y ∈S y ∈S y ∈S

where J n,0 = 2n 2 d 2 Z 1 Z 1 v Z x ≺y

x,y ∈S

d e −n[u1−kxk d +v1− k y k d ] d x dy du dv, J n, ℓ = n 2 d 2 Z 1 Z 1 Z x i y i ,1 ≤i≤ℓ x i y i , ℓi≤d

x,y ∈S

d e −n[u1−kxk d +v1− k y k d ] e nu ∧v1− k x ∨y k ∗ d − 1 d x dy du dv, J n,d = 2n 2 d 2 Z 1 Z 1 v Z y ≺x

x,y ∈S

d e −n[u1−kxk d +v1− k y k d ] e nu ∧v1− k x ∨y k ∗ d − 1 d x dy du dz, for 1 ≤ ℓ ≤ d − 1. Consider first J n, ℓ , 1 ≤ ℓ d. We proceed by four sets of changes of variables to simplify the integral starting from x i 7→ ξ i , y i 7→ ξ i 1 − η i , for 1 ≤ i ≤ ℓ; x i 7→ ξ i 1 − η i , y i 7→ ξ i , for ℓ i ≤ d, which leads to J n, ℓ = nd 2 Z 1 Z 1 Z S d Z [0,1] d e −n h u € 1 − P ξ i + P ′′ ξ i η i Š d +v € 1 − P ξ i + P ′ ξ i η i Š d i ×  e nu ∧v 1 − P ξ i d − 1 ‹ Y ξ i dξ dη du dv, where P ξ i := P d i=1 ξ i , Q ξ i := Q d i=1 ξ i , P ′ x i := P ℓ i=1 x i and P ′′ x i := P d i= ℓ+1 x i . Next, by the change of variables ξ i 7→ 1 d − ξ i n −1d , η i 7→ dη i n −1d , we have J n, ℓ = d 2 Z 1 Z 1 Z S d n Z [0,n 1 d d] d e − h u €P ξ i + P ′′ η i 1 −dξ i n −1d Š d +v €P ξ i + P ′ η i 1 −dξ i n −1d Š d i ×  e u∧v P ξ i d − 1 ‹ Y € 1 − dξ i n −1d Š dξ dη du dv, where S d n = {ξ : ξ i ≤ n 1 d d and ξ 0}. We then perform the change of variables η i 7→ η i € 1 − dξ i n −1d Š , 1870 and obtain J n, ℓ = d 2 Z 1 Z 1 Z S d n Z [0,n 1 d d] d e − h u €P ξ i + P ′′ η i Š d +v €P ξ i + P ′ η i Š d i ×  e u∧v P ξ i d − 1 ‹ dξ dη du dv. Finally, we “linearize” the integrals by the change of variables x 7→ X ξ i , y 7→ X ′′ η i , w 7→ X ′ η i , and get J n, ℓ ≃ d · d d − ℓ − 1ℓ − 1 n 1 −1d Z 1 Z 1 Z ∞ Z ∞ Z ∞ y d −ℓ−1 w ℓ−1 e −u x+ y d −vx+w d × e u∧vx d − 1 dw d y dx du dv, since the change of variables produces the factors n 1 −1d d − 1 , y d −ℓ−1 d − ℓ − 1 and w ℓ−1 ℓ − 1 . Now by symmetry of u and v, we have X 1 ≤ℓd d ℓ J n, ℓ ≃ X 1 ≤ℓd d ℓ 2d · d d − ℓ − 1ℓ − 1 n 1 −1d Z 1 Z 1 v Z ∞ Z ∞ Z ∞ y d −ℓ−1 w ℓ−1 × e −u x+ y d −vx+w d e v x d − 1 dw d y dx du dv. Proceeding in a similar manner for J n,d , we deduce that J n,d = 2d 2 Z 1 Z 1 v Z S d n Z [0,n 1 d d] d e − h u P ξ i d +v P ξ i + P η i d i  e u∧v P ξ i d − 1 ‹ dξ dη du dv. By the change of variables x 7→ P ξ i , w 7→ P η i , we have J n,d ≃ 2d 2 d − 1 2 n 1 −1d Z 1 Z 1 v Z ∞ Z ∞ w d −1 e −ux d −vx+w d e u∧vx d − 1 dw dx du dv = 2d 2 n 1 −1d Z 1 Z 1 v Z ∞ Z ∞ w d −1 e −ux d −vx+w d e v x d − 1 dw dx du dv. Similarly, for J n,0 , we get J n,0 = 2d 2 Z 1 Z 1 v Z S d n Z [0,n 1 d d] d e − h u P ξ i + P η i d +v P ξ i d i dξ dη du dv. The change of variables x 7→ P ξ i , y 7→ P η i then yields J n,0 ≃ 2d 2 n 1 −1d Z 1 Z 1 v Z ∞ Z ∞ y d −1 e −u x+ y d −v x d d y dx du dv. This completes the proof of the theorem. 1871 Remark. By the same arguments, we derive the following asymptotic estimates for the number of maxima in S d . E [M n ] ≃ X ≤ jd d − 1 j −1 j Γ j + 1 d n d−1− jd , V [M n ] = ˜ v d + o1 n 1 −1d , where ˜ v d = Γ 1 d + X 1 ≤kd d k d d d − k − 1k − 1 × Z ∞ Z ∞ Z ∞ y d −k−1 w k −1 e − x+ y d −x+w d e x d − 1 dw d y dx − 2d 2 Z ∞ Z ∞ y d −1 e −x d −x+ y d dx d y. 4 Theorem 2. The number of Pareto records in iud samples from d-dimensional simplex is asymptotically normal with a rate given by sup x P   X n − E[X n ] p V [X n ] x   − Φx = O € n −d−14d log n 2 + n −1d log n 1 d Š , 5 where Φx denotes the standard normal distribution. Proof. Define the region D n := x, z : x ∈ S d , z ∈ [0, 1] and z 1 − kxk d ≤ 2 log n n . Let X n denote the number of maxima in D n and e X n the number of maxima of a Poisson process on D n with intensity dn. Then P   X n − E[X n ] p V [X n ] x   − Φx ≤ P   X n − E[X n ] p V [X n ] x   − P   X n − E[X n ] p V [X n ] x   + P   X n − E[X n ] p V [X n ] x   − P   e X n − E[X n ] p V [X n ] x   + P   e X n − E[ e X n ] p V [ e X n ] y   − Φy + Φ y − Φx , 6 for x ∈ R, where y = x È V [X n ] V [ e X n ] + E [X n ] − E[ e X n ] p V [ e X n ] . 1872 We prove that the four terms on the right-hand side of 6 all satisfy the O-bound in 5. For the first term, we consider the probability P € X n 6= X n Š ≤ nP q 1 ∈ D n and q 1 is a maximum = nd Z S d ×[0,1]−D n € 1 − z1 − kxk d Š n −1 d x dz ≤ nd Z S d ×[0,1]−D n 1 − 2 log n n n −1 d x dz ≤ On −1 . For the second term on the right-hand side of 6, we use a Poisson process approximation sup t P € X n t Š − P € e X n t Š ≤ O € D n Š = O € n −1d log n 1 d Š . To bound the third term, we use Stein’s method similar to the proof for the case of hypercube given in [1] and deduce that sup y P   e X n − E[ e X n ] p V [ e X n ] y   − Φy = O ‚ E[ e X n ] 1 2 Q n V[ e X n ] 3 4 Œ = O € n −d−14d log n 2 Š , where Q n is the error term resulted from the dependence between the cells decomposed and Q n = O € log n 2 Š . Finally, the last term in 6 is bounded above as follows. Φ y − Φx = O     p V [X n ] − p V [ e X n ] + E[X n ] − E[ e X n ] p V [ e X n ]     = O € n −d+12d Š . This proves 6 and thus 5. The Berry-Esseen bound in 5 is expected to be non-optimal and one naturally anticipates an opti- mal order of the form On −12−12d , but we were unable to prove this. Remark. By defining D n := x : x ∈ S d and 1 − kxk d ≤ 2 log n n instead and by applying the same arguments, we deduce the Berry-Esseen bound for the number of maxima in iud samples from S d sup x P   M n − E[M n ] p V [M n ] x   − Φx = O € n −d−14d log n + n −1d log n 1 d Š . 1873 3 Numerical evaluations of the leading constants The leading constants v d see 3 and ˜ v d see 4 appearing in the asymptotic approximations to the variance of X n and to that of M n are not easily computed via existing softwares. We discuss in this section more effective means of computing their numerical values to high degree of precision. Our approach is to first apply Mellin transforms see [12] and derive series representations for the integrals by standard residue calculation and then convert the series in terms of the generalized hypergeometric functions p F q α 1 , . . . , α p ; β 1 , . . . , β q ; z := Γβ 1 · · · Γβ q Γα 1 · · · Γα p X j ≥0 Γ j + α 1 · · · Γ j + α p Γ j + β 1 · · · Γ j + β q · z j j . The resulting linear combinations of hypergeometric functions can then be computed easily to high degree of precision by any existing symbolic softwares even with a mediocre laptop. The leading constant v d of the asymptotic variance of the d-dimensional Pareto records. We consider the following integrals C d = X 1 ≤md d m d − 1 m − 1d − 1 − m × Z 1 Z 1 v Z ∞ Z ∞ Z ∞ y d −1−m w m −1 e −ux+ y d −vx+w d e v x d − 1 dw d y dx du dv + Z 1 Z 1 v Z ∞ Z ∞ w d −1 e −ux d −vx+w d e v x d − 1 dw dx du dv − Z 1 Z 1 v Z ∞ Z ∞ y d −1 e −ux+ y d −v x d d y dx du dv =: d − 1 X 1 ≤md d m d − 2 m − 1 I d,m + I d,d − I d,0 . Then C d is related to v d by v d = d d −1 Γ 1 d + 2d 2 C d . We start from the simplest one, I d,0 and use the integral representation for the exponential function e −t = 1 2 πi Z c Γst −s ds, where c 0, ℜt 0 and the integration path R c is the vertical line from c − i∞ to c + i∞. Substituting this representation into I d,0 , we obtain I d,0 = 1 2 πi Z c Γs Z 1 Z 1 v Z ∞ Z ∞ u −s x + y −ds y d −1 e −v x d d y dx du dv ds. 1874 Making the change of variables y 7→ x y yields I d,0 = 1 2 πi Z c Γs Z 1 Z 1 v Z ∞ Z ∞ u −s x d1 −s 1 + y −ds y d −1 e −v x d d y dx du dv ds = 1 2 πi Z c Γs Z 1 Z 1 v u −s ‚Z ∞ y d −1 1 + y −ds d y Œ‚Z ∞ x d1 −s e −v x d dx Œ du dv ds = dΓd − 1 2 πi Z c ΓsΓds − dΓ1 + 1 d − s Γdsds − 1 ds, where 1 c 1 + 1 d . Moving the integration path to the right, one encounters the simple poles at s = 1 + 1 d + j for j = 0, 1, . . . . Summing over all residues of these simple poles and proving that the remainder integral tends to zero, we get I d,0 = Γd − 1 X j ≥0 −1 j Γ j + 1 + 1 d Γd j + 1 j + 1Γd j + d + 1 , where the terms converge at the rate j −d−1+ 1 d . This can be expressed easily in terms of the general- ized hypergeometric functions. An alternative integral representation can be derived for I d,0 as follows. I d,0 = Γd − 1 Γd X j ≥0 Γ j + 1 + 1 d Γ j + 2 −1 j Z 1 1 − x d −1 x d j dx = Γ 1 d d − 1 Z 1 1 − x d −1 1 − 1 + x d − 1 d x d dx, which can also be derived directly from the original multiple integral representation and successive changes of variables first u, then x, then v, and finally y. In particular, for d = 2, I 2,0 = p π €p 2 − 1 + log 2 + log p 2 − 1 Š . Now we turn to I d,d . I d,d = Z 1 Z 1 v Z ∞ Z ∞ w d −1 e −ux d −vx+w d e v x d − 1 dw dx du dv. By the same arguments used above, we have I d,d = Γd 2d πi Z c ΓsΓds − dΓ1 + 1 d − s Γds I ′ d,d ds, where c 1 and I ′ d,d := Z 1 v −s Z 1 v u − v s −1− 1 d − u s −1− 1 d du dv. 1875 To evaluate I ′ d,d , assume first that 1 d ℜs 1, so that I ′ d,d = Z 1 v −s Z 1 −v u s −1− 1 d du − Z 1 u s −1− 1 d Z u v −s dv du = d d − 1 Γ1 − sΓs − 1 d Γ1 − 1 d − 1 1 − s . Now the right-hand side is well-defined for 1 d ℜs 2. Substituting this into I d,d , we obtain I d,d = Γd − 1 2 πi Z c ΓsΓds − dΓ1 + 1 d − s Γds Γ1 − sΓs − 1 d Γ1 − 1 d − 1 1 − s ds, where 1 c 1 + 1 d . For computational purpose, we use the functional equation for Gamma function Γ1 − sΓs = π sin πs , so that I d,d = Γd − 1 2 πi Z c πΓds − d Γds sinπs − 1 d π Γ1 − 1 d sinπs + Γs − 1 ΓdsΓs − 1 d ds. In this case, we have simples poles at s = j + 1 d for both integrands and s = j for the first integrand to the right of ℜs = 1 for j = 2, 3, . . . . Thus summing over all the residues and proving that the remainder integral goes to zero, we obtain I d,d = Γd − 1Γ 1 d X j ≥2 Γd j − d jΓd j − Γd − 1 X j ≥2 −1 j Γ j − 1 + 1 d Γd j − d + 1 Γ jΓd j + 1 . A similar argument as that used for I d,0 gives the alternative integral representation I d,d = Γ 1 d dd − 1 −1 + Z 1 1 − t 1 d d −1 t 1 d −1 1 + t − 1 d + − log1 − t − t t 2 dt . In particular, for d = 2, I 2,2 = p π € 2 − p 2 − 2 log 2 + log p 2 + 1 Š . Now we consider I d,m for 1 ≤ m d. I d,m := Z 1 Z 1 v Z ∞ Z ∞ Z ∞ y d −1−m w m −1 e −ux+ y d −vx+w d e v x d − 1 dw d y dx du dv, which by the same arguments leads to I d,m = 1 2 πi Z c Γs Z 1 Z 1 v u −s ‚Z ∞ y d −1−m 1 + y −ds d y Œ × Z ∞ w m −1 ‚Z ∞ x d1 −s e −v x d 1+w d −1 − e −v x d 1+w d dx Œ dw du dv ds = dΓd − m 2d − 1πi Z c ΓsΓds − d + mΓ1 + 1 d − s Γdsds − 1 W m s ds, 1876 where 1 c 1 + 1 d and W m s := Z ∞ w m −1 € 1 + w d − 1 Š s −1− 1 d − 1 + w ds −d−1 dw = 1 d Z 1 t −s t − 1 d − 1 m −1 1 − t s −1− 1 d − 1 dt = 1 d X ≤ℓm m − 1 ℓ −1 m −1−ℓ πΓs − 1 d Γ1 − ℓ+1 d Γs + ℓ d sinπs + ℓ d − 1 1 − ℓ d − s , for 1 d ℜs 2 − m − 1d. Note that each term has no pole at s = 1 − ℓ d . Thus I d,m = Γd − m d − 1 X ≤ℓm m − 1 ℓ −1 m −1−ℓ I d,m, ℓ , where I d,m, ℓ := 1 2 πi Z c ΓsΓds − d + mΓ1 + 1 d − s Γdsds − 1 × πΓs − 1 d Γ1 − ℓ+1 d Γs + ℓ d sinπs + ℓ d − 1 1 − ℓ d − s ds. We then deduce that the integral equals the sum of the residues at s = j + 1 d and s = j + 1 − ℓ d I d,m, ℓ = − Γ ℓ+1 d d X j ≥1 Γ j + 1 + 1 d Γd j + m + 1 j + 1Γd j + d + 1Γ j + 1 + ℓ+1 d + 1 d X j ≥1 −1 j Γ j + 1 + 1 d Γd j + m + 1 j + 1Γd j + d + 1 j + ℓ+1 d + Γ ℓ+1 d X j ≥1 Γ j + 1 − ℓ d Γd j + m − ℓ jΓd j + d − ℓd j + d − ℓ − 1 = I [1] d,m, ℓ + I [2] d,m, ℓ + I [3] d,m, ℓ . It follows that C d − I d,d + I d,0 = d − 1 X 1 ≤md d m d − 2 m − 1 I d,m = X 1 ≤md d m d − 2 m − 1 X ≤ℓm m − 1 ℓ −1 m −1−ℓ I [1] d,m, ℓ + I [2] d,m, ℓ + I [3] d,m, ℓ =: C [1] d + C [2] d + C [3] d . 1877 For further simplification of these sums, we begin with C [2] d . Note first that X ≤ℓm m − 1 ℓ −1 m −1−ℓ I [2] d,m, ℓ = 1 d X j ≥1 −1 j Γ j + 1 + 1 d Γd j + m + 1 j + 1Γd j + d + 1 X ≤ℓm m − 1 ℓ −1 m −1−ℓ 1 j + ℓ+1 d = −1 m −1 m − 1 X j ≥1 −1 j Γ j + 1 + 1 d Γd j + 1 j + 1Γd j + d + 1 . Thus C [2] d = X 1 ≤md d m d − 2 m − 1 X ≤ℓm m − 1 ℓ −1 m −1−ℓ I [2] d,m, ℓ = d − 2 X 1 ≤md d m −1 m −1 X j ≥1 −1 j Γ j + 1 + 1 d Γd j + 1 j + 1Γd j + d + 1 = 1 + −1 d I d,0 − Γ1 + 1 d dd − 1 . Accordingly, C [2] d = 0 for odd values of d. For the other two sums containing I [1] d,m, ℓ and I [3] d,m, ℓ , we use the identity X ℓmd N + m−1 m md − mm − 1 − ℓ = −1 d N d − 1 − ℓ N + 1 + ℓ d − N + d d . Then C [1] d = X 1 ≤md d m d − 2 m − 1 X ≤ℓm m − 1 ℓ −1 m −1−ℓ I [1] d,m, ℓ = d − 2 X ≤ℓ≤d−2 d ℓ −1 ℓ Γ ℓ+1 d d X j ≥1 Γ j + 1 + 1 d j + 1Γd j + d + 1Γ j + 1 + ℓ+1 d × X ℓmd Γd j + m + 1−1 m md − mm − 1 − ℓ = −1 d dd − 1 X ≤ℓ≤d−2 d − 1 ℓ −1 ℓ Γ ℓ+1 d X j ≥1 Γ j + 1 + 1 d j + 1Γ j + 1 + ℓ+1 d   d j+ ℓ+1 d d j+d d − 1   . Note that d j+ ℓ+1 d d j+d d − 1 = O j −1 0 ≤ ℓ ≤ d − 2, for large j, so that the series is absolutely convergent. 1878 Similarly, C [3] d = X 1 ≤md d m d − 2 m − 1 X ≤ℓm m − 1 ℓ −1 m −1−ℓ I [3] d,m, ℓ = −1 d d − 1 X ≤ℓ≤d−2 d − 1 ℓ −1 ℓ−1 Γ ℓ+1 d X j ≥1 Γ j + 1 − ℓ d jd j + d − ℓ − 1   d j d d j+d −ℓ−1 d − 1   . Since v d = d d −1 Γ 1 d + 2d 2 C d , we obtain, by converting the series representations into hypergeomet- ric functions, the following approximate numerical values of v d . v 2 ≈ 2.86126 35493 11178 82531 14379, v 3 ≈ 3.22524 36444 05576 89660 59392, v 4 ≈ 3.97797 27442 19455 29292 64760, v 5 ≈ 4.84527 39171 62611 42226 50057, v 6 ≈ 5.76349 95321 96568 64812 77416, v 7 ≈ 6.70865 12250 86590 36364 34742, v 8 ≈ 7.66955 04435 24665 04704 24808, v 9 ≈ 8.64032 79742 08287 24931 00067, v 10 ≈ 9.61764 75521 13755 73944 20940, v 11 ≈ 10.59949 78766 56951 63098 76869, v 12 ≈ 11.58460 78314 60409 77794 37163. In particular, v 2 has a closed-form expression v 2 = 2 3 p π € 2 π 2 − 9 − 12 log 2 Š . The leading constant ˜ v d of the asymptotic variance of the d-dimensional maxima. Let J d,0 := 2d 2 Z ∞ Z ∞ y d −1 e −x d −x+ y d dx d y, and J d,k := d d d − k − 1k − 1 Z ∞ Z ∞ Z ∞ y d −k−1 w k −1 e − x+ y d −x+w d e x d − 1 dw d y dx. Then see 4 ˜ v d = Γ 1 d + X 1 ≤kd d k J d,k − J d,0 . 1879 Consider first J d,0 . By expanding 1 + x d −1− 1 d , interchanging and evaluating the integrals, we obtain J d,0 = 2Γ 1 d Z 1 1 − x d −1 1 + x d 1+ 1 d dx = 2d X j ≥0 Γ j + 1 + 1 d Γd j + 1 Γ j + 1Γd j + d + 1 −1 j , the general terms converging at the rate O j −d− 1 d . The convergence rate can be accelerated as follows. J d,0 = 2Γ 1 + 1 d Z 1 x 1 d −1 1 − x 1 d d −1 1 + x −1− 1 d dx = 2Γ 1 + 1 d X r ≥0 2 −r−1− 1 d Z 1 1 − x r x 1 d −1 1 − x 1 d d −1 dx = Γ 1 + 1 d 2 − 1 d X ≤ℓd d − 1 ℓ −1 ℓ Γ ℓ + 1 d X j ≥0 Γ j + 1 + 1 d Γ j + 1 + ℓ+1 d 2 − j , the convergence rate being now exponential. In terms of the generalized hypergeometric functions, we have J d,0 = Γ 1 d 2 − 1 d X ≤ℓd d − 1 ℓ −1 ℓ ℓ + 1 2 F 1 1 + 1 d , 1; 1 + ℓ + 1 d ; 1 2 . The integrals J d,k can be simplified as follows. J d,k+1 = d 2 d − 1 d − 2 k Z ∞ e x d − 1 Z ∞ x e − y d × Z ∞ x y − x d −2−k z − x k e −z d dz d y dx = 2d 2 d − 1 d − 2 k Z ∞ e − y d Z y e −z d × Z z e x d − 1 y − x d −2−k z − x k dx dz d y = 2d − 1Γ 1 d d − 2 k Z 1 1 − x k Z 1 1 − xz d −2−k z k+1 × 1 1 + z d − x d z d 1+ 1 d − 1 1 + z d 1+ 1 d dz dx = J ′ d,k+1 + J ′′ d,k+1 . 1880 By the same proof used for J d,0 , we have J ′′ d,k+1 = −2d − 1Γ 1 d d − 2 k Z 1 1 − x k × Z 1 1 − xz d −2−k z k+1 1 + z d −1− 1 d dz dx = −1 k+1 2 − 1 d Γ 1 d X k jd d − 1 j −1 j j + 1 2 F 1 1 + 1 d , 1; 1 + j + 1 d ; 1 2 . Similarly, J ′ d,k+1 = 2d − 1Γ 1 d d − 2 k Z 1 1 − x k × Z 1 1 − xz d −2−k z k+1 1 + z d − x d z d −1− 1 d dz dx = 2Γ 1 d d − 1 X ≤ j≤d−2−k −1 j jd − 2 − k − j × X ≤ℓ≤k −1 ℓ ℓk − ℓ · 3 F 2 1 + 1 d , k+ j+2 d , 1; 1 + ℓ+ j+1 d , 1 + k+ j+2 d ; −1 ℓ + j + 1k + j + 2 . Thus we obtain the following numerical values for the limiting constant ˜ v d of V[M n ]n d−1d ˜ v 2 ≈ 0.68468 89279 50036 17418 09957, ˜ v 3 ≈ 1.48217 31873 40583 68601 11369, ˜ v 4 ≈ 2.35824 37612 02486 93742 28054, ˜ v 5 ≈ 3.27773 90059 79491 26684 80858, ˜ v 6 ≈ 4.22231 09450 77067 79998 34338, ˜ v 7 ≈ 5.18220 76686 16078 48517 29967, ˜ v 8 ≈ 6.15196 29023 77474 45508 28039, ˜ v 9 ≈ 7.12835 13658 43360 52793 29089, ˜ v 10 ≈ 8.10938 23221 15849 82527 77117, ˜ v 11 ≈ 9.09377 74697 86680 89694 70616, ˜ v 12 ≈ 10.0806 86465 19733 08113 16376. In particular, ˜ v 2 = p π2 log 2 − 1; see [2]. Yet another constant in [8]. A similar but simpler integral to 4 appeared in [8], which is of the form K d := Z ∞ Z ∞ Z ∞ u + w d −2 e −u+x d +x d −w+x d dx du dw, 1881 this K d is indeed their K d −1 . By Mellin inversion formula for e −t , we obtain K d = 1 2 πi Z c Γs Z ∞ Z ∞ Z ∞ u + w d −2 u + x −ds e −w+x d +x d dx du dw ds = 1 2d πi Z c ΓsΓ1 + 1 d − s × Z ∞ Z ∞ u + w d −2 1 + u −ds € 1 + w d − 1 Š s −1− 1 d du dw ds. Expanding the factor u + w d −2 , we obtain K d = P ≤m≤d−2 d −2 m K d,m , where K d,m := 1 2d πi Z c ΓsΓ1 + 1 d − s ‚Z ∞ u m 1 + u −ds du Œ × ‚Z ∞ w d −2−m € 1 + w d − 1 Š s −1− 1 d Œ dw = 1 2d πi Z c ΓsΓ1 + 1 d − sBm + 1, ds − m − 1U m s ds. 7 Here U m s := Z ∞ w d −2−m € 1 + w d − 1 Š s −1− 1 d dw = 1 d Z 1 t −s 1 − t s −1− 1 d t − 1 d − 1 d −2−m dt = 1 d X ≤ℓ≤d−2−m d − 2 − m ℓ −1 d −2−m−ℓ B1 − s − ℓ d , s − 1 d . Thus we obtain K d = 1 d 2 X ≤m≤d−2 d − 2 m X ≤ℓ≤d−2−m d − 2 − m ℓ −1 d −2−m−ℓ mΓ ℓ+1 d × X j ≥0 Γ j + 1 − ℓ d Γd j + d − ℓ − m − 1 jΓd j + d − ℓ − Γ j + 1 + 1 d Γd j + d − m Γ j + 1 + ℓ+1 d Γd j + d + 1 . 1882 This readily gives, by converting the above series into hypergeometric functions, the numerical values of the first few K d , K 2 ≈ 0.30714 28473 56944 02518 48954, K 3 ≈ 0.21288 24684 73220 99693 80676, K 4 ≈ 0.19494 67028 23033 18190 40460, K 5 ≈ 0.20723 21512 99671 45854 93769, K 6 ≈ 0.24331 17024 51836 72554 88428, K 7 ≈ 0.30744 56566 07893 22242 37300, K 8 ≈ 0.41127 01058 90385 83873 59349, K 9 ≈ 0.57571 68456 67243 64328 08087, K 10 ≈ 0.83615 82236 77116 00233 16115, K 11 ≈ 1.25179 63251 14070 86480 31485, K 12 ≈ 1.92201 04035 18847 36012 85304. These are consistent with those given in Chiu and Quine 1997. In particular, K 2 = 1 4 p π log 2. Further simplification of this formula can be obtained as above, but the resulting integral expression is not much simpler than Γ 1 d d 4 Z 1 Z 1 u − 1 d + v − 1 d − 2 d −2 u −1− 1 d v −1− 1 d € u −1 + v −1 − 1 Š −1− 1 d du dv. 4 Asymptotics of the number of chain records We consider in this section the number of chain records of random samples from d-dimensional simplex; the tools we use are different from [17] and apply also to chain records for hypercube random samples, which will be briefly discussed. For other types of results, see [17].

4.1 Chain records of random samples from d-dimensional simplex