Expected distance between two points on the Surface of a sphere

Problems in Geometric Probability are always fascinating. In this post, we going to see two of the well known and challenging problems in this topic.

The first problem is about the expected distance between two points selected at random from a surface of an $n$-sphere. To clarify, I'm considering a circle to be a $2$-sphere, a sphere to be a $3$-sphere and so on.

This problem is relatively easy as can seen from this stackexchange post. Let $S_n(R)$ and $V_n(R)$ denote the surface area and the volume of an $n$-sphere with radius $R$. We know that,

$S_n(R)=s_nR^{n-1}$ and $V_n(R)=v_nR^n$

where $s_n=\dfrac{2\pi^{n/2}}{\Gamma(n/2)}$ and $v_n=\dfrac{2}{n}\dfrac{\pi^{n/2}}{\Gamma(n/2)}$

For the problem under consideration, we can choose one of the points to be on the south pole of the $n$-sphere. Consider one of the axis that connects this point to the origin. A hyperplane perpendicular to this axis cuts the surface of the $n$-sphere to give a $(n-1)$-spherical surface. 

If we consider multiple such hyperplanes parallel to one another along the axis, the $n$-spherical surface is sliced into multiple rings. The picture below shows this in case of a $3$-sphere (taken from this Stack Exchange post).

Because second point is uniformly distributed in the $n$-spherical surface, it is equally likely to be present in any of these rings.

Let $\theta$ be the angle subtended between the radius that contains the second point and radius containing the first point. Then the distance between the points is $2\sin(\theta/2)$.

If we consider these rings to be infinitesimally small, then the density function of the second point lying in any of these rings is equal to the area of the ring. But this is easy. This is just the surface area of the  sliced $(n-1)$-hypersphere times $rd\theta$ (obviously $r=1$ for the unit sphere; used here for illustrative purposes). Therefore,

$\displaystyle \mathbb{E}(L)=\frac{1}{s_n}\int_0^\pi 2\sin(\theta/2)\cdot S_{n-1}(r\sin\theta)\cdot r\,d\theta=\frac{s_{n-1}}{s_n}\int_0^\pi 2\sin(\theta/2)\cdot \sin^{n-2}\theta \,d\theta$

Using Wolfram Alpha, we then see that

$\displaystyle \mathbb{E}(L)=\frac{\Gamma(n/2)}{\sqrt{\pi}\Gamma((n-1)/2)}\cdot \frac{2\sqrt{\pi}\Gamma(n-1)}{\Gamma(n-1/2)}=2\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}\cdot \frac{\Gamma(n-1)}{\Gamma(n-1/2)}$

Now, the second problem asks for the expected distance between a point randomly selected on the $n$-spherical surface and a point selected on the $n$-spherical volume. This problem is more involved and requires a little more trickery. We'll see this in the next post.

Until then

Yours Aye

Me