Moments of volumes of lower-dimensional random simplices are not monotone

Abstract

In a d-dimensional convex body K, for $n \le d+1$, random points $X_0, \dots , X_{n-1}$ are chosen according to the uniform distribution in K. Their convex hull is a random $(n-1)$-simplex with probability 1. We denote its $(n-1)$-dimensional volume by $V_{K[n]}$. The k-th moment of the $(n-1)$-dimensional volume of a random $(n-1)$-simplex is monotone under set inclusion, if $K \subseteq L$ implies that the k-th moment of $V_{K[n]}$ is not larger than that of $V_{L[n]}$. Extending work of Rademacher (Mathematika 58:77–91, 2012) and Reichenwallner and Reitzner (Mathematika 62:949–958, 2016), it is shown that for $n \le d$, the moments of $V_{K[n]}$ are not monotone under set inclusion. As a consequence, the nonmonotonicity of the expected surface area of the convex hull of $n\le d+1$ uniform random points in a d-dimensional convex body follows.

1 Introduction

For a d-dimensional convex body K and $n \le d+1$, we denote the $(n-1)$-dimensional volume of the convex hull of n independently and uniformly distributed random points $X_0, \ldots , X_{n-1}$ in K by $V_{K[n]}$. Since the n points are not contained in any $(n-2)$-dimensional hyperplane with probability 1, their convex hull is almost surely an $(n-1)$-simplex. In 2006, Meckes [5] asked whether for a pair of convex bodies $K, L \subseteq \mathbb {R}^d$, $K \subseteq L$ would imply

$$\begin{aligned} \mathbb {E}V_{K[d+1]} \le \mathbb {E}V_{L[d+1]}. \end{aligned}$$

(1)

Interest in this question comes from the fact that it would imply a positive solution to the slicing problem, see e.g. [7]. For a more general statement of the conjecture we refer to [9].

In 2012, Rademacher [7] showed that Meckes’ conjecture (1) is not true in general, since there exist counterexamples for each dimension $d \ge 4$. His results were generalized to higher moments of $V_{K[d+1]}$ by Reichenwallner and Reitzner [8], and the expected value in dimension three was handled in [4]:

Theorem 1

Let $k \in \mathbb {N}$ and $d \ge 2$. For two-dimensional convex bodies K and L, $K \subseteq L$ implies that $\mathbb {E}V_{K[d+1]} \le \mathbb {E}V_{L[d+1]} \ \text { and } \ \mathbb {E}V_{K[d+1]}^2 \le \mathbb {E}V_{L[d+1]}^2.$ Unless $d = 2$ and $k \in \{1,2\}$, there exist two d-dimensional convex bodies K, L satisfying $K \subseteq L$ and $\mathbb {E}V_{K[d+1]}^k > \mathbb {E}V_{L[d+1]}^k$.

Note that the problem is trivial in dimension one. In this paper, we perform another extension by considering convex hulls of $n \le d+1$ random points in d-dimensional convex bodies. In order to keep the statement of the result simple, we restrict to $n \le d$ and refer to the theorem above for the case $n=d+1$.

Theorem 2

Let $d,n,k \in \mathbb {N}$ with $d \ge 2$ and $2 \le n \le d$. Then there exist two d-dimensional convex bodies K, L satisfying $K \subseteq L$ and $\mathbb {E}V_{K[n]}^k > \mathbb {E}V_{L[n]}^k$.

Due to Rademacher [7], it is already known that the moments of $V_{K[n]}$ are not monotone under set inclusion in general. However, it might still come as a surprise that for $n \le d$, there exist counterexamples for any moment in any dimension, particularly in dimension two, where the first two moments of the area of a random triangle are monotone.

Theorem 1 is shown using an equivalence statement, first stated in Lemma 12 in [7]. In consideration of Proposition 16 ibidem, it is clear that the lemma can be generalized in the following way:

Lemma 1

For $d,n,k \in \mathbb {N}$ with $2 \le n \le d+1$, monotonicity under inclusion of the map $K \mapsto \mathbb {E}V_{K[n]}^k,$ where K ranges over all d-dimensional convex bodies, holds if and only if we have for each convex body $K \subseteq \mathbb {R}^d$ and for each $x \in {\text {bd}}K$ that $\mathbb {E}V_{K[n]}^k \le \mathbb {E}V_{K[n],x}^k.$

Here, $V_{K[n],x}$ denotes the $(n-1)$-dimensional volume of a random $(n-1)$-simplex, which is the convex hull of a fixed point x and $n-1$ independently and uniformly distributed random points in K. The lemma allows us to consider one convex body K, rather than a pair of convex bodies, and compute two different moments: the moment of the $(n-1)$-dimensional volume of a random $(n-1)$-simplex in K as well as the same, but fixing one of the n points to be a point on the boundary of K, denoted by ${\text {bd}}K$.

For the case $n=d+1$, Rademacher chooses K to be a d-dimensional halfball and x the midpoint of the base which is a $(d-1)$-dimensional ball. This pair forms the counterexample in Theorem 1 to the monotonicity of all moments for $d \ge 4$, and to the monotonicity of the k-th moment for $k \ge 11$ in dimension two and for $k \ge 2$ in dimension three (see [8]; in fact, in dimension two, it does so for $k \ge 8$). Additional counterexamples are given by the triangle, together with the midpoint of one of its edges, for $d=2$ and $k \ge 3$ in [8] and the tetrahedron, together with the centroid of one of its facets, for $d=3$ and $k=1$ in [4].

For our generalization, we use the following lemma, which allows us to create counterexamples inductively:

Lemma 2

Let $d, n, k \in \mathbb {N}$ with $2 \le n \le d+1$. If there exist two d-dimensional convex bodies K, L satisfying $K \subseteq L$ and $\mathbb {E}V_{K[n]}^k > \mathbb {E}V_{L[n]}^k$, there also exist two $(d+1)$-dimensional convex bodies $K^\prime , L^\prime$ satisfying $K^\prime \subseteq L^\prime$ and $\mathbb {E}V_{K^\prime [n]}^k > \mathbb {E}V_{L^\prime [n]}^k$.

We use the counterexamples mentioned above as well as Lemma 2 to get results on the monotonicity of the moments of $V_{K[n]}$, where K ranges over all d-dimensional convex bodies, $d \ge 2$ and $2 \le n \le d$. Since, obviously, monotonicity holds in dimension 1, in order to show that the moments of the length of a random chord in d-dimensional convex bodies are not monotone for $d > 1$, we create another counterexample by computing the moments of the length of a random chord inside a rectangular triangle where both points are randomly chosen as well as the same where one point is fixed to be the midpoint of the hypotenuse.

Proposition 1

For the triangle $T_2$ with vertices (0, 0), (1, 0) and (0, 1), and $c_2=(1/2,1/2)$ the midpoint of the edge $\{(x,y): x,y \ge 0, x+y=1\}$, we have that $\mathbb {E}V_{T_2[2],c_2}^k < \mathbb {E}V_{T_2[2]}^k$ for each $k \in \mathbb {N}$.

Concerning the case $n=3$, $k=1$, we need to construct a pair of three-dimensional convex bodies, or one three-dimensional convex body and a suitable point on its boundary, leading to a counterexample. Already the computation of the expected volume of a random tetrahedron in dimension three is a hard problem, and the same is supposed to be true for the expected area of a random triangle in dimension three.

The only three-dimensional convex sets where the expected volume of a random full-dimensional simplex is known are the ball [6], the cube [12] and the tetrahedron [1], and the ball is the only one for which the expected area of a random triangle inside is known. Since numerical simulations suggest that the ball cannot form a counterexample in our concerns, and neither can the cube, we consider random triangles in a tetrahedron, which nothing is known about. A suitable choice for x should be the center c of one of the facets, but we have to take into account that the considered moments depend on the shape of T as well as of the facet which c lies in. Additionally, already the determination of the expected volume of a random tetrahedron in a tetrahedron T, done by Buchta and Reitzner [1], was extremely hard, and it seems that a computation of $\mathbb {E}V_{T[3]}$ and $\mathbb {E}V_{T[3],c}$ is out of reach. Neverless we will prove the following proposition.

Proposition 2

For the tetrahedron $T_3$ with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1), and $c_3=(1/3,1/3,1/3)$ the centroid of the facet $\{(x,y,z): x,y,z \ge 0, x+y+z=1\}$, we have that $\mathbb {E}V_{T_3[3],c_3} < \mathbb {E}V_{T_3[3]}.$

A combination of this result and a direct computation of $\mathbb {E}V_{T_3[3]}^2$ and $\mathbb {E}V_{T_3[3],c_3}^2$ with Rademacher’s Lemma 1 yields Theorem 2. Proposition 2 is obtained by a combination of methods from stochastic geometry with results from approximation theory. In the background, we use a result about the precise approximation of the square root function by suitable polynomials.

A next step might naturally be given by a generalization of Theorems 1 and 2 to intrinsic volumes, among them the surface area. While a derivation of results on its higher moments seems to be rather involved, it appears that the nonmonotonicity of the expected surface area of a random simplex, eventually not of full dimension, is a direct consequence of Theorem 2. For a d-dimensional convex body K and $2 \le n \le d+1$, let $S_{K[n]}$ denote the surface area of the convex hull of n points that are chosen according to uniform distribution in K. Since the (relative) boundary of an $(n-1)$-simplex is given by the union of n facets and the expected value is additive, the following corollary holds true:

Corollary 1

Let $d,n \in \mathbb {N}$ with $d \ge 2$ and $2 \le n \le d+1$. Then there exist two d-dimensional convex bodies K, L satisfying $K \subseteq L$ and $\mathbb {E}S_{K[n]} > \mathbb {E}S_{L[n]}$.

More generally, it shall be mentioned that the expected value of the sum of the m-dimensional volumes of all m-faces of the convex hull of $n \ge m+2$ uniform random points in a convex body K of dimension $d \ge n-1$ is not monotone under set inclusion.

Finally, we consider a similar problem: Assume that n points are uniformly distributed on the boundary of a convex body K and let $\mathcal {V}_{K[n]}$ be the $(n-1)$-dimensional volume of their convex hull. Then we can use a technique similar to that in the background of Lemma 2 to show that, in general, monotonicity under set inclusion does not hold for the moments of $(n-1)$-dimensional volumes of this type of random $(n-1)$-simplices either. In detail, we show the following:

Theorem 3

Let $d,n,k \in \mathbb {N}$ with $d \ge 3$ and $2 \le n \le d$. Unless $d=3$, $n=3$ and $k \in \{1,2\}$, there exist two d-dimensional convex bodies K, L satisfying $K \subseteq L$ and $\mathbb {E}\mathcal V_{K[n]}^k > \mathbb {E}\mathcal V_{L[n]}^k$.

This paper is organized in the following way. We prove Lemma 2 in Sect. 2. In Sect. 3, we compute the moments of the lengths of two types of random chords inside a triangle as described above, and thereafter we give bounds for the expected areas of random triangles in a tetrahedron. These results will be used in Sect. 5 for the proof of Theorem 2. Finally, Theorem 3 will be proved in Sect. 6.

As a general reference for the tools and results we need in the following, we refer to the book on Stochastic and Integral Geometry by Schneider and Weil [10]. More recent surveys on random polytopes are due to Hug [3] and Reitzner [9].

2 Proof of Lemma 2

Let $d, n,k \in \mathbb {N}$ with $2\le n \le d+1$ and assume that there exist two d-dimensional convex bodies K and L satisfying $K \subseteq L$ and $\mathbb {E}V_{K[n]}^k > \mathbb {E}V_{L[n]}^k$. Our goal is to construct two $(d+1)$-dimensional convex bodies $K^\prime$, $L^\prime$ satisfying similar conditions. For $\varepsilon > 0$, we define two $(d+1)$-dimensional convex bodies by

$$\begin{aligned} K_{\varepsilon } :=K \times [0,\varepsilon ], \quad L_{\varepsilon } :=L \times [0,\varepsilon ]. \end{aligned}$$

Then, $K_\varepsilon \subseteq L_\varepsilon$ for all $\varepsilon$, and if $\mathbb {E}V_{K_\varepsilon [n]}^k$ and $\mathbb {E}V_{L_\varepsilon [n]}^k$ converge to $\mathbb {E}V_{K[n]}^k$ and $\mathbb {E}V_{L[n]}^k$, resp., for $\varepsilon$ tending to zero, there exists an $\varepsilon _0 > 0$ such that $\mathbb {E}V_{K_{\varepsilon _0}[n]}^k > \mathbb {E}V_{L_{\varepsilon _0}[n]}^k$.

It remains to proof the continuity of the map $\varepsilon \mapsto \mathbb {E}V_{K_\varepsilon [n]}^k$ in $\varepsilon =0$. We express the $(n-1)$-dimensional volume of an $(n-1)$-simplex with vertices $x_0,\ldots ,x_{n-1}$ using a matrix that contains the vectors spanning the $(n-1)$-simplex in its columns:

$$\begin{aligned}&{\text {vol}}_{n-1}{\text {conv}}(x_0,x_1,\ldots ,x_{n-1}) \\&\quad = \frac{1}{(n-1)!} \sqrt{\det \left( (x_1 - x_0, \ldots , x_{n-1} - x_0)^T (x_1 - x_0, \ldots , x_{n-1} - x_0)\right) }. \end{aligned}$$

Then we have for $\varepsilon > 0$:

$$\begin{aligned}&\mathbb {E}V_{K_\varepsilon [n]}^k = \frac{1}{{\text {vol}}K_\varepsilon ^n} \int _{K_\varepsilon ^n} {\text {vol}}_{n-1} {\text {conv}}(x_0,\ldots ,x_{n-1})^k\; d(x_0,\ldots ,x_{n-1})\\&\quad = \frac{1}{\varepsilon ^n{\text {vol}}_d K^n} \frac{1}{((n-1)!)^k} \int _{K^n} \int _{[0,\varepsilon ]^n} \det B^{k/2}\; d(z_0,\ldots ,z_{n-1})\; d(x_0,\ldots ,x_{n-1}), \end{aligned}$$

where $B = (b_{ij})_{1 \le i,j \le n-1}$ with

$$\begin{aligned} b_{ij}&= \sum _{m=1}^d \left( x_i^{(m)}-x_0^{(m)}\right) \left( x_j^{(m)}-x_0^{(m)}\right) + (z_i-z_0) (z_j-z_0), \end{aligned}$$

writing $x_i = (x_i^{(1)}, \ldots , x_i^{(d)}, z_i).$ Now, fix the first d coordinates of the points $x_0,\ldots , x_{n-1}$ and let $\zeta :=\min _{z_i} \det B$ and $Z :=\max _{z_i} \det B$, where the minimum and the maximum are taken over all possible choices of $(z_0,\ldots ,z_{n-1}) \in [0,\varepsilon ]^n$. The extrema exist since the determinant is continuous and the intervall $[0,\varepsilon ]$ is compact. We have:

$$\begin{aligned}&\frac{1}{{\text {vol}}_d K^n} \frac{1}{((n-1)!)^k} \int _{K^n} \zeta ^{k/2}\; d(x_0,\ldots ,x_{n-1}) \le \mathbb {E}V_{K_\varepsilon [n]}^k \\&\quad \le \frac{1}{{\text {vol}}_d K^n} \frac{1}{((n-1)!)^k} \int _{K^n} Z^{k/2}\; d(x_0,\ldots ,x_{n-1}). \end{aligned}$$

If $\varepsilon$ tends to zero, the coordinates $z_i$ do so as well, and all summands of $\zeta$ and Z containing a factor $z_i$ vanish in the limit. Therefore, we get that

$$\begin{aligned}&\mathbb {E}V_{K[n]}^k \le \lim _{\varepsilon \rightarrow 0} \mathbb {E}V_{K_\varepsilon [n]}^k \le \mathbb {E}V_{K[n]}^k, \end{aligned}$$

which implies that $\lim _{\varepsilon \rightarrow 0} \mathbb {E}V_{K_\varepsilon [n]}^k = \mathbb {E}V_{K[n]}^k$. $\square$

3 Random chords in a triangle

Essential for our investigations of the case $n=2$ are the moments of two different types of random distances in a given triangle T. In detail, we need the moments of the length of a random chord in a right-angled triangle, which arises as the segment between two uniformly chosen random points in this triangle, as well as the moments of the distance of a random point in this triangle from the midpoint of its hypotenuse, which is equivalent to the length of a random chord where one point is fixed to be this midpoint.

3.1 Moments of the distance of a random point from a point on the boundary

We first prove a lemma that will be useful for the computation of the moments of the distance of a random point from a fixed point on the boundary, since we can dissect a triangle by a line through this fixed point.

Lemma 3

Let T be a triangle with vertices A, B and C. Denote the lengths of the edges opposite to A, B and C by a, b and c, resp., and the angles in these corners by $\alpha$, $\beta$ and $\gamma$, resp. Then it holds for the moments of the distance of a random point in T from A:

$$\begin{aligned} \mathbb {E}V_{T[2],A}^k&= \frac{2(c \sin \beta )^{k+1}}{(k+2)a} \left( \cos \beta \; _2F_1\left( \frac{1}{2},\frac{k+3}{2}; \frac{3}{2}; \cos \beta ^2\right) \right. \\&\quad \left. - \cos (\alpha +\beta ) \; _2F_1\left( \frac{1}{2},\frac{k+3}{2}; \frac{3}{2}; \cos (\alpha +\beta )^2\right) \right) . \end{aligned}$$

Here, the function

$$\begin{aligned} _2F_1\left( a_1,a_2; b; x\right) = \sum \limits _{k=0}^\infty \frac{(a_1)_k (a_2)_k}{(b)_k} \frac{x^k}{k!} \end{aligned}$$

with $(a)_k = a(a+1) \cdots (a+k-1)$ is a hypergeometric function.

Proof

Since the expected length of a random chord in a triangle does not depend on the position of the latter, we build our coordinate system ensuring that A is identical to the origin o and the edge c lies on the x-axis. Then:

$$\begin{aligned} \mathbb {E}V_{T[2],o}^k&= \frac{1}{{\text {vol}}T}\int \limits _T ||x||^k \; dx. \end{aligned}$$

Note that ${\text {vol}}T = a c \sin \beta /2$. Using polar coordinates, we get that

$$\begin{aligned} \mathbb {E}V_{T[2],o}^k&= \frac{2}{a c \sin \beta } \int \limits _0^\alpha \int \limits _0^{l(\varphi )} r^{k+1} \; dr\; d\varphi = \frac{1}{k+2} \frac{2}{a c \sin \beta } \int \limits _0^\alpha l(\varphi )^{k+2}\; d\varphi , \end{aligned}$$

where $l(\varphi )$ denotes the length of the intersection of the triangle with the line whose angle to the x-axis is $\varphi$. By the sine rules, $l(\varphi ) = c \sin \beta /\sin (\beta + \varphi )$, and

$$\begin{aligned} \mathbb {E}V_{T[2],o}^k&= \frac{2(c \sin \beta )^{k+1}}{(k+2)a} \int \limits _0^{\alpha } \frac{1}{\sin (\beta + \varphi )^{k+2}}\; d\varphi . \end{aligned}$$

With

$$\begin{aligned} \int \frac{1}{\sin \varphi ^{k+2}}\; d\varphi = -\cos \varphi \; _2F_1\left( \frac{1}{2},\frac{k+3}{2}; \frac{3}{2}; \cos \varphi ^2\right) , \end{aligned}$$

we get that

$$\begin{aligned} \mathbb {E}V_{T[2],o}^k&= \frac{2(c \sin \beta )^{k+1}}{(k+2)a} \left( \cos \beta \; _2F_1\left( \frac{1}{2},\frac{k+3}{2}; \frac{3}{2}; \cos \beta ^2\right) \right. \\&\quad \left. - \cos (\alpha +\beta ) \; _2F_1\left( \frac{1}{2},\frac{k+3}{2}; \frac{3}{2}; \cos (\alpha +\beta )^2\right) \right) . \end{aligned}$$

$\square$

Given this lemma, we can compute the moments of the distance of a random point from a point on the boundary of a triangle.

Proposition 3

Let T be a triangle with vertices A, B and C. Denote the lengths of the edges opposite to A, B and C by a, b and c, resp., and the angles in these corners by $\alpha$, $\beta$ and $\gamma$, resp. Furthermore, let D be a point lying on the edge between A and B and $c_1$ the distance of D from A. Then it holds for the moments of the distance of a random point in T from D:

$$\begin{aligned} \mathbb {E}V_{T[2],D}^k&= \frac{2}{(k+2) bc \sin \alpha } \left( (c_1 \sin \alpha )^{k+2} \cos \alpha \; _2F_1\left( \frac{1}{2},\frac{k+3}{2}; \frac{3}{2}; \cos \alpha ^2\right) \right. \\&\quad \left. - (c_1 \sin \alpha )^{k+2} \cos (\alpha +\delta ) \; _2F_1\left( \frac{1}{2},\frac{k+3}{2}; \frac{3}{2}; \cos (\alpha +\delta )^2\right) \right. \\&\quad \left. + ((c-c_1) \sin \beta )^{k+2} \cos \beta \; _2F_1\left( \frac{1}{2},\frac{k+3}{2}; \frac{3}{2}; \cos \beta ^2\right) \right. \\&\quad \left. + ((c-c_1) \sin \beta )^{k+2} \cos (\beta -\delta ) \; _2F_1\left( \frac{1}{2},\frac{k+3}{2}; \frac{3}{2}; \cos (\beta -\delta )^2\right) \right) , \end{aligned}$$

where

$$\begin{aligned} \delta = \arcsin \left( b\sin \alpha / \sqrt{c_1^2 + b^2 - 2c_1 b \cos \alpha }\right) . \end{aligned}$$

Proof

We split T into two parts by the line segment between C and D. Denote the length of this line segment by d and the angle between AD and DC by $\delta$. Then, by cosine and sine rule,

$$\begin{aligned} d = \sqrt{c_1^2 + b^2 - 2c_1 b \cos \alpha } \quad \text { and } \quad \delta = \arcsin (b \sin \alpha /d). \end{aligned}$$

If we denote the triangle ACD by $T_1$ and the triangle BCD by $T_2$, we have:

$$\begin{aligned} \mathbb {E}V_{T[2],D}^k&= \frac{{\text {vol}}T_1}{{\text {vol}}T}\, \mathbb {E}V_{T_1[2],D}^k + \frac{{\text {vol}}T_2}{{\text {vol}}T}\, \mathbb {E}V_{T_2[2],D}^k. \end{aligned}$$

The previous lemma yields the result. $\square$

In order to find a counterexample to the monotonicity of the moments of the length of random chords, it suffices to consider one specific triangle. We choose a right-angled, isosceles one and apply Proposition 3 to this triangle.

Corollary 2

For the triangle $T_2$ with vertices (0, 0), (0, 1) and (1, 0), and $c_2 = (1/2,1/2)$, we have:

$$\begin{aligned} \mathbb {E}V_{T_2[2],c_2}^k = \frac{2^{1/2-k}}{k+2} \; _2F_1\left( \frac{1}{2},\frac{k+3}{2}; \frac{3}{2}; \frac{1}{2}\right) . \end{aligned}$$

We give some values of $\mathbb {E}V_{T_2[2],c_2}^k$ for this specific triangle:

$$\begin{aligned}&\mathbb {E}V_{T_2[2],c_2} = \frac{1}{6\sqrt{2}}(2 + \sqrt{2} {\text {arcsinh}}1) = 0.3825\ldots ,\\&\mathbb {E}V_{T_2[2],c_2}^2 = \frac{1}{6} = 0.1666\ldots ,\\&\mathbb {E}V_{T_2[2],c_2}^3 = \frac{1}{160\sqrt{2}} (14 + 3 \sqrt{2} {\text {arcsinh}}1) = 0.0783\ldots ,\\&\mathbb {E}V_{T_2[2],c_2}^4 = \frac{7}{180} = 0.0388\ldots \end{aligned}$$

3.2 Moments of the length of a random chord

In this subsection, we complete our investigations of the triangle by a computation of the moments of the length of a random chord in a triangle, where both endpoints of the chord are uniformly distributed points in the triangle.

Proposition 4

Let T be a triangle with vertices $E_1$, $E_2$ and $E_3$. Denote the length of the edge opposite to $E_i$ by $e_i$ and the angle in this corner by $\eta _i$, $i = 1,2,3$. Then it holds for the moments of the distance of two random point in T:

$$\begin{aligned} \mathbb {E}V_{T[2]}^k&= \frac{8}{(k+2)(k+3)(k+4)} \frac{1}{(e_1 e_3 \sin \eta _2)^2} \sum _{i \ne j} \sin \eta _i^{k+3} e_j^{k+4} \\&\quad \times \left[ \cos \eta _i \cos \eta _j\; _2F_1\left( \frac{1}{2},\frac{k+3}{2};\frac{3}{2};\cos \eta _j^2\right) \right. \\&\quad + \cos \eta _i^2\; _2F_1\left( \frac{1}{2},\frac{k+3}{2};\frac{3}{2};\cos \eta _i^2\right) \\&\quad \left. + \frac{\sin \eta _i}{k+2} \left( \frac{1}{\sin \eta _j^{k+2}} - \frac{1}{\sin \eta _i^{k+2}}\right) \right] .\end{aligned}$$

Proof

As before, the area of the triangle is given by $e_1 e_3 \sin \eta _2/2$. Using the affine Blaschke-Petkantschin formula—see e.g. [10]—we integrate over all lines H intersecting the triangle:

$$\begin{aligned} \mathbb {E}V_{T[2]}^k&= \frac{1}{{\text {vol}}T^2} \int \limits _{T^2} ||x_0 - x_1||^k\; d(x_0,x_1)\\&= \frac{4}{(e_1 e_3 \sin \eta _2)^2} \int \limits _{A(2,1)}\int \limits _{(H\cap T)^2} ||x_0 - x_1||^k\; d(x_0,x_1)\; dH, \end{aligned}$$

where A(2, 1) denotes the affine Grassmannian of lines in $\mathbb {R}^2$. We represent a line H by its unit normal vector $u \in S_1$ and its distance $t>0$ from the origin and we therefore denote the line by

$$\begin{aligned} H_{t,u} = \{x \in \mathbb {R}^2: \langle x,u\rangle = t\}. \end{aligned}$$

We choose the normalization of the Haar measure dH in such a way that $dH = dt\; du,$ where dt and du correspond to Lebesgue measures in $\mathbb {R}$ and $S_1$. We write the appearing integral as an expectation to get

$$\begin{aligned} \mathbb {E}V_{T[2]}^k&= \frac{4}{(e_1 e_3 \sin \eta _2)^2} \int \limits _{S_1} \int \limits _0^{\infty } \int \limits _{(H_{t,u}\cap T)^2} ||x_0-x_1||^{k+1}\; d(x_0,x_1)\; dt\; du\\&= \frac{4}{(e_1 e_3 \sin \eta _2)^2} \int \limits _{S_1} \int \limits _0^{\infty } {\text {vol}}(H_{t,u}\cap T)^2\; \mathbb {E}V_{H_{t,u}\cap T}^{k+1}\ dt\; du. \end{aligned}$$

The $(k+1)$-st moment of the distance of two random points in an interval I of length l has already been computed by Solomon [11], amongst others, to be given by

$$\begin{aligned} \mathbb {E}V_I^{k+1} = \frac{2\, l^{k+1}}{(k+2)(k+3)}. \end{aligned}$$

Hence, we obtain

$$\begin{aligned} \mathbb {E}V_{T[2]}^k&= \frac{8}{(k+2)(k+3)} \frac{1}{(e_1 e_3 \sin \eta _2)^2} \int \limits _{S_1} \int \limits _0^{\infty } {\text {vol}}(H_{t,u}\cap T)^{k+3}\; dt\; du. \end{aligned}$$

A line $H_{t,u}$ that intersects the triangle T a.s. meets exactly two edges of T. It splits T into a triangle and a quadrangle. We say that $H_{t,u}$ cuts off a vertex $E_i$ from T if $E_i$ is contained in the triangular part. Furthermore, we write

$$\begin{aligned} \mathcal {I}^{(i)} = \int \limits _{S_1} \int \limits _0^{\infty } \mathbbm {1}(H_{t,u} \text { cuts off } E_i \text { from } T)\;{\text {vol}}(H_{t,u}\cap T)^{k+3}\; dt\; du \end{aligned}$$

for $i = 1,2,3$, which gives

$$\begin{aligned} \mathbb {E}V_{T[2]}^k = \frac{8}{(k+2)(k+3)} \frac{1}{(e_1 e_3 \sin \eta _2)^2} \Big ( \mathcal {I}^{(1)} + \mathcal {I}^{(2)} + \mathcal {I}^{(3)} \Big ). \end{aligned}$$

We state the following lemma, which will be proved right after the end of the proof of this proposition:

Lemma 4

It holds for $i = 1,2,3$:

$$\begin{aligned} \mathcal {I}^{(i)}&= \frac{\sin \eta _i^{k+3}}{k+4} \left( e_j^{k+4} \left[ \cos \eta _i \cos \eta _j\; _2F_1\left( \frac{1}{2},\frac{k+3}{2};\frac{3}{2};\cos \eta _j^2\right) \right. \right. \\&\quad + \cos \eta _i^2\; _2F_1\left( \frac{1}{2},\frac{k+3}{2};\frac{3}{2};\cos \eta _i^2\right) \\&\quad \left. \left. + \frac{\sin \eta _i}{k+2} \left( \frac{1}{\sin \eta _j^{k+2}} - \frac{1}{\sin \eta _i^{k+2}}\right) \right] \right. \\&\quad + \left. e_l^{k+4} \left[ \cos \eta _i \cos \eta _l\; _2F_1\left( \frac{1}{2},\frac{k+3}{2};\frac{3}{2};\cos \eta _l^2\right) \right. \right. \\&\quad + \cos \eta _i^2\; _2F_1\left( \frac{1}{2},\frac{k+3}{2};\frac{3}{2};\cos \eta _i^2\right) \\&\quad \left. \left. + \frac{\sin \eta _i}{k+2} \left( \frac{1}{\sin \eta _l^{k+2}} - \frac{1}{\sin \eta _i^{k+2}}\right) \right] \right) . \end{aligned}$$

This lemma immediately yields the result. $\square$

Proof of Lemma 4

Let $i \in \{1,2,3\}$. We substitute $u = (\cos \varphi , \sin \varphi )$ such that the line $\varphi = 0$ contains the edge between $E_i$ and $E_j$ for $j \ne i$ fixed, and by the well-known fact that $du = d\varphi$ and with $H_{t,\varphi } = \{(x,y) \in \mathbb {R}^2: x\cos \varphi + y\sin \varphi = t\}$, we get the integral

$$\begin{aligned} \mathcal {I}^{(i)} = \int \limits _{-\pi }^\pi \int \limits _0^{\infty } \mathbbm {1}(H_{t,\varphi } \text { cuts off } E_i \text { from } T)\;{\text {vol}}(H_{t,\varphi }\cap T)^{k+3}\; dt\; d\varphi . \end{aligned}$$

Let $l \in \{1,2,3\}{\setminus }\{i,j\}$. A line $H_{t,\varphi }$ cuts off $E_i$ from T if and only if $\varphi \in [\eta _i-\pi /2,\pi /2]$ and

• (i) $0 \le t \le e_j\cos (\eta _1-\varphi )$,   if $\varphi \le \frac{\pi }{2} - \eta _j,$

• (ii) $0 \le t \le e_l\cos \varphi$,   if $\varphi \ge \frac{\pi }{2} - \eta _j$.

We use the notation $l(t,\varphi )={\text {vol}}(H_{t,\varphi }\cap T)$ for the length of the intersection of T with a line $H_{t,\varphi }$. Basic trigonometric considerations show that

$$\begin{aligned} l(t,\varphi ) = t (\tan (\eta _i-\varphi ) + \tan \varphi ) = t \frac{\sin \eta _i}{\cos \varphi \cos (\eta _i-\varphi )}. \end{aligned}$$

This leads to the expression

$$\begin{aligned} \mathcal {I}^{(i)}&= \int \limits _{\eta _i-\pi /2}^{\pi /2-\eta _j} \left( \frac{\sin \eta _i}{\cos \varphi \cos (\eta _i-\varphi )}\right) ^{k+3} \int \limits _0^{e_j\cos (\eta _i-\varphi )} t^{k+3}\; dt\; d\varphi \\&\quad + \int \limits _{\pi /2-\eta _j}^{\pi /2} \left( \frac{\sin \eta _i}{\cos \varphi \cos (\eta _i-\varphi )}\right) ^{k+3} \int \limits _0^{e_l\cos \varphi } t^{k+3}\; dt\; d\varphi \\&= \frac{\sin \eta _i^{k+3}}{k+4} \left( e_j^{k+4} \int \limits _{\eta _i-\pi /2}^{\pi /2-\eta _j} \frac{\cos (\eta _i-\varphi )}{\cos \varphi ^{k+3}} \; d\varphi + e_l^{k+4} \int \limits _{\pi /2-\eta _j}^{\pi /2} \frac{\cos \varphi }{\cos (\eta _i-\varphi )^{k+3}}\; d\varphi \right) . \end{aligned}$$

With a substitution $\psi = \eta _i-\varphi$ and with $\cos (\eta _i-\varphi ) = \cos \eta _i \cos \varphi + \sin \eta _i \sin \varphi$, we can compute the outer integral:

$$\begin{aligned} \mathcal {I}^{(i)}&= \frac{\sin \eta _i^{k+3}}{k+4} \left( e_j^{k+4} \left[ \cos \eta _i \cos \eta _j\; _2F_1\left( \frac{1}{2},\frac{k+3}{2};\frac{3}{2};\cos \eta _j^2\right) \right. \right. \\&\quad + \cos \eta _i^2\; _2F_1\left( \frac{1}{2},\frac{k+3}{2};\frac{3}{2};\cos \eta _i^2\right) \\&\quad \left. \left. + \frac{\sin \eta _i}{k+2} \left( \frac{1}{\sin \eta _j^{k+2}} - \frac{1}{\sin \eta _i^{k+2}}\right) \right] \right. \\&\quad + \left. e_l^{k+4} \left[ \cos \eta _i \cos \eta _l\; _2F_1\left( \frac{1}{2},\frac{k+3}{2};\frac{3}{2};\cos \eta _l^2\right) \right. \right. \\&\quad + \cos \eta _i^2\; _2F_1\left( \frac{1}{2},\frac{k+3}{2};\frac{3}{2};\cos \eta _i^2\right) \\&\quad \left. \left. + \frac{\sin \eta _i}{k+2} \left( \frac{1}{\sin \eta _l^{k+2}} - \frac{1}{\sin \eta _i^{k+2}}\right) \right] \right) . \end{aligned}$$

$\square$

Again, we apply this proposition to the right-angled, isosceles triangle.

Corollary 3

For the triangle $T_2$ with vertices (0, 0), (0, 1) and (1, 0), we have:

$$\begin{aligned} \mathbb {E}V_{T_2[2]}^k = \frac{2^{(5-k)/2}+2^{7/2}}{(k+2)(k+3)(k+4)} \; _2F_1\left( \frac{1}{2},\frac{k+3}{2}; \frac{3}{2}; \frac{1}{2}\right) . \end{aligned}$$

As before, we give some values of $\mathbb {E}V_{T_2[2]}^k$ for this triangle:

$$\begin{aligned}&\mathbb {E}V_{T_2[2]} = \frac{1+2\sqrt{2}}{30}(2 + \sqrt{2} {\text {arcsinh}}1) = 0.4142\ldots ,\\&\mathbb {E}V_{T_2[2]}^2 = \frac{2}{9} = 0.2222\ldots ,\\&\mathbb {E}V_{T_2[2]}^3 = \frac{1+4\sqrt{2}}{840}(14 + 3\sqrt{2} {\text {arcsinh}}1) = 0.1405\ldots ,\\&\mathbb {E}V_{T_2[2]}^4 = \frac{1}{10} = 0.1. \end{aligned}$$

4 Random triangles in a tetrahedron

In this section, our aim is to show that for a specific tetrahedron T and the center c of one of its facets, we have

$$\begin{aligned} \mathbb {E}V_{T[3]} > \mathbb {E}V_{T[3],c}. \end{aligned}$$

(2)

Therefore, we want to find polynomials $P(x) = \sum _{i=0}^n a_i x^{i}$ and $Q(x) = \sum _{i=0}^l b_i x^{i}$ such that $P(x) \le |x|$ and $Q(x) \ge |x|$ for relevant ranges of values $x \in \mathbb {R}$ and consequently

$$\begin{aligned} \mathbb {E}V_{T[3]} \ge \mathbb {E}P(V_{T[3]}) > \mathbb {E}Q(V_{T[3],c}) \ge \mathbb {E}V_{T[3],c}, \end{aligned}$$

which suffices in order to show that the pair (T, c) is a counterexample to monotonicity of the expected area of a random triangle in dimension three. Simulations suggest that

$$\begin{aligned} \mathbb {E}V_{T[3]} = 0.0592... \ \text { and } \ \mathbb {E}V_{T[3],c} = 0.0466... \end{aligned}$$

(3)

and conjecture (2) should hold true.

4.1 A lower bound for the expected area of a random triangle in a tetrahedron

Let T be a tetrahedron. For random points $X_0,X_1,X_2 \in T$, we write $X_i = (x_i,y_i,z_i)$. The volume of the triangle with vertices $X_0,X_1$ and $X_2$ is given by

$$\begin{aligned} V_{T[3]}&= \frac{1}{2}\sqrt{\det \left( \begin{pmatrix} x_1-x_0 &{} y_1-y_0 &{} z_1-z_0\\ x_2-x_0 &{} y_2-y_0 &{} z_2-z_0 \end{pmatrix} \begin{pmatrix} x_1-x_0 &{} x_2-x_0 \\ y_1-y_0 &{} y_2-y_0 \\ z_1-z_0 &{} z_2-z_0 \end{pmatrix} \right) }\\&\quad = \frac{1}{2}\, D(x_0,\ldots ,z_2)^{1/2}, \end{aligned}$$

and this is the square root of a polynomial of degree four in the coordinates of $X_0,X_1$ and $X_2$. For even moments of $V_{T[3]}$, we get rid of the square root:

$$\begin{aligned} \mathbb {E}V_{T[3]}^{2k}&= {\text {vol}}T^{-3}\, 2^{-2k} \int \limits _T \int \limits _T \int \limits _T \ D(x_0, \dots , z_2)^{k} \ d(x_0,y_0,z_0)\, d(x_1,y_1,z_1)\, d(x_2,y_2,z_2). \end{aligned}$$

(4)

Note that the expectation $\mathbb {E}V_{T[3]}$ depends on the shape of the tetrahedron T. We consider the specific tetrahedron

$$\begin{aligned} T_3 = \{(x,y,z)\in \mathbb {R}^3: x,y,z \ge 0, x+y+z\le 1\}, \end{aligned}$$

i.e., that with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1). Expanding the determinant, the polynomial D can be written as

$$\begin{aligned} D(x_0,\ldots ,z_2)&= \sum _{i=1}^{63} s_i \ \text { with } \ s_i = \varepsilon _i\, 2^{c_i}\, x_0^{e_{i0}}\cdots z_2^{e_{i8}}, \end{aligned}$$

where $\varepsilon _i \in \{-1,1\}$, $c_i \in \{0,1\}$ and $e_{ij} \ge 0$, $j=0,\ldots ,8$, are explicitly given constants and $\sum _j e_{ij} = 4$ for $i=1,\ldots ,63$. By the Multinomial Theorem, and using the multinomial coefficient

$$\begin{aligned} \left( {\begin{array}{c}k\\ k_1,\ldots ,k_{63}\end{array}}\right) = \frac{k!}{k_1! \cdots k_{63}!}, \end{aligned}$$

the k-th power of it can be rewritten as

$$\begin{aligned} D(x_0,\ldots ,z_2)^k&= \sum _{\sum _1^{63}k_i =k} \left( {\begin{array}{c}k\\ k_1,\ldots ,k_{63}\end{array}}\right) \, \prod _{i=1}^{63} s_i^{k_i} \nonumber \\&= \sum _{\sum _1^{63}k_i =k} (-1)^{k^\prime } 2^{k^{\prime \prime }} \left( {\begin{array}{c}k\\ k_1,\ldots ,k_{63}\end{array}}\right) \prod _{i=0}^2 x_i^{l_1} y_i^{m_1} z_i^{n_1}. \end{aligned}$$

(5)

For completeness, we list the exponents appearing in (5):

$$\begin{aligned} k^\prime&= k_{19} + \cdots + k_{37} + k_{40} + k_{44} + k_{47} + k_{51} + k_{54} + k_{55} + k_{59} + k_{63},\\ k^{\prime \prime }&= k_{19}+\cdots + k_{63},\\ l_0&= 2 k_1 + \cdots + 2 k_4 + 2 k_{19} + 2 k_{20} + k_{27} + k_{29} + k_{33} + k_{35} + k_{37} + \cdots + k_{48}, \\ m_0&= 2 k_5 + 2 k_9 + 2 k_{13} + 2 k_{14} + k_{21} + k_{23} + 2 k_{25} + 2 k_{26} + k_{34} + k_{36} + \cdots + k_{38} \\&\quad + k_{43} + k_{44} + k_{49} + k_{50} + k_{55} + \cdots + k_{60}, \\ n_0&= 2 k_7 + 2 k_{11} + 2 k_{15} + 2 k_{17} + k_{22} + k_{24} + k_{28} + k_{30} + 2 k_{31} + 2 k_{32} + k_{40} + k_{41} \\&\quad + k_{46} + k_{47} + k_{52} + k_{53} + k_{55} + k_{56} + k_{58} + k_{59} + k_{61} + k_{62}, \\ l_1&= 2 k_5 + \cdots + 2 k_8 + 2 k_{21} + 2 k_{22} + k_{25} + k_{29} + k_{31} + k_{35} + k_{37} + \cdots + k_{42}\\&\quad + k_{49} + \cdots + k_{54}, \\ m_1&= 2 k_1 + 2 k_{10} + 2 k_{15} + 2 k_{16} + k_{19} + k_{23} + 2 k_{27} + 2 k_{28} + k_{32} + k_{36} + k_{37} + k_{39} \\&\quad + k_{43} + k_{45} + k_{49} + k_{51} + k_{55} +\cdots + k_{57} + k_{61} + \cdots + k_{63}, \\ n_1&= 2 k_3 + 2 k_{12} + 2 k_{13} + 2 k_{18} + k_{20} + k_{24} + k_{26} + k_{30} + 2 k_{33} + 2 k_{34} + k_{40} + k_{42} \\&\quad + k_{46} + k_{48} + k_{52} + k_{54} + k_{55} + k_{57} + k_{58} + k_{60} + \cdots + k_{63}, \\ l_2&= 2 k_9 + 2 k_{10} + 2 k_{11} + 2 k_{12} + 2 k_{23} + 2 k_{24} + k_{25} + k_{27} + k_{31} + k_{33} + k_{43}\\&\quad \ + \cdots + k_{54}, \\ m_2&= 2 k_2 + 2 k_6 + 2 k_{17} + 2 k_{18} + k_{19} + k_{21} + 2 k_{29} + 2 k_{30} + k_{32} + k_{34} + k_{38} + k_{39} \\&\quad + k_{44} + k_{45} + k_{50} + k_{51} + k_{58} + \cdots + k_{63}, \\ n_2&= 2 k_4 + 2 k_8 + 2 k_{14} + 2 k_{16} + k_{20} + k_{22} + k_{26} + k_{28} + 2 k_{35} + 2 k_{36} + k_{41} + k_{42} \\&\quad + k_{47} + k_{48} + k_{53} + k_{54} + k_{56} + k_{57} + k_{59} + k_{60} + k_{62} + k_{63}. \end{aligned}$$

Integration of the monomials over the tetrahedron $T_3$ gives with the substitution $z=t$, $y=s\,(1-t)$, $x=r\,(1-s)\,(1-t)$ that

$$\begin{aligned} \int \limits _{T_3} x^{l_i}\, y^{m_i}\, z^{n_i}\ d(x,y,z)= & {} \int \limits _0^1 r^{l_i}\; dr \int \limits _0^{1} s^{m_i}\, (1-s)^{l_i+1}\,ds \int \limits _0^{1} t^{n_i}\, (1-t)^{l_i+m_i+2}\,dt \nonumber \\= & {} \frac{1}{l_i+1}\, B(m_i+1,l_i+2)\, B(n_i+1, l_i+m_i+3)\nonumber \\= & {} \frac{l_i!\,m_i!\,n_i!}{(l_i+m_i+n_i+3)!}. \end{aligned}$$

(6)

Combining this with Eqs. (4) and (5) gives

$$\begin{aligned} \mathbb {E}V_{T_3[3]}^{2k}&= 6^3\, 2^{-2k} \sum _{\sum _1^{63}k_i=k} (-1)^{k^\prime } 2^{k^{\prime \prime }} \left( {\begin{array}{c}k\\ k_1,\ldots ,k_{63}\end{array}}\right) \, \prod _{i=0}^2 \frac{l_i!\,m_i!\,n_i!}{(l_i+m_i+n_i+3)!}. \end{aligned}$$

We list the first five even moments of the area of a random triangle in our tetrahedron $T_3$:

$$\begin{aligned} \mathbb {E}V_{T_3[3]}^2&= \frac{9}{1\,600} = 5.625\cdot 10^{-3},\\ \mathbb {E}V_{T_3[3]}^4&= \frac{27}{196\,000} \approx 1.37755\cdot 10^{-4},\\ \mathbb {E}V_{T_3[3]}^6&= \frac{3\,161}{379\,330\,560} \approx 8.3331\cdot 10^{-6},\\ \mathbb {E}V_{T_3[3]}^8&= \frac{93\,957}{106\,247\,680\,000} \approx 8.8432\cdot 10^{-7},\\ \mathbb {E}V_{T_3[3]}^{10}&= \frac{209\,022\,679}{1\,551\,386\,124\,288\,000} \approx 1.34733\cdot 10^{-7}. \end{aligned}$$

We have seen that the (2k)-th moment of $V_{T_3[3]}$ can be computed, with fast increasing complexity in k. Also note that the area of a triangle in $T_3$ is not larger than $\sqrt{3}/2$, which is the area of the facet $\{(x,y,z)\in \mathbb {R}^3: x,y,z\ge 0, x+y+z= 1\}$. Hence, we want to approximate the square root function in the interval [0, 3 / 4] by a polynomial

$$\begin{aligned} P(x) = \sum _{i=0}^n a_i x^{i} \end{aligned}$$

of degree n for some $n\in \mathbb {N}$ such that $P(x) \le \sqrt{x}$ for all $x \in [0,\frac{3}{4}]$. This way, we can approximate $\mathbb {E}V_{T_3[3]}$ from below by the polynomial P:

$$\begin{aligned} \mathbb {E}V_{T_3[3]} \ge \mathbb {E}P(V_{T_3[3]}). \end{aligned}$$

Moreover, the best polynomial for fixed $n\in \mathbb {N}$ can be found via the linear optimization problem

$$\begin{aligned} \max _P \ \mathbb {E}P(V_{T_3[3]})&= \max _{a_i}\ \sum _{i=0}^n a_{i}\, \mathbb {E}P(V_{T_3[3]})^{2i} \quad \text {s.t.} \quad P\left( x^2\right) \le x, \; x\in \left[ 0,\frac{\sqrt{3}}{2}\right] . \end{aligned}$$

This constraint is infinite dimensional, but we get an upper bound on $\mathbb {E}P(V_{T_3[3]})$ via the finite dimensional linear program

$$\begin{aligned} \max _P \mathbb {E}P(V_{T_3[3]})= & {} \max _{a_i} \sum _{i=0}^n a_{i}\, \mathbb {E}V_{T_3[3]}^{2i} \quad \text {s.t.} \quad P\left( x_{\ell }^2\right) \le x_{\ell }, \; x_{\ell }\in \left[ 0,\frac{\sqrt{3}}{2}\right] ,\\&\ell =0,\ldots ,L. \end{aligned}$$

For $n=6$ and $L=200$ equidistant points $x_{\ell }\in [0,\frac{\sqrt{3}}{2}]$, we numerically compute via Matlab and the optimization toolbox CVX [2] that

$$\begin{aligned} \mathbb {E}P(V_{T_3[3]}) < 0.04647, \end{aligned}$$

and in consideration of the simulations in (3), this yields that we do not get a sufficiently precise estimate using only $n=6$ even moments.

We solve the above linear program for $n=7$ and $L=1\,000$, get the squares of three interpolation nodes with the square root function in $(0,\frac{\sqrt{3}}{2})$ and rationalize these points to

$$\begin{aligned} \{x_1,x_2,x_3\} = \left\{ \frac{2}{19}, \frac{4}{15}, \frac{8}{17}\right\} . \end{aligned}$$

Additionally, we take the points $x_0 = 0$ and $x_4 = \frac{47}{54} > \frac{\sqrt{3}}{2}$ as interpolation nodes of our desired polynomial. We want to use these five points to get a polynomial of degree 7. Consider a variation of Lemma 2 in [4] with two conditions

$$\begin{aligned} P(x_j^2)=x_j \ \text { and } \ P^\prime (x_j^2) = \frac{1}{2x_j} \end{aligned}$$

for $j = 1,2,3$ and the single conditions $P(x_0^2)=P(0)=0$ and $P(x_4^2)=x_4$. Then we have with $f(x)=\sqrt{x}$ that

$$\begin{aligned} f(x)-P(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!} x\,(x-x_4^2)\prod _{j=1}^3 (x-x_j^2)^2 \ge 0 \end{aligned}$$

for some $\xi \in [0,x_4^2]$, since $f^{(n+1)}(x)<0$, $x \ge 0$ and $x \le x_4$ for all $x \in [0,\frac{3}{4}]$.

We solve the interpolation problem and finally get a polynomial $P_\mathrm{cert}(x)= \sum _{i=1}^{7} a_i x^{i}$ of degree 7 with explicitly given rational coefficients $a_1, \dots , a_7$ and the property $P_\mathrm{cert} \le \sqrt{x}$, and this polynomial yields with usage of the even moments computed above that

$$\begin{aligned}&\mathbb {E}V_{T_3[3]}\ge \mathbb {E}P_\mathrm{cert}(V_{T_3[3]})\nonumber \\&\quad = \frac{91\,096\,443\,868\,688\,279\,627\,145\,529\,306\,423\,486\,222\,767\,653\,883\,418\,444\,589}{1\,940\,591\,908\,276\,502\,542\,424\,150\,723\,573\,750\,536\,040\,094\,315\,231\,313\,920\,000}\nonumber \\&\quad > 0.046942. \end{aligned}$$

(7)

4.2 An upper bound for the expected area of a random triangle in a tetrahedron where one point is fixed

Now we consider the convex hull of two random points $X_1, X_2$ and one fixed point in a tetrahedron T. We write again $X_i = (x_i,y_i,z_i)$ and choose the centroid $c = (x_c,y_c,z_c)$ of one of the facets of T as third point. The volume of the triangle with vertices $X_1,X_2$ and c is given by

$$\begin{aligned} V_{T[3],c}&= \frac{1}{2}\sqrt{\det \left( \begin{pmatrix} x_1-x_c &{} y_1-y_c &{} z_1-z_c\\ x_2-x_c &{} y_2-y_c &{} z_2-z_c \end{pmatrix} \begin{pmatrix} x_1-x_c &{} x_2-x_c \\ y_1-y_c &{} y_2-y_c \\ z_1-z_c &{} z_2-z_c \end{pmatrix} \right) } \\&\quad = \frac{1}{2}\, D(x_0,\ldots ,z_c)^{1/2}, \end{aligned}$$

and this is again the square root of a polynomial of degree four in the coordinates of $X_1,X_2$ and c. As before, we have:

$$\begin{aligned} \mathbb {E}V_{T[3],c}^{2k}&= {\text {vol}}T^{-2}\, 2^{-2k} \int \limits _T \int \limits _T \ D(x_1, \dots , z_c)^{k} \ d(x_1,y_1,z_1)\, d(x_2,y_2,z_2). \end{aligned}$$

(8)

The expectation $\mathbb {E}V_{T[3],c}$ particularly depends on the facet whose centroid we choose to be the fixed vertex. We take again the tetrahedron $T_3$ as defined in Sect. 4.1 and choose $c_3=(1/3,1/3,1/3)$, the centroid of the facet $\{(x,y,z)\in \mathbb {R}^3: x,y,z\ge 0, x+y+z= 1\}$, to be the fixed vertex of the random triangle.

Expanding the determinant and inserting the coordinates of $c_3$, the polynomial D can be displayed as

$$\begin{aligned} D(x_1,\ldots ,z_{c_3})&= \frac{1}{9} \sum _{i=1}^{54} s_i \quad \text { with } \quad s_i = \varepsilon _i\, 2^{c_i}\, 3^{d_i}\, x_1^{e_{i0}}\cdots z_2^{e_{i5}}, \end{aligned}$$

where $\varepsilon _i \in \{-1,1\}$, $c_i, d_i \in \{0,1,2\}$ and $e_{ij} \ge 0$, $j=0,\ldots ,5$, are explicitly given constants and $\sum _j e_{ij} \le 4$ for $i=1,\ldots ,54$. Again with usage of the Multinomial Theorem, the series representation of the k-th power of it is given by

$$\begin{aligned} D(x_1,\ldots ,z_{c_3})^k&= 3^{-2k} \sum _{\sum _1^{54}k_i =k} \left( {\begin{array}{c}k\\ k_1,\ldots ,k_{54}\end{array}}\right) \, \prod _{i=1}^{54} s_i^{k_i} \nonumber \\&= 3^{-2k} \sum _{\sum _1^{54}k_i =k} (-1)^{k^\prime } 2^{k^{\prime \prime }} 3^{k^{\prime \prime \prime }} \left( {\begin{array}{c}k\\ k_1,\ldots ,k_{54}\end{array}}\right) \prod _{i=1}^2 x_i^{l_1} y_i^{m_1} z_i^{n_1}, \end{aligned}$$

(9)

where we use abbreviations analogously as in the previous proof:

$$\begin{aligned} k^\prime&= k_7 + \cdots + k_{10} + k_{12} + k_{15} + k_{17} + k_{18} + k_{21} + \cdots + k_{33} + k_{52} + \cdots + k_{54},\\ k^{\prime \prime }&= k_1+\cdots k_6 + 2k_7 + \cdots + 2k_9 + k_{10} + \cdots + k_{45} + k_{52} + \cdots + k_{54},\\ k^{\prime \prime \prime }&= k_{22} + \cdots + k_{45} + 2k_{46} + \cdots + 2k_{54},\\ l_1&= 2k_1 + k_7 + k_{10} + \cdots + k_{13} + 2k_{22} + 2k_{23} + k_{28} + k_{32} + k_{34} + \cdots + k_{39}\\&\quad + 2k_{46} + 2k_{47}+ k_{52}+ k_{53},\\ m_1&= 2k_3 + k_8 + k_{10} + k_{14} + k_{18} + k_{19} + k_{24} + 2k_{26} + 2k_{27} + k_{33} + k_{34}\\&\quad + k_{38} + k_{40} + k_{42} + \cdots + k_{44} + 2k_{48} + 2k_{50}+ k_{52}+ k_{54},\\ n_1&= 2k_5 + k_9 + k_{12} + k_{16} + k_{18} + k_{20} + k_{25} + k_{29} + 2k_{30} + 2k_{31} + k_{36} \\&\quad + k_{39} + k_{41} + k_{42} + k_{44} + k_{45} + 2k_{49}+ 2k_{51}+ k_{53}+ k_{54},\\ l_2&= 2k_2 + k_7 + k_{14} + \cdots + k_{17} + 2k_{24} + 2k_{25} + k_{26} + k_{30} + k_{34} \\&\quad + \cdots + k_{37} + k_{40} + k_{41} + 2k_{48} + 2k_{49}+ k_{52}+ k_{53},\\ m_2&= 2k_4 + k_8 + k_{11} + k_{15} + k_{20} + \cdots + k_{22} + 2k_{28} + 2k_{29} + k_{31} + k_{35} \\&\quad + k_{38} + k_{40} + k_{42} + k_{43} + k_{45} + 2k_{46}+ 2k_{51}+ k_{52}+ k_{54},\\ n_2&= 2k_6 + k_9 + k_{13} + k_{17} + k_{19} + k_{21} + k_{23} + k_{27} + 2k_{32} + 2k_{33} + k_{37} \\&\quad + k_{39} + k_{41} + k_{43} + k_{44} + k_{45} + 2k_{47}+ 2k_{50}+ k_{53}+ k_{54}. \end{aligned}$$

Now, Eqs. (6), (8) and (9) give

$$\begin{aligned} \mathbb {E}V_{T_3[3],{c_3}}^{2k} = 6^{2-2k} \sum _{\sum _1^{54}k_i=k} (-1)^{k^\prime } 2^{k^{\prime \prime }} 3^{k^{\prime \prime \prime }} \left( {\begin{array}{c}k\\ k_1,\ldots ,k_{54}\end{array}}\right) \, \prod _{i=0}^2 \frac{l_i!\,m_i!\,n_i!}{(l_i+m_i+n_i+3)!}. \end{aligned}$$

Again, we list the first five even moments of the area of a random triangle in our tetrahedron $T_3$, where one point lies in $c_3$:

$$\begin{aligned} \mathbb {E}V_{T_3[3],c_3}^2&= \frac{7}{2\, 400} \approx 2.91667\cdot 10^{-3},\\ \mathbb {E}V_{T_3[3],c_3}^4&= \frac{11}{529\,200} \approx 2.07861\cdot 10^{-5},\\ \mathbb {E}V_{T_3[3],c_3}^6&= \frac{2\,839}{10\,973\,491\,200} \approx 2.58714\cdot 10^{-7},\\ \mathbb {E}V_{T_3[3],c_3}^8&= \frac{29\,419}{6\,224\,027\,040\,000} \approx 4.72668\cdot 10^{-9},\\ \mathbb {E}V_{T_3[3],c_3}^{10}&= \frac{4\,134\,139}{36\,352\,301\,290\,905\,600} \approx 1.13724\cdot 10^{-10}. \end{aligned}$$

Note that the area of a triangle in $T_3$, where one vertex is fixed to lie in $c_3$, is not larger than $\sqrt{3}/6$. Hence, we want to approximate the square root function in the interval [0, 1 / 12] by a polynomial

$$\begin{aligned} Q(x) = \sum _{i=0}^l b_i x^{i} \end{aligned}$$

of degree $l\in \mathbb {N}$ such that $Q(x) > \sqrt{x}$ for all $x \in [0,1/12]$. Then we have that

$$\begin{aligned} \mathbb {E}V_{T_3[3],c_3} \le \mathbb {E}Q(V_{T_3[3],c_3}). \end{aligned}$$

We use Lemma 2 in [4], which yields that for $m\in \mathbb {N}$, $l=2m+1$ and $0<x_0<\ldots <x_m$, the system of equations

$$\begin{aligned} Q(x_j^2)=x_j \text { and } Q'(x_j^2)=\frac{1}{2 x_j} \quad \text { for } j=0,\ldots ,m \end{aligned}$$

determines uniquely a polynomial $Q(x)=\sum _{i=0}^l b_i x^{i}$ with the property $Q(x)\ge \sqrt{x}$ for all $x\in \mathbb {R}$. We get a lower bound on $\mathbb {E}Q(V_{T_3[3],c_3})$ via the finite dimensional linear program

$$\begin{aligned} \min _Q \mathbb {E}Q(V_{T_3[3],c_3})= & {} \min _{b_i} \sum _{i=0}^l b_{i}\, \mathbb {E}V_{T_3[3],c_3}^{2i} \quad \text {s.t.} \quad Q\left( x_{\ell }^2\right) \ge x_{\ell }, \; x_{\ell }\in \left[ 0,\frac{\sqrt{3}}{6}\right] ,\\&\ell =0,\ldots ,L. \end{aligned}$$

For $l=14$ and $L=200$ equidistant points $x_{\ell }\in [0,\frac{\sqrt{3}}{6}]$, we numerically compute via Matlab and CVX that

$$\begin{aligned} \mathbb {E}Q(V_{T_3[3],c_3}) > 0.04699, \end{aligned}$$

and this approximation is not sufficient with consideration of (7).

For $l=15$ and $L=1\,000$, we solve the above linear program, compute the interpolation nodes with the square root function numerically, and rationalize these points to

$$\begin{aligned} \{x_0,\ldots ,x_7\} = \left\{ \frac{1}{45}, \frac{1}{17}, \frac{1}{11}, \frac{1}{8}, \frac{1}{6}, \frac{1}{5}, \frac{3}{13}, \frac{7}{27}\right\} . \end{aligned}$$

Using these points, we solve the interpolation problem from Lemma 2 in [4] and finally get a polynomial $Q_\mathrm{cert}(x)= \sum _{i=0}^{15} b_i x^{i}$ of degree 15 with explicitly given rational coefficients $b_0, \dots , b_{15}$ and the property $Q_\mathrm{cert} \ge \sqrt{x}$. Inserting the even moments computed above, we see that

5 Proof of Theorem 2

5.1 Some basic considerations

In the first part of the proof, we use the results on the monotonicity of the moments of volumes of random full-dimensional simplices, which have already been stated in Thorem 1. It follows with Lemma 2 that, for $d \ge 2$ and $2\le n \le d$, monotonicity of the map $K \rightarrow \mathbb {E}V_{K[n]}^k$ for K ranging over all d-dimensional convex bodies K does not hold unless $n = 2$ and $k \in \mathbb {N}$ or $n =3$ and $k \in \{1,2\}$.

5.2 Proof of Proposition 1

Now we give a counterexample for $d \ge 2$ and $n=2$. Here we prove that random chords in a specific right-angled triangle are suitable for our purposes by giving the explicit values.

We consider the ratio $r(k):=\mathbb {E}V_{T_2[2],c_2}^k/\mathbb {E}V_{T_2[2]}^k$ for the triangle $T_2$ with vertices (0, 0), (0, 1) and (1, 0) and $c_2 = (1/2,1/2)$ the midpoint of its hypotenuse. It holds:

$$\begin{aligned} r(k)&= \frac{(k+3)(k+4)}{2^{k+3} + 2^{k/2+2}}. \end{aligned}$$

The ratio r(k) is clearly monotonely decreasing, and r(1) is smaller than 1. Therefore, r(k) is smaller than 1 for each $k \in \mathbb {N}$ and Proposition 1 holds.

5.3 Proof of Proposition 2

We can settle the case $n=3, k=2$ by comparing the second moments of the areas of our two different types of random triangles in a tetrahedron directly. We computed in Sect. 4:

$$\begin{aligned} \mathbb {E}V_{T_3[3],c_3}^2 = \frac{7}{2\, 400} = 0.002916... < 0.005625 = \frac{9}{1\,600} = \mathbb {E}V_{T_3[3]}^2. \end{aligned}$$

Consequently, with Lemmata 1 and 2, we have shown that the map $K \mapsto \mathbb {E}V_{K[3]}^2$ is not monotone if K ranges over all d-dimensional convex bodies for $d \ge 3$.

In the same section, we have shown that

$$\begin{aligned} \mathbb {E}V_{T_3[3],c_3}< 0.046942 < \mathbb {E}V_{T_3[3]}. \end{aligned}$$

Hence, we also have a counterexample for $k=1$, $n=3$ and $d \ge 3$, and the proof of Proposition 2 as well as that of Theorem 2 are completed.

6 Proof of Theorem 3

Let $d\ge 3$, $2 \le n \le d$ and $k \in \mathbb {N}$. If $d>3$, $n \ne 3$ or $k > 2$, there exists a pair of convex bodies $K, L \subseteq \mathbb {R}^{d-1}$ such that $K \subset L$, but $\mathbb {E}V_{K[n]}^k > \mathbb {E}V_{L[n]}^k$. As in the proof of Lemma 2, for $\varepsilon > 0$, we define two d-dimensional convex bodies by

$$\begin{aligned} K_{\varepsilon } :=K \times [0,\varepsilon ], \quad L_{\varepsilon } :=L \times [0,\varepsilon ]. \end{aligned}$$

In the sequel, we will show that $\mathbb {E}\mathcal V_{K_\varepsilon [n]}^k$ and $\mathbb {E}\mathcal V_{L_\varepsilon [n]}^k$ converge to $\mathbb {E}V_{K[n]}^k$ and $\mathbb {E}V_{L[n]}^k$, resp., for $\varepsilon$ tending to zero. Therefore, there exists an $\varepsilon _0 > 0$ such that $\mathbb {E}\mathcal V_{K_{\varepsilon _0}[n]}^k > \mathbb {E}\mathcal V_{L_{\varepsilon _0}[n]}^k$, but $K_{\varepsilon _0} \subseteq L_{\varepsilon _0}$.

Let $X_0,\ldots , X_{n-1}$ be random points in ${\text {bd}}K_\varepsilon$ and

$$\begin{aligned} N_0 = \#\{X_i \in K\times \{0,\varepsilon \}\}, \quad N_1 = \#\{X_i \in {\text {bd}}K\times [0,\varepsilon ]\}. \end{aligned}$$

Then $N_0+N_1=n$. It holds for $0\le i \le n-1$:

$$\begin{aligned} \mathbb {P}(X_i \in K\times \{0,\varepsilon \})&= \frac{2{\text {vol}}_{d-1} K}{2{\text {vol}}_{d-1} K + S(K)\,\varepsilon }, \\ \mathbb {P}(X_i \in {\text {bd}}K\times [0,\varepsilon ])&= \frac{S(K)\,\varepsilon }{2{\text {vol}}_{d-1} K + S(K)\,\varepsilon }, \end{aligned}$$

if S(K) is the surface area of the boundary of K. Therefore, we have that

$$\begin{aligned} \mathbb {E}\mathcal {V}_{K_\varepsilon [n]}^k&= \mathbb {P}(N_1=0)\, \mathbb {E}\mathcal {V}_{K\times \{0,\varepsilon \}[n]}^k + \mathbb {P}(N_1>0)\, \mathbb {E}_{N_1> 0} \mathcal {V}_{K_\varepsilon [n]}^k \nonumber \\&= \frac{2^n {\text {vol}}_{d-1} K^n}{(2{\text {vol}}_{d-1} K + S(K)\varepsilon )^n}\, \mathbb {E}\mathcal {V}_{K\times \{0,\varepsilon \}[n]}^k\nonumber \\&\quad + \left( 1 - \frac{2^n {\text {vol}}_{d-1} K^n}{(2{\text {vol}}_{d-1} K + S(K)\varepsilon )^n}\right) \mathbb {E}_{N_1 > 0} \mathcal {V}_{K_\varepsilon [n]}^k. \end{aligned}$$

(10)

If $\varepsilon$ tends to zero, $\mathbb {P}(N_1>0)$ and hence the second summand in (10) tend to zero. Now, an argument similar to that in the proof of Lemma 2 yields that $\mathbb {E}\mathcal {V}_{K\times \{0,\varepsilon \}[n]}^k$ converges to $\mathbb {E}V_{K[n]}^k$, and the same is true for K replaced by L, which completes the proof.