On Wendel’s equality for intersections of balls

Abstract

We study the analogue of Wendel’s equality in random polytope models in which the hull of the random points is formed by intersections of congruent balls, called the spindle (or hyper-) convex hull. According to the classical identity of Wendel the probability that the origin is contained in the (linear) convex hull of n i.i.d. random points distributed according to an origin symmetric probability distribution in the d-dimensional Euclidean space $\mathbb {R}^{d}$ that assigns measure zero to hyperplanes is a constant depending only on n and d. While in the classical convex case one gets nonzero probabilities only for $n\ge d+1$ points in $\mathbb {R}^{d}$, for the spindle convex hull this happens for all $n\ge 2$. We study this question for the uniform and normally distributed random models.

1 Introduction and results

Wendel’s equality [10] is one of the classical results in geometric probability: it states that if $x_1, \ldots , x_n$ are i.i.d. random points in $\mathbb {R}^d$ whose distribution is (centrally) symmetric with respect to the origin o, and the probability measures of hyperplanes are 0, then the probability that o is not contained in the convex hull $[x_1,\ldots , x_n]$ is

$$\begin{aligned} \mathbb {P}(o\notin [x_1,\ldots , x_n])=\frac{1}{2^{n-1}}\sum _{i=0}^{d-1}{n-1\atopwithdelims ()i}. \end{aligned}$$

(1.1)

One can find a simple proof of (1.1) in Bárány [1, pp. 94–95], which is independent of the distribution (under the above conditions).

It was proved by Wagner and Welzl [9], that o-symmetric distributions are extremal in this sense. For more information, see also [8, Section 8.1.2].

Recently, Kabluchko and Zaporozhets [3] investigated the related problem of finding the probability that the convex hull of n i.i.d. normally distributed random points in $\mathbb {R}^d$ contains a fixed points of space; they called these absorption probabilities. For a general introduction to random polytopes we refer to the recent survey paper by Schneider [7] and the book by Schneider and Weil [8].

We denote the d-dimensional origin centered unit radius closed ball by $B^d$ and its boundary by $S^{d-1}$. The symbol $\kappa _d$ denotes the volume (Lebesgue measure) of $B^d$, and $\omega _d$ is the surface volume of $B^d$. For general information on convex sets, see the monograph [6] by Schneider.

In this paper we study the following spindle convex variant of the above problems. Let $x,y\in \mathbb {R}^d$ be two points and $\varrho >0$. If $|x-y|\le 2\varrho$, then let the spindle $[x,y]_\varrho$ determined by x and y be the intersection of all radius $\varrho$ closed balls that contain both x and y. If $|x-y|>2\varrho$, then let $[x,y]_\varrho =\mathbb {R}^d$. We say that a convex body $K\subset \mathbb {R}^d$ (compact convex set with non-empty interior) is spindle convex with radius $\varrho$, or $\varrho$-spindle convex if together with any two points $x,y\in K$, it contains the spindle $[x,y]_\varrho$. It is known ([2]) that if a convex body $K\subset \mathbb {R}^d$ is spindle convex with radius $\varrho$, then K is the intersection of all radius $\varrho$ closed balls that contain K. This latter property is called radius $\varrho$ ball-convexity.

Let $X\subset \mathbb {R}^d$. If $X\subset \varrho B^d+v$ for some $v\in \mathbb {R}^d$, then the radius $\varrho$ spindle convex hull $[X]_\varrho$ of K is defined as the intersection of all radius $\varrho$ closed balls containing X. If $X\not \subset \varrho B^d+v$ for any $v\in \mathbb {R}^d$, then let $[X]_\varrho =\mathbb {R}^d$. If $K\subset \mathbb {R}^d$ is spindle convex with radius $\varrho$, and $X\subset K$, then $[X]_\varrho \subset K$. For more information on spindle convexity, see, for example, the paper [2] by Bezdek et al. and the book [4] by Martini, Montejano and Oliveros and the references therein.

First, we describe the $\varrho$-spindle convex uniform model. Let $\varrho >0$, and let $K\subset \mathbb {R}^d$ be an o-symmetric convex body that is $\varrho$-spindle convex. Let $x_1,\ldots , x_n$ be i.i.d. uniform random points from K. We denote the radius $\varrho$ spindle convex hull of $x_1,\ldots , x_n$ by $K_{(n)}^\varrho =[x_1,\ldots , x_n]_\varrho$. By the $\varrho$-spindle convexity of K, the random ball-polytope $K_{(n)}^\varrho$ is contained in K. We ask the same question as in the classical convex case: what is the probability that $o\in K_{(n)}^\varrho$? We note that in this model we may always achieve by scaling (simultaneously K and radius $\varrho$ circles) that $\varrho =1$. Henceforth, in the following two theorems we assume that $\varrho =1$.

We study the special case when $K=rB^d$ with $0<r\le 1$. Then K is clearly spindle convex with radius $\varrho =1$. We wish to determine the probability

$$\begin{aligned} P(d,r,n):=\mathbb {P}(o\in [x_1,\ldots , x_n]_1). \end{aligned}$$

In Sect. 2 we prove the following theorem:

Theorem 1.1

Let $K=r B^d$. Then

$$\begin{aligned} P(d,r,2)=\frac{\omega _{d-1}\omega _d}{(r^d\kappa _d)^2}\int _0^r \int _{0}^r\int _{0}^{\varphi (r_1, r_2)} r_1^{d-1}r_2^{d-1} \sin ^{d-2} \varphi \, d\varphi dr_2 d r_1, \end{aligned}$$

where $\varphi (r_1, r_2)=\arcsin (r_1/2)+\arcsin (r_2/2)$. In particular,

$$\begin{aligned} P(2,1,2)&=\frac{\sqrt{3}}{\pi }-\frac{1}{3}= 0.2179\ldots ,\\ P(3,1,2)&=\frac{1}{64}(23+12\sqrt{3}\pi -8\pi ^2)= 0.1459\ldots . \end{aligned}$$

Furthermore, for the case of three points, we prove the following statement in Sect. 3.

Theorem 1.2

Let $K=B^2$. Then

$$\begin{aligned} P(2,1,3)=\frac{-84\pi ^2-477+360\sqrt{3}\pi }{144\pi ^2}= 0.4594\ldots . \end{aligned}$$

Finally, in Sect. 4, we study the Gaussian $\varrho$-spindle convex model. Let $x_1, \ldots , x_n$ be i.i.d. random points from $\mathbb {R}^d$ distributed according to the standard normal distribution. The question is the same, what is the probability that $o\in K_{(n)}^\varrho$? We note that in this second case, it may, and does, happen that $K_{(n)}^\varrho =\mathbb {R}^d$. We give an integral formula for the probability that a Gaussian unit radius spindle contains the origin and evaluate it numerically in the plane.

2 Proof of Theorem 1.1

Note that it is the simplest case of the model when $n=2$, and $K=r B^d$, where $0<r\le 1$ is a fixed number. This, of course, is of no interest in the classical version of Wendel’s problem as $\mathbb {P}(o\in [x_1, x_2])=0$ since $[x_1, x_2]$ is a segment.

Let us examine what it means geometrically that $o\in [x_1, x_2]_1$. Let $M( x_1)$ denote the union of all open unit balls that contain o and $x_1$ on their boundary. Let $K(d, r, x_1)$ be the part of $rB^d{\setminus }M(x_1)$ that is in the closed half-space bounded by the hyperplane through o and orthogonal to $x_1$ which does not contain $x_1$. We depicted this region in Fig. 1 when $d=2$. We will only use $K(2,r,x_1)$ in our calculations, so, in order to simplify notation, we will denote it by $K(r, x_1)=K(2,r,x_1)$.

Fig. 1

The region $K(r,x_1)$

In order to evaluate P(d, r, 2), we use the linear Blaschke-Petkantschin formula. Let G(d, 2) denote the Grassmannian manifold of 2-dimensional linear subspaces of $\mathbb {R}^d$, and $\nu _2$ be the unique rotation invariant Haar probability measure on G(d, 2). The 2-dimensional special case of the linear Blaschke-Petkantschin formula (see, for example, [8, Theorem 7.2.1 on p. 271]) says the following: If $f:(\mathbb {R}^d)^2\rightarrow \mathbb {R}$ is a non-negative measurable function, then

$$\begin{aligned} \int _{(\mathbb {R}^d)^2} f\, d\lambda ^2=\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _{G(d,2)}\int _{L^2} f(x_1, x_2)\nabla ^{d-2}_2(x_1, x_2)\, d\lambda _L^2\, \nu _2 (dL), \end{aligned}$$

(2.1)

where $\nabla _2$ denotes the area of the parallelogram spanned by the vectors $x_1, x_2$ in L. The symbol $\lambda$ denotes the Lebesgue measure in $\mathbb {R}^d$, and $\lambda _L$ the (2-dimensional) Lebesgue measure in L.

Next, using polar coordinates for $x_1, x_2\in L$, that is, $x_1=r_1 u_1$, $x_2=r_2 u_2$, where $u_1, u_2\in S^{1}$, $r_1, r_2\in \mathbb {R}_+$, we may write the right-hand-side of (2.1) as follows.

$$\begin{aligned}&\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _{G(d,2)}\int _{L^2} f(x_1,x_2)\nabla ^{d-2}_2(x_1,x_2)\, d\lambda _L^2\, \nu _2 (dL)\nonumber \\&\quad =\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _{G(d,2)}\int _{(S^1\times \mathbb {R})^2} f(r_1u_1, r_2u_2)\nonumber \\&\qquad \times \nabla ^{d-2}_2(r_1u_1, r_2u_2)\, r_1 r_2 dr_1 du_1 d r_2 du_2\, \nu _2 (dL)\nonumber \\&\quad =\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _{G(d,2)}\int _{(S^1\times \mathbb {R})^2} f(r_1u_1, r_2u_2)r_1^{d-1} r_2^{d-1}\times \nonumber \\&\qquad \times |u_1\times u_2|^{d-2} dr_1 du_1 d r_2 du_2\, \nu _2 (dL). \end{aligned}$$

(2.2)

Now, from (2.2) we obtain that

$$\begin{aligned} P(d,r,2)&=\frac{1}{(r^d\kappa _d)^2}\int _{rB^d}\int _{rB^d} {\mathbf {1}}(o\in [x_1, x_2]_1)\, dx_1 d x_2\\&=\frac{1}{(r^d\kappa _d)^2}\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _{G(d,2)}\int _{S^1}\int _0^r\int _{S^1}\int _{0}^r {\mathbf {1}}(o\in [r_1u_1, r_2u_2]_1) r_1^{d-1} r_2^{d-1}\\&\qquad \times |u_1\times u_2|^{d-2} dr_1 du_1 d r_2 du_2\, \nu _2 (dL)\\&=\frac{1}{(r^d\kappa _d)^2}\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _{S^1}\int _0^r\int _{S^1}\int _{0}^r {\mathbf {1}}(o\in [r_1u_1, r_2u_2]_1) r_1^{d-1} r_2^{d-1}\\&\qquad \times |u_1\times u_2|^{d-2} dr_1 du_1 d r_2 du_2\\&=\frac{1}{(r^d\kappa _d)^2}\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _{S^1}\int _0^r\int _{S^1}\int _{0}^r {\mathbf {1}}(x_2\in K(r, x_1)) r_1^{d-1} r_2^{d-1}\\&\qquad \times |u_1\times u_2|^{d-2} dr_2 du_2 d r_1 du_1. \end{aligned}$$

By the rotational symmetry of $rB^d$, integration with respect to $u_1$ is a multiplication by $2\pi$. Hence, from now on, we fix $u_1=(0,1)$. Let $\varphi$ be the angle of $u_2$ and $-u_1$, as shown on Fig. 1, and let

$$\begin{aligned} \varphi (r_1, r_2)=\arcsin (r_1/2)+\arcsin (r_2/2). \end{aligned}$$

Then

$$\begin{aligned} P(d,r,2)&=\frac{2\pi }{(r^d\kappa _d)^2}\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _0^r\int _{0}^r \int _{-\varphi (r_1,r_2)}^{\varphi (r_1,r_2)} r_1^{d-1}r_2^{d-1}|\sin \varphi |^{d-2}\, d\varphi dr_2 d r_1\\&=\frac{4\pi }{(r^d\kappa _d)^2}\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _0^r \int _{0}^r\int _{0}^{\varphi (r_1, r_2)} r_1^{d-1}r_2^{d-1} \sin ^{d-2} \varphi \, d\varphi dr_2 d r_1\\&=\frac{\omega _{d-1}\omega _d}{(r^d\kappa _d)^2}\int _0^r \int _{0}^r\int _{0}^{\varphi (r_1, r_2)} r_1^{d-1}r_2^{d-1} \sin ^{d-2} \varphi \, d\varphi dr_2 d r_1. \end{aligned}$$

The above integral can be evaluated for any specific value of d using multiple integration by parts. In particular,

$$\begin{aligned} P(2,r,2)&=\frac{4}{\pi r^4}\int _0^r\int _{0}^r \int _{0}^{\varphi (r_1, r_2)} r_2r_1 d\varphi dr_2 d r_1\nonumber \\&=\frac{4}{\pi r^4}\int _0^r\int _{0}^r r_2r_1 (\arcsin (r_1/2)+\arcsin (r_2/2))\, dr_2 d r_1\nonumber \\&=\frac{4}{\pi r^4}\left( \frac{r^2}{4}(r\sqrt{4-r^2}+2(r^2-2)\arcsin (r/2))\right) \nonumber \\&=\frac{1}{\pi r^2}\left( r\sqrt{4-r^2}+2(r^2-2)\arcsin (r/2)\right) , \end{aligned}$$

(2.3)

and

$$\begin{aligned} P(3,r,2)&=\frac{9}{2r^6}\int _0^r\int _{0}^r \int _{0}^{\varphi (r_1, r_2)} r_2^2 r_1^2 \sin \varphi \, d\varphi dr_2 d r_1\\&=\frac{9}{2r^6}\left( \frac{r^2}{288}(-72+90r^2-4r^4+9r^6)\right. \\&\quad \left. +\frac{1}{4}\arcsin (r/2)(R\sqrt{4-r^2}(r^2-2)+4\arcsin (r/2))\right) . \end{aligned}$$

In particular,

$$\begin{aligned} P(2,1,2)&=\frac{\sqrt{3}}{\pi }-\frac{1}{3}= 0.2179\ldots ,\\ P(3,1,2)&=\frac{1}{64}(23+12\sqrt{3}\pi -8\pi ^2)= 0.1459\ldots . \end{aligned}$$

This finishes the proof of Theorem 1.1. $\square$

We conclude this section with the following statements.

Corollary 2.1

For any fixed $d\ge 2$, it holds that

$$\begin{aligned} \lim _{r\rightarrow 0^+} P(d,r,2)=0. \end{aligned}$$

Furthermore, for any fixed $0<r\le 1$, it holds that

$$\begin{aligned} \lim _{d\rightarrow \infty } P(d,r,2)=0. \end{aligned}$$

Proof

Note that, using $\arcsin x\le \pi x/2$ for $x\in [0,\pi /2]$ and $\sin x\le x$ for $x\in [0, \pi /2]$, we get that

$$\begin{aligned} P(d,r,2)&\le \frac{C(d)}{r^{2d}}\int _0^r \int _{0}^r\int _{0}^{r_1+r_2} r_1^{d-1}r_2^{d-1} (r_1+r_2)^{d-2}\, d\varphi dr_2 d r_1\\&\le \frac{2^{d-1}C(d)}{r^{2d}}\int _0^r \int _{r_1}^r\int _{0}^{2r_2} r_2^{3d-4}\, d\varphi dr_2 d r_1\\&=\frac{2^{d}C(d)}{r^{2d}}\int _0^r \int _{0}^r r_2^{3d-3}\, dr_2 dr_1\\&=\frac{2^{d}C(d)}{r^{2d}}\frac{r^{3d-1}}{3d-2}, \end{aligned}$$

where the constant C(d) depends only on the dimension d. From this it follows that

$$\begin{aligned} \lim _{r\rightarrow 0^+}P(d,r,2)=0 \end{aligned}$$

for $d\ge 2$, as claimed.

In the proof of the second statement we use the fact that $\varphi (r_1, r_2)\le \pi /3$. Thus

$$\begin{aligned} P(d,r,2)&\le \frac{\omega _{d-1}\omega _d}{r^{2d}\kappa _d^2}\int _0^r\int _0^r r_1^{d-1}r_2^{d-1}\left( \frac{\sqrt{3}}{2}\right) ^{d-1} \, dr_2dr_1\\&=\frac{\omega _{d-1}\omega _d}{d^2\kappa _d^2}\left( \frac{\sqrt{3}}{2}\right) ^{d-1}=\frac{d-1}{d}\frac{\kappa _{d-1}}{\kappa _d}\left( \frac{\sqrt{3}}{2}\right) ^{d-1}. \end{aligned}$$

From $\kappa _{d-1}/\kappa _d\sim c\cdot \sqrt{d}$ as $d\rightarrow \infty$, it follows that $P(d,r,2)\rightarrow 0$ as $d\rightarrow \infty$. $\square$

3 Proof of Theorem 1.2

The case when $n=3$, can be treated, at least in the plane, as follows. We only consider when $r=1$, that is, $K=B^2$. Let $x_1, x_2, x_3$ be i.i.d. uniform random points from $B^2$. Let

$$\begin{aligned} P(2,1,3):&=\mathbb {P}(o\in [x_1, x_2, x_3]_1)\\&=\mathbb {P}(o\in [x_1, x_2]_1)+\mathbb {P}(o\notin [x_1, x_2]_1\text { and }o\in [x_1, x_2, x_3]_1)\\&=P(2,1,2)+\mathbb {P}(o\notin [x_1, x_2]_1\text { and }o\in [x_1, x_2, x_3]_1). \end{aligned}$$

Let

$$\begin{aligned} \overline{P}(2,1,3):=\mathbb {P}(o\notin [x_1, x_2]_1\text { and }o\in [x_1, x_2, x_3]_1). \end{aligned}$$

Due to the rotational symmetry of $B^2$, we may assume that $x_1=(0,r_1)$. Let $x_2=r_2u_2$, where $\varphi$ is the angle of $u_2$ and the negative half of the y-axis. Making use of the previously introduced notation, we write $K(x_1)=K(1,x_1)$ and, similarly, $K(x_2)=K(1,x_2)$. The ray $ox_i$ divides $K(x_i)$ into two congruent parts. The part that is on the positive side of $ox_i$ is denoted by $K^+(x_i)$, and the negative part is $K^-(x_i)$, as shown in Fig. 2.

Fig. 2

.

Let $V^+(x_i)=V_2(K^+(x_i))$ and $V^-(x_i)=V_2(K^-(x_i))$ for $i=1,2$. Then it holds that

$$\begin{aligned} V^+(x_i)&=V^-(x_i)=\int _0^1\int _{0}^{\varphi (r_i, r)} r\, d\varphi dr =\int _0^1 (\arcsin (r_i/2)+\arcsin (r/2))r\, dr\\&= \frac{1}{12} \left( 3\sqrt{3}-\pi +6\arcsin (r_i/2)\right) . \end{aligned}$$

We distinguish four cases according to the relative position of $x_1$ and $x_2$.

Case 1. $r_2\le r_1$ and $x_2\notin [x_1, o]_1$.

In this case, $\varphi \in [\varphi (r_1, r_2), \pi -\arcsin (r_1/2)+\arcsin (r_2/2)]$. Then

$$\begin{aligned} P_1&:=\mathbb {P}(o\notin [x_1, x_2]_1\text { and }o\in [x_1, x_2, x_3]_1\text { and }x_2\notin [x_1, o]_1\text { and } r_1\ge r_2)\\&=\frac{2\pi }{\pi ^3}\int _0^1\int _{0}^{r_1}\int _{\varphi (r_1,r_2)}^{\pi -\arcsin (r_1/2)+\arcsin (r_2/2)}\\&\quad \times \left( V^+(x_1)+V^-(x_2)+\frac{\pi -\varphi }{2}\right) r_1r_2 d\varphi dr_2 dr_1\\&=\frac{1}{\pi ^2}\int _0^1\int _{0}^{r_1}\int _{\varphi (r_1,r_2)}^{\pi -\arcsin (r_1/2)+\arcsin (r_2/2)} \left( \sqrt{3}-\frac{\pi }{3}+\arcsin (r_1/2)\right. \\&\qquad \left. +\arcsin (r_2/2)+\frac{\pi -\varphi }{2}\right) r_1r_2 d\varphi dr_2 dr_1\\&=-\frac{5}{72}-\frac{1}{\pi ^2}+\frac{5}{4\sqrt{3} \pi }. \end{aligned}$$

Case 2.  $r_2\ge r_1$ and $x_1\notin [x_2, o]_1$. By the symmetry of $x_1$ and $x_2$,

$$\begin{aligned} P_2&:=\mathbb {P}(o\notin [x_1, x_2]_1\text { and }o\in [x_1, x_2, x_3]_1\text { and }x_1\notin [x_2, o]_1\text { and }r_1\le r_2)\\&=\mathbb {P}(o\notin [x_1, x_2]_1\text { and }o\in [x_1, x_2, x_3]_1\text { and }x_2\notin [x_1, o]_1\text { and } r_1\ge r_2)\\&=-\frac{5}{72}-\frac{1}{\pi ^2}+\frac{5}{4\sqrt{3} \pi }. \end{aligned}$$

Case 3.  $x_2\in [x_1, o]_1$.

In this case $r_1\ge r_2$ and $\varphi \in [\pi -\arcsin (r_1/2)+\arcsin (r_2/2),\pi ]$. Then $K(x_2)\subset K(x_1)$, thus

$$\begin{aligned} P_3&:=\mathbb {P}(o\notin [x_1, x_2]_1\text { and }o\in [x_1, x_2, x_3]_1\text { and }x_2\in [x_1, o]_1)\\&=\frac{2\pi }{\pi ^3}\int _0^1\int _{0}^{r_1}\int _{\pi -\arcsin (r_1/2)+\arcsin (r_2/2)}^\pi V(x_1) r_1r_2 d\varphi dr_2 dr_1\\&=\frac{1}{\pi ^2}\int _0^1\int _{0}^{r_1}\int _{\pi -\arcsin (r_1/2)+\arcsin (r_2/2)}^\pi \left( \frac{\sqrt{3}}{2}-\frac{\pi }{6}+\arcsin (r_1/2)\right) r_1r_2 d\varphi dr_2 dr_1\\&=\frac{99-24\sqrt{3}\pi +4\pi ^2}{576 \pi ^2}. \end{aligned}$$

Case 4.  $x_1\in [x_2, o]_1$. Again, by the symmetry of $x_1$ and $x_2$,

$$\begin{aligned} P_4&=\mathbb {P}(o\notin [x_1, x_2]_1\text { and }o\in [x_1, x_2, x_3]_1\text { and }x_1\in [x_2, o]_1)\\&=\mathbb {P}(o\notin [x_1, x_2]_1\text { and }o\in [x_1, x_2, x_3]_1\text { and }x_2\in [x_1, o]_1)\\&=\frac{99-24\sqrt{3}\pi +4\pi ^2}{576 \pi ^2}. \end{aligned}$$

Thus, considering the symmetry with respect to the line $ox_1$, we obtain that

$$\begin{aligned} \overline{P}(2,1,3)&=2(P_1+P_2+P_3+P_4)=\frac{-36\pi ^2-477+216\sqrt{3}\pi }{144\pi ^2}. \end{aligned}$$

Thus,

$$\begin{aligned} P(2,1,3)=P(2,1,2)+\overline{P}(2,1,3)=\frac{-84\pi ^2-477+360\sqrt{3}\pi }{144\pi ^2}= 0.4594\ldots . \end{aligned}$$

We note that the actual calculation can be carried out, at least numerically, for any $0<r\le 1$. Furthermore, the cases of $n=4, 5,\ldots$ are essentially similar, although the case analysis grows significantly more complicated as n increases.

Finally, we note that according to Wendel’s equality (1.1),

$$\begin{aligned} \mathbb {P}(0\in [x_1,x_2,x_3])=\frac{1}{4}<P(2,1,3). \end{aligned}$$

4 The case of normally distributed random points

In this subsection we consider the model in which $\varrho =1$ and $x_1,\ldots , x_n$ are i.i.d. random points in $\mathbb {R}^d$ that are distributed according to the standard normal distribution with density function

$$\begin{aligned} f(x)=\frac{1}{(2\pi )^\frac{d}{2}}e^{-\frac{|x|^2}{2}}, \; x\in \mathbb {R}^d. \end{aligned}$$

Here we need to use the part of the definition of the spindle convex hull that normally does not come into play when the random points are chosen from a convex body that is spindle convex with radius less than or equal to 1. Namely, if $x,y\in \mathbb {R}^d$ are such that $|x-y|>2$, then $[x,y]_1:=\mathbb {R}^d$.

We are interested in the following probability

$$\begin{aligned} P_N(d,1,n):=\mathbb {P}(o\in [x_1,\ldots , x_n]_1). \end{aligned}$$

It is clear that

$$\begin{aligned} \mathbb {P}(o\in [x_1,\ldots , x_n])\le \mathbb {P}(o\in [x_1,\ldots , x_n]_1) \end{aligned}$$

as $[X]\subset [X]_1$ for any $X\subset \mathbb {R}^d$.

Let E be the event that $|x_1-x_2|\le 2$. Then

$$\begin{aligned} P_N(d,1,2)=\mathbb {P}(o\in [x_1, x_2]_1 \text { and } E)+\mathbb {P}(E^c), \end{aligned}$$

where $E^c$ is the complement of E, as $E^c$ automatically implies that $o\in [x_1, x_2]_1$.

Let l denote the length of the random segment $[x_1x_2]$. It is known (see [5, p. 438] and the historical references therein) that the density of $s:=l^2/4$ is

$$\begin{aligned} g(s)=\frac{s^{\frac{d}{2}-1}e^{-s}}{\Gamma (d/2)}, \; 0< s<\infty . \end{aligned}$$

(4.1)

Thus,

$$\begin{aligned} \mathbb {P}(E^c)=\int _1^\infty g(s)\, ds&=\frac{\gamma (d/2, 1)}{\Gamma (d/2)}, \end{aligned}$$

where $\Gamma (\cdot )$ is Euler’s gamma function, and $\gamma (d/2, x)$ denotes the lower incomplete gamma function.

Using the linear Blaschke–Petkantschin formula (2.2) and the rotational invariance of the standard normal distribution we obtain that

$$\begin{aligned} \mathbb {P}&(o\in [x_1, x_2]_1 \text { and }E)\\&=\frac{1}{(2\pi )^d}\int _{\mathbb {R}^d}\int _{\mathbb {R}^d} {\mathbf {1}}(o\in [x_1, x_2]_1 \text { and }E)\,e^{-\frac{|x_1|^2+|x_2|^2}{2}}\, dx_1 d x_2\\&=\frac{1}{(2\pi )^d}\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _{G(d,2)}\int _{L^2} {\mathbf {1}}(o\in [x_1, x_2]_1\,\text { and }E)\\&\quad \times \Delta ^{d-2}(x_1, x_2)\,e^{-\frac{|x_1|^2+|x_2|^2}{2}} dx_1 d x_2 \nu _2 (dL)\\&=\frac{1}{(2\pi )^d}\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _{L^2} {\mathbf {1}}(o\in [x_1, x_2]_1\text { and }E)\Delta ^{d-2}(x_1, x_2)\,e^{-\frac{|x_1|^2+|x_2|^2}{2}} dx_1 d x_2.\\ \end{aligned}$$

In order to evaluate the above integral, we use polar coordinates $x_1=r_1 u_1$ and $x_2=r_2u_2$, $r_1, r_2\ge 0$, $u_1, u_2\in S^1$. Let $\varphi$ be the angle of $-u_1$ and $u_2$, as before. For $2-r_1\le r_2\le \sqrt{4-r_1^2}$, let

$$\begin{aligned} \psi (r_1, r_2)=\pi -\arccos \left( \frac{r_1^2+r_2^2-4}{2r_1r_2}\right) . \end{aligned}$$

We distinguish two cases according to $r_2$. When $0\le r_2\le 2-r_1$, then $-\varphi (r_1, r_2)\le \varphi \le \varphi (r_1, r_2)$, and when $2-r_1\le r_2\le \sqrt{4-r_1^2}$, then $-\varphi (r_1, r_2)\le \varphi \le -\psi (r_1, r_2)$ and $\psi (r_1, r_2)\le \varphi (r_1, r_2)$, see Fig. 3.

Fig. 3

Integration bounds in $\varphi$ according to $r_2$

By the rotational symmetry of the normal distribution, integration with respect to $u_1$ is just a multiplication by $2\pi$. Then, w obtain that

$$\begin{aligned}&\mathbb {P}(o\in [x_1, x_2]_1\text { and } E)\\&\quad =\frac{2}{(2\pi )^{d-1}}\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _0^2\int _0^{2-r_1} \int _{0}^{\varphi (r_1, r_2)}r_1^{d-1} r_2^{d-1}\sin ^{d-2}(\varphi )\,e^{-\frac{r_1^2+r_2^2}{2}} d\varphi dr_2 dr_1 \\&\qquad +\frac{2}{(2\pi )^{d-1}}\frac{\omega _{d-1}\omega _d}{\omega _1\omega _2}\int _0^2\int _{2-r_1}^{\sqrt{4-r_1^2}} \int _{\psi (r_1, r_2)}^{\varphi (r_1, r_2)}r_1^{d-1} r_2^{d-1}\\&\qquad \times \sin ^{d-2}(\varphi )\,e^{-\frac{r_1^2+r_2^2}{2}} d\varphi dr_2 dr_1 . \end{aligned}$$

The above integrals can be evaluated, at least numerically, for any specific value of d. In particular, for $d=2$, we obtain for the first integral

$$\begin{aligned}&\frac{1}{\pi }\int _0^2 \int _{0}^{2-r_1} \int _{0}^{\varphi (r_1, r_2)} r_1r_2 \,e^{-\frac{r_1^2+r_2^2}{2}}\, d\varphi dr_2 d r_1\\&\quad =\frac{1}{\pi }\int _0^2 \int _{0}^{2-r_1} (\arcsin (r_1/2)+\arcsin (r_2/2)) r_1r_2 \,e^{-\frac{r_1^2+r_2^2}{2}}\, dr_2 d r_1\\&\quad = 0.079214\ldots . \end{aligned}$$

The second integral is

$$\begin{aligned}&\frac{1}{\pi }\int _0^2 \int _{2-r_1}^{\sqrt{4-r_1^1}} \int _{\psi (r_1, r_2)}^{\varphi (r_1, r_2)} r_1r_2 \,e^{-\frac{r_1^2+r_2^2}{2}}\, d\varphi dr_2 d r_1\\&\quad =\frac{1}{\pi }\int _0^2 \int _{2-r_1}^{\sqrt{4-r_1^2}} \left( \arcsin (r_1/2)+\arcsin (r_2/2)-\pi +\arccos \left( \frac{r_1^2+r_2^2-4}{2r_1r_2}\right) \right) \\&\qquad \times r_1r_2 \,e^{-\frac{r_1^2+r_2^2}{2}}\, dr_2 d r_1\\&\quad = 0.01866\ldots . \end{aligned}$$

For $d=2$,

$$\begin{aligned} \mathbb {P}(E^c)=\frac{\gamma (1, 1)}{\Gamma (1)}=\frac{1}{e}= 0.367879\ldots , \end{aligned}$$

thus, in summary,

$$\begin{aligned} P_N(2,1,2)= 0.465753\ldots . \end{aligned}$$