The degree profile of random Pólya trees

Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

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Journal of Combinatorial Theory, Series A www.elsevier.com/locate/jcta

The degree profile of random Pólya trees ✩ Bernhard Gittenberger a , Veronika Kraus b a b

Institute of Discrete Mathematics and Geometry, TU Wien, Wiedner Hauptstr. 8-10/104, A-1040 Wien, Austria Institute of Bioinformatics and Translational Research, UMIT, Eduard-Wallnoefer Zentrum 1, 6020 Hall in Tyrol, Austria

a r t i c l e

i n f o

Article history: Received 21 April 2011 Available online 9 May 2012 Keywords: Unlabelled trees Profile Nodes of fixed degree Brownian excursion Local time

a b s t r a c t We investigate the profile of random Pólya trees of size n when only nodes of degree d are counted in each level. It is shown that, as in the case where all nodes contribute to the profile, the suitably normalized profile process converges weakly to a Brownian excursion local time. Moreover, we investigate the joint distribution of the number of nodes of degrees d1 and d2 on the same level of the tree. © 2012 Elsevier Inc. All rights reserved.

1. Introduction Consider the size of level k in a rooted tree, i.e., the number of nodes at distance k from the root. The sequence of level sizes of the tree is commonly called the profile of the tree. First investigations on the profile of random trees of given size (i.e., their number of nodes) seem to go back to Stepanov [54]. A first distributional result as well as the relation to Brownian excursion local time was achieved by Kolchin [37] for the family trees of a Galton–Watson branching process conditioned on the total progeny. Different representations of the same result have been obtained later by several authors by approaching the problem from very different directions (random walks, queuing theory, general theory of stochastic processes, random trees), see [24,32,55,56]. The relation between trees and diffusion processes has also been studied by probabilistic methods. In this context, first and foremost the seminal papers of Aldous [1–3] in which he developed the theory of continuum trees should be mentioned. Here, a metric space called continuum tree is identified as the limit of several classes of random trees with respect to the Gromov–Hausdorff topology. This theory was further elaborated by Marckert and Miermont [41] and Haas and Miermont [30]. There is also a continuous limit of more complicated discrete structures like planar maps. A comprehensive analysis has been carried out by Le Gall [39,40]. Links to fractal structures were explored by Duquesne ✩

This research has been supported by FWF (Austrian Science Foundation), National Research Network S9600, grant S9604.

0097-3165/$ – see front matter © 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jcta.2012.04.007

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and Le Gall [20]. For getting an overview of these topics we recommend Le Gall’s very well-written survey [38]. These approaches tell us a lot about the behaviour of large random trees. They imply limit theorems for functionals which can be regarded as a continuous mapping. But they do neither provide limit theorems for functionals which depend on some local structures of the tree, like the profile, nor always convergence of moments. A different approach to the relation between trees and diffusion process was pursued in [46]. For a general overview on the relation between stochastic processes of combinatorial and therefore discrete origin and their continuous counterparts can be found in [47]. Not only the relation between trees and diffusion processes attracted much attention, but the processes themselves are of interest in their own right. A thorough overview on Brownian local times and related processes can be found in [50]. Explicit representations for the moments and the density of the one-dimensional projections of the local time of a Brownian excursion and related processes have been derived by Takács [58,57,59]. Multi-dimensional analogues can be found in [28,29]. For results on density representations for related processes such as occupation times we refer to [15,35,33]. The profile of random trees has recently attracted the attention of numerous authors. A survey on the theory of random trees in general and the profile in particular can be found in [14]. Roughly speaking, the tree classes which have been studied can √ be divided into trees of height  log n ( meaning the order of magnitude) and trees of height  n where n is the size of the tree. Trees of logarithmic height are for instance binary search trees and recursive trees and variations or generalizations thereof. First investigations on the profile of binary search trees have been done Chauvin et al. [7,8]. A thorough analysis was done later by Fuchs, Hwang and Neininger [23]. See also [19] and Schopp [52] for related work on search trees. For an analysis of the profile of tree classesrelated to recursive trees we refer to [18,34,23]. The maximum of the profile, commonly called the width of a tree, in logarithmic trees was analyzed by Devroye and Hwang [12]. A related structure appearing in the theory of data structures are so-called tries. Their profile was examined in [11,43,45]. √ Trees of height  n are for instance trees obtained from conditioned Galton–Watson branching processes or Pólya trees, i.e., rooted unlabelled trees. The analysis of the profile of Galton–Watson trees was done in [15]. The start of the profile, i.e., the behaviour close to the root, for trees as well as forests was investigated in [25,26]. Binary Pólya trees have been studied by Broutin and Flajolet [5,6] and general Pólya trees in [17]. The joint distribution of two level sizes was addressed explicitely by van der Hofstad et al. [60] for recursive trees and in [28] for Galton–Watson trees. Note that this question also arises implicitely when proving a functional limit theorem by showing convergence of the finite-dimensional projections as well as tightness of the profile, albeit the proofs of tightness utilizes the moments of the difference of the two level sizes. Recently, also patterns in random trees were investigated. Here, the question whether certain given trees occur as a substructure of a large tree, and for the number of such occurrences, is addressed. The simplest pattern is the star graph. This is equivalent to counting the number of nodes of a given degree in random trees. First investigations in this directions were performed by Robinson and Schwenk [51]. In [16] it was shown for several tree classes (certain classes of Galton–Watson trees as well as Pólya trees) that the number of nodes of given degree is asymptotically normally distributed, as long as the degree is fixed. If the given degree grows with the tree size, then the limiting distribution may change. This depends on the rate of the growth. The change occurs if the expectation does no longer tend to infinity, see [42] for Galton–Watson trees and [27] for Pólya trees. The analogues for general patterns instead of star graphs was carried out by Chyzak et al. [9]. Note that all results on patterns in trees tell us something about the number of occurrences of a given pattern, but nothing about their location within the large tree. For Galton–Watson trees this question was settled by Drmota [13] for star graphs. The same question for Pólya trees is addressed in this paper. Plan of the paper In the next section we will recall some basic results on Pólya trees and present the main results afterwards. The first result is Theorem 1 which states that the d-profile (number of nodes of degree d at fixed distance from the root), viewed as a stochastic process, weakly converges to Brownian excursion local time. To prove this theorem we will split it into two partial results, the

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convergence of the finite-dimensional projections of the profile (Theorem 2) as well as the tightness of the profile (Theorem 3). To examine the joint behaviour of two different patterns we derive the covariance and the correlation (Proposition 2.1 and Theorem 4). The results on covariance and correlation exhibit a surprisingly regular structure of the limiting object. We show that the correlation coefficient tends to 1 and derive the speed of convergence as well. The proof of these theorems will be carried out by means of generating functions. This will be described in Section 2 as well. This section ends with an introduction of the notation we will use throughout the paper. To proceed we need a singularity analysis of the generating functions together with a kind of transfer of the singular behaviour into the asymptotic behaviour of the coefficients in the sense of [21]. Section 3 provides some a priori estimates which are to be refined later. In Section 4 we first present the proof of the one-dimensional analogue of Theorem 2 which is based on the refinement of the results in Section 3. The rest of the section is devoted to the refined analysis. The generalization to multiple dimensions is done in Section 5. In Section 6 we show tightness. This is usually a very technical matter (cf. the eight-page proof in [17]). Here we offer a considerably shorter proof by showing a more general result using Faà di Bruno’s formula, which generalizes the chain rule to higher derivatives. The final section is devoted to the joint behaviour of the numbers of nodes of two different degrees within one level. 2. Results and notation 2.1. Preliminaries A Pólya tree is an unlabelled rooted tree and thus it can be viewed as a root with a set of Pólya trees attached to it. By the machinery of symbolic transfers described in [22], we easily obtain that the generating function y (x) of unlabelled rooted trees fulfills the functional equation

y (x) =



yn xn = x exp

n0



 



y xi / i ,

(2.1)

i 1

where yn denotes the number of trees with n vertices. This result goes back to Pólya [48] (cf. also [49]), who also showed that y (x) has exactly one singularity ρ on the disk of convergence and that ρ ≈ 0.3383219. Around its singularity, y (x) has the local expansion

y (x) = 1 − b

√

√

ρ − x + c (ρ − x) + d ρ − x3 + · · · ,

(2.2)

with b ≈ 2.6811266, as Otter [44] showed, and y (ρ ) = 1. From the expansion, asymptotic estimations for yn can be derived by transfer lemmas (cf. [22]):

√ ρ yn ∼ √ b

2

1

π n3/2 ρ n

.

(2.3)

Remark. The constants ρ , b and C in Theorem 1 below have been computed numerically by Pólya [48], Otter [44] and Schwenk [53], respectively. Nowadays, these constants can be easily computed using computer algebra systems: Note that the functional equation (2.1) can be written in the form y (x) = Ψ ( y (x)) := xe y (x) A (x) where A (x) is an analytic function. Since the operator Ψ ( y (x)) is a contraction w.r.t. the formal metric (cf. [22, p. 731]), it is an easy exercise to obtain a large number of coefficients of y (x) exactly, yielding a fairly good approximation of y (x). Since ρ is the solution of the equation 1 = xe y (x) A (x), the approximation and standard numerical procedures can be used to obtain √ easily a good numerical approximation for ρ . If we set F (x, y ) = xe y A (x) − y, then b = 2ρ F x (ρ , 1), where F x = ∂ F /∂ x, gives an approximation for b. Similar arguments apply to the computation of C .

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2.2. Main results (d)

We denote by L n (k) the number of nodes of degree d at distance k from the root in a randomly chosen unlabelled rooted tree of size n. By linear interpolation, we create a continuous stochastic (d) (d) (d) process L n (t ) = (t + 1 − t ) L n (t )+(t −t ) L n (t + 1), t  0. This simplifies the proof of tightness (Theorem 3) since it allows us to work with the discrete process but nevertheless use the tightness conditions for continuous processes. (d)

(d)

√

√

Theorem 1. Let ln (t ) = L n (t n )/ n and l(t ) denote the local time of a standard Brownian excursion. Then (d)

ln (t ) converges weakly to the local time of a Brownian excursion, i.e., we have

 √

 (d)

 b ρ Cd ρd w ln (t ) t 0 − → ·l √ t √ 2ρ b 2 2



(2.4) t 0



in C[0, ∞) as n → ∞, where C d = C + O (dρ d ) with C = exp(

ρ i )/ρ i − 1)/i ) ≈ 7.7581604.

i 1 ( y (

Remark. In [17] it is shown that the general profile of an unlabelled rooted random tree converges to Brownian excursion local time, i.e.





ln (t ) t 0 →

 √  √  b ρ b ρ √ l √ t 2 2

2 2

,

t 0

where ln (t ) is the profile process L n (t ), rescaled√by normalizing constant in Theorem 1 equals

b

√

n analogously to the definition in Theorem 1. The

ρ

μd 2√2 , where μd n is asymptotically equal to the expected

value of nodes of degree d in trees of size n, with

μd =

2C d b2 ρ

ρ d , see for example [16]. To prove the

above statement, weak convergence of the finite-dimensional distributions and tightness have to be shown: Theorem 2. For any choice of fixed numbers t 1 , . . . , tm and for large d

 √

 (d)

√

 w Cd ρd b ρ b ρ (d) l ln (t 1 ), . . . , ln (tm ) − →√ √ t 1 , . . . , √ tm 2ρ b 2 2 2 2 as n → ∞.



Remark. We will show this theorem by proving the convergence of the corresponding characteristic functions. From a result in [10] the following representation for the characteristic function of √ d

b

ρ

Cd ρ √ l( √ t ) at level 2ρ b

κ can be derived.  Cd ρd φ(t ) = 1 + √ ψκ (x, it )e −x dx, 2 2

ib

where

ρπ

(2.5)

γ

γ = (c − i ∞, c + i ∞) with some arbitrary c < 0 and √ √ t −se −κ b −ρ s/2 √ ψκ ( s , t ) = √ √ √ . −se κ b −ρ s/2 − tC d ρ d sinh( 12 κ b −ρ s )/(b ρ )

(2.6)

A sequence of stochastic processes might not converge in C[0, ∞) even if the sequence of their images with respect to every finite-dimensional projection does. Roughly speaking, in order to guarantee convergence in the sense of stochastic processes (i.e., when constructing a sequence by applying an arbitrary continuous bounded functional to the corresponding probability measures, this sequence must converge) the sample paths of the processes must not fluctuate too wildly. Tightness is a technical property of stochastic processes which guarantees this. The next theorem states a technical condition for the profile process which implies tightness (cf. [4] and [36] for the general theory).

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Theorem 3. There exists a constant c > 0 such that for all positive integers r , h, n the inequality

 4 E Ln (r ) − Ln (r + h)  ch2n

(2.7)

holds.

√

Remark. According to [4, Theorem 12.3] the inequality E| L n (r ) − L n (r + h)|α = O (hβ ( n )α −β ) implies tightness of the process ln (t ) if α > 0 and β > 1. In the theorem above we have α = 4 and β = 2 and thus ln (t ) is √ tight. We remark here that in [17, remark on p. 2050] the authors erroneously stated the bound O ((h n )β ). In order to examine the dependence of the numbers of nodes for two different degrees, say d1 and d2 (at the same level k), we will compute the covariance and the correlation. (d )

(d )

(d )

(d )

Proposition 2.1. The covariance Cov( X n 1 (k), X n 2 (k)) of random √ variables X n 1 (k) and X n 2 (k) counting vertices of degrees d1 and d2 , with d1 = d2 fixed, at level k = κ n in a random Pólya tree of size n is asymptotically given by

 (d )  (d ) Cov Xn 1 (k), Xn 2 (k)      2  1 = C d1 C d2 ρ d1 +d2 n 2 e −ω/4 + e −ω − κ 2 e −ω/2 1 + O √ , b

ρ

n

(2.8)

as n tends to infinity, with ω = κ 2 b2 ρ . (d )

(d )

) be the random variables counting the number of vertices of degrees d1 Theorem 4. Let X n 1 (k) and X n 2 (k√ and d2 , respectively, on a level k = κ n in a Pólya tree of size n. Then the correlation coefficient is asymptoti√ cally equal to 1 as n tends to infinity. The speed of convergence is of order 1/ n. 2.3. Description of the problem with generating functions In order to prove Theorem 2, we will start with the one-dimensional case and then extend results (d) to multiple dimensions. Therefore, we introduce generating functions yk (x, u ), which represent trees where all nodes of degree d on level k are marked and counted by u. Note that we consider planted trees instead of ‘ordinary’ rooted trees, that is, we assume that the root node is adjacent to an additional node which is not counted. This assumption does not alter the tree structure, but allows us to treat the root vertex like a normal vertex, that is, a root of degree d has in-degree 1 and out-degree (d) d − 1. Refining the decomposition of trees along their root, the yk (x, u ) fulfill a recursion:



 





y 0 (x, u ) = y (x) + (u − 1)xZ d−1 y (x), y x2 , . . . , y xd−1 , (d)

(d) yk+1 (x, u ) = x exp



(d)  i

yk

i





x , u /i ,

(2.9)

i 1

where Z d (s1 , s2 , . . . , sd ) is the cycle index of the symmetric group Sd on d elements, given by

1

|Sd |

d 

λ

si i ,

π ∈Sd i =1

where λi is the number of cycles of length i in the permutation π . (d) Examining two levels k and k + h simultaneously, we use the generating function yk,h (x, u 1 , u 2 ) where all nodes of degree d on level k are marked by u 1 and nodes of degree d on level k + h are marked by u 2 . As before, x marks the total size of the tree. We get the recursive relation

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(d)



(d)

(d)  d−1

(d)

y 0,h (x, u 1 , u 2 ) = y h (x, u 2 ) + (u 1 − 1)xZ d−1 y h (x, u 2 ), . . . , y h (d)

yk+1,h (x, u 1 , u 2 ) = x exp



(d) 





x

1533

 , ud2−1 ,

yk,h xi , u 1i , u 2i /i .

(2.10)

i 1

In general, observing levels k1 , . . . , km such that k2 = k1 + h1 , . . . , km = km−1 + hm−1 , we get (d)

y 0,h ,...,h (x, u 1 , . . . , um ) 1 m −1 (d)

= yh1 ,...,hm−1 (x, u 2 , . . . , um )  (d)   (d) d −1 + (u − 1)xZ d−1 yh1 ,...,hm−1 (x, u 2 , . . . , um ), . . . , yh1 ,...,hm−1 xd−1 , ud2−1 , . . . , um ,    i i  (d) (d) i /i . yk+1,h ,...,h (x, u 1 , . . . , um ) = x exp yk,h ,...,h x , u 1 , . . . , um 1 m −1 1 m −1 i 1

(d)

These functions are related to the process L n (t ) by



  (d) (d) (d) P Ln (k) = 1 , Ln (k + h1 ) = 2 , . . . , Ln k + h i = m (d)

=

m [xn u 11 u 22 · · · um ] y 0,h1 ,...,hm−1 (x, u 1 , . . . , um )

[xn ] y (x)

.

For the computation of the covariance of the numbers of nodes of degrees d1 and d2 we will utilize the functions

     (x, u , v ) = y (x) + (u − 1)xZ d1 −1 y (x), y x2 , . . . , y xd1 −1      + ( v − 1)xZ d2 −1 y (x), y x2 , . . . , y xd2 −1 ,    (d ,d ) (d ,d )  yk+11 2 (x, u , v ) = x exp y k 1 2 xi , u i , v i / i . (d ,d2 )

y0 1

(2.11)

i 1

2.4. Notations For proving the theorems we will carry out a singularity analysis of the generating functions. All generating functions are in some sense close to the tree function y (x) from (2.1). Therefore we will use the differences to y (x) and related functions: Set (d)

(d)

(d)



w k (x, u ) = yk (x, u ) − y (x),

Σk (x, u ) =

(d) 

wk

 i

xi , u / i ,

(d1 ,d2 )

(x, u , v ) = yk(d1 ,d2 ) (x, u , v ) − y (x),  (d ,d )   (d ,d ) Σk 1 2 (x, u , v ) = w k 1 2 xi , u i , v i / i ,

wk

i 2

i 2 2

∂ (d) ∂ (d ,d ) y (x, u ), y 1 2 (x, u , v ), γ˜k(d1 ,d2 ) (x, u , v ) = ∂u k ∂ u∂ v k ∂2 γk(d)[2] (x, u ) = 2 yk(d) (x, u ). ∂u

γk(d) (x, u ) =

(2.12)

We further introduce some domains, depicted in Fig. 1:

  = (η, θ) = z ∈ C | z| < ρ + η, |arg( z − ρ )| > θ ,

  =  (θ) = z ∈ C | z − ρ | <  , |arg( z − ρ )| > θ ,

 Θ = Θ(η) = z ∈ C | z| < ρ + η, |arg( z − ρ )| = 0 ,

 Ξk = Ξk (η˜ ) = v ∈ C | v |  1, k| v − 1|  η˜ ,

with  , η, η˜ > 0 and 0 < θ < π2 .

(2.13) (2.14) (2.15) (2.16)

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Fig. 1. The regions used for the proofs.

In all the proofs in the subsequent sections we will assume (even without explicitely mentioning) that η, θ,  are sufficiently small for all arguments to be valid. (d1 ,d2 )

(d)

3. The local behaviour of y k (x, u ) and y k

(x, u , v ) – A priori bounds

In order to analyze the local behaviour of the generating functions we will first derive a priori (d) (d ,d ) estimates for w k , w k 1 2 and the related functions which will be used frequently in the following to derive the needed refinements. Lemma 3.1. There exists a constant L with 0 < L < 1 and a positive constant D such that for |x|  ρ 2 + ε ,

ε > 0 and sufficiently small, and for |u |  1 the estimate | w k(d) (x, u )|  D |u − 1| · |x|d · Lk holds.

Proof. We will only provide a short sketch, since the proof is similar to that of [17, Lemma 2]. Since |x|  ρ 2 + ε < ρ < 1, for k = 0 we have

(d)

   

w (x, u ) = |u − 1| · |x| · Z d−1 y (x), y x2 , . . .  |u − 1| · D · |x|d . 0

Since the cycle index can be bounded by a constant times | y (x)|d−1 which itself is bounded by D · |x|d . (d) The result for general w k (x, u ) follows by induction. Starting with the recurrence relation





(d)

(d)

w k+1 (x, u ) = y (x) exp w k (x, u ) +



(d)  i

wk





x , u i /i



−1

i 2

we use the trivial estimate | w k (x, u )|  2 y (|x|) which is valid for |x|  ρ and |u |  1, the convexity of y (x)/x on the positive reals, and some elementary estimates for e x . For the precise details see [17]. 2 Corollary 3.2. There is a positive constant C˜ such that for |u |  1 and |x|  ρ + ε (and for all k  0, d  1) (d) |Σk (x, u )|  C˜ |u − 1| L k .

Proof. Same as proof of Corollary 1 in [17]. Corollary 3.3. Let u ∈ Ξk and x ∈ Θ . Then Proof. As i  2, the functions (d)

Γk (x, u ) := (d)



i 2



i 2

2

γk(d) (xi , u i ) = O( Lk ), as k → ∞, uniformly w.r.t. x and u.

γk(d) (xi , u i ) are analytic in the whole region Θ . Moreover, note that

γk(d) (xi , u i ) =



n,m

(d)

(d)

ynmk xn ym and all the coefficients yknm are positive, which im-

(d)

plies |Γk (x, u )|  Γk (|x|, |u |) and the right-hand side is monotone in |x| and |u |. Now let x  0 and 0 < u < 1. Using Taylor’s theorem and the above argument we get (d)

(d) 

Σk (x, u ) = (u − 1)Γk



(d)

x, 1 + ϑ(u − 1)  Γk (x, u ),

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

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where 0 < ϑ < 1. In view of Corollary 3.2 this implies for all x ∈ Θ and u ∈ Ξk the estimate |Γk(d) (x, u )|  Γk(d) (|x|, |u |)  C L k for some positive constant L < 1. 2 (d1 ,d2 )

In a similar fashion we obtain the analogous results for w k

:

Lemma 3.4. If |x|  ρ 2 + ε for sufficiently small ε , |u |  1, and | v |  1, then there exist a constant L with 0 < L < 1 and a positive constant D (both independent of x, u , v) such that

(d1 ,d2 )

 

w (x, u , v )  D Lk |u − 1| · |x|d1 + | v − 1| · |x|d2 . k

Corollary 3.5. There is a positive constant C˜ such that for |u |  1, | v |  1 and |x|  ρ + ε (and for all k  0, (d1 ,d2 )

d  1) we have |Σk

(x, u , v )|  C˜ (|u − 1| + | v − 1|) L k .

Corollary 3.6. There is a positive constant C¯ such that for u ∈ Ξk , v ∈ Ξk , and x ∈ Θ







γ˜k(d1 ,d2 ) xi , u i , v i  C¯ Lk .

(3.1)

i 2

4. The one-dimensional case Our main goal is to prove the following theorem from where the main result follows by integration. Theorem 5. Let x = ρ (1 + ns ), u = e

√it

n

, k = κ

√

n and d be a fixed integer. Moreover, assume that |arg s|  (d)

ϑ > 0 and, as n → ∞, we have |s| = O(log2 n), whereas κ and t are fixed. Then, w k (x, u ) admits the local

representation

Cd ρd (d) w k (x, u ) ∼ √ · ψκ (s, it ), n

(4.1)

where ψκ (s, t ) is given in (2.6). The one-dimensional limiting distribution Let us first assume that Theorem 5 holds. Then, to prove Theorem 2 in one dimension, we need to determine the characteristic function (d)

φk,n (t ) =



1  n  (d)  1 √it  x yk x, e n = yn 2π iyn

(d) 

yk

x, e

√it

n

 ds

(4.2)

xn+1

Γ

where the contour Γ = γ ∪ Γ consists of the line





γ = x=ρ 1−

1 + iτ n

 

2 2

− D log n  τ  D log n

with an arbitrarily chosen constant D > 0 and Γ is a circular arc centered at the origin and closing the curve. The contribution of Γ is negligible since for x ∈ Γ we have (d)

1 |x−(n+1) | yn

3

2

= O(n 2 e − log n )

√it n

on the one hand whereas on the other hand | yk (x, e )| is bounded. If x ∈ γ the local expansion (4.1) is valid and thus, inserting into (4.2) leads to

1 (d) lim φ (t ) = lim n→∞ k,n n→∞ 2π iy n



√

C d ρ d+n n 2

2 −1 + i log n

√

ψκ (s, it )

ib 2ρπ 2

−1−i log n

= φ(t ), where φ(t ) is the function given by (2.5).

1

ρnn

e

−s

 ds + Γ





ds

y (x)

xn+1





=2π iyn

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Now let us turn back to the proof of Theorem 5. (d)

4.1. The local behaviour of yk (x, u )-refined analysis Now we will refine the a priori estimates of the previous section. First we show that the first (d) derivative γk (x, 1) is almost a power of y (x). Afterwards we will derive estimates for the second (k)

derivative and then obtain a power-like representation for w k (x, u ). Finally, utilizing the recurrence (k)

relation for w k (x, u ) we will arrive at the desired result (4.1). (d)

Lemma 4.1. For x ∈ Θ (where η > 0 is sufficiently small) the functions γk (x) can be represented as

γk(d) (x) := γk(d) (x, 1) = C k(d) (x) y (x)k+d , (d)

where the functions C k (x), k = 0, 1, 2, . . . , are analytic and the sequence converges uniformly (for x ∈ Θ ) to (d)

an analytic limit function C (d) (x) with convergence rate C k (x) = C (d) (x) + O ( L k ) for some 0 < L < 1, and further C (d) (ρ ) = C d ρ d , where C d is the constant given in (2.4).

Proof. Let C k (x) := γk (x)/ y (x)k+d . We prove the analyticity of the functions (d)

(d)

Since the function

(d)

d−1

γ0 (x) = xZ d−1 ( y (x), y (x ), . . . , y (x 2

γk(d) (x) by induction:

)) is obviously analytic in Θ and bounded

by D |xy (x)d−1 | for some D > 0, so is



C 0 (x)  D x/ y (x)d  D . (d)

(4.3)

The step of induction works like in [17], as the

γk(+d)1 (x, u ) =

γk(d) fulfill the same recursion as the γk :

 (d)   ∂ i1 y (d) (xi ,u i ) k = yk(d+)1 (x, u ) γk xi , u i u i−1 , xe ∂u

(4.4)

i 1

and for u = 1

γk(+d)1 (x) = y (x)γk(d) (x) + y (x)Γk(d) (x), (d)



(4.5)

√

γk(d) (xi ), which is analytic for |x|  ρ and hence in Θ . Applying the induction (d) hypothesis, this proves the analyticity of γk (x). Solving the recurrence (4.5), we obtain

with Γk (x) =

i 2

γk(d) (x) = y (x)k γ0(d) (x) +

k −1 

y (x)k− Γ (x) (d)

=0

and hence, the analyticity of

γk(d) (x) implies the analyticity of the functions C k(d) (x) in Θ . (d)

We now have to show that the functions (C k (x))k0 have a uniform limit C (d) (x) but this works analogously to [17, Lemma 3]. Finally, note that



γk(d) (x, 1) =

k0



(d)

dn xn = D (d) (x),

k0

(d)

where dn is the total number of vertices of degree d in all trees of size n, and D (d) (x) is the corresponding generating function, introduced e.g. in [51]. On the other hand,

 k0

γk(d) (x, 1) =

 k0

 

C (d) (x) + O L k

y (x)k =

C (d) (x) y (x)d 1 − y (x)

+ O(1)

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

1537

and therefore C (d) (ρ ) = limx→ρ (1 − y (x)) D (d) (x)/ y (x)d , as x → ρ . We know that D (d) (x) =  ( y (x) i 2 D (d) (xi )+ xZ d−1 ( y (x), . . . , y (xd−1 )))/(1 − y (x)) (cf. [51, Eq. (36)]). Schwenk [53, Lemma 4.1] computed the limit of the cycle index in the numerator. In his proof he provides the speed of convergence as well. In fact, [53, Eq. (32)] says that

  d   d   

 1 y (x) 1 y (x) y (xd )

d +1 ,..., d −1 x exp λ − exp

Zd

x i x i x i =1

i =1

with λ = sup0xρ ( y (x)/x − 1)/x = (1 − ρ )/ρ 2 . Thus xZ d−1 ( y (x), . . . , y (xd−1 )) is of the form xd F (x)+ O(dx2d+1 ). Note further that D (d) (x) = O(xd+1 ) for |x|  ρ < 1 since there are no nodes of degree d in trees of size less than d + 1. This implies C (d) (ρ ) = C d ρ d as d → ∞ with C d = C + O (dρ d ) and C as in Theorem 1. 2 (d) Lemma 4.2. There exist constants  (d) , θ (d) , η˜ (d) > 0 and θ (d) < π2 such that |γk (x, u )| = O (| y (x)|k+d ), as k → ∞, uniformly for x ∈  and u ∈ Ξk . (d) (d) Proof. For   k we set C¯  = sup x∈ |γ (x, u ) y (x)−−d |. First we derive the following inequality, u ∈Ξk

(d)

using the recurrence for yk :

 

(d)



 1 (d)  i i 

y (x, u )  y (x) exp w (d) (x, u ) +

w x , u   +1 i 2

i



(d) 

  y (x) exp γ x, 1 + ϑ(u − 1) |u − 1|    C¯  |u −1| (d)

  |u i − 1| (d)    i i

+ γ x ,1 + ϑ u − 1 i 2



i

with 0 < ϑ < 1 and thus 1 + ϑ(u i − 1) ∈ Ξk . To get an estimate for the second term, we use that

|u i − 1| = |1 + u + · · · + u i −1 ||u − 1|  i |u − 1| as |u |  1 and hence (d) use |γ (xi , 1 + ϑ(u i

|u i −1| i

 |u − 1|  2. Further we

(d) − 1))|  |γ (xi , 1)|, | y (x)|  1 and Corollary 3.3 to obtain

(d)



  

y (x, u )  y (x) exp C¯ (d) |u − 1| + O L  .  +1 Using recurrence (4.4) leads to



(d)



y (x, u )

γ(d) (x, u ) + i 2 γ(d) (xi , u i )u i −1

¯C (d) = sup +1 +1

y (x)

y (x)+d x∈ u ∈Ξk

¯ (d) η˜ +O ( L  )  ¯ (d) k C

  + O L   ¯ (d) η˜  = C¯ (d) e C  k 1 + O L  , as  → ∞,  eC

where we used Lemma 4.1 to get |

i 2



γ(d) (xi , u i )u i−1 | = O(

i 2 | y (x

i

)|+d ) and hence



+d    

i 2 γ(d) (xi , u i )u i −1   y ( xi )

= sup O = O L ,

+ d y (x) y (x)

sup

x∈ u ∈Ξk



(4.6)

i 2

as  → ∞.

1538

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

We now set c 0 =



j 0 (1 + O ( L

j

)). Note that

(d)



   

γ (x, u ) y (x)−d = xZ d−1 y (x), y x2 , . . . , y (x)d−1 y (x)−d  c x/ y (x)  c˜ 0

for x ∈  (c and c˜ are suitable positive constants). Hence C¯ 0 = sup |γ0 (x, u )/ y (x)d | is bounded, (d)

¯ (d)

too. Thus we can choose η˜ > 0 such that e 2C 0 2C¯ 0 c 0 c k (d)

c 0 η˜

(d)

(d) (d)  2. For fixed k we get: C¯   C¯ 0

¯ (d)

O( L ))e  2C 0 c 0 . The second estimate is clear by the choice of inequality can be obtained from (4.6) by induction: j

(d) 

C¯ 1  C¯ 0 (d)

 

¯ (d) η

1 + O L0 eC0

k



 C¯ 0(d)

 

¯ (d) c 0 η 1

1 + O L j e 2C 0

k



j < (1

+

η and by   k. The first

,

j <1

(d) η

(d)

C¯ +1  C¯  e k C  (d)

=





 

1 + O L

 j 

1+O L

  

1+O L

 (d)  ¯ (d) 2C¯ 0 c0 η k

C0 e

exp

k

j <

 C¯ 0



 j 

1+O L

e

1 2C¯ 0 c 0 η + k

η ¯ (d) C0

j <

 e

2C¯0

(d)

c 0 η k







2 k 2

2

.

j <+1

(d)

For the second derivatives with respect to u of yk (x, u ),

γk(d)[2] (x, u ), we find that

Lemma 4.3. Suppose that |x|  ρ − η for some η > 0 and |u |  1. Then

 

k+d 

γk(d)[2] (x, u ) = O y |x|

as k → ∞,

,

(4.7) (d)[2]

uniformly w.r.t. x and u. There also exist constants  , θ, η˜ such that γk holds uniformly for u ∈ Ξk and x ∈  .

(x, u ) = O(ky (|x|)k+d ), as k → ∞,

Proof. Derivation of (4.4) leads to the recurrence



γk(+d)[1 2] (x, u ) = yk(d+)1 (x, u ) +

(d)  i

γk





x ,u u

i 1

(d) yk+1 (x, u )

i

i −1

2 (d)

+ yk+1 (x, u )

(d)  i

(i − 1)γk

i



x ,u u



(d)[2]  i

i γk



x , u i u 2 ( i −1 )

i 1

i −2

i 2

γ0(d)[2] (x, u ) = 0. (d)[2] (d)[2] (x, u )|  γk (|x|, 1). Thus, in this For |x| < ρ − η , for some η > 0 and for |u |  1 we have |γk case we can restrict ourselves to non-negative real x  ρ − η .  (d) (d) (d) i i k By using the bounds γk (x, 1)  C k y (x)k+d from Lemma 4.2, i 2 γk (x , u ) = O ( L ) from (d)[2] (d) (d) k+d Corollary 3.3 and the induction hypothesis γk (x, 1)  D k y (x) with D k = O (1), we can derive with initial condition

the following upper bound from the above:



γk(+d)[1 2] (x) = y (x)

i 1

(d)  i 

γk

x

2 + y (x)



(d)[2]  i 

i γk

x + y (x)

i 1



(d)  i 

(i − 1)γk

i 2

x

      k 2  i k+d  (d) (d) (d)  i k+d k+d + +O L + Dk iy x Ck y x  y (x) C k y (x) i 1

i 2

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

1539

 (d)2 (d)    y (x)k+d+1 C k y (x)k+d + C k O Lk     y (xi )k+d  y (xi )k+d i C ( i − 1) + + D k(d) 1 + y (x)k+d y (x)k+d i 2 i 2        y (x)k+d+1 C k(d)2 y (ρ − η)k+d + D k(d) 1 + O Lk + O Lk .

γk(+d)[12]  D k(d+)1 y (x)k+d+1 , with

Consequently,



(d)

(d)2

D k +1 = C k

(d) 

y (ρ − η)k+d + D k

 

1 + O Lk

  + O Lk = O(1),

To prove the second property we use the same constants (d)[2] supx∈ , u ∈Ξk |γ (x, u ) y (x)ell−d |.

as k → ∞.

 , θ, η˜ as in Lemma 4.2 and set D¯ (d) =

We use the already known bound |γ (x, u )|  C¯ (d) | y (x)k+d | and by similar considerations as in the proof of Lemma 4.2 we get (d)

(d)

y +1 (x, u )

y (x)−(+d)

y (x)

¯ D +1 = sup (d)

x∈ u ∈Ξk

 

2  

(d)  i i  i −1 (d)[2]  i i  2(i −1) (d)  i i  i −2

× γ x , u u + i γ x ,u u + (i − 1)γ x , u u i 1

 α(d) D¯ (d) with

¯ (d) η˜

α(d) = e C

(d)

=

k −1 



i 2

¯ (d) η˜

(d)

(1 + O( L  )) and β = C (d)2 e C

k

¯ α D k k −1 (d)

i 1

+ β(d) 





k



+ O( L  ). Thus, as k → ∞,



αk(d−)2 . . . α0(d) D 0 + β0(d) . . . βk(d−)2 + βk(d−)1 −1 k

(d)

βj

j =0

αi(d) + α0(d) D¯ (0d)

i = j +1

(d) C¯ (d) η

 k max β j e j



 

1 + O Li

= O(k),

i 0

2

which completes the proof of the lemma. (d)

Lemma 4.4. Let  , θ, η˜ and C k (x) be as in Lemma 4.1 and Lemma 4.2. Then







w k (x, u ) = C k (x)(u − 1) y (x)k+d 1 + O k|u − 1| , (d)

(d)

as k → ∞,

(4.8)

uniformly for x ∈  and u ∈ Ξk . Furthermore we have for |x|  ρ + η and |u |  1

 k+d   k+d  , Σk(d) (x, u ) = C˜ k(d) (x)(u − 1) y x2 + O |u − 1|2 y |x|2

as k → ∞,

(4.9)

˜ (d)

where the analytic functions C k (x) are given by (d) C˜ (x) =



k

i 2

(d)  i 

Ck

x



y ( xi )

k+d

y (x2 )

and have a uniform limit C˜ (d) (x) with convergence rate C˜ k (x) = C˜ (d) (x) + O ( L k ), for some constant L with 0 < L < 1. (d)

1540

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

(d)

Proof. To prove the first statement, we expand w k (x, u ) into a Taylor polynomial of degree 2 around u = 1 and apply Lemmas 4.1 and 4.3. To prove the second statement, we again use Taylor series. Note that for i  2 we have |xi | < ρ − η if |x| < ρ + η and η is sufficiently small. We get (d)  i



(d)  i  i

x , ui = Ck

wk

  k+d

u − 1 y xi

x

2  k+d   + O u i − 1 y x i

and consequently (d)

Σk (x, u ) =

1 i 2

i

(d)  i  i

  k+d

u − 1 y xi

x

Ck

2  k+d   + O u 2 − 1 y x2

2  k+d   (d)    k+d = (u − 1)C˜ k x2 y x2 , + O u 2 − 1 y x2 where we used the property that



− 1 y (xi )k+d  (d)  i  (1 + u + · · · + u i −1 ) y (xi )k+d = Ck x i (u − 1) y (x2 )k+d i y (x2 )k+d i 2  (d)  (d) = C˜ k (x) + O C˜ k (x)(u − 1)

i (d)  i  u

Ck

x

i 2

(d) represents an analytic function in x and u, and thus its leading term, as u → 1, is our function C˜ k (x).

Finally, since C k (x) = C (d) (x) + O ( L k ) it follows that C˜ k (x) has a limit C˜ (d) (x) with the same order of convergence. 2 (d)

(d)

Lemma 4.5. For x ∈  and u ∈ Ξk (with the constants  , θ, η˜ as in Lemma 4.2) we have

(u − 1) y (x)k+d C k (x) (d)

(d)

w k (x, u ) =

1−

1 2

1− y (x)k

(d)

y (x)d C k (x) · (u − 1) 1− y (x) + O (|u − 1|)

.

The constant hidden in the error term may depend on d, but not on x, u , k. (d)

Proof. Recall that by (2.9), w k (x, u ) satisfies the recursive relation (d)

w k+1 (x, u ) = x exp

= x exp



(d)  i





x , u i /i

yk

− y (x)

i 1





(d)  i

wk



 



x , u i + y xi / i

− y (x)

i 1

    = y (x) exp w k(d) (x, u ) + Σk(d) (x, u ) − 1 , (d)

(d)

(d)

and further, since by Lemma 4.4 it follows that Σk (x, u ) = O ( w k (x, u ) L k ) = O ( w k (x, u )), as k → ∞ and uniformly in x and u (for brevity, we omit the variables now),

 (d) w k +1

=y



(d)

(d) 

w k + Σk

(d)

+

(d)

( w k + Σk )2 2





(d)

(d) 3 

+ O w k + Σk

     2  ( w k(d) + Σk(d) ) = y w k(d) + Σk(d) 1 + + O w k(d) + Σk(d)  (d)

= y wk

2

(d) 

1+

Σk

(d)

wk

(d)

1+

wk 2

  (d)   (d) 2  , + O Σk + O w k

as k → ∞ and uniformly in x and u. From there, we obtain

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

y (d)

w k +1

 (d)  Σ · 1 + k(d) =



1 (d)

wk

wk



1

=

(d)

1 wk

2

    + O Σk(d) + O w k(d)2

wk 2



1

−

(d)

 (d)   (d) 2  + O Σk + O w k

wk

(d)

1−

wk

=

(d)

1+

2

+O

(d) 

Σk

(d)

wk

1541

 −1



 (d)  + O wk .

This leads us to a recursion

y k +1 (d) w k +1

=

yk (d)

wk

−

Σk(d) · y (x)k+1 (d)

(d) w k w k +1

−

1 2

 yk + O

Σk(d) · yk



(d)

wk

  + O w k(d) yk

which we can solve to

yk (d)

wk

=

=

1 (d)

w0 1

(d)

w0

+O

− 

k −1  Σ (d) · y +1

 (d) (d) =0 w  w +1 (d)

1 − w0

 

 (d) (d) w =0  w +1



(d)

w

=0

k (d) 1 1 − y

− w0

2 1− y

+O

(d)

k (d) 1 − L

+ O w0

(4.10)

,

(d)

(d)

(d)

(d)

(d)

 k −1 xZ d−1  y (x)d

k−1 ˜ (d)  C (u − 1) y (x2 )+d + O (|u − 1|2 y (|x|2 )+d )

 (d) (d) 2(+d)+1 (u =0 C  C +1 y (x)

C˜ 

(d)

y (x2 )+d

(d) (d) =0 C  C +1

y (x)+d





+



− 1|))  k −1  O(|u − 1|2 y (|x|2 )+d )  =0



− 1)2 (1 + O(|u

(d) (d) C  C +1 y (x)+d



y +1





1 + O |u − 1|



=O (|u −1|2 L  )

  (d) = c k + O | u − 1| ,

denotes the first term in the parenthesis times the prefactor. Note that xZ d−1 / y (x)d is (d) 1− y 2k 1− y 2

bounded. Now turn back to (4.10) and observe that w 0

= O(k|u − 1| y (x)d ) = O( y (x)d ) = O(1) (d)

if k|u − 1|  η˜ . Thus, we obtain the following representation for w k (x, u ): (d)

(d)

= O( y w k−1 ) = O( yk w 0 ).

w  w +1

=O ( L  )

(d)

1−L



= (u − 1)xZ d−1

where ck



k −1  Σ (d) · y (x)+1

=0

=

 (d) w  y

=0



where we used that Σ y  / w  = O ( L  ) and that by Lemma 4.4 w k Again we apply Lemma 4.4 and (2.9) to obtain

w0

 k −1 



1 − y2

(d)

 k−1 (d)   Σ · y

=0

k −1 (d)  Σ · y +1

2k (d) 2 1 − y

w0

k −1

1  − y +O 2



(d)

w k = w 0 yk

−1



(d)

(d)

1 − ck (x) −

w 0 1 − yk 2

1− y

   + O | u − 1| .

1542

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557



We use the expressions C k+1 = i 1 C k ( y (xi )/ y (x))k+d and C˜ k = are consequences of Lemma 4.1, Eq. (4.4) and Lemma 4.4, to obtain (d)



(d)



C˜ k (x) = C k+1 (x) − C k (x) (d)

(d)

(d)



(d)

y (x)



i 2

C k ( y (xi )/ y (x2 ))k+d , which (d)

k+d (4.11)

.

y (x2 )

This provides the telescope sum: (d)

ck =

=0

(d)

and hence, since C 0 = the result.



(d) k−1 (d) xZ d−1  C +1 − C  xZ d−1 = (d) (d) y (x)d y (x)d C C

2



x Z d −1 , y (x)d

1

−

(d)

C0

+1



1

(4.12)

(d)

Ck

we get 1 − ck (x) = xZ d−1 ( y (x), . . . , y (xd−1 )) y (x)−d C k (x), which yields (d)

(d)

It is now easy to prove Theorem 5. With x = ρ (1 + ns ), u = e (2.2) we obtain the expansions for n → ∞:



it u−1∼ √ , n

√it

n

, d and t = 0 fixed, k = κ

√

y (x)k ∼ e −κ b −ρ s ,

1 − y (x) ∼ b −ρ s/n,

√

n and

y (x)d ∼ 1.

(d)

Since the functions C k (x) are continuous and uniformly convergent to C (d) (x), they are also uni(d)

formly continuous and thus C k (x) ∼ C (d) (ρ ) = C d ρ d . This leads to √ √it C d ρ d e −κ b −ρ s n

(d)

w k (x, u ) ∼

1− 1

−κ b √it C d ρ d ( 1 1−e 2

n

√

=√ ·√ n

√ −ρ s

−ρ s

b

)

n

−sitC d ρ d e −κ b

−s −

√

−ρ s

√ itC d √ (1 − e −κ b −ρ s )

ρd

2b

Cd ρd

= √

n

ρ

· ψκ (s, it ).

5. Finite-dimensional limiting distributions First we consider the case m = 2. The computation of the two-dimensional limiting distribution shows the general method of the proof. Iterative applications of the arguments will eventually prove Theorem 2. Theorem 6. Let x =

ρ (1 + ns ), u 1 = e

it

√1

n

, u2 = e

it

√2

n

, k = κ

√

√

n and h = η n. Moreover, assume that

|arg s|  Θ > 0 and, as n → ∞, we have |s| = O(log2 n), whereas κ , t 1 and t 2 are fixed. Then, for n → ∞, (d) w k,h (x, u ) admits the local representation

  Cd ρd (d) w k,h (x, u , v ) ∼ √ · ψκ s, it 1 + ψη (s, it 2 ) , n

(5.1)

with ψη (s, t ), ψκ (s, t ) given in (2.6). The two-dimensional limiting distribution Let us assume first that Theorem 6 holds. The characteristic function of the two-dimensional distribution is given by (d)

φk,k+h,n (t 1 , t 2 ) =

it it 1  n  (d)  1 √1 √2  x yk,h x, e n , e n = 1 + yn 2π iyn



(d) 

w k,h x, e

it

√1

n

it

,e

√2

n

 ds xn+1

.

(5.2)

Γ

We use the same contour as in the one-dimensional case. With the same arguments, only the (d) integral over γ hence the representation (5.1) of w k,h leads to

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

1543

2 −1 + i log n

  n→∞ ψκ s, it 1 + ψη (s, it 2 ) e −s ds − −−→ φκ ,η (t 1 , t 2 ),

d

Cd ρ φk(d,h),n (t 1 , t 2 ) ∼ 1 + √ b ρπ i

−1−i log2 n √

d

√

b ρ b ρ C ρ where φκ ,η (t 1 , t 2 ) is the characteristic function of the random variable √d (l( √ κ , √ η)). 2ρ b 2 2 2 2

(d)

5.1. The local behaviour of yk,h (x, u 1 , u 2 ) (d)

(d)

(d)

(d)

Note that yk,h (x, u 1 , 1) = yk (x, u 1 ) and yk,h (x, 1, u 2 ) = yk+h (x, u 2 ). Considering the first derivative, let

γk(,dh)[u i ] (x, u 1 , u 2 ) :=

∂ (d) y (x, u 1 , u 2 ), ∂ u i k,h

i = 1, 2

and by simple induction, we observe that

γk(,dh)[u1 ] (x, 1, 1) = γk(d) (x, 1) = γk(d) (x) = C k(d) (x) y (x)k+d , γk(,dh)[u2 ] (x, 1, 1) = γk(+d)h (x, 1) = γk(+d)h (x) = C k(d+)h (x) y (x)k+h+d . (d)[u i ]

As |γk,h

(d)[u i ]

(x, u 1 , u 2 )|  γk,h

(d)[u ] |γk,h 1 (x, u 1 , u 2 )|

(5.3)

(x, 1, 1) for i = 1, 2; u 1 ∈ Ξk , u 2 ∈ Ξk+h , and |x|  ρ it follows that (d)[u 2 ]

= O( y (x)k+d ) and γk,h

(x, u 1 , u 2 )  O( y (x)k+h+d ) in the same regions. To be more precise, we can prove the following analogue to Lemma 4.2. Note. In the remainder of the paper, unless otherwise mentioned, we will use the assumptions of Theorem 6 on k and h. In particular, error terms are meant as n → ∞. Lemma 5.1. There exist constants

 , θ, η1 , η2 , such that, uniformly for x ∈  , u 1 ∈ Ξk (η1 ) and u 2 ∈

Ξk+h (η2 )





k+d γk(,dh)[u1 ] (x, u 1 , u 2 ) + γk(,dh)[u2 ] (x, u 1 , u 2 ) = O y (x) .

Proof. Set (d)[u ] C ,h 1

(d)[u 1 ]

γ,h (x, u 1 , u 2 )

, = sup

y (x)k+d x,u ,u 1

2

(d)[u ] C ,h 2

(d)[u 2 ]

γ,h (x, u 1 , u 2 )

, = sup

y (x)k+h+d x,u ,u 1

2

where here and in the following supx,u 1 ,u 2 f (x, u 1 , u 2 ) means the supremum taken over the domain x ∈  , u 1 ∈ Ξk , u 2 ∈ Ξk+h . As in the proof of Lemma 4.3, we apply Taylor’s theorem (in two variables) to get

 



 | w (d) (xi , u 1i , u 2i )|

= y (x) exp w (d) (x, u 1 , u 2 ) + ( x , u , u ) 1 2  +1,h

(d)

y

i

i 2



   y (x) exp γ,(dh)[u 1 ] x, 1 + ϑ1 (u 1 − 1), 1 + ϑ2 (u 2 − 1) (u 1 − 1) (d)[u 2 ] 

+ γ,h +





x, 1 + ϑ1 (u 1 − 1), 1 + ϑ2 (u 2 − 1) (u 2 − 1)









γ,(dh)[u1 ] xi , 1 + ϑ1 u 1i − 1 , 1 + ϑ2 u 2i − 1

 u 1i − 1 i

i 2

(d)[u 2 ]  i

+ γ,h







x , 1 + ϑ1 u 1i − 1 , 1 + ϑ2 u 2i − 1

 u 2i − 1 i



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B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

and this implies

(d)

y



  

 y (x) exp C (d)[u 1 ] |u 1 − 1| y (x)+d + C (d)[u 2 ] |u 2 − 1| y (x)+d + O L  ,

+1,h (x, u 1 , u 2 )

,h

,h

where we use that, for i  2, and j = 1, 2

(d)[u j ]  i     (d)[u j ]  i 

γ x , 1 + ϑ1 u 1i − 1 , 1 + ϑ2 u 2i − 1  γ,h x , 1, 1 . ,h

By using recursion (2.10) and Lemma 4.2, we obtain (d)[u ] C +1,h1



(d)

(d)[u ] (d)[u ]

y +1,h (x, u 1 , u 2 )

γ,h 1 (x, u 1 , u 2 ) + i 2 γ,h 1 (xi , u 1i , u 2i )



= sup

y (x) y (x)k+d x,u 1 ,u 2      (d)[u 1 ]   (d)[u 1 ] η1 (d)[u 2 ] η2  exp C , + C , + O L C ,h + O L h h k k     (d)[u ] (d)[u ] η1 (d)[u ] η2  = C ,h 1 exp C ,h 1 + C ,h 2 1 + O L , k

and analogously (d)[u ]

(d)[u 2 ]

C +1,h2  C ,h We choose η1 and for j = 1, 2, (d)[u j ]

C ,h



(d)[u 1 ] η1

exp C ,h

η2 such that

(d)[u j ] 

 C 0,h

k

k

(d)[u 2 ] η2

+ C ,h

 

k

(d)[u 1 ] (d)[u ] 2c (C η1 +C 0,h 2 η2 ) e o 0,h

 

1 + O Lr e

(d)[u 1 ]

2co (C 0,h

 

1 + O L .

 2. Then, by induction we get, as  → ∞ and (d)[u 2 ]

η1 +C 0,h

η2 )/k

(d)[u j ]

 2C 0,h

c 0 = O (1).

r <

Note therefore that (d)[u 1 ]

C 0,h

(d)[u 2 ]

C 0,h

 (d)  (d)  = sup xZ d−1 yh (x, u 2 ), . . . , yh xd−1 , ud2−1 y (x)−d = O(1), x,u 1 ,u 2

= sup

γ0(,dh)[u 2 ] (x, u 1 , u 2 ) x,u ,u 1

2

   ∂ Z d−1 y h (x, u 2 ), . . . , y h xd−1 , ud2−1 y (x)−h−d

+ ( u 1 − 1) x ∂ u2

(d)[u 2 ]

 = O sup γ (x, u 1 , u 2 ) y (x)−h−d = O(1). 2 x,u 1 ,u 2

0,h

Let (d)[u i u j ]

γk,h

(x, u 1 , u 2 ) :=

∂2 (d) y (x, u 1 , u 2 ), ∂ u i ∂ u j k,h

(d)[u 21 ]

γk(,dh)[2] (x, u 1 , u 2 ) := γk,h

(d)[u 22 ]

(x, u 1 , u 2 ) + γk,h

i , j ∈ {1, 2},

(x, u 1 , u 2 ) + γk(,dh)[u 1 u 2 ] (x, u 1 u 2 ).

Lemma 5.2. Uniformly for |u 1 |  1, |u 2 |  1 and for |x|  ρ − η with some η > 0

  k+d  , (x, u 1 , u 2 ) = O y |x| 2    (d)[u ] k+h+d  γk,h 2 (x, u 1 , u 2 ) = O y |x|

 

(d)[u 21 ]

uniformly. Furthermore, uniformly for x ∈  , u 1 ∈ Ξk and u 2 ∈ Ξk+h



k+h+2d−1 

γk(,dh)[u1 u2 ] (x, u 1 , u 2 ) = O y |x|

γk,h



γk(,dh)[2] (x, u 1 , u 2 ) = O (k + h) y (x)k+d .

,

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

1545

Proof. The proof of the first statement is identical to the one of Lemma 4.3, as we can derive identical recursive relations for

γk(,dh)[u1 u2 ] (x, u 1 , u 2 ):

(d)[u 21 ]

γk,h

(d)[u 22 ]

(x, u 1 , u 2 ) and γk,h



γk(+d)[1,uh1 u2 ] (x, u 1 , u 2 ) = yk(d+)1,h (x, u 1 , u 2 ) +



2   j =1

(x, u 1 , u 2 ) and a similar one for

 (d)[u ]  γk,h j xi , u 1i , u 2i u ij−1

i 1

 (d)[u u ]  i γk,h 1 2 xi , u 1i , u 2i (u 1 u 2 )i −1



 .

i 1

We then prove the statement inductively with the following initial conditions (note therefore that ∂ 1 ∂ s Z n (s1 , . . . , sn ) = i Z n−i (s1 , . . . , sn−i ) (cf. [31, Eq. (8.5.3)] for i = 1 and [16, Eq. (2.5)]): i

(d)[u 21 ]

γ0,h

(x, u 1 , u 2 ) = 0,

(5.4)



 γ0,h (x, u 1 , u 2 )  γh(d)[2] (x, u 2 ) = O y (x)h+d ,  (d)  ∂ (d)  γ0(,dh)[u1 u2 ] (x, u 1 , u 2 ) = x Z d−1 y h (x, u 2 ), . . . , y h xd−1 , ud2−1 ∂ u2

d −1

   1

= γh(d) xr , ur2 rur2−1 Z d−r −1 (s1 , . . . , sd−1−r )

r i i (d)[u 22 ]

r =1





s i = y h ( x ,u 2 )

(d)  d−2

(d)

= O Z d−2 yh (x, u 2 ), . . . , yh   = O y (x)h+2d−2 .

(d) For the proof of the second statement we define D ,h

x

 (d)  , ud2−2 γh (x, u 2 ) (5.5)

(d)[2] supx,u 1 ,u 2 |γ,h (x, u 1 , u 2 )/ y (x)+d |, for

=   k, as in the proof of the second part of Lemma 4.3. We use the estimate

(d)

y



   

 y (x) exp C (d)[u 1 ] η1 + C (d)[u 2 ] η2 k−1 + O L  ,

+1,h (x, u 1 , u 2 )

,h

,h

(5.6)

which we obtained in the proof of Lemma 5.1. From the recursive description, we can derive the following by applying known bounds from Lemma 5.1, the previous statement, (5.3) and (5.6), similar to the proof of Lemma 4.3.





(d) (d) D +1,h = sup y +1,h (x, u 1 , u 2 )

y (x)−1 x,u ,u 1

2

2  2  2  

     

× γ,(dh)[ur ] xi , u 1i , u 2i uri−1 + γ,(dh)[ur ] xi , u 1i , u 2i uri−1

r =1 r =1 i 1 i 1

2   (d)[2]

 i −2

(d)[ur ]  i i i + γ,h (x, u 1 , u 2 ) + (i − 1)γ,h x , u 1 , u 2 u r

y (x)−(+d)

i 1



r =1 i 2

  (d)[u 1 ] 2 (d)[u ] (d)[u ]  C ,h y (x)k+d  exp C ,h 1 η1 + C ,h 2 η2 k−1 + O L     (d)[u ] 2 (d)[u 1 ] (d)[u 2 ] (d) + C , C ,h y (x)k+h+d + C ,h 2 y (x)k+2h+d + D ,h + O L  h  (d)[u ]    (d) (d)[u ]   D ,h exp C ,h 1 η1 + C ,h 2 η2 k−1 1 + O L   (d)[u ]  (d)[u ]  + exp C ,h 1 η1 + C ,h 2 η2 k−1    (d)[u ] 2  (d)[u ] 2 (d)[u ] (d)[u ] × C ,h 1 + C ,h 1 C ,h 2 + C ,h 2 + O L  (d) (d) (d) = D ,h α,h + β,h .

1546

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

k−1

(d)

(d) k−1

(d)

(d)

As in the proof of Lemma 4.3 we get D k,h  α0,h D 0,h + j =0 β j ,h i = j +1 α j ,h = O (k) + O ( D 0,h ). It (d) remains to prove that D 0,h = O (h): First note that (5.4) and (5.5) are still valid for x ∈  and the (d)[u 22 ]

(x, u 1 , u 2 ) = O(hy (x)h+d ) can be shown in a similar fashion as (5.5). Hence D 0,h  supx,u 1 ,u 2 O (hy (x)h+d + y (x)h+2d−2 ) = O (h). 2

estimate

(d)

γ0,h

Remark. For x ∈  and u 1 ∈ Ξk , u 2 ∈ Ξk+h a similar statement holds for the partial derivatives:

    (d)[u 2 ] (x, u 1 , u 2 ) = O ky (x)k+d , γk,h 2 (x, u 1 , u 2 ) = O (k + h) y (x)k+h+d ,   γk(,dh)[u1 u2 ] (x, u 1 , u 2 ) = O (k + h) y (x)k+d . (d)[u 21 ]

γk,h

Lemma 5.3. For x ∈  , u 1 ∈ Ξk and u 2 ∈ Ξk+h , with the same constants as in the previous lemmata, we can approximate

w k,h (x, u 1 , u 2 ) = C k (x)(u 1 − 1) y (x)k+d + C k+h (x)(u 2 − 1) y (x)k+h+d (d)

(d)

(d)

   + O (k + h) y (x)k+d |u 1 − 1|2 + |u 2 − 1|2 .

Furthermore

   k+d  k+h+d Σk(,dh) (x, u 1 , u 2 ) = C˜ k(d) x2 (u 1 − 1) y x2 + C˜ k(d+)h (x)(u 2 − 1) y x2   k  k+h  + O y |x|2 |u 1 − 1|2 + y |x|2 |u 2 − 1|2 . (d)

Proof. For the first statement, we expand w k,h (x, u 1 , u 2 ) into a Taylor polynomial of degree 2 around u 1 = u 2 = 1 and obtain (d)

(d)

(d)

w k,h (x, u 1 , u 2 ) = γk (x)(u 1 − 1) + γk+h (x)(u 2 − 1) + R with

|R| 

 1  (d)[u 21 ]  γ x, 1 + ϑ1 (u 1 − 1), 1 + ϑ2 (u 2 − 1) (u 1 − 1)2 2 k,h (d)[u 1 u 2 ] 

+ 2γk,h

(d)[u 22 ] 

+ γk,h



x, 1 + ϑ1 (u 1 − 1), 1 + ϑ2 (u 2 − 1) (u 1 − 1)(u 2 − 1)





x, 1 + ϑ1 (u 1 − 1), 1 + ϑ2 (u 2 − 1) (u 2 − 1)2 .

Hence,

w k,h (x, u 1 , u 2 ) = C k (x)(u 1 − 1) y (x)k+d + C k+h (x)(u 2 − 1) y (x)k+h+d (d)

(d)

(d)

   + O (k + h) y (x)k+d |u 1 − 1|2 + |u 2 − 1|2 ,

where we can neglect the mixed derivatives as either (u 1 − 1)2 or (u 2 − 1)2 will determine the dominant part. For the second part we again use a Taylor polynomial, using the fact that |xi | < ρ < 1 and |u ri − 1|  i |u r − 1| for i > 2, r = 1, 2. Hence the result follows immediately. 2 Note that the √terms u 1 − 1 and u 2 − 1 are asymptotically proportional: In fact u 2 /u 1 = √ (e it1 / n − 1)/(e it2 /√n − 1) ∼ t 2 /t 1 . Moreover, note that y (x2 )k+h+d is exponentially smaller than y (x2 )k+d as h = η n.

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

(d)

1547

(d)

Lemma 5.4. There exist constants such that w k,h = w k,h (x, u 1 , u 2 ) is given by (d)

w 0,h y (x)k

(d)

w k,h =

(d)

1 − fk

− ( w (0d,h) /2) · (1 − y (x)k )/(1 − y (x)) + O(|u 1 − 1| + |u 2 − 1|) (d)

for u 1 ∈ Ξk , u 2 ∈ Ξk+h and x ∈  , where f k (d)

fk

(d)

is given by

(d)

= f k (x, u 1 , u 2 ) = w 0,h (x, u 1 , u 2 )

k −1 

Σl,h (x, u 1 , u 2 ) y (x)l+1 . (d) l=0 w l,h (x, u 1 , u 2 ) w l+1,h (x, u 1 , u 2 )

(5.7)

Proof. We can argue similarly as in the proof of Lemma 4.5 and derive the recursive description

 (d)

(d) 

(d)

w k+1,h = y w k,h 1 +

Σk,h

(d)

1+

(d)

w k,h

w k,h 2

  (d)2   (d)  + O w k,h + O Σk,h ,

and equivalently

y (d)

w k+1,h

  Σk(,dh) · 1 + (d) =

1

−

(d)

w k,h

w k,h

1 2

 (d)  Σk,h   + O w k(d,h) + O . (d) w k,h

Further we get

y k +1 (d)

w k+1,h

=

yk (d)

w k,h

−

Σk(,dh) · y (x)k+1 (d)

(d)

w k,h w k+1,h

−



1





(d)

y (x)k + O w k,h yk + O 2

Σk(,dh) · yk

 .

(d)

w k,h

Solving the recurrence leads to

yk (d)

w k,h

=

=

1 (d)

w 0,h

−

(d) k −1  Σl,h · y (x)l+1 l =0



1 (d) w 0,h

1−

(d)

(d)

w l,h w l+1,h

−

1 1 − yk 2 1− y

k −1 

Σl(,dh) y (x)l+1 (d) w 0,h (d) (d) l=0 w l,h w l+1,h







−

+O

 k −1  l =0

(d) w 0,h 1 − yk

1− y

2

 (d) w ,h y 

+O

   =O ( w 0,h

 k−1 (d) l   Σl,h · y l =0

y 2 )

(d)

w l,h

   =O ( L l )

 + O w (0d,h)2

1− y

2k 

1 − y2

(d)

 +O

= f k (x,u 1 ,u 2 )



k

(d) 1 − L w 0,h 1−L

.

Observe that (d)



(d)

(d)  i

(d)

w 0,h = y h (x, u 2 ) + (u 1 − 1)xZ d−1 y h (x, u 2 ), . . . , y h (d)



(d)

(d)  i

= w h + (u 1 − 1)xZ d−1 yh (x, u 2 ), . . . , yh

x

, u 2i

x , u 2i





− y (x)

  ∼ C h(d) (x)(u 2 − 1) y (x)h+d + (u 1 − 1) y (x)d = O |u 1 − 1| + |u 2 − 1| . (d)

2 (d)

In the following, we denote by U := (u 1 − 1) y (x)d and W := w h (x, u 2 ). Note that w 0,h ∼ U + W . (d) (d) (d) By Lemma 5.3 we obtain for w ,h (note that C +h (x) = C h (x)(1 + L  ))

1548

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

w ,h (x, u 1 , u 2 ) = C  (x)(u 1 − 1) y (x)+d + C +h (x)(u 2 − 1) y (x)+h+d (d)

(d)

(d)

   + O ( + h) y (x) |u 1 − 1|2 + |u 2 − 1|2  = y (x) C (d) U + C h(d) (x)(u 2 − 1) y (x)h+d    + O ( + h) |u 1 − 1|2 + |u 2 − 1|2  (d)     = y (x) C  (x)U + W 1 + O h |u 1 − 1| + |u 2 − 1| .

We use the representation (4.11) for C˜  (x), which we already used in the proof of Lemma 4.5, and omit all error terms, to obtain by telescoping (d)

(d)

f k (x, u 1 , u 2 ) ∼ w 0,h (x, u 1 , u 2 )

k −1 

∼

(d)

 (d) =0 y (x) (C  (x)U

∼ U w (0d,h) (x, u 1 , u 2 ) (d) w 0,h (x, u 1 , u 2 )

y (x)+1 (C˜  (x)(u 1 − 1) y (x2 )k+d ) (d)

+ W ) y (x)+1 (C +1 (x)U + W )

k −1 

C˜ l (x)( y (x2 )/ y (x))l+d

l =0

(C l (x)U + W )(C l+1 (x)U + W )

(d)

(d)

(d)

(d) (d) k −1  (C l+1 (x)U + W ) − (C l (x)U + W ) (d)

(d)

(C l (x)U + W )(C l+1 (x)U + W )  1 1 − . (d) (d) C 0 (x)U + W C k (x)U + W

l =0

 (d) ∼ w 0,h (x, u 1 , u 2 ) (d)

As we know from (4.3), C 0 = xZ d−1 / y (x)d = O (1) near u 1 = 1 (analytic), hence



(U + W )

(d)

f k (x, u 1 , u 2 ) ∼ 1 −

(d)

C k (x)U + W



. √

(d) it 1 Using C k (x) ∼ C d ρ d , (u 1 − 1) ∼ √ , y (x)k ∼ e −κ b −ρ s , 1 − y (x) ∼ b



n

W + U , we can derive

ρs n

(d)

, and w 0,h (x, u 1 , u 2 ) ∼

(d)

w 0,h y (x)k

(d)

w k,h ∼

(U + W ) Cd ρd U +W

−

(d)

w 0,h 1− y (x)k 2 1− y (x)

√ √ √ (d) (C d ρ d it 1 + w h (x, u 2 ) n ) −se −κ b −ρ s ∼√ √ √ √ n −s − (C d ρ d it 1 + w (d) (x, u 2 ) n )(1 − e −κ b −ρ s )/2b√ρ h

1

Cd ρd

= √

n

  ψκ it 1 + ψη (s, it 2 ) (d)

with the expansion (4.1) of w h (x, u 2 ) with u 2 = e

it

√2

n

√

and h = η n, given by Theorem 5.

6. Tightness We must show the estimate (2.7) in Theorem 3. The fourth moment in (2.7) can be obtained by applying the operator (u ∂∂u )4 and setting u = 1 afterwards. Hence, using the transfer lemma of Flajolet and Odlyzko [21] it turns out that it suffices to show that



      ∂ ∂4 ∂2 ∂3 h2 + 7 2 + 6 3 + 4 y˜ r ,h x, u , u −1 =O ∂u 1 − | y (x)| ∂u ∂u ∂u u =1

uniformly for x ∈  and r  1 (see [17, p. 2046] for the detailed argument).

(6.1)

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

1549

Set



γk(d)[ j] (x) =

∂ j yk (x, u ) ∂u j

 and u =1

γk(,dh)[ j] (x) =



∂ j y˜ r ,h (x, u , u1 ) ∂u j

 . u =1

(d)[ j ]

The left-hand side of (6.1) is a linear combination of γk,h (x) for j = 1, 2, 3, 4. Therefore we need a bound for those quantities. We will derive upper bounds for all j since this more general result is easier to achieve. We start with an auxiliary result. Lemma 6.1. Let j be a positive integer. Under the assumption that for all i  j the bound

O(|x/ρ |k ) holds uniformly for |x|  for x ∈ .

γk(d)[i] (x) =

ρ , we have [( ∂∂u ) j Σk(d) ]u=1 = O(|x2 /ρ |k ), as k → ∞ and uniformly

Proof. By Faà di Bruno’s formula we have



∂ ∂u

j

 (d)

Σk

u =1

  1  ∂  j (d)   = w k xi , u i i ∂u u =1 i 2

=



1 i 2



j!

i j

m=1 mνm = j

ν1 ! · · · ν j !

γk(d)[ν1 +···+νm ] xi , 1

 j



λ=1

1

λ!



∂ ∂u

λ 

νλ

ui

. u =1

γk(d)[ν1 +···+νm ] (xi , 1) = O(|xi /ρ |k ) for x ∈ , since x ∈  implies |xi |  ρ  for i  2. The product is essentially a derivative of order j = λνλ of u i andcan therefore be j j −1 ik x /ρ k = estimated by O (i ). So the whole expression is bounded by a constant times i 2 i 2 k O((|x |/ρ ) i ). 2 By our assumption we have

Exactly the same line of arguments yield the analogous result for two levels: Lemma 6.2. Let j be a positive integer and set

Σ˜ k(,dh) =

1 i 2

i

(d) 



w k,h xi , u i , u −i .

(6.2) (d)[i ]

Under the assumption that for all i  j the bound γk,h

(x) = O(|x/ρ |k ) holds uniformly for |x|  ρ we have

(d) [( ∂∂u ) j Σ˜ k,h ]u =1 = O(|x2 /ρ |k ) uniformly for x ∈ . (d)[ j ]

(d)[ j ]

(x) and γk,h (x) rely on Faà di Bruno’s formula and involve therefore Our estimate for γk typically shaped sums. To cope with these we provide a technical lemma. Lemma 6.3. Let k, , λ1 , . . . , λ−1 denote integers satisfying k > 0,   3, λ j  0 (for j = 1, . . . ,  − 1) and

−1 j =1

j λ j = . Moreover, let

A k := k

−1

j =1 ( j −1)λ j

and

B k := k

−1

j =2 ( j −2)λ j

−1 λ  / 1 − y (x) j=2 j ,

˜ k := k−2 , B˜ k := k−3 /1 − | y (x)|. Then min( A k , B k )  min( A˜ k , B˜ k ) holds for x ∈ . and A

(6.3)

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Proof. First note that the exponent in A k is

−1

˜ := k−2 . AA

j =1 ( j − 1)λ j

= −

−1

j =1 λ j

and that

−1

j =1 λ j

 2. Thus

To complete the proof we have to exclude that B˜ k < A k and B˜ k < B k are both true. We will show ˜ k and this in turn is equivalent to that B˜ k < A k implies B k  B˜ k . Observe that B˜ k < A k implies B˜ k < A 1/(1 − | y (x)|) < k. Case 1.

−1

j =2 λ j > 1. 1+α

(1 − | y (x)|) < k B k  B˜ k . Case 2.

−1

j =2 λ j

In this case, a routine calculation shows that B˜ k  B k is equivalent to 1/

with

−1

α=(

j =1 λ j

− 1)/(−1 +

−1

j =2 λ j )

> 0. Hence B˜ k < A˜ k indeed implies

= 1. In this case we must have λ1 =  − j 0 , λ j 0 = 1 for some j 0 with 2  j 0   − 1

and λ j = 0 for j = j 0 and j = 1. Thus B k = k j 0 −2 /(1 − | y (x)|)  B˜ k . Case 3.

−1

j =2 λ j

= 0. In this case the only nonzero λ j is λ1 = . Thus B k = 1  k−3  B˜ k . 2

With these auxiliary lemmas we can easily get bounds for

γk(d)[ j] (x) and γk(,dh)[ j] (x).

Lemma 6.4. As k → ∞, we have

γk(d)[1] (x) =



uniformly for x ∈ ,

O(1)

(6.4)

O(|x/ρ | ) uniformly for |x|  ρ k

and for any fixed integer  > 1 −2

k O(min(k−1 , 1−| )) uniformly for x ∈ , y (x)|

γk(d)[] (x) =

O(|x/ρ |k )

Proof. The estimate (6.4) essentially follows from Lemma 4.1: We know k+d

(6.5)

uniformly for |x|  ρ .

γk(d)[1] (x) = C (d) (x) ×

y (x) (1 + O( L )) = O(1) with some 0 < L < 1 and | y (x)|  1 and this is sufficient to show the first part of (6.4). If |x|  ρ we can exploit the convexity of y (x) on the positive real line to get | y (x)|  |x/ρ |. This k

(d)[1]

(x) = O(|x/ρ |k+d ), an even better bound than stated in the assertion. implies γk Now we are left with the induction step. Again we use Faà di Bruno’s formula and the fact that (d) (d) w k (x, 1) = 0 and Σk (x, 1) = 0 and obtain



     (d)  ∂  (d) ∂ (d) w ( x , u ) = y ( x ) exp w ( x , u ) + Σ ( x , u ) k −1 k −1 ∂ u k ∂ u u =1 u =1  λ j − 1 1  ∂  j  (d)   ! (d) = w k−1 (x, u ) + Σk−1 (x, u ) λ1 ! · · · λ−1 ! j! ∂u −1 u =1

γk(d)[] (x) =

i =1

i λi =

j =1

i =1

i λi =

j =1

    (d)  ∂ (d) + y (x) w k−1 (x, u ) + Σk−1 (x, u ) ∂u u =1  (d)[ j ] − 1 (d)[ j ]  γ ( x ) + Γ (x) λ j ! k −1 k −1 = λ1 ! · · · λ−1 ! j! −1  (d)[] (d)[]  + y (x) γk−1 (x) + Γk−1 (d)[]

where Γk−1

(d)

= [( ∂∂u ) Σk−1 (x, u )]u =1 .

(6.6)

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

1551

Consider the case |x|  ρ . The product in (6.6) comprises only terms which essentially have the

γk(−d)[1 j] (x) + Γk(−d)[1 j] (x) with j < . Since  is fixed we may disregard the factorials. The induction (d)[ j ] hypothesis yields γk−1 (x) = O (|x2 /ρ |k ). Therefore the assumption of Lemma 6.1 is satisfied and we (d)[ j ] obtain Γk−1 (x) = O (|x2 /ρ |k ) and hence the factors as a whole are bounded by C · |x/ρ |k . Since

form

−1

j =1 λ j

 2 we get







γk(d)[] (x) = y (x) γk(−d)[] (x) + Γk(−d)[] + O |x/ρ |2k . 1 1 So we finally get the desired estimate by induction on k and Lemma 6.1, starting with



γ0(d)[] =

xZ d−1 ( y (x), y (x2 ), . . . , y (xd−1 ))

if  = 1,

0

otherwise.

(6.7)

Now let us turn to general x ∈ . Like before we focus first on the terms of the product of (6.6). Again the induction hypothesis (6.5) (actually its second part) guarantees that the assumption of (d)[ j ]

Lemma 6.1 is satisfied and so Γk−1 (x) = O (|x2 /ρ |k ). Furthermore, the induction hypothesis implies (d)[1] k j −2 γk(−d)[1 j] (x) = O(min(k j−1 , 1−| )). Since γk−1 (x) = O(1) this implies y (x)| − 1



j =1



λ j

γk(−d)[1 j] (x) + Γk(−d)[1 j] (x) / j !

  = O min( Ak , B k ) (d)[]

where A k , B k are given in (6.3). Set ak := γk (x). Then (6.6) is of the form ak = y (x)ak−1 + y (x)αk with αk = O (min( A k , B k )). Solving this recurrence relation gives

 

1 − y (x)k

· min( Ak , B k ) . ak = y (x) a0 + O y (x) 1 − y (x) k

1− y (x)k

If  = 2 we obtain min( A k , B k ) = 1 and since | y (x) 1− y (x) |  k and a0 y (x)k = O ( y (x)k+d ) = O (1) we get the desired bound for ak . ˜ k and B˜ k of Lemma 6.3 and these two quantities are If   3, then A k and B k can be replaced by A precisely the arguments of minimum in (6.5). 2 Lemma 6.5. We have

γk(,dh)[1] (x) =

h O ( k+ ) h

uniformly for x ∈ ,

O(|x/ρ | ) uniformly for |x|  ρ , k

(6.8)

and for any fixed  > 1 −2

γk(,dh)[] (x) =

h O(min(h−1 , 1−| )) uniformly for x ∈ , y (x)|

O(|x/ρ |k )

uniformly for |x|  ρ .

Proof. As the bounds are precisely the same as in the previous lemma, the induction step works in an analogous way, using Lemma 6.2 instead of Lemma 6.1. Thus we only have to show the initial step of the induction, Eq. (6.8). We can use a similar reasoning as in the proof of [17, Lemma 7]. Indeed, by applying the opera (d)[1] (d)[1] tor [ ∂∂u ·]u =1 to (2.10) we obtain the recurrence relation γk+1,h (x) = y (x) i 1 γk,h (xi ) with initial

γ0(,dh)[1] (x) = xZ d−1 ( y (x), y (x2 ), . . . , y (xd−1 )) − γh (x). Induction on k gives the representation (d)[1] γk,h (x) = γk(d)[1] (x) − γk(+d)[h 1] (x) and using γk(d)[1] (x) = C (d) (x) y (x)k+d (1 + O( Lk )) from Lemma 4.1

value

we obtain

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B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557









γk(,dh)[1] (x) = O sup y (x)k+d 1 − y (x)h + Lk = O



 ,

h k+d+h

x∈

2

and the proof is complete.

Now, applying Lemma 6.5 to (6.1) proves tightness and Theorem 3 after all. 7. The joint distribution of two degrees We want√to gain knowledge on the correlation between two different degrees d1 , d2 in a certain level k = κ n. (d1 )

7.1. The covariance Cov( X n

(d2 )

(k), Xn

(k))

The covariance of two random variables X and Y is given by

Cov( X , Y ) = E( X Y ) − E( X )E(Y ). In this section, we will prove the result on the covariance function of the two random variables (d ) (d ) X n 1 (k) and X n 2 (k), counting the vertices of degrees d1 and d2 , respectively, on level k, given in Proposition 2.1. (d ) (d ) To compute E( X n 1 (k) · X n 2 (k)) we need to determine yn−1 [xn ][∂ 2 yk (x, u , v )/∂ u ∂ v ]u = v =1 , while (d1 )

E( Xn

(d2 )

(k)) and E( Xn

(k)) are given by yn−1 [xn ]γk

(d1 )

(d1 )

(x) and yn−1 [xn ]γk

(d2 )

(x), respectively.

(d2 )

(x, u , v ) = ∂ yk (x, u , v )/∂ u, γk (x, u , v ) = ∂ yk (x, u , v )/∂ v as well as (d ,d ) (d ,d ) (d ,d ) (recall (2.12)) γ˜k 1 2 (x, u , v ) = ∂ 2 yk (x, u , v )/∂ u ∂ v and γ˜k 1 2 (x) = γ˜k 1 2 (x, 1, 1). We use the notations

γk

Lemma 7.1. There exist constants  and θ such that for z ∈ (η, θ)

γ˜k(d1 ,d2 ) (x) = C (d1 ) (x) · C (d2 ) (x) y (x)k+d1 +d2

k −1  

 

y (x) + O L  ,

=0

where C (d1 ) (x) and C (d2 ) (x) are given in Lemma 4.1. Proof. We use the recursive representation (4.4) for gives

γk(+d11) (x, u , v ) = yk(d+)1 (x, u , v )





γ (d1 ) (x, u , v ) with the additional variable v. This 

γk(d1 ) xi , u i , v i u i−1 .

i 1

Derivating with respect to v gives

γ˜k(+d11,d2 ) (x, u , v ) = γ (d2 ) (x, u , v )

 i 1

+ yk+1 (x, u , v )





γk(d1 ) xi , u i , v i u i−1



(d1 ,d2 )  i

i γ˜k



x , u i , v i u i −1 v i −1

i 1

        = yk+1 (x, u , v ) γk(d1 ) xi , u i , v i u i−1 γk(d2 ) xi , u i , v i v i−1 i 1

+ yk+1 (x, u , v )



(d1 ,d2 )  i

i γ˜k

i 1

(d1 ,d2 )

with γ˜0

(x) = 0. Setting u = v = 1 we obtain

i

i



x ,u , v u

i 1 i −1 i −1

v

,

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

1553

Fig. 2. The integration contour δ .

       (d1 ,d2 ) (d1 )  i  (d2 )  i  (d1 ,d2 )  i  γ˜k+1 (x) = y (x) γk x γk x + i γ˜k x i 1

i 1

i 1

    = y (x) γk(d1 ) (x) + Γk(d1 ) (x) γk(d2 ) (x) + Γk(d2 ) (x) + Γ˜k(d1 ,d2 ) (x) + γ˜k(d1 ,d2 ) (x) ,   (d ) (d ) (d ) (x) = i 2 γk 1 (xi ) and Γk 2 (x) = i 2 γk 2 (xi ) as in the proof  ( d , d ) ( d , d ) 1 2 1 2 i of Lemma 4.1, and Γ˜k (x) = i 2 i γ˜k (x ). Solving the recurrence, we get (d1 )

where we use the notations Γk

γ˜k(d1 ,d2 ) (x) =

k −1 

y (x)k−









γ(d1 ) (x) + Γ(d1 ) (x) γ(d2 ) (x) + Γ(d2 ) (x) + Γ˜(d1 ,d2 ) (x) .

(7.1)

=1 (d1 )

From Corollary 3.3 we know that Γ with Eq. (3.1) we have k −1 

γ˜k(d1 d2 ) (x) =

(x) = O( L  ) and Γ(d2 ) (x) = O( L  ) in Θ for u = v = 1. Together



 

y (x)k− C (d1 ) (x) y (x)+d1 + O L 

 

C (d2 ) y (x)+d2 + O L 

  + O L ,

=1

and the result follows.

2

To extract coefficients we will use Cauchy’s formula

 n x

γ˜ (d1 ,d2 ) (x) =

1



γ˜ (d1 ,d2 ) (x)

2π i

1 xn+1

dx,

δ

where δ is the truncated contour δ = δ1 ∪ δ2 ∪ δ3 ∪ δ4 given by

 δ1 = x = a +



ρ i

n

2

ρ  a  ρ + η log n/n ,





δ2 = x = ρ 1 −

eiϕ n

  π

π ,

− ϕ 2

δ3 = δ¯1

2

(7.2)

and δ4 being a circular arc closing the contour, cf. Fig. 2. It can be shown that the contribution of the circular arc δ4 is exponentially small and √thus asymptotically negligible. Near ρ , more precisely for x = ρ (1 + ns ) ∈ δ1 ∪ δ2 ∪ δ3 and k = κ n, we have

√

y (x)d1 +d2 = 1 + O ( |s/n| ),

1 − y (x) ∼ b

√ 



ρ −s/n 1 + O 



C (d1 ) (x) ∼ C d1 ρ d1 + O |s/n| , Hence, we obtain



 |s/n| ,

   y (x)k ∼ e −κ b −ρ s 1 + O |s/n| , √





C (d2 ) (x) ∼ C d2 ρ d2 + O |s/n| .

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B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

Fig. 3. The covariance for

 n x

1

γ˜ (d1 ,d2 ) (x) ∼ C d1 C d2 ρ d1 +d2



2π i

b

√

κ ∈ [0, √1n (E( H n ) + 3 Var( H n ))].

√ n √ √

ρ

δ

k−1

  1 e −κ b −ρ s 1 − e −κ b −ρ s e −s ρ −n ds, n −s √

√

!

√ μ −α −s−s ds can be easily transformed into δ (−s) e √ b ρ Hankel’s representation of the Gamma-function and together with yn ∼ √ ρ −n n−3/2 (cf. Eq. (2.3))

as

 =0 y (x)

and

=

1− y (x)k 1− y (x)

. Integrals of the shape

√

2

α = κ b ρ we obtain

π

  √ 2  κ 2 b2 ρ 2 2  E Xn(d1 ) (k) · Xn(d2 ) (k) = C d1 C d2 ρ d1 +d2 2 n e − 4 + e −κ b ρ + O ( n ). b

ρ

From this the representation of the covariance given in Proposition 2.1 easily follows (see Fig. 3). 7.2. The correlation coefficient To obtain more information on the correlation of two degrees d1 and d2 on the same level k = √ κ n, we compute the correlation coefficient, given by



(d1 )

Cor X n

(d2 )

(k), Xn

(d ) (d )  Cov( Xn 1 (k), Xn 2 (k))  . (k) =  Var( Xn(d1 ) (k)) Var( Xn(d2 ) (k)) (d1 )

For the computation, it remains to compute the variance Var( X n

(k)), given by



  2    2 Var Xn(d1 ) (k) = E Xn(d1 ) (k) − E Xn(d1 ) (k) . (d1 )

We need to determine E(( X n

(k))2 ), which can be done very similarly to the previous part.

    (d ) 2  1  n ∂ ∂ E Xn 1 (k) = x u . yk (x, u , 1) yn ∂u ∂u u =1 (d )

(d )

1 1 Proposition 7.2. The variance √ Var( Xn (k)) of the random variable Xn (k) counting vertices of degree d1 , with d1 fixed, at level k = κ n in a random Pólya tree of size n is asymptotically given by

  Var Xn(d1 ) (k) = C d1 C d2 ρ 2d1 n



b

κ 2 b2 ρ 2  − κ 2 b2 ρ 2 2  4 e + e −κ b ρ − κ 2 e − 2 2

ρ



√ + O ( n ),

(7.3)

as n tends to infinity. We proceed analogously to the computation of the variance, and obtain the following auxiliary result.

B. Gittenberger, V. Kraus / Journal of Combinatorial Theory, Series A 119 (2012) 1528–1557

1555

Lemma 7.3. There exist constants  and θ such that for z ∈ (η, θ)



2

γ˜k(d1 [2]) (x) = C (d1 ) (x) y (x)k+2d1

1 − y (x)k 1 − y (x)

+ C (d1 ) (x) y (x)k+d1 ,

where C (d1 ) (x) is given in Lemma 4.1. Proof. The proof of this lemma is analogous to the proof of Lemma 7.1. Derivating recurrence (4.4) a second time, we obtain an additional summand C (d1 ) (x) y (x)k+d1 which origins in derivating twice with respect to the same variable u. 2 (d1 [2])

Note that the additional summand C (d1 ) (x) y (x)k+d1 in Lemma 7.3, where γ˜k (d )

(x) and γ˜k(d1 ,d2 ) (x)

differ from each other, is equal to the expected value E( X n 1 (k)) when extracting coefficients √ 1 n ( d1 ) k+d1 [x ]C (x) y (x) . As this is of order n, while the coefficient of the other terms will be of yn order n, this term is negligible, and we obtain

 (d ) 2  2  κ 2 b2 ρ 2 2  E Xn 1 (k) = C d1 C d2 ρ 2d1 2 n e − 4 + e −κ b ρ b

ρ

(7.4)

by using Cauchy’s formula and the integration contour δ given in (7.2). Applying the known estimate (d ) for E( X n 1 (k)) we obtain the representation given in Proposition 7.2, and with Proposition 2.1 the result given in Theorem 4 follows immediately. Acknowledgments The authors thank two anonymous referees for pointing out numerous imprecisions in the manuscript as well as drawing our attention to further references. References [1] D. Aldous, The continuum random tree. I, Ann. Probab. 19 (1) (1991) 1–28. [2] D. Aldous, The continuum random tree. II. An overview, in: Stochastic Analysis, Durham, 1990, in: London Math. Soc. Lecture Note Ser., vol. 167, Cambridge Univ. Press, Cambridge, 1991, pp. 23–70. [3] D. Aldous, The continuum random tree. III, Ann. Probab. 21 (1) (1993) 248–289. [4] P. Billingsley, Convergence of Probability Measures, John Wiley & Sons Inc., New York, 1968. [5] N. Broutin, P. Flajolet, The height of random binary unlabelled trees, Discrete Math. Theor. Comput. Sci. Proc. AI (2008) 121–134. [6] N. Broutin, P. Flajolet, The distribution of height and diameter in random non-plane binary trees, Random Structures Algorithms, http://dx.doi.org/10.1002/rsa.20393, in press. [7] B. Chauvin, M. Drmota, J. Jabbour-Hattab, The profile of binary search trees, Ann. Appl. Probab. 11 (4) (2001) 1042–1062. [8] B. Chauvin, T. Klein, J.-F. Marckert, A. Rouault, Martingales and profile of binary search trees, Electron. J. Probab. 10 (12) (2005) 420–435 (electronic). [9] F. Chyzak, M. Drmota, T. Klausner, G. Kok, The distribution of patterns in random trees, Combin. Probab. Comput. 17 (1) (2008) 21–59. [10] J.W. Cohen, G. Hooghiemstra, Brownian excursion, the M / M /1 queue and their occupation times, Math. Oper. Res. 6 (4) (1981) 608–629. [11] L. Devroye, Laws of large numbers and tail inequalities for random tries and PATRICIA trees, in: Probabilistic Methods in Combinatorics and Combinatorial Optimization, J. Comput. Appl. Math. 142 (1) (2002) 27–37. [12] L. Devroye, H.-K. Hwang, Width and mode of the profile for some random trees of logarithmic height, Ann. Appl. Probab. 16 (2) (2006) 886–918. [13] M. Drmota, On nodes of given degree in random trees, in: Probabilistic Methods in Discrete Mathematics, Petrozavodsk, 1996, VSP, Utrecht, 1997, pp. 31–44. [14] M. Drmota, Random Trees, Springer, Wien/New York, 2009. [15] M. Drmota, B. Gittenberger, On the profile of random trees, Random Structures Algorithms 10 (4) (1997) 421–451. [16] M. Drmota, B. Gittenberger, The distribution of nodes of given degree in random trees, J. Graph Theory 31 (3) (1999) 227–253. [17] M. Drmota, B. Gittenberger, The shape of unlabeled rooted random trees, European J. Combin. 31 (2010) 2028–2063. [18] M. Drmota, H.-K. Hwang, Profiles of random trees: correlation and width of random recursive trees and binary search trees, Adv. in Appl. Probab. 37 (2) (2005) 321–341.

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