beatricalwu

2022-07-17

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

uavklarajo

Beginner2022-07-18Added 17 answers

Step 1

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 $\mathrm{\Delta}$ of this simplex are

$E\left[{\mathrm{\Delta}}^{k}\right]=\phantom{\rule{0ex}{0ex}}\frac{1}{r{!}^{k}}{\left(\frac{n}{n+k}\right)}^{i}\frac{\mathrm{\Gamma}(\frac{1}{2}(r+1)(n+k)-j+1)}{\mathrm{\Gamma}(\frac{1}{2}[(r+1)n+rk]-j+1)}{\left(\frac{\mathrm{\Gamma}\left(\frac{1}{2}n\right)}{\mathrm{\Gamma}(\frac{1}{2}[n+k])}\right)}^{r}\prod _{l=1}^{r-1}\frac{\mathrm{\Gamma}(\frac{1}{2}[n-r+k+l])}{\mathrm{\Gamma}(\frac{1}{2}[n-r+l])}\phantom{\rule{thickmathspace}{0ex}}.$

Step 2

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

${A}_{n}=\frac{1}{n!}\frac{\mathrm{\Gamma}(\frac{1}{2}{n}^{2}+\frac{1}{2})}{\mathrm{\Gamma}\left(\frac{1}{2}{n}^{2}\right)}{\left(\frac{\mathrm{\Gamma}\left(\frac{1}{2}n\right)}{\mathrm{\Gamma}(\frac{1}{2}n+\frac{1}{2})}\right)}^{n}\prod _{l=1}^{n-1}\frac{\mathrm{\Gamma}(\frac{1}{2}l+\frac{1}{2})}{\mathrm{\Gamma}\left(\frac{1}{2}l\right)}\phantom{\rule{thickmathspace}{0ex}}.$

With

$\begin{array}{r}\mathrm{\Xi}(n):=\frac{\mathrm{\Gamma}(n+\frac{1}{2})}{\mathrm{\Gamma}(n)}\end{array}$

this becomes

${A}_{n}=\frac{1}{n!}\mathrm{\Xi}\left(\frac{{n}^{2}}{2}\right)\mathrm{\Xi}{\left(\frac{n}{2}\right)}^{-n}\prod _{l=1}^{n-1}\mathrm{\Xi}\left(\frac{l}{2}\right)\phantom{\rule{thickmathspace}{0ex}}.$

Thus, with

$\begin{array}{ccccccc}n& \frac{1}{2}& 1& \frac{3}{2}& 2& \frac{9}{2}& 8\\ \mathrm{\Xi}(n)& \frac{1}{\sqrt{\pi}}& \frac{\sqrt{\pi}}{2}& \frac{2}{\sqrt{\pi}}& \frac{3\sqrt{\pi}}{4}& \frac{128}{35\sqrt{\pi}}& \frac{6435\sqrt{\pi}}{4096}\end{array}$

we find

${A}_{2}=\frac{1}{2}\frac{\mathrm{\Xi}(2)\mathrm{\Xi}\left(\frac{1}{2}\right)}{\mathrm{\Xi}(1)\mathrm{\Xi}(1)}=\frac{3}{2\pi}$

and

${A}_{3}=\frac{1}{3!}\frac{\mathrm{\Xi}\left(\frac{9}{2}\right)\mathrm{\Xi}\left(\frac{1}{2}\right)\mathrm{\Xi}(1)}{\mathrm{\Xi}\left(\frac{3}{2}\right)\mathrm{\Xi}\left(\frac{3}{2}\right)\mathrm{\Xi}\left(\frac{3}{2}\right)}=\frac{4\pi}{105}\phantom{\rule{thickmathspace}{0ex}},$

in agreement with the MathWorld values, and also

${A}_{4}=\frac{1}{4!}\frac{\mathrm{\Xi}(8)\mathrm{\Xi}\left(\frac{1}{2}\right)\mathrm{\Xi}(1)\mathrm{\Xi}\left(\frac{3}{2}\right)}{\mathrm{\Xi}(2)\mathrm{\Xi}(2)\mathrm{\Xi}(2)\mathrm{\Xi}(2)}=\frac{6435}{31104{\pi}^{2}}\phantom{\rule{thickmathspace}{0ex}}.$

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 $\mathrm{\Delta}$ of this simplex are

$E\left[{\mathrm{\Delta}}^{k}\right]=\phantom{\rule{0ex}{0ex}}\frac{1}{r{!}^{k}}{\left(\frac{n}{n+k}\right)}^{i}\frac{\mathrm{\Gamma}(\frac{1}{2}(r+1)(n+k)-j+1)}{\mathrm{\Gamma}(\frac{1}{2}[(r+1)n+rk]-j+1)}{\left(\frac{\mathrm{\Gamma}\left(\frac{1}{2}n\right)}{\mathrm{\Gamma}(\frac{1}{2}[n+k])}\right)}^{r}\prod _{l=1}^{r-1}\frac{\mathrm{\Gamma}(\frac{1}{2}[n-r+k+l])}{\mathrm{\Gamma}(\frac{1}{2}[n-r+l])}\phantom{\rule{thickmathspace}{0ex}}.$

Step 2

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

${A}_{n}=\frac{1}{n!}\frac{\mathrm{\Gamma}(\frac{1}{2}{n}^{2}+\frac{1}{2})}{\mathrm{\Gamma}\left(\frac{1}{2}{n}^{2}\right)}{\left(\frac{\mathrm{\Gamma}\left(\frac{1}{2}n\right)}{\mathrm{\Gamma}(\frac{1}{2}n+\frac{1}{2})}\right)}^{n}\prod _{l=1}^{n-1}\frac{\mathrm{\Gamma}(\frac{1}{2}l+\frac{1}{2})}{\mathrm{\Gamma}\left(\frac{1}{2}l\right)}\phantom{\rule{thickmathspace}{0ex}}.$

With

$\begin{array}{r}\mathrm{\Xi}(n):=\frac{\mathrm{\Gamma}(n+\frac{1}{2})}{\mathrm{\Gamma}(n)}\end{array}$

this becomes

${A}_{n}=\frac{1}{n!}\mathrm{\Xi}\left(\frac{{n}^{2}}{2}\right)\mathrm{\Xi}{\left(\frac{n}{2}\right)}^{-n}\prod _{l=1}^{n-1}\mathrm{\Xi}\left(\frac{l}{2}\right)\phantom{\rule{thickmathspace}{0ex}}.$

Thus, with

$\begin{array}{ccccccc}n& \frac{1}{2}& 1& \frac{3}{2}& 2& \frac{9}{2}& 8\\ \mathrm{\Xi}(n)& \frac{1}{\sqrt{\pi}}& \frac{\sqrt{\pi}}{2}& \frac{2}{\sqrt{\pi}}& \frac{3\sqrt{\pi}}{4}& \frac{128}{35\sqrt{\pi}}& \frac{6435\sqrt{\pi}}{4096}\end{array}$

we find

${A}_{2}=\frac{1}{2}\frac{\mathrm{\Xi}(2)\mathrm{\Xi}\left(\frac{1}{2}\right)}{\mathrm{\Xi}(1)\mathrm{\Xi}(1)}=\frac{3}{2\pi}$

and

${A}_{3}=\frac{1}{3!}\frac{\mathrm{\Xi}\left(\frac{9}{2}\right)\mathrm{\Xi}\left(\frac{1}{2}\right)\mathrm{\Xi}(1)}{\mathrm{\Xi}\left(\frac{3}{2}\right)\mathrm{\Xi}\left(\frac{3}{2}\right)\mathrm{\Xi}\left(\frac{3}{2}\right)}=\frac{4\pi}{105}\phantom{\rule{thickmathspace}{0ex}},$

in agreement with the MathWorld values, and also

${A}_{4}=\frac{1}{4!}\frac{\mathrm{\Xi}(8)\mathrm{\Xi}\left(\frac{1}{2}\right)\mathrm{\Xi}(1)\mathrm{\Xi}\left(\frac{3}{2}\right)}{\mathrm{\Xi}(2)\mathrm{\Xi}(2)\mathrm{\Xi}(2)\mathrm{\Xi}(2)}=\frac{6435}{31104{\pi}^{2}}\phantom{\rule{thickmathspace}{0ex}}.$

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