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
−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
Assume that p
1
, . . . , p
n
are iud in the d-dimensional simplex S
d
. Let Y
n
denote the number of chain records of this sample. Then Y
n
satisfies the recurrence Y
n d
= 1 + Y
I
n
n ≥ 1, 8
with Y := 0, where
P I
n
= k = π
n,k
= d n
− 1 k
Z
1
t
kd
1 − t
d n
−1−k
1 − t
d −1
dt, for 0
≤ k n. An alternative expression for the probability distribution π
n,k
is π
n,k
= n
− 1 k
X
≤ jd
d − 1
j −1
j
Γn − kΓ k +
j+1 d
Γ n +
j+1 d
,
1883
which is more useful from a computational point of view. Let
z + 1 · · · z + d − d = z Y
1 ≤ℓd
z − λ
ℓ
, where the
λ
ℓ
’s are all complex 6∈ R, except when d is even in that case, −d − 1 is the unique real
zero among {λ
1
, . . . , λ
d −1
}. Interestingly, the same equation also arises in the analysis of random increasing k-trees see [9], in some packing problem of intervals see [4], and in the analysis of
sorting and searching problems see [7].
Theorem 3. The number of chain records Y
n
for random samples from d-dimensional simplex is asymp- totically normally distributed in the following sense
sup
x ∈R
P
Y
n
− µ
s
log n σ
s
p log n
x
− Φx = O
log n
−12
,
9 where
µ
s
:= 1 d H
d
and σ
s
:= Æ
H
2 d
d H
3 d
. The mean and the variance are asymptotic to E
[Y
n
] = H
n
d H
d
+ c
1
+ On
−ǫ
, 10
V [Y
n
] = H
2 d
d H
3 d
H
n
+ c
2
+ On
−ǫ
, 11
respectively, for some ǫ 0, where
c
1
= 1
d H
d
X
1 ≤ℓd
ψ −
λ
ℓ
d − ψ
ℓ d
, c
2
= 1
6 +
π
2
6d
2
H
2 d
− 2H
3 d
3H
3 d
+ H
2 d
2
2H
4 d
+ 1
d
2
H
2 d
X
1 ≤ℓd
ψ
′
− λ
ℓ
d − ψ
′
ℓ d
+ c
1
H
2 d
H
2 d
− 2d
H
d
X
j ≥1
d j + 1 · · · d j + dH
d j+d
− H
d j
d j + 1 · · · d j + d − d
2
. Here
ψx denotes the derivative of log Γx. The error terms in 10 and 11 can be further refined, but we content ourselves with the current
forms for simplicity.
Expected number of chain records. We begin with the proof of 10. Consider the mean
µ
n
:= E
[Y
n
]. Then µ = 0 and, by 8,
µ
n
= 1 + X
≤kn
π
n,k
µ
k
n ≥ 1. 12
Let ˜ f z := e
−z
P
n ≥0
µ
n
z
n
n denote the Poisson generating function of µ
n
. Then, by 12, ˜
f z + ˜ f
′
z = 1 + d Z
1
˜ f t
d
z1 − t
d −1
dt. 1884
Let ˜ f z =
P
n ≥0
˜ µ
n
z
n
n. Taking the coefficients of z
n
on both sides gives the recurrence ˜
µ
n
+ ˜ µ
n+1
= d
d n + 1 · · · dn + d ˜
µ
n
n ≥ 1. Solving this recurrence using ˜
µ
1
= 1 yields ˜
µ
n
= −1
n −1
Y
1 ≤ jn
1 −
d d j + 1 · · · d j + d
n ≥ 1. It follows that for n
≥ 1 µ
n
= X
1 ≤k≤n
n k
˜ µ
k
= X
1 ≤k≤n
n k
−1
k −1
Y
1 ≤ jk
1 −
d d j + 1 · · · d j + d
. 13
This is an identity with exponential cancelation terms; cf. [17]. In the special case when d = 2, we have an identity
µ
n
= H
n
+ 2 3
. No such simple expression is available for d
≥ 3 since there are complex-conjugate zeros; see 14.
Exact solution of the general recurrence. In general, consider the recurrence
a
n
= b
n
+ X
≤kn
π
n,k
a
k
n ≥ 1, with a
= 0. Then the same approach used above leads to the recurrence ˜
a
n+1
= − 1
− d
d n + 1 · · · dn + d ˜
a
n
+ ˜b
n
+ ˜b
n+1
, which by iteration gives
˜ a
n+1
= X
≤k≤n
−1
k
˜b
n −k
+ ˜b
n −k+1
Y
≤ jk
1 −
d dn − j + 1 · · · dn − j + d
, by defining b
= ˜b = 0. Then we obtain the closed-form solution
a
n
= X
1 ≤k≤n
n k
˜ a
k
. A similar theory of “d-analogue” to that presented in [13] can be developed by replacing 2
d
j
d
there by d d j + 1 · · · d j + d.
However, this type of calculation becomes more involved for higher moments.
1885
Asymptotics of µ
n
. We now look at the asymptotics of
µ
n
. To that purpose, we need a better expression for the finite product in the sum-expression 13.
In terms of the zeros λ
j
’s of the equation z + 1 · · · z + d − d, we have
Y
1 ≤ jn
1 −
d d j + 1 · · · d j + d
= Q
1 ≤ jn
d j
Q
1 ≤ℓd
d j − λ
ℓ
Q
1 ≤ jn
d j + 1 · · · d j + d =
1 n
Y
1 ≤ℓd
Γ n
−
λ
ℓ
d
Γ
1 +
ℓ d
Γ
n +
ℓ d
Γ
1 −
λ
ℓ
d
=: φn. 14
The zeros λ
j
’s are distributed very regularly as showed in Figure 4.
−1.25 −1
−0.75 −0.5
−0.25
−0.5 −0.3
0.3 0.5
Figure 4: Distributions of the zeros of z + 1 · · · z + d − d = 0 for d = 3, . . . , 50. The zeros approach,
as d increases, to the limiting curve |z
−z
z + 1
1+z
| = 1 the blue innermost curve. Now we apply the integral representation for the n-th finite difference called Rice’s integrals; see
[14] and obtain µ
n
= − 1
2 πi
Z
1 2
+i∞
1 2
−i∞
Γn + 1Γ−s Γn + 1 − s
φs ds. Note that
φs is well defined and has a simple pole at s = 0. The integrand then has a double pole at s = 0; standard calculation moving the line of integration to the left and summing the residue of
the pole encountered then leads to
µ
n
= 1
d H
d
H
n
+ X
1 ≤ℓd
ψ −
λ
ℓ
d − ψ
ℓ d
+ O
n
−ǫ
,
1886
where the O-term can be made more explicit if needed. Note that to get this expression, we used the identity
z + 1 · · · z + d − d z
= X
1 ≤ j≤d
dΓz + d − j + 1
d − j + 1Γz + 1 .
The probability generating function. Let P
n
y := E[ y
Y
n
]. Then P y = 1 and for n ≥ 1
P
n
y = y X
≤kn
π
n,k
P
k
y. The same procedure used above leads to
P
n
y = X
≤k≤n
n k
−1
k
Y
≤ jk
1 −
d y d j + 1 · · · d j + d
= 1 + y − 1 X
1 ≤k≤n
n k
−1
k −1
Y
1 ≤ jk
1 −
d y d j + 1 · · · d j + d
n ≥ 0. Let now
| y − 1| be close to zero and z + 1 · · · z + d − d y =
Y
1 ≤ℓ≤d
z − λ
ℓ
y .
Note that the λ
ℓ
’s are analytic functions of y. Let λ
d
y denote the zero with λ
d
1 = 0. Then we have
P
n
y = 1 − y
− 1 2
πi Z
1 −ǫ+i∞
1 −ǫ−i∞
Γn + 1Γ−s Γn + 1 − s
φs, y ds, where
ǫ 0 and φs, y =
Γ s
−
λ
d
y d
Γs + 1Γ 1
−
λ
d
y d
Y
1 ≤ℓd
Γ s
−
λ
ℓ
y d
Γ
1 +
ℓ d
Γ
s +
ℓ d
Γ
1 −
λ
ℓ
y d
. Note that for y
6= 1, φ0, y = 1 − y. When y ∼ 1, the dominant zero is λ
d
y, and we then deduce that
P
n
y = Q yn
λ
d
yd
+ O|1 − y|n
−ǫ
, where
Q y := d y
− 1 λ
d
yΓ1 +
λ
d
y d
Y
1 ≤ℓd
Γ
λ
d
y−λ
ℓ
y d
Γ
1 +
ℓ d
Γ
λ
d
y+ℓ d
Γ 1
−
λ
ℓ
y d
. By writing z + 1
· · · z + d − d y = 0 as 1 + z · · ·
1 +
z d
− 1 = y − 1,
and by Lagrange’s inversion formula, we obtain λ
d
y = y
− 1 H
d
− H
2 d
− H
2 d
2H
3 d
y − 1
2
+ O
| y − 1|
3
.
1887
From this we then get Q1 = 1 + O | y − 1| and
λ
d
e
η
= η
H
d
+ H
2 d
2H
3 d
η
2
− 2H
d
H
3 d
− 3H
2 d
2
6H
5 d
η
3
+ O|η|
4
, for small
|η|. This is a typical situation of the quasi-power framework see [15, 22], and we deduce 10, 11 and the Berry-Esseen bound 9. The expression for c
2
is obtained by an ad-hoc calculation based on computing the second moment the expression obtained by the quasi-power
framework being less explicit. When d = 2, a direct calculation leads to the identity
V [Y
n
] = 5
27 H
n
+ 2
π
2
27 +
H
2 n
9 −
26 27
− 2
9 X
j ≥1
2 j
− 1 j
2 n+ j n
− 2 j
j +
1 2
2 n+ j+
1 2
n
, for n
≥ 1, which is also an asymptotic expansion. This is to be contrasted with E[Y
n
] = H
n
+ 23.
4.2 Chain records of random samples from hypercubes
