Expected distance between two points inside a Sphere

This post picks up where we left earlier. As discussed before, we want to find the expected distance between two points randomly selected in an $n$-sphere.

Let the two points be $P$ and $Q$, and $O$ be the centre. The first step is to apply the technique of projective Reduction with $O$ as the scaling point. This shows that the expected length between two points inside a $n$-sphere is $2n/(2n+1)$ times the expected length between a point on the surface of $n$-sphere and another point inside the $n$-sphere.

WLOG, let $P$ be the point on the surface and let's orient the sphere so that point $P$ becomes the south pole. The next step is apply another Projective Reduction with the south pole as the scaling point. This step gives another factor of $n/(n+1)$ for the expected length. Note that while the first reduction preserves the uniformity of the points on the respective 'surfaces', it is not the case with the second reduction. This is what makes the problem tricky.

To find the density of $Q$, consider a random chord from point $P$ (the south pole) along with the axis on which $P$ lies. Let $\theta$ be the angle between the chord and the axis.

Rotating the chord along this axis cuts out a (for a lack of a better word) spherical-cone which contains a cone with its apex at the south pole and a spherical cap at the top (Imaging the other end of the chord to be on the opposite side of $P$). With simple geometry, it is also apparent that the angle between $OQ$ and the axis is $2\theta$.

If we let $V_{\text{sp. cone}}(\theta, r)$ be the volume of the spherical cone from a $n$-sphere of radius $r$ and apex angle $\theta$, then we have,

$\begin{align} \displaystyle V_{\text{sph. cone}}(\theta, r) &= V_{\text{cone}} + V_{\text{sph. cap}}\\&=  \frac{1}{n}V_{n-1}(r \sin 2\theta)(r+r\cos2\theta)+\int_{0}^{2\theta}V_{n-1}(r \sin t) \,d(-r\cos t)\\ &= r^n\frac{v_{n-1}}{n}\sin^{n-1}2\theta \cdot 2  \cos^2\theta + r^nv_{n-1}\int_0^{2\theta}\sin^nt\,dt \\ &=\frac{2^nr^nv_{n-1}}{n}\sin^{n-1}\theta\cos^{n+1}\theta + + r^nv_{n-1}\int_0^{2\theta}\sin^nt\,dt \end{align}$

We have used the fact that the area of an $n$-dimensional cone is $1/n$ times the volume of the circumscribing 'cylinder'. For the volume of the cap, we have used that the volume is the sum of the areas of $(n-1)$ dimensional 'circles' with infinitesimal thickness (see here for more). 

Now, $\frac{\partial V_{\text{sph. cone}}}{\partial r \partial \theta}$ gives volume between $V_{\text{sph. cone}}(\theta + d\theta, r+dr)$ and $V_{\text{sph. cone}}(\theta, r)$. When $Q$ lies in this infinitesimal volume, the distance between $P$ and $Q$ is just $2\cos\theta$.

Because we are only interested in distribution of $Q$ on the surface of the $n$-sphere, we can integrate out the $r$. Therefore, the density is given by,

$\displaystyle f(\theta)=\frac{1}{v_n}\int_{0}^1\frac{\partial V_{\text{sph. cone}}}{\partial r \partial \theta}\,dr$

We now get,

$\displaystyle f(\theta)=\frac{2^nv_{n-1}}{nv_n}((n-1)\cos^{n+2}\theta\sin^{n-2}\theta-(n+1)\sin^n\theta\cos^n\theta)+\frac{v_{n-1}}{v_n}\cdot 2\sin^n2\theta$

where we have used Leibniz differential under integral sign for the last term. Further simplification then gives,

$\displaystyle f(\theta)=\frac{2^nv_{n-1}}{nv_n}(n-1)\sin^{n-2}\theta\cos^n\theta$

This nice simplification allows us to express the expected length $e_n$ in terms of Beta function.

$\displaystyle e_n=\int_0^{\pi/2}2\cos\theta\cdot f(\theta)\,d\theta=\frac{2^n}{n}\frac{v_{n-1}}{v_n}(n-1)B(n/2-1/2, n/2+1)$

Finally, we see that the expected length $L$ between two points randomly selected inside an $n$-sphere is,

$\displaystyle \mathbb{E}(L)=\frac{2n}{2n+1}\frac{n}{n+1}\frac{2^n(n-1)}{n}\frac{B(n/2-1/2,n/2+1)}{B(n/2+1/2,1/2)}$

Interestingly, at $n\to \infty$, irrespective of whether we choose points on the surface or the inside, the expected length goes to $\sqrt{2}$. I can't clearly see why but if you have an idea, please do share it with us.

UPDATE 3 Jul 2023: Generalising this, we can see that

$\displaystyle \mathbb{E}(L^k)=\frac{2n}{2n+k}\frac{n}{n+k}\frac{2^{n+k-1}(n-1)}{n}\frac{B(n/2-1/2,n/2+k/2+1/2)}{B(n/2+1/2,1/2)}$

Using this, we can see, using WA, that

$\displaystyle \lim_{n \to \infty}\mathbb{E}(L^k)=\sqrt{2}^k$

Also, using the limit definition of the logarithm function,

$\displaystyle \mathbb{E}(\ln L)=-\frac{1}{n}+\frac{n}{2}\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k(n+k)}$

This shows us the geometric mean of distance between two points chosen uniformly randomly from a unit circle is $e^{-1/4}$ and that of the sphere is $2e^{-3/4}$

Until then

Yours Aye

Me