aa r X i v : . [ m a t h . HO ] A p r Steven Finch
April 24, 2019
Abstract.
Let the Euclidean plane be simultaneously and independentlyendowed with a Poisson point process and a Poisson line process, each of unitintensity. Consider a triangle T whose vertices all belong to the point process.The triangle is 0-pierced if no member of the line process intersects any side of T . Our starting point is Ambartzumian’s 1982 joint density for angles of T ;our exposition is elementary and raises several unanswered questions. A triangle with angles α , β , γ is acute if max { α, β, γ } < π/ well-conditioned if min { α, β, γ } > π/
6. Given a random mechanism for generating triangles in theplane, we dutifully calculate corresponding probabilities out of sheer habit and forthe sake of numerical concreteness.Beginning with a Poisson point process of unit intensity, let us form a triangleby taking the convex hull of three particles (members of the process). The triangleis if no other particles are contained in the convex hull. Study of suchconfigurations is complicated by the prevalence of long, narrow triangles with anglestypically ≈ ≈ π . We defer discussion of these until later.Beginning with a Poisson point process and a Poisson line process, also of unitintensity and independent, let us form a triangle as before. The triangle is if the intersection of each line with the convex hull is always empty. Nothing ispresumed about the existence or number of other interior particles; there may be 0or 1 or 2 or many more. Since the angles satisfy α + β + γ = π , we can eliminate γ from consideration and write the joint density for α , β as [1, 2, 3, 4]42 π sin( x ) sin( y ) sin( x + y )[sin( x ) + sin( y ) + sin( x + y )] where x > y > x + y < π . This is a remarkable result, owing to the scatteredcomplexity of particles overlaid with lines. Integrating out y , we obtain the marginaldensity for α : f ( x ) = 212 π (7 − x )) (1 + cos( x )) + 4 (5 − cos( x )) (1 − cos( x )) ln (cid:0) sin( x ) (cid:1) (1 + cos( x )) Copyright c (cid:13) -Pierced Triangles within a Poisson Overlay E ( α ) = π . ..., E ( α ) = 1310 − = 1 . .... Corresponding to the density for max { α, β, γ } , the expression 3 f ( x ) holds when π/ < x < π ; the expression when π/ < x < π/ f ( x ) where˜ f ( x ) = 212 π (7 − x )) (1 − x )) − − cos( x )) (1 − cos( x )) ln (cid:0) x ) (cid:1) (1 + cos( x )) . It thus follows that P (a typical 0-pierced triangle is acute) = 96 −
132 ln(2) − π π = 0 . .... Corresponding to the density for min { α, β, γ } , the expression − f ( x ) holds when0 < x < π/ P (a typical 0-pierced triangle is well-conditioned) = − √ − √
3) ln(2) + (4380 − √
3) ln( − √
3) + (71 + 41 √ π √ π = 0 . .... From V ( α + β + γ ) = 0, we deduce that E ( αβ ) − E ( α ) E ( β ) = − (1 / V ( α ) andtherefore E ( α β ) = − − + π . .... Simulation provides compelling evidence that Ambartzumian’s [1, 2] joint density isvalid – see Figure 1 – although it does not provide insight leading to an actual proof.
1. Related Expressions
We turn attention to the bivariate densities C j sin( x ) sin( y ) sin( x + y )[sin( x ) + sin( y ) + sin( x + y )] j where x > y > x + y < π and C = 412 − π , C = 1( − π , C = 8 . The case j = 1 appears in [5, 6] with regard to cells of a Goudsmit-Miles tessellation(sampled until a triangle emerges); the case j = 3 appears in [7, 8] with regard to -Pierced Triangles within a Poisson Overlay j = 2,we obtain the univariate density for α : g ( x ) = 12 ( − π ( π − x ) sin( x ) + 4 (1 − cos( x )) ln (cid:0) sin( x ) (cid:1) x )and E ( α ) = 4 ( π −
12 ln(2)) ln(2) − ζ (3)6 ( − . ... where ζ (3) is Ap´ery’s constant [9].Corresponding to the density for max { α, β, γ } , the expression 3 g ( x ) holds when π/ < x < π ; the expression when π/ < x < π/ g ( x ) where˜ g ( x ) = 12 ( − π (3 x − π ) sin( x ) − − cos( x )) ln (cid:0) x ) (cid:1) x ) . It thus follows that P (a typical such triangle is acute) = −
24 ln(2) + (4 − π + 12 G − π = 0 . ... -Pierced Triangles within a Poisson Overlay G is Catalan’s constant [10]. Corresponding to the density for min { α, β, γ } , theexpression − g ( x ) holds when 0 < x < π/ P (a typical such triangle is well-conditioned) = 0 . ... (exact expression omitted for reasons of length). As before, we deduce that E ( α β ) = 2 ( π + 24 ln(2)) ln(2) − π + 3 ζ (3)12 ( − . .... What’s missing, of course, is a natural procedure for generating (not necessarilyplanar) triangles whose angles α , β , γ obey the distributional law prescribed by j = 2.
2. 0-Filled Triangles
The phrase “0-filled” first appeared in [11, 12]. Let us initially discuss the simulationunderlying 0-pierced triangles. Given a parameter value λ >
0, we generated data( α , β ), ( α , β ), . . . , ( α n , β n ) via Poisson overlays in the planar disk of radius λ centered at the origin. Our goal was to verify a probability theoretic expression: P ( x < α ≤ x + dx, y < β ≤ y + dy ) = 42 π sin( x ) sin( y ) sin( x + y )[sin( x ) + sin( y ) + sin( x + y )] dx dy as λ → ∞ . This was done simply by histogramming the data, given large enough n and λ .For 0-filled triangles, however, we face a situation where the goal is less tangible.Ambartzumian’s measure theoretic expression [2]: M ( x < α ≤ x + dx, y < β ≤ y + dy ) = 2sin( x ) sin( y ) sin( x + y ) dx dy cannot be normalized to give a probability density (that is, encompassing unit area).It follows that [13, 14] M ( x < max { α, β, γ } ≤ x + dx ) = (cid:26) −
12 csc( x ) ln (2 cos( x )) if π/ ≤ x < π/ , ∞ if π/ ≤ x < π and, for 0 < x < π/ M ( x < min { α, β, γ } ≤ x + dx ) = 12 csc( x ) ln (2 cos( x ))– see Figures 2 and 3 – but verification is problematic. As before, we can generate -Pierced Triangles within a Poisson Overlay φ ( x ) = 4 (3 x − π + 3 cot( x ) ln(2 cos( x ))) for π/ < x < π/ φ ( x ) = 4 (3 x − π + 3 cot( x ) ln(2 cos( x ))) for 0 < x < π/ -Pierced Triangles within a Poisson Overlay λ .data over disks of increasing radius λ . Figure 4 provides histograms of α for λ =2 , , ,
5; Figures 5 and 6 do likewise for max { α, β, γ } and min { α, β, γ } . Clearlylim λ →∞ P (a typical 0-filled triangle is acute) = 0 , lim λ →∞ P (a typical 0-filled triangle is well-conditioned) = 0on empirical grounds. Unfortunately we do not know how to confirm theoreticalpredictions stemming from [13, 14]: M (a typical 0-filled triangle is acute) = 2 π ≈ . , M (a typical 0-filled triangle is well-conditioned) = 2 (cid:16) √ − π (cid:17) ≈ . λ → ∞ , would be welcome.
3. Process Intensities
We report here on work in [15]. Given a Poisson overlay Ω, define a T · process tobe the set of all 0-pierced triangles within Ω. Let the intensity i of the process bethe mean number of triangles per unit area. It is known that i ( T · ) = 2 π
21 = 0 . ... = 16 (5 . ... ) . -Pierced Triangles within a Poisson Overlay λ .Figure 6: Histograms for minimum angle in 0-filled triangles, for increasing λ . -Pierced Triangles within a Poisson Overlay / i ( T · ) = ∞ . Most interesting, however,is the set of all triangles that are both 0-filled and 0-pierced: i ( T ) = 0 . ... = 16 (3 . ... )= π √ π Z ξ ( x ) η ( x ) p b ( x ) ( a ( x ) − c ( x )) sin (cid:16) x (cid:17) dx where a ( x ) = 23 (cid:16) cos (cid:16) x (cid:17) + 1 (cid:17) , b ( x ) = 23 (cid:18) cos (cid:18) x − π (cid:19) + 1 (cid:19) , c ( x ) = 23 (cid:18) cos (cid:18) x + 2 π (cid:19) + 1 (cid:19) ,q ( x ) = s a ( x ) ( b ( x ) − c ( x )) b ( x ) ( a ( x ) − c ( x )) , h ( x ) = 2(27) / cos (cid:16) x (cid:17) − / ,ξ ( x ) = 2 (cid:0) h ( x ) (cid:1) − √ π (cid:0) h ( x ) (cid:1) h ( x ) exp (cid:18) h ( x ) (cid:19) erfc (cid:18) h ( x )2 (cid:19) ,η ( x ) = (cid:18) c ( x ) − a ( x ) (cid:19) E ( q ( x )) + (cid:18) a ( x ) − (cid:19) K ( q ( x )); K ( y ) = π/ Z p − y sin( θ ) dθ, E ( y ) = π/ Z p − y sin( θ ) dθ are complete elliptic integrals of the first and second kind; anderf( z ) = 2 √ π z Z exp( − τ ) dτ = 1 − erfc ( z )is the error function. Formulas (7) and (8) in [15], devoted to a more general scenario T kℓ than our T , specialize to ∞ Z s exp (cid:0) − s − t √ s (cid:1) ds = 18 (cid:20) (cid:0) t (cid:1) − √ π (cid:0) t (cid:1) t exp (cid:18) t (cid:19) erfc (cid:18) t (cid:19)(cid:21) for t > D − (cid:0) t/ √ (cid:1) which is lessfamiliar).Theory fails for T – we do not possess density predictions for the histogramsin Figure 7 – nor do we know exact probabilities that a such a triangle is acute orwell-conditioned. -Pierced Triangles within a Poisson Overlay T triangles. References [1] R. V. Ambartzumian, Random shapes by factorization,
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