## Geometric Probability of Subspaces Intersecting with a Convex Body in $\mathbb{R}^n$

Zhao Jiangfu,

 基金资助: 福建省中青年教师教育科研项目基金(科技类).  JT180585福建江夏学院科研培育人才项目基金.  JXZ2019016

 Fund supported: the Educational Research Project Fund of Young and Middle-Aged Teachers of Fujian Province.  JT180585the Project Fund for Scientific Research and Cultivation of Talents of Fujian Jiangxia University.  JXZ2019016

Abstract

The probability that three independent random subspaces in $\mathbb{R}^n$ intersecting a convex body K have their common point intersecting K is found by using of the mean curvature integral of convex sets. Then we focus on the particular case of hyperplanes. On the base of this, we state the geometric probability of hyperplanes that intersect a ball, a cube or a right parallelepiped having an intersection inside the same ball, the cube or the right parallelepiped respectively. Finally, the monotonicity, convergence, and size relationship of the geometric probabilistic sequence are discussed.

Keywords： Mean curvature integral ; Geometric probability ; Buffon needle throwing ; Elementary symmetric function ; Hyperplanes ; Subspaces

Zhao Jiangfu. Geometric Probability of Subspaces Intersecting with a Convex Body in $\mathbb{R}^n$. Acta Mathematica Scientia[J], 2021, 41(3): 770-782 doi:

## 1 引言

$$$p = \frac{\pi^4V}{M^3},$$$

## 2 预备知识

$$$e_k(x) = e_k(x_1, x_2, \cdots , x_n) = \sum\limits_{1\leq i_1<i_2< \cdots<i_k\leq n} x_{i_1}x_{i_2}\cdots x_{i_k}$$$

$k$阶初等对称函数, 并补充规定$e_0(x) = e_0(x_1, x_2, \cdots , x_n) = 1$.

$1$阶初等对称函数$e_1(x_1, x_2, \cdots , x_n) = \sum\limits_{i = 1}^{n}x_i$; 第$2$阶初等对称函数$e_2(x_1, x_2, \cdots , x_n) = \sum\limits_{1\leq i<j \leq n }^{n}x_ix_j$.

$L_{n-r[O]} $${{\Bbb R}} ^n 中过定点 O 且垂直于 L_{r[O]}$$ n-r$维平面, 则由密度公式${\rm d}L_{r[O]} = {\rm d}L_{n-r[O]}$

$$$m(G_{n-r, r}) = \int_{G_{n-r, r}}{\rm d}L_{n-r[O]} = \int_{G_{r, n-r}}{\rm d}L_{r[O]} = \frac{O_{n-1}O_{n-2}\cdots O_{n-r}}{O_{r-1}O_{r-2}\cdots O_{0}} = m(G_{r, n-r}).$$$

## 3 主要结论

$$$m_2 = T^n_pT^n_qT^n_sM_{p-1}(\partial K)M_{q-1}(\partial K)M_{s-1}(\partial K).$$$

$\begin{eqnarray} p_n(K) = \left\{\begin{array}{ll} { } \frac{\pi (n-2)(n-3)(O_{n-2})^3M_{n-4}(\partial K)}{6(n-1)^2O_{n-4}(M_{n-2}(\partial K))^3}, &n\geq 4, \\ { } \frac{\pi ^4V(K)}{M^3}, &n = 3. \end{array}\right. \end{eqnarray}$

$K$为一些特殊凸体时, 将对应的平均曲率积分代入(3.10)式, 可得到推论3.1–3.3.

$\begin{eqnarray} p_n(C_e) = \left\{\begin{array}{ll} { } \frac{\pi (n-1)(n-2)}{6n^2}, &n\geq 4, \\ { } \frac{\pi}{27}, &n = 3. \end{array}\right. \end{eqnarray}$

$\begin{eqnarray} p_n(C) = \left\{\begin{array}{ll} { } \frac{\pi e_3(a_1, a_2, \cdots, a_n)}{(e_1(a_1, a_2, \cdots, a_n))^3}, &n\geq 4, \\ { } \frac{\pi a_1a_2a_3}{27(a_1+a_2+a_3)^3}, &n = 3. \end{array}\right. \end{eqnarray}$

$\begin{eqnarray} p_n(B) = \left\{\begin{array}{ll} { } \frac{\pi (n-2)\Gamma ^2(\frac{n}{2})}{3(n-1)^2\Gamma ^2(\frac{n-1}{2})}, &n\geq 4, \\ { } \frac{\pi ^2}{48}, &n = 3. \end{array}\right. \end{eqnarray}$

(1) 当$q\geq n-r$时,

$$$M^{(n)}_q(\partial K^r) = \frac{(^{\ \, r-1}_{q-n+r})O_q}{(^{n-1}_{\ \; q})O_{q-n+r}}M^{(r)}_{q-n+r}(\partial K^r);$$$

(2) 当$q = n-r-1$时,

$$$M^{(n)}_{n-r-1}(\partial K^r) = \frac{O_{n-r-1}}{(^{\ \, n-1}_{n-r-1})}V_r(K^r),$$$

(3) 当$q<n-r-1$时,

$$$M^{(n)}_q(\partial K^r) = 0.$$$

$$$\int_{L_r\cap K\neq \varnothing}M^{(r)}_q(\partial (L_r\cap K)){\rm d}L_r = \frac{O_{n-2}O_{n-3}\cdots O_{n-r}O_{n-q}}{O_{r-2}O_{r-3}\cdots O_0O_{r-q}}M_q(\partial K).$$$

$$$m_1 = \int_{(L_p\cap L_q \cap L_s\cap K \neq\varnothing)}{\rm d}L_p\wedge {\rm d}L_q \wedge {\rm d}L_s = \int_{L_s\cap K\neq\varnothing}\bigg(\int_{(L_p\cap L_q \cap (L_s\cap K) \neq\varnothing}{\rm d}L_p\wedge {\rm d}L_q\bigg){\rm d}L_s.$$$

(1) 当$p+q\geq n+1$时, 由引理$4.1$

$\begin{eqnarray} & & \int_{(L_p\cap L_q \cap (L_s\cap K) \neq\varnothing}{\rm d}L_p\wedge {\rm d}L_q \\ & = & \frac{2\pi(p-1)!(q-1)!O_{p-1}O_{q-1}O_{2n-p-q+1}}{(n-1)!(p+q-n-1)!O_{n-p+1}O_{n-q+1}O_{p+q-n-1}}T^n_pT^n_qM_{p+q-n-1}(\partial (L_s\cap K)). \end{eqnarray}$

1) 当$p+q+s\geq 2n+1$时, 即$p+q-n-1\geq n-s$, 由$(\label(4.3)$式得

$\begin{eqnarray} M_{p+q-n-1}(\partial (L_s\cap K))& = & M^{(n)}_{p+q-n-1}(\partial K^s)\\ & = & \frac{\left(^{{\qquad} s-1}_{p+q+s-2n-1}\right)O_{p+q-n-1}}{\left(^{{\quad} n-1}_{p+q-n-1}\right) O_{p+q+s-2n-1}}M^{(s)}_{p+q+s-2n-1}(\partial K^s). \end{eqnarray}$

$\begin{eqnarray} \label{eq:4.11} & &\int_{L_s\cap K\neq\varnothing}M^{(s)}_{p+q+s-2n-1}(\partial K^s){\rm d}L_s\\ & = & \frac{O_{n-2}O_{n-3}\cdots O_{n-s}O_{3n-p-q-s+1}}{O_{s-2}O_{s-3}\cdots O_0O_{2n+1-p-q}}M_{p+q+s-2n-1}(\partial K)\\ & = & \frac{2\pi O_{s-1}O_{3n-p-q-s+1}}{O_{n-s+1}O_{2n+1-p-q}}T^n_sM_{p+q+s-2n-1}(\partial K). \end{eqnarray}$

2) 当$p+q+s = 2n$时, 即$p+q-n-1 = n-s-1$, 由(4.4)式得

$$$M_{p+q-n-1}(\partial (L_s\cap K)) = M^{(n)}_{p+q-n-1}(\partial K^s) = \frac{O_{n-s-1}} {\left(^{\ n-1}_{n-s-1}\right)}V_s(K^s) = \frac{O_{n-s-1}}{\left(^{\ n-1}_{n-s-1}\right)}\sigma_s(L_s\cap K).$$$

$$$\int_{L_s\cap K\neq\varnothing}\sigma_s(L_s\cap K){\rm d}L_s = \frac{O_{n-1}O_{n-2}\cdots O_{n-s}}{O_{s-1}O_{s-2}\cdots O_0}V(K) = \frac{2\pi O_{n-1}}{O_{n-s+1}}T^n_sV(K).$$$

3) 当$p+q+s<2n$时, 即$p+q-n-1<n-s-1$, 由(4.5)式得

$$$M_{p+q-n-1}(\partial (L_s\cap K)) = 0;$$$

(2) 当$p+q = n$时, 由引理$4.1$

$$$\int_{L_p\cap L_q \cap (L_s\cap K) \neq\varnothing}{\rm d}L_p\wedge {\rm d}L_q = \frac{p!q!O_{n-1}}{(n-1)!}T^n_pT^n_qV(L_s\cap K),$$$

$$$m_r\{L_r:L_r\cap K\neq \varnothing\} = \int_{L_r\cap K\neq \varnothing}{\rm d}L_r = \frac{O_{n-2}O_{n-3}\cdots O_{n-r-1}}{(n-r)O_{r-1}O_{r-2}\cdots O_0}M_{r-1}(\partial K).$$$

由引理4.5得

### 4.3 定理3.5的证明

$p_n(C_e)$显然收敛于$\frac{\pi}{6}$, 因此定理3.5只需证明以下三个结论:

(1) $p_n(C_e)$关于维数$n$严格单调递增;

(2) $p_n(B)$关于维数$n$严格单调递增;

(3) $\lim \limits_{n\rightarrow \infty}p_n(B) = \frac{\pi}{6}$.

(1) 证明$p_n(C_e)$严格单调递增.

$n = 3$时, $p_{4}(C_e) = \frac{\pi}{16}>\frac{\pi}{27} = p_3(C_e)$;

$n\geq4$时, 因为

(2) 证明$p_n(B)$严格单调递增.

$n = 3$时, $p_3(B) = \frac{\pi^2}{48}<\frac{8}{27} = p_4(B);$

$n\geq4$时, 因为

(3) 证明$\lim \limits_{n\rightarrow \infty}p_n(B) = \frac{\pi}{6}$.

(1) 当$n = 2m$时, 由Wallis's公式可得偶子列的极限为

(2) 当$n = 2m+1$时, 由Wallis's公式可得奇子列的极限为

### 4.4 定理3.6的证明

(1) 首先证明$p_n(C)\leq p_n(C_e)$.

(i) 当$n = 3$时, 由$a_1, a_2, a_3>0$可得

(ii) 当$n\geq4$时, 令$A = \sum\limits_{i = 1}^na^3_i, B = \sum\limits_{i\neq j}^na_ia^2_j, C = \sum\limits_{1\leq i<j\leq n}^na_ia_ja_k$, 由$a_i, a_j, a_k>0$

$(n-2)B\geq6C,$所以

$$$B\geq\frac{6C}{n-2} \; \; {\rm{(当且仅当}}| a_1 = a_2 = \cdots = a_n {\rm{时等号成立)}}.$$$

$$$\frac{(n-1)(n-2)}{2}A+3(n-2)B\geq21C.$$$

(2) 证明$p_n(C_e)< p_n(B)$. 这里需要用到以下两个不等式[19]

$$$\frac{(2m)!!^2}{(2m-1)!!^2}>\frac{\pi (4m+1)}{4}, m = 1, 2, \cdots,$$$

$$$\frac{(2m-1)!!^2}{(2m)!!^2}>\frac{4m+3}{\pi(2m+1)^2}, m = 1, 2, \cdots.$$$

1) 当$n = 3$时, $p_3(C_e) = \frac{\pi}{27}<\frac{\pi^2}{48} = p_3(B)$;

2) 当$n\geq4$时, 令

(i) 当$n = 2m$时, 由(4.20)式得

(ii) 当$n = 2m+1$时, 由(4.21)式得

## 参考文献 原文顺序 文献年度倒序 文中引用次数倒序 被引期刊影响因子

Xie F F , Li D Y .

On generalized Buffon needle problem for lattices

Acta Mathematica Scientia, 2011, 31B (1): 303- 308

E3中有界网格的Buffon问题

Xie F F , Hu S G .

On Buffon problem for bounded lattices in E3

Acta Math Sci, 2011, 31A (2): 410- 414

Li D Y , Jiang H F , Xiong G .

The reference point of matching ahapes

J of Math, 2005, 25 (3): 336- 340

Li S G , Han H F , YANG P P .

The average distance from the interior points to the boundary points of a parallelogram

Journal of Wuhan University of Science and Technology, 2011, 34 (5): 376- 380

Li D Y , Yang P P , Li T .

Chord length distribution of rectangles

Journal of Wuhan University of Science and Technology, 2011, 34 (5): 381- 383

Huang Z X .

Study on Buffon throwing problem

Journal of Jimei University (Natural Science), 2005, 10 (4): 381- 384

Wang Y Y , Li D Y , Li X C , et al.

Buffon needle problem in regula tri-prism

J of Math, 2013, 33 (5): 887- 890

Zou M T , Li S G , Chen L L .

A special class of the geometric probability

J of Math, 2014, 34 (2): 374- 378

Ren D L . Topics in Integral Geometry. Singapore: World Scientific, 1994

Zhang G Y .

Dual kinematic formulas

Transactions of the American Mathematical Society, 1999, 351 (3): 985- 995

Santalo L A . Integral Geometry and Geometric Probability. London: Addidion-Wesley, 1976

Merca M .

New convolutions for complete and elementary symmetric functions

Integral Transforms and Special Functions, 2016, 27 (12): 965- 973

Macdonald I G . Symmetric Functions and Hall Polynomials. 2nd ed Oxford: Clarendon Press, 1995

Zeng C N , Jiang D S .

Some notes on mean curvature integral of convex sets

Journal of Chongqing Normal University (Natural Science), 2013, 30 (5): 62- 65

Zeng C N , Bo S K .

Some inequalities on mean curvature integral of convex sets (Ⅱ)

Journal of Chongqing Normal University (Natural Science), 2017, 34 (6): 57- 60

Zeng C N , Ma L , Xia Y W .

On mean curvature integrals of the outer parallel body of the projection of a convex body

Journal of Inequalities and Applications, 2014, Article number, 415

$\mathbb{S}.4$中的Willmore泛函与包含问题

Zhou J Z .

Willmore functional and inclusion problems in $\mathbb{S}.4$

Science in China, 2007, 37 (2): 249- 256

Zhang G Y, Zhou J Z. Containment Measures in Integral Geometry//Grinberg E, Li S G, Zhang G Y, Zhou J Z. Integral Geometry and Convexity. Singapore: World Scientific, 2005: 153-168

John G .

On Wallis formula

The American Mathematical Monthly, 1956, 63 (9): 643- 645

Guo B N , Q F .

On the increasing monotonicity of sequence originating from computation of the probability of intersecting between a plane couple and a convex body

Turkish Journal of Analysis and Number Theory, 2015, 3 (1): 21- 23

/

 〈 〉