface of a sphere. What is the average area? (Each point is independently chosen relative to a uniform distribution on the sphere) Also, what would be the average area if I choose three points on other types of surfaces? Such as 2-dimensional square/circle or 1-dimensional line.
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$\begingroup$ Are you talking about the area of the resulting spherical triangle, or the triangle created inside of the sphere? And I assume the answer will be vastly different for different surfaces, but symmetry can help and you can ask for a general form in terms of integrals. $\endgroup$– anonAug 23, 2011 at 16:12
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$\begingroup$ @anon: not inside the sphere, the area of the triangle on the surface (spherical triangle). $\endgroup$– MarkAug 23, 2011 at 16:13
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$\begingroup$ @anon: do you think the ways to exploit the symmetry on these surfaces are similar? $\endgroup$– MarkAug 23, 2011 at 16:47
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3$\begingroup$ On the unit sphere, the area is equal to the sum of the angles minus $\pi$. The expected value for one of the angles is $\pi/2$, so the expected value of the area is $3*\pi/2 - \pi = \pi/2$, which is one eigth of the sphere. $\endgroup$– Willie WongAug 23, 2011 at 17:03
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$\begingroup$ @Willie: Do you mean the angle for the triangle? Why is it equal to the sum of the angles minus $\pi$?. The link from wolfram doesn't say much... $\endgroup$– MarkAug 23, 2011 at 17:08
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An other way to see the result is by a symmetry argument.
For any given spherical triangle, you may toggle each point with their diametrical opposite to obtain a total of $2^3 = 8$ triangles. The key property to see here is that the union of all these triangles is equal to the sphere, and that the intersection of two of these triangles has an area equal to $0$ (it is at most a common side). So they basically form a partition of the sphere (area-wise, at least).
This means that the expected area of a random spherical triangle has to be an eighth that of the total sphere, ie $\pi R^2/2$.
Edit: To clarify the last step, since these $8$ triangles have equal probability, their average area is the sum of the areas ($4\pi R^2$) divided by $8$. Then you need to see that each triangle is part of such a set of $8$ triangles, with no set more probable than any other.
The way I see it: choosing a triangle at random is the same as choosing a set of $8$ triangles (ie, choosing three great circles at random, which relates to the angle solution), and then choosing one of the $8$ triangles uniformly at random.
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$\begingroup$ "...The key property to see here is that the union of all these triangles is equal to the sphere, and that the intersection of two of these triangles has an area equal to 0 ..." How do I prove this part? $\endgroup$– MarkAug 24, 2011 at 1:58
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$\begingroup$ "...For any given spherical triangle, you may toggle each point with their diametrical opposite to obtain a total of $2^3=8$ triangles..." I still don't feel clear on how to construct these eight triangles. Suppose I just happen to choose all three points at one point. It doesn't seem to work in this case. So, are you talking about the average area or any three points on the surface of a sphere? $\endgroup$– MarkAug 24, 2011 at 2:17
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$\begingroup$ Lastly, suppose the statement is true what would be the procedure to obtain these eight triangles? $\endgroup$– MarkAug 24, 2011 at 2:18
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$\begingroup$ @Mark (1) The sphere's center, along with any pair of the three points, determines a plane. Use this plane to divide the sphere into two parts. That gives three planes--one for each point pair--and therefore 2^3 = 8 parts of the sphere are created. When no two points coincide, each of these parts is nonempty and each part is a triangle. (2) The chance that two or more of the points coincide equals zero, so you don't have to consider this possibility. $\endgroup$– whuberAug 24, 2011 at 4:13
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$\begingroup$ @whuber: Ok I am convinced that they cover the sphere's surface. But "...This means that the expected area of a random spherical triangle has to be an eighth that of the total sphere, ie $πR^2/2$.." is still not clear to me. How should I prove this part? $\endgroup$– MarkAug 24, 2011 at 6:54