Large Cells in Poisson-Delaunay Tessellations

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1 Large Cells in Poisson-Delaunay Tessellations Daniel Hug an Rolf Schneier Mathematisches Institut, Albert-Luwigs-Universität, D Freiburg i. Br., Germany aniel.hug, Abstract. It is prove that the shape of the typical cell of a Delaunay tessellation, erive from a stationary Poisson point process in -imensional Eucliean space, tens to the shape of a regular simplex, given that the volume of the typical cell tens to infinity. This follows from an estimate for the probability that the typical cell eviates by a given amount from regularity, given that its volume is large. As a tool for the proof, a stability result for simplices is establishe. 1 Introuction an main result Voronoi tessellations (also calle Voronoi mosaics or Voronoi iagrams) an their uals, Delaunay tessellations, are a thoroughly stuie subject of iscrete geometry. The book by Okabe, Boots, Sugihara an Chiu [11] gives an impression of the richness of the theory of these tessellations an of the variety of their applications. If the iscrete point set in R from which such a tessellation is erive is ranom, one gets a ranom tessellation. Important examples are the Poisson-Voronoi an Poisson-Delaunay tessellations, which are erive from (stationary) Poisson point processes. We refer to Chapter 5 of [11], Chapter 10 of Stoyan, Kenall an Mecke [14], an Chapter 6 of Schneier an Weil [13] for introuctions to ranom tessellations. A conjecture of D. G. Kenall initiate the stuy of limit shapes of large cells in special ranom mosaics. In the early 1940s, Kenall conjecture (as ocumente in the introuction to the first eition of [14]) that the shape of the zero cell of the ranom tessellation generate by a stationary an isotropic Poisson line process in the plane tens to circular shape given that the area of the zero cell tens to infinity. Contributions to this problem were mae by Miles [9] an Golman [1], an Kenall s conjecture was finally prove by Kovalenko [3], [5]. In [], the limit shape for zero cells an typical cells of not necessarily isotropic, stationary Poisson hyperplane tessellations was foun, an the probability of large eviations from the limit shape was estimate. Kovalenko 1

2 [4] treate an analogue of Kenall s problem for the typical cell of a stationary Poisson- Voronoi tessellation in the plane. Again, the shape tens to circularity, given that the area tens to infinity. The present paper is in a similar spirit. We consier the typical cell of a stationary Poisson-Delaunay mosaic in -imensional space an prove, as a consequence of a precise estimate, that its shape tens to that of a regular simplex, given that the volume tens to infinity. Let X be a stationary Poisson point process with intensity λ > 0 in -imensional Eucliean space R ( ). Let Y enote the Poisson-Delaunay tessellation erive from X. The typical cell of Y (as efine in [13, 6.]) is enote by Z (explanations are given below in Section ). Almost surely, Z is a simplex which is inscribe to a sphere centere at the origin. For the formulation of our result, we nee a measure for the eviation of the shape of a simplex from the shape of a regular simplex. For -simplices S 1, S, we efine η(s 1, S ) as the smallest number η with the property that for each vertex p of one of the simplices there is a vertex q of the other such that p q η (here enotes the Eucliean norm). Note that δ(s 1, S ) η(s 1, S ), where δ enotes the Hausorff metric (cf. [1, 1.8]). For a -simplex S, let z be the center an r the raius of the sphere through the vertices of S, an set ρ(s) := minη(r 1 (S z), T ) : T T }, where T enotes the set of regular simplices inscribe to the unit sphere S 1. By P we enote the unerlying probability, an P( ) is a conitional probability. We write V for volume in R. Theorem 1. Let Y enote the Poisson-Delaunay tessellation erive from a stationary Poisson process with intensity λ > 0 in R ; let Z be its typical cell. There is a constant c 0 epening only on such that the following is true. If ɛ (0, 1) an I = [a, b) is any interval (possibly b = ) with aλ σ 0 > 0, then P(ρ(Z) ɛ V (Z) I) c exp c 0 ɛ aλ }, where c is a constant epening only on, ɛ, σ 0. As a consequence, we have for any fixe ɛ > 0. lim P(ρ(Z) ɛ V (Z) a) = 0 a Using similar arguments as in [], one can also euce a corresponing result for the zero cell Z 0 (the cell containing the origin of R ) of Y. This will not be carrie out here, since the proceure is clear from []. The proof of Theorem 1 is base on a geometric stability result for simplices (Theorem in Section 3) an on two estimates, provie by Lemmas an 3 in Section 4. The erivation of these estimates is facilitate by an explicit formula for the istribution of Z, which is ue to R. E. Miles.

3 The typical cell of a Poisson-Delaunay tessellation We recall briefly the notion of a Poisson-Delaunay tessellation an its typical cell (etails can be foun, e.g., in [13]). Let X be a stationary Poisson process in R, with intensity λ > 0. With probability one, no + 1 points of X are in a hyperplane, an no + points lie on a sphere. If + 1 points x 1,..., x +1 of X lie on a sphere that contains no point of X in its interior, then the convex hull of x 1,..., x +1 is calle a cell. The set Y of all such cells is a tessellation of R by simplices, the Poisson-Delaunay tessellation erive from X. (This construction of a Delaunay tessellation is equivalent to the usual one as the ual of a Voronoi tessellation.) Since Y can be consiere as a stationary particle process (of intensity λ = [( + 1)a()] 1 λ, where a() is given by (1) below), one can associate with it a shape istribution (cf. [13, 4.]). It can be escribe as follows. For a -simplex S, we enote by z(s) the center of the sphere through the vertices of S. Let 0 be the set of all -simplices S in R with z(s) = 0. Let C be the cube [ 1/, 1/]. The shape istribution of Y is the probability measure Q 0 on 0 with the property that Q 0 (A) = 1 λ E cars Y : z(s) C, S z(s) A} for Borel sets A 0 ; here E enotes mathematical expectation. The typical cell of the Poisson-Delaunay tessellation Y is efine as a ranom polytope with istribution Q 0. A more intuitive interpretation of this istribution is possible ue to the fact that stationary Poisson-Delaunay tessellations are mixing an hence ergoic ([13, Satz 6.4.]). This entails that, for A as before, hols with probability one. Q 0 (A) = lim r cars Y : z(s) rc, S z(s) A} cars Y : z(s) rc } For example, suppose we are intereste in P(V (Z) a), the probability that the typical cell has volume at least a > 0. Then we can take an arbitrary realization of the tessellation Y an a large number r an consier, among the cells S of the realization with center z(s) in the cube rc, the relative frequency of those with volume at least a. This proportion will almost surely be a goo approximation to the probability P(V (Z) a). We shall make use of the explicit integral representation of the istribution Q 0 given by Lemma 1. It is ue to Miles [8, formula (76)]; the proof can also be foun in [10, Theorem 7.5] an [13, Satz 6..10]. Let σ enote the spherical Lebesgue measure on the unit sphere S 1, an let κ be the volume of the -imensional unit ball. Lemma 1. Let Y be the Delaunay tessellation erive from a stationary Poisson process of intensity λ > 0 in R, an let Q 0 be the istribution of its typical cell. Let A 0 be a Borel set. Then Q 0 (A) = a()λ λκr 1 A (convru 0,..., ru })e r 1 0 S 1 S 1 V (convu 0,..., u }) σ(u 0 ) σ(u ) r 3

4 with a() := +1 π 1 Γ ( ) Γ ( +1 ) [ ( Γ +1 ) ] Γ ( + 1). (1) 3 A stability result for simplices For a -imensional simplex S R we say that S is inscribe to the unit sphere S 1 if the vertices of S lie on S 1. Let S be such a simplex an suppose that it has maximal volume among all simplices inscribe to S 1. Then it is easy to see that S is a regular simplex (e.g., [7, p. 317]). Here we nee an improve version of such a uniqueness result, in the form of a stability estimate. In the following, T is a regular simplex inscribe to S 1. Theorem. There is a positive constant c() such that the following is true for any ɛ [0, 1]. If S is a simplex inscribe to S 1 an if ρ(s) ɛ, then V (S) (1 c()ɛ )V (T ). () Proof. First we consier the case =, where we show that c() = 1/1 is a possible choice. Let S be a triangle inscribe to S 1 an satisfying V (S) > (1 ɛ /1)V (T ); here V (T ) = 3 3/4. Then 0 int S. Let α, β, γ be the angles at 0 spanne by the eges of S. Then V (S) = sin α cos α + sin β cos β + sin γ cos γ an α+β+γ = π. We can choose the notation in such a way that the angles ϕ := α π/3 an ψ := β π/3 are either both non-negative or both non-positive. An elementary calculation gives V (S) V (T ) = 3 [cos ϕ + cos ψ ] sin(ϕ + ψ)[sin ϕ sin β + sin ψ sin α]. Since either ϕ 0, ψ 0 or ϕ 0, ψ 0 (an ϕ + ψ < π), we get We euce that hence, observing that π/3 ϕ π/6, sin(ϕ + ψ)[sin ϕ sin β + sin ψ sin α] 0. ɛ < V (S) V (T ) 1 4 (cos ϕ 1), 1 1 ϕ cos ϕ > ɛ. 4

5 Thus ϕ < ɛ/, an similarly ψ < ɛ/, hence α π/3 < ɛ an β π/3 < ɛ. Let p be the vertex of S common to the eges spanning the angles α an β. Let T be the regular triangle inscribe to S 1 with one vertex at p. Then η(s, T ) < ɛ. This proves the assertion for =. Now let 3, an assume that the assertion has been prove in imension 1. Let S be a -simplex inscribe to S 1 an satisfying V (S) > (1 α)v (T ) for some given number α > 0. First we assume that 0 int S. The inraius r(s) of S satisfies r(s) 1/ (see, e.g., [6, Satz 1], or [7, Lemma 13..]). Hence, there is at least one facet of S, say F, which has a istance at most 1/ from 0. Therefore, there are a vector u S 1 an a number t (0, 1/] such that aff F = H(u, t) := z R : u, z = t}. Let p be the vertex of S not in F, an let q be the point in S 1 with maximal istance from H(u, t). Let a be the istance between the hyperplanes through p an q parallel to H(u, t). The ( 1)-volume of F is less than or equal to the ( 1)-volume of a regular ( 1)-simplex inscribe to H(u, t) S 1. Moreover, the function x (1 x ) 1 (1 + x) attains a unique maximum on the interval [0, 1] at 1/. All this implies V (S) = 1 V 1(F )(1 + t a) 1 V 1(T 1 )(1 t ) 1 (1 + t a) 1 V 1(T 1 )(1 t ) 1 (1 + t) 1 V 1(T 1 ) (1 1 ) 1 ( ) = V (T ) < V (S) + αv (T ). From this chain of inequalities, we raw three conclusions. In this proof, c 1, c,... enote positive constants epening only on the imension. The first conclusion is that 1 V 1(T 1 )(1 t ) 1 a < αv (T ). Here t 1/, hence a < c 1 α. Since p q = a, we get p q < c α. (3) 5

6 Let The secon conclusion is that [ ( 1 V 1(T 1 ) 1 1 ) 1 ( ) ] (1 t ) 1 (1 + t) < αv (T ). (4) The first two erivatives are given by an g(x) := (1 x ) 1 (1 + x) for x [0, 1]. g (x) = (1 x ) 3 (x + ( 1)x 1) g (x) = ( 1)(1 x ) 5 (x 3 + ( )x 3x 1), for x [0, 1). Since g (1/) = 0, we get g(x) = g ( ) ( g (ξ) x ) 1, x [0, 1/], (5) with a suitable ξ [x, 1/]. We estimate g from above by a negative constant. For this, we set f(x) := x 3 + ( )x 3x 1. An elementary iscussion shows that f is strictly ecreasing in [0, 1/]. In particular, we euce that f(x) f(0) = 1 for x [0, 1/]. This shows that, for x [0, 1/], g (x) ( 1)(1 x ) 5 ( 1), 3, 4}, ( 1)(1 ) 5, 5, hence 1 g (x) c 3 for x [0, 1/] with c 3 > 0. Now (5) gives an from (4) we conclue that Our thir conclusion is that 1 c 3 (t 1/) g(1/) g(t), [V 1 (T 1 )(1 t ) 1 V 1 (F ) t 1/ < c 4 α. (6) Here 1 + t a > 1 + 1/ c 4 α c1 α > 1, if we assume that ] (1 + t a) < αv (T ). (7) c 4 α + c1 α < 1/. (8) 6

7 The ( 1)-simplex F := (1 t ) 1/ (F tu) is inscribe to H(u, 0) S 1 an, as a consequence of (7) an of t 1/, satisfies V 1 (F ) > V 1 (T 1 ) αv (T )(1 ) 1 = (1 c 5 α)v 1 (T 1 ). (9) Let c( 1) be the constant appearing in the inuction hypothesis. We assume that c 5 α/c( 1) 1 (10) an put c 5 α/c( 1) =: γ. Then (9) an the inuction hypothesis imply that η(t 1, F ) < γ for a suitably chosen regular ( 1)-imensional simplex T 1 H(u, 0) S 1. Let T := (1 1 ) 1 T u an T := conv (T q}). Then T is a regular -simplex, inscribe to S 1. The two ( 1)-simplices F an T have the property that to each vertex v of one of them there is a vertex w of the other such that v w < (1 t )γ + 1 t 1/ 1 1/ c 5 α/c( 1) + c 6 α c7 α by (6). Together with (3), this shows that η(s, T ) < c 8 α. (11) Now we choose c() > 0 so small that c 8 c() 1 an that α c() implies (8) an (10). Let ɛ (0, 1] be given, an put α := c()ɛ. Then V (S) > ( 1 c()ɛ ) V (T ) implies η(s, T ) < ɛ. Finally, we can ecrease c(), if necessary, so that 0 / int S implies V (S) (1 c())v (T ). The inuction step is now finishe, hence the proof of Theorem is complete. Remark. The estimate () is of optimal orer, that is, ɛ cannot be replace by a smaller power of ɛ. This is easily seen by appropriately moving one vertex of a regular simplex. 7

8 4 Proof of Theorem 1 In the following, we set := V (T ). By c 1,..., c 5 we enote positive constants epening only on or on an ɛ, as inicate. We write c 1 = c 1 () for the constant c() in Theorem an c = c () for the constant a() of Lemma 1 in Section. Lemma. For each ɛ (0, 1), there is a constant c 3 = c 3 (, ɛ) such that, for 0 < h h 0 := (c 1 /(c 1 + 1))ɛ an aλ > 0, P(V (Z) a[1, 1 + h]) c 3 h(aλ) exp κ ( 1 + (c1 /4)ɛ ) } aλ. Proof. Let ɛ (0, 1), h (0, h 0 ] an a > 0, λ > 0 be given. For x 0,..., x R, we set V (x 0,..., x ) := V (convx 0,..., x }). From Lemma 1, we obtain P(V (Z) a[1, 1 + h]) = c λ 0 S 1 Substituting s = λκ r, we get S 1 1 } r V (u 0,..., u ) a[1, 1 + h] exp λκ r } r 1 V (u 0,..., u ) σ(u 0 ) σ(u ) r. P(V (Z) a[1, 1 + h]) = c κ S 1 S s aλκ /V (u 0,..., u )[1, 1 + h]} e s s 1 s V (u 0,..., u ) σ(u 0 ) σ(u ). For fixe u 0,..., u S 1 in general position, we apply to the inner integral the mean value theorem for integrals. This gives the existence of some such that P(V (Z) a[1, 1 + h]) = c κ 1 c κ 1 ξ(u 0,..., u ) aλκ /V (u 0,..., u )[1, 1 + h] (1) haλ S 1 exp ξ(u 0,..., u )}ξ(u 0,..., u ) 1 σ(u 0 ) σ(u ) S 1 haλ } } R(,ɛ) exp ξ(u 0,..., u )}ξ(u 0,..., u ) 1 σ(u 0 ) σ(u ), 8

9 where R(, ɛ) := (u 0,..., u ) (S 1 ) +1 : V (u 0,..., u ) ( 1 + (c 1 /1)ɛ ) 1 τ }. For (u 0,..., u ) R(, ɛ) we can estimate an ξ(u 0,..., u ) aλκ / ξ(u 0,..., u ) (1 + h 0 )(1 + (c 1 /1)ɛ )aλκ /. Since σ +1 (R(, ɛ)) epens only on an ɛ, this gives Now hence the assertion follows. P(V (Z) a[1, 1 + h]) c 3 (, ɛ)h(aλ) exp κ (1 + h 0 ) ( 1 + (c 1 /1)ɛ ) } aλ. (1 + h 0 ) ( 1 + (c 1 /1)ɛ ) 1 + (c 1 /4)ɛ, Lemma 3. For each ɛ (0, 1), there is a constant c 5 = c 5 (, ɛ) such that, for aλ > 0 an h > 0, P(V (Z) a[1, 1 + h], ρ(z) ɛ) c 5 haλ exp κ ( 1 + (c1 /)ɛ ) } aλ. Proof. As in the proof of Lemma, we get P(V (Z) a[1, 1 + h], ρ(z) ɛ) = c λ 0 S 1 S 1 1 1ρ(convu 0,..., u }) ɛ} } r V (u 0,..., u ) a[1, 1 + h] exp λκ r } r 1 V (u 0,..., u ) σ(u 0 ) σ(u ) r = c κ S 1 S s aλκ /V (u 0,..., u )[1, 1 + h]} e s s 1 s 1ρ(convu 0,..., u }) ɛ}v (u 0,..., u ) σ(u 0 ) σ(u ) = c κ 1 haλ S 1 exp ξ(u 0,..., u )}ξ(u 0,..., u ) 1 S 1 1ρ(convu 0,..., u }) ɛ} σ(u 0 ) σ(u ), 9

10 where again (1) hols. By Theorem, the inequality ρ(convu 0,..., u }) ɛ implies V (u 0,..., u ) (1 c 1 ɛ ). Hence, if 1ρ(convu 0,..., u }) ɛ) 0, then ξ(u 0,..., u ) κ aλ ( 1 c 1 ɛ ) 1 τ 1 Moreover, there is a constant c 4 = c 4 (, ɛ) such that, for ξ 0, ( exp( ξ)ξ 1 c 1 c 4 exp 1 (1 + c 1 ) ɛ κ (1 + c 1 ɛ )aλ. (13) The estimates (13) an (14) imply, for ξ = ξ(u 0,..., u ) as above, that exp( ξ)ξ 1 c 4 exp κ ( ) } (1 + c 1 ɛ c 1 ) 1 (1 + c 1 ) ɛ aλ Therefore, as asserte. c 4 exp κ ( 1 + (c1 /)ɛ ) } aλ. P(V (Z) a[1, 1 + h], ρ(z) ɛ) c κ 1 ) ξ haλ(κ ) +1 c 4 exp κ ( 1 + (c1 /)ɛ ) } aλ = c 5 (, ɛ)haλ exp κ ( 1 + (c1 /)ɛ ) } aλ, }. (14) The proof of Theorem 1 is now similar to the final argument in []. Let ɛ (0, 1), a > 0 an λ > 0 with aλ σ 0 > 0 be given, an let h 0 be as in Lemma. Let I = [a, b) be a given interval. The constants c 6,..., c 9 below epen only on, ɛ, σ 0. If h 0 > (b a)/a, we put h 1 := (b a)/a, then a[1, 1 + h 1 ) = [a, b) = I. Lemma gives P(V (Z) I) c 6 (, ɛ, σ 0 )h 1 aλ exp Aaλ} with A := κ (1 + (c 1 /4)ɛ ), an Lemma 3 gives P(V (Z) I, ρ(z) ɛ) c 5 (, ɛ)h 1 aλ exp Baλ} with B := κ (1 + (c 1 /)ɛ ) Both estimates together yiel P(ρ(Z) ɛ V (Z) I) c 7 (, ɛ, σ 0 ) exp (B A)aλ} 10

11 with B A = κ τ 1 (c 1/4)ɛ. Suppose now that h 0 (b a)/a. Then 1 + h 0 b/a an a[1, 1 + h 0 ) [a, b). Lemma gives P(V (Z) I) c 6 (, ɛ, σ 0 )h 0 aλ exp Aaλ}. (15) For i N 0, Lemma 3 (together with a(1 + h 0 ) i λ σ 0 ) gives P(V (Z) a(1 + h 0 ) i [1, 1 + h 0 ], ρ(z) ɛ) c 5 (, ɛ)h 0 (1 + h 0 ) i aλ exp Ba(1 + h 0 ) i λ} = c 5 (, ɛ)h 0 (1 + h 0 ) i aλ exp Aa(1 + h 0 ) i λ} exp (B A)a(1 + h 0 ) i λ} c 5 (, ɛ)h 0 aλ exp Aaλ} exp ((B A)/)aλ} (1 + h 0 ) i exp ((B A)/)σ 0 (1 + h 0 ) i }. From [a, b) i=0 a(1 + h 0) i [1, 1 + h 0 ] we now get P(V (Z) I, ρ(z) ɛ) c 5 (, ɛ)h 0 aλ exp Aaλ} exp ((B A)/)aλ} (1 + h 0 ) i exp ((B A)/)σ 0 (1 + h 0 ) i } i=0 = c 8 (, ɛ, σ 0 )h 0 aλ exp Aaλ} exp ((B A)/)aλ}. Together with (15), this gives P(ρ(Z) ɛ V (Z) I) c 9 (, ɛ, σ 0 ) exp ((B A)/)aλ}. This completes the proof of Theorem 1. References [1] A. Golman, Sur une conjecture e D. G. Kenall concernant la cellule e Crofton u plan et sur sa contrepartie brownienne, Ann. Probab. 6 (1998), [] D. Hug, M. Reitzner, an R. Schneier, The limit shape of the zero cell in a stationary Poisson hyperplane tessellation. Ann. Probab. (to appear). [3] I. N. Kovalenko, A proof of a conjecture of Davi Kenall on the shape of ranom polygons of large area, (Russian) Kibernet. Sistem. Anal. 1997, 3 10, 187; Engl. transl. Cybernet. Systems Anal. 33 (1997),

12 [4] I. N. Kovalenko, An extension of a conjecture of D. G. Kenall concerning shapes of ranom polygons to Poisson Voronoï cells, In: Engel, P. et al. (es.), Voronoï s impact on moern science. Book I. Transl. from the Ukrainian. Kyiv: Institute of Mathematics. Proc. Inst. Math. Natl. Aca. Sci. Ukr., Math. Appl. 1 (1998), [5] I. N. Kovalenko, A simplifie proof of a conjecture of D. G. Kenall concerning shapes of ranom polygons, J. Appl. Math. Stochastic Anal. 1 (1999), [6] K. Leichtweiß, Über ie affine Exzentrizität konvexer Körper, Arch. Math. 10 (1959), [7] J. Matoušek, Lectures on Discrete Geometry, Grauate Texts in Mathematics 1, Springer, New York, 00. [8] R. E. Miles, A synopsis of Poisson flats in Eucliean spaces, Izv. Aka. Nauk Arm. SSR, Mat. 5 (1970), 63 85; reprinte in Stochastic Geometry (E. F. Haring, D. G. Kenall, es.) Wiley, New York, 1974, pp [9] R. E. Miles, A heuristic proof of a long-staning conjecture of D. G. Kenall concerning the shapes of certain large ranom polygons, Av. in Appl. Probab. 7 (1995), [10] J. Møller, Ranom tessellations in R, Av. in Appl. Probab. 1 (1989), [11] A. Okabe, B. Boots, K. Sugihara, an S. N. Chiu, Spatial Tessellations; Concepts an Applications of Voronoi Diagrams., n e., Wiley, Chichester, 000. [1] R. Schneier, Convex Boies: the Brunn-Minkowski Theory, Encyclopeia of Mathematics an Its Applications 44, Cambrige University Press, Cambrige, [13] R. Schneier an W. Weil, Stochastische Geometrie, Teubner Skripten zur Mathematischen Stochastik, Teubner, Stuttgart, 000. [14] D. Stoyan, W. S. Kenall, an J. Mecke, Stochastic Geometry an its Applications, n e., Wiley, Chichester,