Some geometric probability problems involving the Eulerian numbers

Transcription

1 Some geometrc probablty problems nvolvng the Euleran numbers Frank Schmdt Rodca Smon Department of Mathematcs The George Washngton Unversty Washngton, DC Dedcated to Herb Wlf on the occason of hs sxty-ffth brthday Abstract We present several problems nvolvng geometrc probablty. Each s related to the dvson of a smplex or cube by a famly of hyperplanes. Both the classcal Euleran numbers and ther analogue for the hyperoctahedral group arse n the solutons. 0. Introducton Consder the followng general type of problem: From a convex polytope P R n, select a pont x = (x 1, x 2,..., x n ) at random accordng to a certan fxed dstrbuton. Gven a functon f: P R and a sequence of functons ρ 0, ρ 1,..., ρ m : P R, satsfyng ρ 0 (x) ρ 1 (x)... ρ m (x) and ρ 0 (x) f(x) ρ m (x), what s the probablty that ρ 1 (x) f(x) ρ (x)? For example, what s the probablty that the average of the coordnates of x s at most 1 2, f x s selected unformly at random from P = [0,1]n, the n-dmensonal unt cube? Ths arses upon choosng f(x) = x 1+x x n n, and the constant functons ρ 0 (x) = 0, ρ 1 (x) = 1 2, and ρ 2 (x) = 1. Partally supported through NSF grant DMS

2 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 2 Here we present several problems of ths type, n whch the selecton of x s done unformly at random, and the functons f, ρ 0, ρ 1,..., ρ m are lnear. Hence, the problems can be reformulated geometrcally as follows. Let H 1, H 2,..., H m be a sequence of affne hyperplanes n R n. For each, let H + and H be the two closed half-spaces n R n determned by H. In terms of the orgnal formulaton of the problem, the hyperplane H has equaton f(x) = ρ (x), and H + = {x R n : f(x) ρ (x)}. If a pont x s selected unformly at random from the polytope P, what s the probablty that x les n P H 1 H+? Snce the selecton of x s done unformly at random, ths geometrc probablty can be expressed n terms of the (n-dmensonal) volume of the regon P H 1 H+. Thus, we are led to consder problems n whch the goal s to fnd the volumes of the regons nto whch a polytope s dvded by a famly of hyperplanes. When the choce of polytope s P = [0, 1] n, the selecton of a pont x unformly at random corresponds to the selecton of x 1, x 2,..., x n from [0, 1] ndependently at random, accordng to the unform dstrbuton. A related probablstc queston s to consder order statstcs. That s, after selectng x = (x 1, x 2,..., x n ) as above, frst reorder ts coordnates n weakly ncreasng order, and then apply the functons f and ρ to the ncreasngly ordered n-tuple x R n thus obtaned. Clearly, x les n the n-dmensonal smplex n := {(x 1, x 2,..., x n ) R n : 0 x 1 x 2... x n 1}. If the selecton s accordng to the unform dstrbuton, then the desred probablty can be obtaned by selectng x unformly drectly from the smplex n. Ths leads to the queston of fndng the volumes of the regons nto whch the smplex n s dvded by a famly of hyperplanes. The specfc problems descrbed here pertan to the cube [0, 1] n and the smplex n, and the answers turn out to nvolve well-known sequences from combnatoral enumeraton. The (n 1)-dmensonal volumes of the sectons [0,1] n H or n H turn out to have closely related expressons as well. As an ntal example, consder the smplex n and the functons f(x) = 0, ρ 0 (x) = α, and ρ (x) = x α, for some real number α (0, 1) and 1 n. If x s chosen unformly at random from n, then the probablty Pr[ρ 1 (x) f(x) ρ (x)] equals Pr[x 1 α x ] (where x 0 = 0). The values of the frst 1 coordnates, selected from [0, 1], are all from [0, α] wth probablty α 1. Smlarly, the values of the th through nth coordnates are from [α,1] wth probablty (1 α) n +1. Fnally, upon orderng the coordnates n ncreasng order, n! we obtan the probablty ( 1)!(n +1)! α 1 (1 α) n +1. Equvalently, the hyperplanes H wth equatons x = α, for 1 n, dssect n nto n + 1 regons wth volumes gven by ( ) = 1 n n! 1) α 1 (1 α) n +1, for 1 n + 1. When α = 1 2, the volumes have especally smple expressons proportonal to bnomal coeffcents: ( ) = 1 n n!2 1). n (Ths corresponds to the case of a far con f the problem s phrased n terms of tossng a con havng probablty α for heads.) The sectons S (n) := n H have (n 1)-dmensonal volumes proportonal to bnomal coeffcents as well (for a gven, we have x = α, and the 2

3 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 3 probablty that x 1,..., x 1 α and x +1,..., x n α can be computed smlarly to the prevous calculaton). The precedng example nvolves a pencl of n hyperplanes passng through the pont (α, α,..., α) R n, whch are parallel to the coordnate hyperplanes, and the volumes of the resultng regons and sectons of n nvolve bnomal coeffcents. In Secton 2 we consder the smplex and a dfferent pencl of n hyperplanes passng through ( 1 2, 1 2,..., 1 2 ). The volumes of the resultng regons and sectons of the smplex turn out to be related to the Euleran numbers. These numbers are well-known n permutaton enumeraton. Frst we solve the problem of fndng the volumes of the regons through a drect geometrc and nductve argument. Then we sketch a second approach, usng tools from probablty theory. The Euleran numbers arse agan n Secton 3, n a problem datng back to Laplace, concernng the cube and a certan famly of parallel hyperplanes. Smlar results are derved for a dfferent famly of parallel hyperplanes and the cube, whch gve rse to the analogue of Euleran numbers for the hyperoctahedral group. Ths follows readly from recurrence relatons gven n [ChLo], where the volumes of regons and sectons are dscussed. We provde an explanaton for ths connecton between geometry and the enumeraton of sgned permutatons, by adaptng to the hyperoctahedral case a result for the symmetrc group (the prevous problem) found by Stanley [St2] n response to a queston posed by Foata [Fo]. The fnal secton ncludes several open problems. 1. Euleran numbers and the smplex Suppose that a pont x s selected unformly at random from the smplex n = {x = (x 1, x 2,..., x n ) R n : 0 x 1 x 2... x n 1}. What s the probablty that the average of 0, 1, and the coordnates of x falls between the ( 1)st and th coordnates of x? Note that Pr[x 1 < x x n+2 n+2 x ] wth x selected from n+2 gves the same dstrbuton, as can be seen by condtonng on the values of x 1 and x n+2. Ths geometrc probablty queston arses from choosng f(x) = x 1+x x n +1 n+2, ρ 0 (x) = 0, and ρ (x) = x for = 1,2,..., n. Thus, we consder the hyperplanes H R n wth equatons x 1 + x x n + 1 = (n + 2)x, for = 1, 2,..., n. These form a pencl of hyperplanes through the pont ( 1 2, 1 2,..., 1 2 ) Rn, and they determne n + 1 regons, R (n) 1, R(n) 2,..., R(n) n+1 n the smplex n : R (n) 1 = {x n : n k=1 x k +1 (n+2)x 1 }, R (n) = {x n : (n+2)x 1 n k=1 x k +1 (n + 2)x } for 2 n, and R (n) n+1 = {x n : (n + 2)x n n k=1 x k + 1}. The queston of fndng Pr [ x 1 < 0 + x 1 + x x n + 1 ] x n + 2 3

4 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 4 s answered then by determnng the sequence of volumes ( ) ) n+1 of these regons. =1 For example, n dmensons n = 1,2, 3, the volumes of the regons are: ( V (1) (R (1) 1 ), V (1) (R (1) 2 ) ) = ( 1 2, ) 1 2, ( V (2) (R (2) 1 ), V (2) (R (2) 2 ), V (2) (R (2) 3 ) ) = ( 1 12, 4 12, 12) 1, ( V (3) (R (3) 1 ), V (3) (R (3) 2 ), V (3) (R (3) 3 ), V (3) (R (3) 4 ) ) = ( 1 144, , , 1 144). These low-dmensonal cases suggest that the volumes are proportonal to the Euleran numbers (see, e.g., [Co], [St1]). A permutaton σ S n has a descent n poston (where 1 n 1) f σ() > σ( + 1), and the Euleran numbers count permutatons accordng to ther number of descents. Theorem 1.1. Let A(m, j) denote the Euleran numbers,.e., the number of permutatons n the symmetrc group S m havng exactly j descents. Then, the hyperplanes wth equatons n k=1 x k +1 = (n + 2)x, for 1 n, dssect the smplex n nto regons R (n) whose volumes are gven by ) = A(n + 1, 1), n!(n + 1)! and the (n 1)-dmensonal volumes of the sectons S (n) = n H for 1 n are gven by V (n 1) (S (n) A(n, 1) ) = c(n) (n 1)!n!, where c(n) = n 2 + 3n/(n + 1). For ease of exposton, we postpone the proof of the theorem untl after three prelmnary results. Lemma 1.2. For each = 1,2,..., n, the secton S (n) has (n +1) vertces x whose coordnates are x 1 = x 2 =... = x s = 0, x s+1 =... = x =... = x r = n r + 1 n r + s + 2, x r+1 =... = x n = 1, where 0 s < r n. In partcular, the pont ( 1 2, 1 2,..., 1 2 ) belongs n the ntersecton of n wth every hyperplane H. Proof. The vertces of n have coordnates of the form (0,0,...,0,1,1,...,1), wth the number of 0 s rangng between zero and n. Hence, an edge of n conssts of ponts havng a 4

5 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 5 certan number, s 0, of ntal coordnates equal to 0, followed by a number, r s 0, of equal coordnates whose common value les n (0,1), followed n turn by n r 0 coordnates equal to 1. To determne such a pont x whch les n H, observe that f x r+1 =... = x n = 1 for some r+1, then the equaton of H requres s 0+(r s)x r +(n r) 1+1 = (n+2) 1, mplyng x r = r+1 r s > 1. Snce such a pont x s not n n, we must have r. Smlarly, f s, then the equaton of H requres (r s)x r +(n r) 1+1 = 0, mplyng that x r < 0, so agan x n. Therefore we must have 0 s < r n, and t follows from the equaton of H that x s+1 =... = x r = n r+1 n r+s+2, and x n H. In the sequel, we wll wrte H (d) to ndcate the dmenson d of the ambent Eucldean space n whch we vew the hyperplane H. Lemma 1.3. For each such that 1 n, the regon R (n 1) S (n) of n. Proof. Frst, we descrbe the vertces of R (n 1) fall nto three classes: those of the secton S (n 1) of n 1. of n 1 s a projecton of the secton when satsfes 2 n 1. These vertces, those of the secton S (n 1) 1, and vertces In the frst two types we have vertces as descrbed n Lemma 1.2 (shftng the dmenson down to n 1). These overlap n the vertces of n 1 H (n 1) H (n 1) 1. There are ( 1)(n ) vertces n ths ntersecton, namely, for each choce of s, r such that 0 s < 1 < r n 1, we obtan a vertex x n 1 H (n 1) H (n 1) 1 whose coordnates are x 1 = x 2 =... = x s = 0, x s+1 =... = x 1 = x =... = x r = n r n r+s+1, x r+1 =... = x n 1 = 1. Only one vertex of n 1 appears n R (n 1). Ths s the vertex whose coordnates are x 1 = x 2 =... = x 1 = 0, x = x +1 =... = x n 1 = 1, and we denote t by v (n 1). In partcular, for 2 n 1, the total number of vertces of R (n 1) s (n ) + ( 1)(n + 1) ( 1)(n ) + 1 = (n + 1), the same (by Lemma 1.2) as the number of vertces of S (n). The regons R (n 1) 1 and R n (n 1) are (n 1)-dmensonal smplces. Indeed, the vertces of 1 as n Lemma 1.2. Smlarly, R (n 1) 1 are v (n 1) 1 = (1, 1,...,1) and the n 1 vertces of S (n 1) has n affnely ndependent vertces (the orgn v (n 1) R (n 1) n vertces of S (n 1) n 1 ). By comparng the above descrpton of the vertces of R (n 1) of the vertces of S (n) (Lemma 1.2), we see that the projecton 5 n = (0, 0,..., 0) and the n 1 wth the earler descrpton

6 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 6 (x 1, x 2,..., x n ) (x 1,..., x 1, x +1,..., x n ) s a bjecton between the vertces of S (n) R (n 1), whch maps S (n) onto R (n 1). and Corollary 1.4. The (n 1)-dmensonal volumes of S (n) and R (n 1) are related by V (n 1) (S (n) n2 + 3n ) = V (n 1) (R (n 1) ). n + 1 Proof. Snce the hyperplane H (n) N (n) = (1,1,...,1, (n+1),1,...,1) n 2 +3n follows from Lemma 1.3. contanng S (n) has unt normal vector (the non-unt coordnate s the th one), the desred relaton Proof of Theorem 1.1. The proof s by nducton on the dmenson n. The result s obvously true for n = 1, where R (1) 1 and R (1) 2 are segments of length 1 2. Consder n 2. For = 2,3,..., n, the regon R (n) s the unon of two pyramds wth apex v (n) = (0,..., 0, 1,..., 1) (the frst unt coordnate s the th one). One pyramd s pyr(v (n), S (n) ) over the secton S (n) and the other s pyr(v (n), S (n) 1 ) over the secton S (n) 1. Ther ntersecton s the (n 1)-dmensonal pyramd wth apex v(n) over the (n 2)-dmensonal ntersecton of S (n) and S (n) 1. The dstance d from v (n) to H (n) s easly calculated as 1 (N (n), v (n) ) N (n), and s equal to d = n2 + 3n. Smlarly, the dstance d 1 between v (n) and H (n) 1 s d 1 = n + 2 n2 + 3n. Together wth Corollary 1.4, ths mples that for 2 n, ) = 1 n d V (n 1) (S (n) ) + 1 n d 1 V (n 1) (S (n) 1 ) = 1 n(n + 1) [ V (n 1) (R (n 1) ) + (n + 2) V (n 1) (R (n 1) 1 )]. 6

7 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 7 By nducton, we obtan ) = 1 [A(n, 1) + (n + 2)A(n, 2)] n!(n + 1)! = 1 A(n + 1, 1). n!(n + 1)! The last equalty follows from the standard recurrence relaton satsfed by the Euleran numbers (see, e.g., [Co]). For = 1, the volume of the smplex R (n) 1 can be computed drectly va a determnant evaluaton or, usng the notaton establshed above, 1 ) = 1 n d 1 V (n 1) (S (n) 1 ), 1 n whch we have d 1 = and, by Corollary 1.4, V (n 1) n (S (n) 2 +3n 1 ) = n 2 +3n n+1 V (n 1) (R (n 1) 1 ). Agan, nductvely we have V (n 1) (R (n 1) 1 ) = A(n,0) (n 1)!n! = 1, and t follows that (n 1)!n! 1 ) = A(n+1,0) n!(n+1)!. We omt the calculaton of n+1 ) whch follows through a smlar drect computaton, or smply by symmetry. Usng Theorem 1.1 we derve an expresson for the partal sums of the Euleran numbers. As before, let x = (x 1,..., x n ) denote a random pont n the smplex n. Successve dfferences x j x j 1 are called spacngs. We may model the spacngs as follows (see [Py]): y x j x j 1 = j y 1 +y y n+1, for 1 j n+1, where each y j s a standard exponental random varable and where we defne x 0 = 0 and x n+1 = 1. Next, rewrtng the probablty n terms of the y s, we get for 1 j n: Pr [ x 1 + x x n + 1 n + 2 x j ] = Pr [ (n j + 1)y j y n 1 + 2y n + y n+1 y 1 + 2y 2 + 3y jy j ]. ( ) Each y j has probablty densty functon (p.d.f.) g(t) = { e t f t 0, 0 f t < 0. For dstnct, postve values c j (1 j k), the p.d.f. for Y = c 1 y 1 + c 2 y c k y k s (see [JK], p.222) k c k 2 g Y (t) = j (c c j ) e t/c, for t 0. =1 7

8 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 8 Let Y 1 = (n j + 1)y j y n 1 + 2y n + y n+1 and Y 2 = y 1 + 2y 2 + 3y jy j. Wth ths notaton, ( ) s gven by where and g Y1 (s) = 0 n j+1 a=1 g Y2 (t) = g Y2 (t) dt t 0 g Y1 (s) ds, ( 1) n+j+1 a a n j 1 (a 1)!(n j + 1 a)! e s/a j =1 ( 1) j j 2 ( 1)!(j )! e t/. Performng the ntegraton gves Pr [ x 1 + x x n + 1 n + 2 x j ] = j =1 n j+1 a=1 ( 1) n+1 +a a n j j ( + a)( 1)!(j 1)!(a 1)!(n j + 1 a)!. Equvalently, we get the followng expresson for the partal sums of the Euleran numbers: j 1 ( ) j n 1 A(n + 1, k) = (n + 1)n j 1 k=0 for 1 j n. =1 n j+1 a=1 ( 1) n+1 a + a ( )( ) j 1 n j a n j j, 1 a 1 2. Euleran numbers and the cube Turnng to a cube as the choce of polytope P, we wll consder two problems. The frst one corresponds to the choce of functons f(x) = x 1 + x x n, ρ 0 (x) = 0 and ρ (x) = for 1 n 1. Thus, ths problem concerns the volumes of the n regons of [0, 1] n determned by the n 1 hyperplanes H : x 1 + x x n =, for 1 n 1. We denote the regons as R (n) = {x [0, 1] n : 1 n k=1 x k }, for 1 n. The soluton to ths problem s mplct n the work of Laplace [Lap] and t appears n [Fo], n the context of results about combnatoral statstcs on permutatons. The Euleran numbers arse agan: the volumes of the regons are gven by ) = 1 A(n, 1), n! and the (n 1)-dmensonal volumes of the sectons are also proportonal to Euleran numbers. The expresson for the volume ) suggests that (up to a set of measure zero) the 8

9 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 9 regon R (n) may be parttoned by the mages, under a measure-preservng transformaton, of A(n, 1) n-dmensonal smplces, each of volume 1 n!. Moreover, these smplces may be n natural correspondence wth the permutatons n S n whch have 1 descents. The exstence of such a map would provde an alternatve proof of the volume formulas for the regons, reflectng the combnatoral nature of the formulas. In reponse to a queston asked by Foata, Stanley [St2] exhbted such a map. The map ϕ whose descrpton follows s essentally that gven n [St2]. Frst, t s well-known that the unt cube, [0,1] n, s dssected by the hyperplanes x = x j, 1 < j n, nto n! smplces, each havng volume 1 n!. For each such smplex, there exsts a permutaton σ S n whch permutes n ncreasng order the coordnates of every pont n the nteror of the smplex. Denote the smplex by σ. Now, the desred map ϕ on [0, 1] n less the measure zero set of ponts havng any equal consecutve coordnates, s defned by ϕ(x) = y [0,1] n, where y n = 1 x n, and y = { x+1 x f x < x +1, 1 + x +1 x f x > x +1, for = 1,2,..., n 1. Note that f x σ, then f(y) = n k=1 y k = des(σ) + 1 x 1, where des(σ) denotes as usual the number of descents of σ. Thus, ϕ( σ ) R (n) des(σ)+1. Note also that ϕ s an affne transformaton on each of the 2 n 1 regons determned by a choce of x +1 > x or x +1 < x for each = 1,2,..., n 1, and that the determnant of the transformaton s equal to ( 1) n. Therefore ϕ s measure-preservng. The nverse of ϕ s defned on [0, 1] n less the set of measure zero {y [0,1] n n : k= y k Z, for some }, and s gven by ϕ 1 (y) = x, where x = 1 + n k= y k n k= y k for each. The remander of ths secton s devoted to analogous results for a second problem nvolvng the cube. Let agan P = [0, 1] n, f(x) = x 1 + x x n, and ρ 0 (x) = 0, and consder the functons ρ (x) = 1 2, for 1 n. These gve rse to n parallel hyperplanes, H : x 1 + x x n = 1 2, for 1 n. The volumes of the resultng n + 1 regons and n sectons were nvestgated by Chakeran and Logothett [ChLo], who establshed recurrence relatons satsfed by the sequence of volumes. We denote the regons by R (n) +1 = {x [0,1]n : 1 2 n k=1 x k }, for 0 n. Proposton 2.1. ([ChLo]) For = 0, 1,..., n, let S(, n ) = 2 n n! +1 ). Then wth S(0, 0) = 1. S(, n ) = (2 + 1)S(, n 1) + (2n 2 + 1)S( 1, n ), 9

10 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 10 It turns out that ths recurrence mples that the volumes are proportonal to the Euleran numbers for sgned permutatons. A sgned permutaton on n letters s a permutaton of {1,2,..., n} n whch each letter may bear a mnus sgn. The sgned permutatons on n letters form the hyperoctahedral group of order 2 n n!. For notatonal convenence, we wll wrte m nstead of m. The noton of descent for sgned permutatons s based on the lnear orderng 1 < 2 <... < n < n < n 1 <... < 2 < 1 of the symbols, together wth the fact that f the last letter n the permutaton s negatve ( barred ), then the last poston contrbutes a descent. For example, the sgned permutaton has 3 descents (occurrng n postons 1, 3, and 6). For 0 n, let A(n, ) denote the number of sgned permutatons on n letters, havng exactly descents. For example, when for n = 2, we have A(2,0) = 1, A(2, 1) = 6, and A(2, 2) = 1. Corollary 2.2. Let A(n, m) denote the Euleran numbers for the hyperoctahedral group on n letters. Then the hyperplanes wth equatons n k=1 x k = 1 2, for 1 n, dssect the unt cube [0, 1]n nto regons R (n) whose volumes are gven by ) = A(n, 1), 1 n n n! Proof. In Proposton 2.1 one recognzes, as explaned below, the recurrence relaton for the hyperoctahedral Euleran numbers, leadng to the concluson S(, n ) = A(n, ) for all 0 n. Indeed, t can easly be checked that A(n, ) = (2 + 1) A(n 1, ) + (2n 2 + 1) A(n 1, 1), by examnng how a sgned permutaton counted by the left-hand-sde can arse from the nserton of ether 1 or 1 nto a sgned permutaton τ on {2, 3,..., n}. Clearly, f 1 s nserted nto τ, then t wll be the absolute mnmum of the resultng sgned permutaton, whle f 1 s nserted nto τ, then t wll be the absolute maxmum. Therefore, f des(τ) =, then one of 1 or 1 should be nserted so that the number of descents wll be preserved. There are precsely such nsertons, of whch 2 are of type () and one s of type (), as follows: () nsert ether 1 or 1 after the larger element of one of the descents of τ; () nsert 1 at the front of τ. Ths gves the frst term on the rght-hand-sde. The second term s obtaned smlarly: f des(τ) = 1, then the number of descents must be ncreased by one. Ths s acheved through each of precsely the followng 2(n )+1 nsertons: () nsert ether 1 or 1 after the smaller element of one of the ascents of τ; () nsert 1 at the front of τ. The ntal condtons are obvous and the desred concluson follows. 10

11 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 11 Next, observe that the hyperplanes x = 1 2, x = x j, and x + x j = 1 for 1, j n dssect the cube [0,1] n nto 2 n 1 n! smplces, each of volume 2 n n!. Now let x be any pont from the nteror of such a smplex. Let x = (x 1, x 2,..., x n), where { x x f x = < 1/2, 1 x f x > 1/2. Clearly, x has dstnct coordnates. Let σ be the (ordnary) permutaton whch orders the coordnates of x ncreasngly. Fnally, we obtan a sgned permutaton σ by placng a bar over σ() f x > 1 2. For example, f x = (0.4,0.8) [0,1]2, then x = (0.4, 0.2), leadng to σ(1)σ(2) = 21 and, fnally, to σ = 21. All ponts from the nteror of a smplex gve rse to the same sgned permutaton, and we denote by σ the smplex correspondng to σ. An adaptaton of the map found by Stanley for the prevous problem plays the analogous role here, mappng the smplex σ to the regon R (n) des(σ)+1. Proposton 2.3. The map ϕ gven by ϕ(x) = y where { 1/2 xn f x y n = n < 1/2, 3/2 x n f x n > 1/2, and, for = 1, 2,..., n 1, y = { x+1 x f x < x +1, 1 + x +1 x f x > x +1, s defned on [0,1] n, measure-preservng, and nvertble on [0,1] n, up to measure-zero sets. Moreover, for each j = 0,1,..., n, the regon R (n) j+1 of [0,1]n s parttoned, up to a set of measure zero, by the mages of the nterors of the smplces σ for those sgned permutatons satsfyng des(σ) = j. Proof. The former part of the concluson follows from arguments as n [St2]. For the latter part, usng the descrpton of x, t s easy to check that the second case n the defnton of y, for all 1 n, corresponds precsely to a descent n poston for the sgned permutaton σ correspondng to x. Thus, n k=1 y k = des(σ) x 1 les n the nterval (des(σ) 1 2,des(σ) ), and ϕ maps each σ as clamed. 3. Remarks and open questons. a) For dmensons n = 1,2,3, the sectons S (n) exhbt the followng property: they form a dssecton of an (n 1)-dmensonal smplex (Secton 1) or parellelpped (Secton 2). In partcular, the unon of the sectons S (n) from Secton 1 s, nformally put, a folded 11

12 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 12 (n 1)-dmensonal smplex, wth the folds occurrng along the ntersectons S (n) S (n) +1, for 1 n 1. Does ths property hold n every dmenson, and under what condtons does t hold for more general choces of the polytope P and the functons f and ρ s? Arguments based on tlng may prove frutful n ths regard. b) In each of the problems dscussed here, the sequence of the volumes of regons turns out to be symmetrc and unmodal; n fact, even logarthmcally concave. Is there a connecton between these sequences and mxed volumes (see, e.g., [Lag, p. 946]), also known to be logarthmcally concave? c) The regons arsng from the dssectons consdered here are themselves convex polytopes, and ther vertces have ratonal coordnates. Upon scalng, they can be transformed nto ntegral polytopes, and thus, the nformaton about the volumes of the regons corresponds to the leadng coeffcents of the Ehrhart polynomals of the regons (expostons on the Ehrhart polynomal appear, e.g., n [St1], [H], n addton to the orgnal treatment [E]). Do the sequences of coeffcents of lower-order terms of the Ehrhart polynomal admt enumeratve nterpretatons as well? d) The symmetrc group and the group of sgned permutatons are, n fact, the Weyl groups for the root systems A n 1 and B n, respectvely. In both cases, the noton of a descent s motvated by the geometrc context of reflecton groups. Is there a unfed approach to the results presented here, n the framework of Coxeter systems? e) A drect combnatoral proof of Theorem 1.1 would be desrable. f) What other combnatoral sequences arse naturally as volumes of regons and sectons of polytopes? Acknowledgment The authors thank the anonymous referee for the careful readng of the manuscrpt. References [ChLo] D. Chakeran and D. Logothett, Cube slces, pctoral trangles, and probablty, Math. Mag. 64 (1991) [Co] L. Comtet, Advanced Combnatorcs, D. Redel, Dordrecht, [E] E. Ehrhart, Polynômes Arthmétques et Méthode des Polyèdres en Combnatore, Brkhäuser, Basel and Stuttgart, [Fo] D. Foata, Dstrbutons eulérennes et mahonennes sur le groupe de permutatons, n Hgher Combnatorcs (M. Agner, ed.), NATO Adv. Study Inst. Seres, Seres C: Mat. and Phys. Sc., D. Redel, Dordrecht, 1977, p

13 the electronc journal of combnatorcs 4 (no. 2) (1997), #R18 13 [H] T. Hb, Algebrac Combnatorcs on Convex Polytopes, Carslaw Publcatons, Glebe, Australa, [JK] N. Johnson and S. Kotz, Contnuous Unvarate Dstrbutons - 1, Houghton-Mffln, Boston, [Lag] J. Lagaras, Pont lattces, n Handbook of Combnatorcs (R. Graham, M. Grötschel and L. Lovász, eds.), MIT Press, Cambrdge, 1995, p [Lap] Marqus de Laplace, Oeuvres complètes, vol. 7, 1820; reprnted by Gauthers-Vllars, Pars, 1886, p. 257 ff. [Py] R. Pyke, Spacngs, Royal Stat. Soc. (B) 27 (1965) [St1] R. Stanley, Enumeratve Combnatorcs, vol. 1, Wadsworth & Brooks/Cole, Monterey, 1986; second edton, Cambrdge Unv. Press, to appear. [St2] R. Stanley, Euleran parttons of a unt hypercube, n Hgher Combnatorcs (M. Agner, ed.), NATO Adv. Study Inst. Seres, Seres C: Mat. and Phys. Sc., D. Redel, Dordrecht, 1977, p