Return to Answer

Let $A_d$ be the area of a triangle whose vertices are chosen uniformly at random from a unit sphere in $\mathbb{R}^d$.

I claim that for $d\ge 2$ and $m\ge 1$:

$$ E(A_d^{2m})=\frac{3}{4^m} \prod _{q=1}^{m-1} \frac{3 d+6 m-2 q-6}{d+2 m-2 q-2}\prod _{q=1}^m \frac{d+2 m-2 q-1}{d+2 m-2 q}\\ = \frac{3\ \Gamma \left(\frac{d}{2}\right)^2 \Gamma \left(\frac{d-1}{2}+m\right) \Gamma \left(\frac{3 d}{2}+3 m-3\right)}{4^m\ \Gamma \left(\frac{d-1}{2}\right) \Gamma \left(\frac{d}{2}+m-1\right) \Gamma \left(\frac{d}{2}+m\right) \Gamma \left(\frac{3 d}{2}+2 m-2\right)} $$

For $d=2$ (and any $m\ge 1$) this simplifies to the established formula:

$$E(A_2^{2m}) = \frac{(3m)!}{16^m\ (m!)^3}$$

For $m=1$ (and any $d\ge 2$) it simplifies to:

$$ E(A_d^2)=\frac{3(d-1)}{4d} $$

To prove the general formula, first note that the squared area of a triangle can be described in terms of a Grammian determinant:

$$ A_d^2 = \frac{1}{4} \det{\left(s_i \cdot s_j\right)} $$

where the triangle has vertices $v_0, v_1, v_2$ and:

$$ s_i = v_i - v_0, \: i=1,2 $$

For $d\ge3$, we can always rotate a triangle with vertices on the sphere into the configuration:

$$\begin{array}{rcl} v_0 & = & e_0 \\ v_1 & = & \cos(\theta_1)\, e_0 + \sin(\theta_1)\, e_1 \\ v_2 & = & \cos(\theta_2)\, e_0 + \sin(\theta_2)\cos(\phi_2)\, e_1 + \sin(\theta_2)\sin(\phi_2)\, e_2 \end{array}$$

The expectation values of the even moments can then be expressed as an integral over three coordinates of a suitably weighted version of the squared area raised to a power:

$$ E(A_d^{2m}) = \frac{(d-2)\ \Gamma \left(\frac{d}{2}\right)}{2 \pi ^{3/2}\ \Gamma \left(\frac{d-1}{2}\right)} \int_{0}^\pi \int_{0}^\pi \int_{0}^\pi (A_d^2)^m \sin(\theta_1)^{d-2} \sin(\theta_2)^{d-2} \sin(\phi_2)^{d-3} \,d\theta_1\,d\theta_2\,d\phi_2 $$

where the vertex $v_1$ is a representative of a $(d-2)$-sphere of radius $\sin(\theta_1)$ over which it can be rotated while keeping $v_0$ fixed, and $v_2$ is a representative of a $(d-3)$-sphere of radius $\sin(\theta_2)\sin(\phi_2)$ over which it can be rotated while keeping $v_0, v_1$ fixed, and the weights incorporate the measures of these spheres with respect to the whole $(d-1)$-sphere.

For $m=1$, the integrand expands as a sum of products of non-negative integer powers of sines, and the integral can be carried out explicitly to obtain:

$$ E(A_d^2)=\frac{3(d-1)}{4d} $$

For $m\ge 2$, we can integrate by parts to obtain the recursion relation:

$$ E(A_d^{2(m+1)}) = \frac{(d-1) (3 d+4 m)}{4 d^2} E(A_{d+2}^{2m}) $$

The general formula then follows by induction.

Let $A_d$ be the area of a triangle whose vertices are chosen uniformly at random from a unit sphere in $\mathbb{R}^d$.

I claim that for $d\ge 2$ and $m\ge 1$:

$$ E(A_d^{2m})=\frac{3}{4^m} \prod _{q=1}^{m-1} \frac{3 d+6 m-2 q-6}{d+2 m-2 q-2}\prod _{q=1}^m \frac{d+2 m-2 q-1}{d+2 m-2 q}\\ = \frac{3\ \Gamma \left(\frac{d}{2}\right)^2 \Gamma \left(\frac{d-1}{2}+m\right) \Gamma \left(\frac{3 d}{2}+3 m-3\right)}{4^m\ \Gamma \left(\frac{d-1}{2}\right) \Gamma \left(\frac{d}{2}+m-1\right) \Gamma \left(\frac{d}{2}+m\right) \Gamma \left(\frac{3 d}{2}+2 m-2\right)} $$

For $d=2$ (and any $m\ge 1$) this simplifies to the established formula:

$$E(A_2^{2m}) = \frac{(3m)!}{16^m\ (m!)^3}$$

For $m=1$ (and any $d\ge 2$) it simplifies to:

$$ E(A_d^2)=\frac{3(d-1)}{4d} $$

To prove the general formula, first note that the squared area of a triangle can be described in terms of a Grammian determinant:

$$ A_d^2 = \frac{1}{4} \det{\left(s_i \cdot s_j\right)} $$

where the triangle has vertices $v_0, v_1, v_2$ and:

$$ s_i = v_i - v_0, \: i=1,2 $$

For $d\ge3$, we can always rotate a triangle with vertices on the sphere into the configuration:

$$\begin{array}{rcl} v_0 & = & e_0 \\ v_1 & = & \cos(\theta_1)\, e_0 + \sin(\theta_1)\, e_1 \\ v_2 & = & \cos(\theta_2)\, e_0 + \sin(\theta_2)\cos(\phi_2)\, e_1 + \sin(\theta_2)\sin(\phi_2)\, e_2 \end{array}$$

The expectation values of the even moments can then be expressed as an integral over three coordinates of a suitably weighted version of the squared area raised to a power:

$$ E(A_d^{2m}) = \frac{(d-2)\ \Gamma \left(\frac{d}{2}\right)}{2 \pi ^{3/2}\ \Gamma \left(\frac{d-1}{2}\right)} \int_{0}^\pi \int_{0}^\pi \int_{0}^\pi (A_d^2)^m \sin(\theta_1)^{d-2} \sin(\theta_2)^{d-2} \sin(\phi_2)^{d-3} \,d\theta_1\,d\theta_2\,d\phi_2 $$

where the vertex $v_1$ is a representative of a $(d-2)$-sphere of radius $\sin(\theta_1)$ over which it can be rotated while keeping $v_0$ fixed, and $v_2$ is a representative of a $(d-3)$-sphere of radius $\sin(\theta_2)\sin(\phi_2)$ over which it can be rotated while keeping $v_0, v_1$ fixed, and the weights incorporate the measures of these spheres with respect to the whole $(d-1)$-sphere.

For $m=1$, the integrand expands as a sum of products of non-negative integer powers of sines, and the integral can be carried out explicitly to obtain:

$$ E(A_d^2)=\frac{3(d-1)}{4d} $$

For $m\ge 2$, we can integrate by parts to obtain the recursion relation:

$$ E(A_d^{2(m+1)}) = \frac{(d-1) (3 d+4 m)}{4 d^2} E(A_{d+2}^{2m}) $$

The general formula then follows by induction.

Although we derived this formula for even moments, it also gives correct values for odd moments using half-integer values for $m$, including the average area if we set $m=1/2$:

$$ E(A_d) = \frac{3\ \Gamma \left(\frac{3 (d-1)}{2}\right) \Gamma \left(\frac{d}{2}\right)^3}{2\ \Gamma \left(\frac{d-1}{2}\right)^2 \Gamma \left(\frac{d+1}{2}\right) \Gamma \left(\frac{3 d}{2}-1\right)} $$

For example:

$$\begin{array}{rcl} E(A_2) & = & \frac{3}{2\pi} \\ E(A_3) & = & \frac{\pi}{5} \end{array}$$

Let $A_d$ be the area of a triangle whose vertices are chosen uniformly at random from a unit sphere in $\mathbb{R}^d$.

I claim that for $d\ge 2$ and $m\ge 1$:

$$ E(A_d^{2m})=\frac{3}{4^m} \prod _{q=1}^{m-1} \frac{3 d+6 m-2 q-6}{d+2 m-2 q-2}\prod _{q=1}^m \frac{d+2 m-2 q-1}{d+2 m-2 q}\\ = \frac{3\ \Gamma \left(\frac{d}{2}\right)^2 \Gamma \left(\frac{d-1}{2}+m\right) \Gamma \left(\frac{3 d}{2}+3 m-3\right)}{4^m\ \Gamma \left(\frac{d-1}{2}\right) \Gamma \left(\frac{d}{2}+m-1\right) \Gamma \left(\frac{d}{2}+m\right) \Gamma \left(\frac{3 d}{2}+2 m-2\right)} $$

For $d=2$ (and any $m\ge 1$) this simplifies to the established formula:

$$E(A_2^{2m}) = \frac{(3m)!}{16^m\ (m!)^3}$$

For $m=1$ (and any $d\ge 2$) it simplifies to:

$$ E(A_d^2)=\frac{3(d-1)}{4d} $$

To prove the general formula, first note that the squared area of a triangle can be described in terms of a Grammian determinant:

$$ A_d^2 = \frac{1}{4} \det{\left(s_i \cdot s_j\right)} $$

where the triangle has vertices $v_0, v_1, v_2$ and:

$$ s_i = v_i - v_0, \: i=1,2 $$

For $d\ge3$, we can always rotate a triangle with vertices on the sphere into the configuration:

$$\begin{array}{rcl} v_0 & = & e_1 \\ v_1 & = & \cos(\theta_1)\, e_0 + \sin(\theta_1)\, e_1 \\ v_2 & = & \cos(\theta_2)\, e_0 + \sin(\theta_2)\cos(\phi_2)\, e_1 + \sin(\theta_2)\sin(\phi_2)\, e_2 \end{array}$$

The expectation values of the even moments can then be expressed as an integral over three coordinates of a suitably weighted version of the squared area raised to a power:

$$ E(A_d^{2m}) = \frac{(d-2)\ \Gamma \left(\frac{d}{2}\right)}{2 \pi ^{3/2}\ \Gamma \left(\frac{d-1}{2}\right)} \int_{0}^\pi \int_{0}^\pi \int_{0}^\pi (A_d^2)^m \sin(\theta_1)^{d-2} \sin(\theta_2)^{d-2} \sin(\phi_2)^{d-3} \,d\theta_1\,d\theta_2\,d\phi_2 $$

where the vertex $v_1$ is a representative of a $(d-2)$-sphere of radius $\sin(\theta_1)$ over which it can be rotated while keeping $v_0$ fixed, and $v_2$ is a representative of a $(d-3)$-sphere of radius $\sin(\theta_2)\sin(\phi_2)$ over which it can be rotated while keeping $v_0, v_1$ fixed, and the weights incorporate the measures of these spheres with respect to the whole $(d-1)$-sphere.

For $m=1$, the integrand expands as a sum of products of non-negative integer powers of sines, and the integral can be carried out explicitly to obtain:

$$ E(A_d^2)=\frac{3(d-1)}{4d} $$

For $m\ge 2$, we can integrate by parts to obtain the recursion relation:

$$ E(A_d^{2(m+1)}) = \frac{(d-1) (3 d+4 m)}{4 d^2} E(A_{d+2}^{2m}) $$

The general formula then follows by induction.

Let $A_d$ be the area of a triangle whose vertices are chosen uniformly at random from a unit sphere in $\mathbb{R}^d$.

I claim that for $d\ge 2$ and $m\ge 1$:

$$ E(A_d^{2m})=\frac{3}{4^m} \prod _{q=1}^{m-1} \frac{3 d+6 m-2 q-6}{d+2 m-2 q-2}\prod _{q=1}^m \frac{d+2 m-2 q-1}{d+2 m-2 q}\\ = \frac{3\ \Gamma \left(\frac{d}{2}\right)^2 \Gamma \left(\frac{d-1}{2}+m\right) \Gamma \left(\frac{3 d}{2}+3 m-3\right)}{4^m\ \Gamma \left(\frac{d-1}{2}\right) \Gamma \left(\frac{d}{2}+m-1\right) \Gamma \left(\frac{d}{2}+m\right) \Gamma \left(\frac{3 d}{2}+2 m-2\right)} $$

For $d=2$ (and any $m\ge 1$) this simplifies to the established formula:

$$E(A_2^{2m}) = \frac{(3m)!}{16^m\ (m!)^3}$$

For $m=1$ (and any $d\ge 2$) it simplifies to:

$$ E(A_d^2)=\frac{3(d-1)}{4d} $$

To prove the general formula, first note that the squared area of a triangle can be described in terms of a Grammian determinant:

$$ A_d^2 = \frac{1}{4} \det{\left(s_i \cdot s_j\right)} $$

where the triangle has vertices $v_0, v_1, v_2$ and:

$$ s_i = v_i - v_0, \: i=1,2 $$

For $d\ge3$, we can always rotate a triangle with vertices on the sphere into the configuration:

$$\begin{array}{rcl} v_0 & = & e_0 \\ v_1 & = & \cos(\theta_1)\, e_0 + \sin(\theta_1)\, e_1 \\ v_2 & = & \cos(\theta_2)\, e_0 + \sin(\theta_2)\cos(\phi_2)\, e_1 + \sin(\theta_2)\sin(\phi_2)\, e_2 \end{array}$$

The expectation values of the even moments can then be expressed as an integral over three coordinates of a suitably weighted version of the squared area raised to a power:

$$ E(A_d^{2m}) = \frac{(d-2)\ \Gamma \left(\frac{d}{2}\right)}{2 \pi ^{3/2}\ \Gamma \left(\frac{d-1}{2}\right)} \int_{0}^\pi \int_{0}^\pi \int_{0}^\pi (A_d^2)^m \sin(\theta_1)^{d-2} \sin(\theta_2)^{d-2} \sin(\phi_2)^{d-3} \,d\theta_1\,d\theta_2\,d\phi_2 $$

where the vertex $v_1$ is a representative of a $(d-2)$-sphere of radius $\sin(\theta_1)$ over which it can be rotated while keeping $v_0$ fixed, and $v_2$ is a representative of a $(d-3)$-sphere of radius $\sin(\theta_2)\sin(\phi_2)$ over which it can be rotated while keeping $v_0, v_1$ fixed, and the weights incorporate the measures of these spheres with respect to the whole $(d-1)$-sphere.

For $m=1$, the integrand expands as a sum of products of non-negative integer powers of sines, and the integral can be carried out explicitly to obtain:

$$ E(A_d^2)=\frac{3(d-1)}{4d} $$

For $m\ge 2$, we can integrate by parts to obtain the recursion relation:

$$ E(A_d^{2(m+1)}) = \frac{(d-1) (3 d+4 m)}{4 d^2} E(A_{d+2}^{2m}) $$

The general formula then follows by induction.

Let $A_d$ be the area of a triangle whose vertices are chosen uniformly at random from a unit sphere in $\mathbb{R}^d$.

As originally conjectured and now proved, for $d=2$ we have:

$$E(A_2^{2m}) = \frac{(3m)!}{16^m\ (m!)^3}$$

For $d=3$, which I’ve checked up to $m=15$, the formula appears to be:

$$E(A_3^{2m})=\frac{6\ (m!)^3\ (6 m+1)!}{(2 m)!\ (3 m)!\ (4 m+3)!}$$

For $d=4$, which I’ve checked up to $m=12$, there appears to be a simple relationship with the $d=2$ case:

$$ E(A_4^{2m})=\frac{8\ (3(m+1))!}{16^{m+1}\ ((m+1)!)^3\ (2 m+3)} =\frac{8\ E(A_2^{2(m+1)})}{2m+3} $$

Further computations up to $d=13$ are all consistent with the following recursion formula, for $d\ge 4$:

$$ E(A_d^{2m})=\frac{4\ (d-2)^2}{(d-3)\ (4m+3d-6)} E(A_{d-2}^{2(m+1)}) $$

Something very similar holds for the even moments for the distance $s$ between two random points on the unit sphere. As discussed in the blog posts linked to in the question, I proved a general formula for the moments of $s_d$:

$$E(s_d^n) = \frac{2^{d+n-2}\ \Gamma \left(\frac{d}{2}\right) \Gamma \left(\frac{1}{2} (d+n-1)\right)}{\sqrt{\pi }\ \Gamma \left(d+\frac{n}{2}-1\right)}$$ For even moments $n=2m$ and dimensions $d=2,4$, this specialises to the central binomial coefficients and the Catalan numbers:

$$E(s_2^{2m})=\binom{2m}{m}$$ $$E(s_4^{2m})=\frac{1}{m+2}\binom{2(m+1)}{m+1} = \frac{E(s_2^{2(m+1)})}{m+2}$$

And in general, we have the relationship:

$$E(s_d^{2m}) = \frac{(d-2)}{2\ (m+d-2)}E(s_{d-2}^{2(m+1)})$$

Let $A_d$ be the area of a triangle whose vertices are chosen uniformly at random from a unit sphere in $\mathbb{R}^d$.

I claim that for $d\ge 2$ and $m\ge 1$:

$$ E(A_d^{2m})=\frac{3}{4^m} \prod _{q=1}^{m-1} \frac{3 d+6 m-2 q-6}{d+2 m-2 q-2}\prod _{q=1}^m \frac{d+2 m-2 q-1}{d+2 m-2 q}\\ = \frac{3\ \Gamma \left(\frac{d}{2}\right)^2 \Gamma \left(\frac{d-1}{2}+m\right) \Gamma \left(\frac{3 d}{2}+3 m-3\right)}{4^m\ \Gamma \left(\frac{d-1}{2}\right) \Gamma \left(\frac{d}{2}+m-1\right) \Gamma \left(\frac{d}{2}+m\right) \Gamma \left(\frac{3 d}{2}+2 m-2\right)} $$

For $d=2$ (and any $m\ge 1$) this simplifies to the established formula:

$$E(A_2^{2m}) = \frac{(3m)!}{16^m\ (m!)^3}$$

For $m=1$ (and any $d\ge 2$) it simplifies to:

$$ E(A_d^2)=\frac{3(d-1)}{4d} $$

To prove the general formula, first note that the squared area of a triangle can be described in terms of a Grammian determinant:

$$ A_d^2 = \frac{1}{4} \det{\left(s_i \cdot s_j\right)} $$

where the triangle has vertices $v_0, v_1, v_2$ and:

$$ s_i = v_i - v_0, \: i=1,2 $$

For $d\ge3$, we can always rotate a triangle with vertices on the sphere into the configuration:

$$\begin{array}{rcl} v_0 & = & e_1 \\ v_1 & = & \cos(\theta_1)\, e_0 + \sin(\theta_1)\, e_1 \\ v_2 & = & \cos(\theta_2)\, e_0 + \sin(\theta_2)\cos(\phi_2)\, e_1 + \sin(\theta_2)\sin(\phi_2)\, e_2 \end{array}$$

The expectation values of the even moments can then be expressed as an integral over three coordinates of a suitably weighted version of the squared area raised to a power:

$$ E(A_d^{2m}) = \frac{(d-2)\ \Gamma \left(\frac{d}{2}\right)}{2 \pi ^{3/2}\ \Gamma \left(\frac{d-1}{2}\right)} \int_{0}^\pi \int_{0}^\pi \int_{0}^\pi (A_d^2)^m \sin(\theta_1)^{d-2} \sin(\theta_2)^{d-2} \sin(\phi_2)^{d-3} \,d\theta_1\,d\theta_2\,d\phi_2 $$

where the vertex $v_1$ is a representative of a $(d-2)$-sphere of radius $\sin(\theta_1)$ over which it can be rotated while keeping $v_0$ fixed, and $v_2$ is a representative of a $(d-3)$-sphere of radius $\sin(\theta_2)\sin(\phi_2)$ over which it can be rotated while keeping $v_0, v_1$ fixed, and the weights incorporate the measures of these spheres with respect to the whole $(d-1)$-sphere.

For $m=1$, the integrand expands as a sum of products of non-negative integer powers of sines, and the integral can be carried out explicitly to obtain:

$$ E(A_d^2)=\frac{3(d-1)}{4d} $$

For $m\ge 2$, we can integrate by parts to obtain the recursion relation:

$$ E(A_d^{2(m+1)}) = \frac{(d-1) (3 d+4 m)}{4 d^2} E(A_{d+2}^{2m}) $$

The general formula then follows by induction.