What is the expected volume of the simplex formed by $n+1$ points independently uniformly distributed on $\mathbb S^{n-1}$?

Question

I was surprised that I couldn’t find this question answered on this site (not for lack of trying). I need the result for answering Probability of random sphere lying inside the unit ball, so I’m posting it as a separate question and answer for future reference.

What is the expected volume of the simplex formed by $n+1$ points independently uniformly distributed on $\mathbb S^{n-1}$?

MathWorld has the answers $\frac3{2\pi}$ for $n=2$ under Circle Triangle Picking and $\frac{4\pi}{105}$ for $n=3$ under Sphere Tetrahedron Picking, but the general result is surprisingly hard to find.

Answer 1

The question is answered in Isotropic Random Simplices by R. E. Miles (Advances in Applied Probability, 3(2), 353–382) in Theorem $2$ on p. $362$. More generally, for $i$ points independently uniformly distributed in the interior of the $n$-ball and $j$ points independently uniformly distributed on its boundary (the sphere $\mathbb S^{n-1}$), with $1\le r:=i+j-1\le n$ so that the points almost surely form an $r$-simplex, the moments of the volume $\Delta$ of this simplex are

$$ E\left[\Delta^k\right] =\\ \frac1{r!^k}\left(\frac n{n+k}\right)^i\frac{\Gamma\left(\frac12(r+1)(n+k)-j+1\right)}{\Gamma\left(\frac12[(r+1)n+rk]-j+1\right)}\left(\frac{\Gamma\left(\frac12n\right)}{\Gamma\left(\frac12[n+k]\right)}\right)^r\prod_{l=1}^{r-1}\frac{\Gamma\left(\frac12[n-r+k+l]\right)}{\Gamma\left(\frac12[n-r+l]\right)}\;. $$

In our case, $i=0$, $j={n+1}$, $r=n$ and $k=1$, so the desired volume is

$$ A_n=\frac1{n!}\frac{\Gamma\left(\frac12n^2+\frac12\right)}{\Gamma\left(\frac12n^2\right)}\left(\frac{\Gamma\left(\frac12n\right)}{\Gamma\left(\frac12n+\frac12\right)}\right)^n\prod_{l=1}^{n-1}\frac{\Gamma\left(\frac12l+\frac12\right)}{\Gamma\left(\frac12l\right)}\;. $$

With

\begin{eqnarray} \Xi(n):=\frac{\Gamma\left(n+\frac12\right)}{\Gamma(n)} \end{eqnarray}

this becomes

$$ A_n=\frac1{n!}\Xi\left(\frac{n^2}2\right)\Xi\left(\frac n2\right)^{-n}\prod_{l=1}^{n-1}\Xi\left(\frac l2\right)\;. $$

Thus, with

\begin{array}{c|cc} n&\frac12&1&\frac32&2&\frac92&8\\\hline \Xi(n)&\frac1{\sqrt\pi}&\frac{\sqrt\pi}2&\frac2{\sqrt\pi}&\frac{3\sqrt\pi}4&\frac{128}{35\sqrt\pi}&\frac{6435\sqrt\pi}{4096}\\ \end{array}

we find

$$ A_2=\frac12\frac{\Xi(2)\Xi\left(\frac12\right)}{\Xi(1)\Xi(1)}=\frac3{2\pi} $$

and

$$ A_3=\frac1{3!}\frac{\Xi\left(\frac92\right)\Xi\left(\frac12\right)\Xi(1)}{\Xi\left(\frac32\right)\Xi\left(\frac32\right)\Xi\left(\frac32\right)}=\frac{4\pi}{105}\;, $$

in agreement with the MathWorld values, and also

$$ A_4=\frac1{4!}\frac{\Xi(8)\Xi\left(\frac12\right)\Xi(1)\Xi\left(\frac32\right)}{\Xi(2)\Xi(2)\Xi(2)\Xi(2)}=\frac{6435}{31104\pi^2}\;. $$