Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2015, Article ID 756548, 9 pages http://dx.doi.org/10.1155/2015/756548

Research Article Normal Limiting Distribution of the Size of Binary Interval Trees

Jie Liu1 and Yang Yang2,3

1 Department of Statistics and Finance, University of Science and Technology of China, Hefei 230026, China 2Department of Statistics, Nanjing Audit University, Nanjing 211815, China 3The Key Lab of Financial Engineering of Jiangsu Province, Nanjing Audit University, Nanjing 211815, China

Correspondence should be addressed to Yang Yang; [email protected]

Received 1 August 2015; Revised 3 September 2015; Accepted 9 September 2015

Academic Editor: Xinguang Zhang

Copyright © 2015 J. Liu and Y. Yang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The limiting distribution of the size of binary interval tree is investigated. Our illustration is based on the contraction method, and it is quite different from the case in one-sided binary interval tree. First, we build a distributional recursive equation ofthe size. Then, we draw the expectation, the variance, and some high order moments. Finally, it is shown that the size (with suitable standardization) approaches the standard normal random variable in the Zolotarev metric space.

1. Introduction a one-sided version of a random tree with a digital flavor. Fill et al. [12] followed with a study of the nonexistence of Random trees are usually generated based on combinatorics limit distribution for the height of the incomplete trie. Itoh and occur also in the context of algorithms from com- and Mahmoud [13] considered five incomplete one-sided puter science. There are many kinds of random trees with variants of binary interval trees and proved that their sizes all different structures, such as recursive trees, search trees, approach some normal random variables. Janson [14] drew binary trees, and interval trees. The asymptotic probability the same result for a larger scale of one-sided interval trees by behavior of random variables in random trees has attracted the renewal theory, and one kind of fragmentation trees was more scholars’ attention and has become a popular research area. Drmota [1] introduced some labelled and unlabelled discussed by Janson and Neininger [15]. Javanian et al. [16] random trees in his book. Devroye and Janson [2] studied investigated the paths in m-ary interval trees. Su et al. [17] the protected nodes in several random trees. Feng and Hu studied the complete binary interval trees and got the Law [3] researched the phase changes of scale-free trees. The of Large Numbers. In addition, Pan et al. [18] considered the limitinglawfortheheight,size,andsubtreeofbinarysearch construction algorithm about binary interval trees. trees was also considered (see [4–6]). There were also some The binary interval tree is a tree associated with repeated researchers investigating the Zagreb index and nodes of divisions of a line interval of length 𝑥. The process of divisions random recursive trees (see [7–9]). is as follows. If 𝑥<1, there is no division in effect; the Binary interval tree is a random structure that underlies associated interval tree consists only of one terminal node. the process of random division of a line interval and parking Supposing that 𝑥≥1,webeginwiththeinterval(0, 𝑥).Divide problems. It has recently been a popular subject. Sibuya the interval (0, 𝑥) into two subintervals by choosing 𝑈𝑥,a andItoh[10]showedthatthenumberofinternaland point uniformly distributed over the interval (0, 𝑥).Then, external nodes in different levels of binary internal tree is we get two intervals, (0, 𝑈𝑥) and (𝑈𝑥,𝑥). Each of the two asymptotically normal, from which the asymptotic normality subintervals is further divided at a uniform point of its length, ofthesizeofthetreecouldnotbeachieveddirectly.Prodinger and two smaller subintervals are got as before. If the length of [11] looked into various parameters of the incomplete trie, the subinterval is less than 1, we stop the division. Repeat this 2 Mathematical Problems in Engineering

0 4

0 2.5 2.5 4

0 0.8 0.8 2.5 2.5 3.2 3.2 4

0.8 1.6 1.6 2.5 (a) (b)

Figure 1: (a) The division process. (b) The binary interval tree.

process until the length of every interval (or subinterval) is From the definition of binary interval tree, it is easy to less than 1. see that 𝑆1 =3and 𝑆𝑥 =1,for𝑥<1.Forourpurposeto We take 𝑥=4, for instance. Figures 1(a) and 1(b) show investigate the case of 𝑥≥1,let𝑈𝑥 denote the point chosen (0, 𝑥) 𝑈 ∼ 𝑈(0, 𝑥) how the above random division process of interval generates uniformly from interval ;hence, 𝑥 .Forany (1) abinaryintervaltree. fixed real number 0<𝑢<𝑥,if𝑈𝑥 =𝑢, we denote 𝑆𝑢 If some different conditions are added and those intervals to be the size of the left subtree associated with the interval (0, 𝑢) 𝑆(2) satisfying the conditions are not allowed to be divided (see . Correspondingly, 𝑥−𝑢 denotes the size of the right subtree associated with the interval (𝑢, 𝑥). According to the [13, 14]), then we can get different incomplete interval trees. In (1) (2) particular, if we only divide one subinterval of every interval, rule of division, we can see that 𝑆𝑢 and 𝑆𝑥−𝑢 are mutually then the interval tree we get is the so-called one-sided interval independent. Thus, we have tree (see [13]). 󵄨 𝑑 𝑆 󵄨 =1+𝑆(1) +𝑆(2) ,∀0<𝑢<𝑥. (1) It is obvious that interval tree could embody many 𝑥󵄨𝑈=𝑢 𝑢 𝑥−𝑢 properties of random division, so it can elicit lots of valuable This formula implies that if 𝑈𝑥 =𝑢is given, 𝑆𝑥 has the same subjects related to probability. For example, for 𝑥>0, (1) (2) distribution as 1+𝑆 +𝑆 .Obviously,wecanrewritethe the height of the interval tree is the greatest level of all 𝑢 𝑥−𝑢 above formula as subintervals after the divisions, denoted by 𝐻𝑥;thetotal number of nodes of an interval tree is the total number of 𝑑 (1) (2) 𝑆𝑥 = 1+𝑆𝑈 +𝑆𝑥−𝑈 . (2) intervalsthatweregotfromtherandomdivisionprocess,and 𝑥 𝑥 so on. Let 𝑆𝑥 bethesizeoftheintervaltrees,thatis,thetotal Define number of nodes of the binary interval trees. Our intention 𝑚1 (𝑥) := E𝑆𝑥; is to investigate the random variable 𝑆𝑥,thesizeofbinary (3) interval trees. 2 𝑚2 (𝑥) := E (𝑆𝑥) . In this paper, the central limit theorem of the size of binary interval trees is investigated. In view of the difficulty It is easy to see that 𝑆 to calculate the moment generating function of 𝑥,the 𝑚 (1) = E𝑆 =3; method we used is completely different from that in the 1 1 case of one-sided interval trees. In Section 2, we build a 𝑚2 (1) =9; (4) distributional recursive equation of 𝑆𝑥 andgivetheexpec- tation, the variance, and some high order moments of 𝑆𝑥. 𝑚1 (𝑥) =𝑚2 (𝑥) =1, 0<𝑥<1. In Section 3, via the contraction method, the limit law of From the distributional recursive equation (2) and the 𝑆𝑥 isshowntoapproachtheuniquesolutionofafixed- point distributional equation in the Zolotarev metric space. above boundary conditions, Su et al. [17] calculated the expectation E𝑆𝑥 and the variance Var 𝑆𝑥,forany𝑥≥0. Finally, we demonstrate that 𝑆𝑥,withsuitablestandardization, converges to a normal limiting random variable, as 𝑥→∞. Lemma 1. Let 𝑆𝑥 bethesizeofabinaryintervaltree.Then E𝑆 =𝑚 (𝑥) =4𝑥−1, 𝑥≥1. 2. The Moments of 𝑆𝑥 𝑥 1 (5) Lemma 2. 𝑆 Compared with the one-sided interval trees, the properties Let 𝑥 bethesizeofabinaryintervaltree.Then of binary interval trees are much more complex. There are 2 {32𝑥 ln 𝑥−16𝑥 +8𝑥+8, 1≤𝑥≤2, a lot of difficulties when it comes to obtaining the moment Var 𝑆𝑥 = { (6) 𝑆 32 2−20 𝑥, 𝑥≥2. generating function of 𝑥. Therefore, the method used in {( ln ) the case of one-sided interval trees (see [13]) is no longer applicable. Here, we build a distributional recursive equation In order to prove that the asymptotic distribution of 𝑆𝑥 is 4 of 𝑆𝑥. We can calculate the expectation and the variance of normal, we also need the order of E(𝑆𝑥 − E𝑆𝑥) as 𝑥→∞. 𝑆𝑥. Furthermore, we find that the order of the fourth central The following proposition shows the fourth central moment 2 moment of 𝑆𝑥 is 𝑂(𝑥 ) as 𝑥 goes to infinity. of 𝑆𝑥. Mathematical Problems in Engineering 3

∗ Proposition 3. Let 𝑆𝑥 bethesizeofabinaryintervaltree.Then But, we can find that, in the set D , there is only one distri- bution, the standard normal N(0, 1),satisfying(10). 4 2 E (𝑆𝑥 − E [𝑆𝑥]) =𝑂(𝑥), 𝑥󳨀→∞. (7) (𝑚) Suppose that 𝑚 is a nonnegative integer. Denote F by Proof. See the appendix. thesetofallrealfunctionsthatare𝑚 times continuous and differentiable, defined on the real line. Let 𝑆 3. The CLT for 𝑥 (𝑚) F𝛼 𝑆 In this section, we will prove the asymptotic normality of 𝑥 󵄨 󵄨 (13) 𝑥→∞ (𝑚) 󵄨 (𝑚) (𝑚) 󵄨 󵄨 󵄨𝛼 as . The main method is the contraction method and := {𝑓 : 𝑓 ∈ F , 󵄨𝑓 (𝑥) −𝑓 (𝑦)󵄨 ≤ 󵄨𝑥−𝑦󵄨 }, some metrics are needed especially the Zolotarev metrics (see [19]). where 0<𝛼≤1is a fixed real number. Let 𝑠=𝑚+𝛼and First we introduce the Zolotarev metrics. Denote the 𝑋 L(𝑋) D distribution of the random variable by .Let be the 𝜁 (𝑋, 𝑌) := 𝜁 (L (𝑋) , L (𝑌)) set of the distributions of all real random variables, and define 𝑠 𝑠 (14) 󵄨 󵄨 (𝑚) = sup {󵄨E𝑓 (𝑋) − E𝑓 (𝑌)󵄨 :𝑓∈F𝛼 }, D∗ ={𝐹:𝐹∈D, ∫ 𝑥𝑑𝐹 (𝑥) =0, ∫ 𝑥2𝑑𝐹 (𝑥) R R (8) and then 𝜁𝑠 is the Zolotarev metrics with order 𝑠 on the set D. =1, ∫ |𝑥|3 𝑑𝐹 (𝑥) <∞}. According to the properties of the Zolotarev metric, we know R

𝜁𝑠 (𝑋, 𝑌) <∞ It can be verified that random variable 𝑍 with L(𝑍) = 2 𝑠 𝑠 N(0, 𝜎 ) satisfies the following formula. For any 𝑢∈[0,1], ⇐⇒ E |𝑋| + E |𝑌| <∞, (15) 𝑘 𝑘 𝑑 E𝑋 = E𝑌 , 𝑘=1,...,𝑚. 𝑍 =𝑍√𝑢+𝑍√1−𝑢, (9)

Therefore, we can choose 𝜁3 as the metric we need on the ∗ and more generally, we have the following lemma. subset D (see [20, 21]); that is, 𝑚=2, 𝛼=1. This is due ∗ ∗ to the fact that, for any L(𝑋) ∈ D and L(𝑌) ∉ D ,wehave Lemma 4. 𝑍 𝑍 ∗ If and are standard normal random variables, 𝜁3(𝑋, 𝑌),butif =∞ L(𝑋), L(𝑌) ∈ D ,then𝜁3(𝑋, 𝑌) <∞. 𝑈 is uniformly distributed over interval [0, 1],and(𝑈, 𝑍, 𝑍) are The metric 𝜁𝑠(𝑋, 𝑌) has several properties as follows (see mutually independent and then one has [20]):

𝑑 𝑐>0 𝑍 =𝑍√𝑈+𝑍√1−𝑈. (10) (1) For any constant , 𝑠 𝜁𝑠 (𝑐𝑋, 𝑐𝑌) =𝑐𝜁𝑠 (𝑋, 𝑌) ; (16) Proof. In fact, for any 𝑢∈[0,1],wehave

E exp {i𝑡(√𝑢𝑍 + √1−𝑢𝑍)} (2) if random variables 𝑌 and (𝑋1,𝑋2) are mutually independent, then = E exp {i𝑡(√𝑢𝑍 + √1−𝑢𝑍)} 𝜁𝑠 (𝑋1 +𝑌,X2 +𝑌)≤𝜁𝑠 (𝑋1,𝑋2); (17) = E𝑒i(𝑡√𝑢)𝑍E𝑒i(𝑡√1−𝑢)𝑍 (11)

𝑢𝑡2 (1−𝑢) 𝑡2 𝑡2 𝑋 {𝑋 , 𝑛=1,2,3,...} = {− } {− }= {− }. (3) for random variables and 𝑛 , exp 2 exp 2 exp 2 𝑑 𝜁𝑠 (𝑋𝑛,𝑋)󳨀→0󳨐⇒𝑋𝑛 →𝑋.󳨀 (18) Therefore,

E exp {i𝑡(√𝑈𝑍 + √1−𝑈𝑍)} Now,webegintoprovethemainresultinthispaper.

1 Theorem 5. Let 𝑆𝑥 be the size of a binary interval tree. Then, = ∫ E exp {i𝑡(√𝑢𝑍 + √1−𝑢𝑍)} 𝑑𝑢 𝑥→∞ 0 (12) as ,

1 2 2 𝑡 𝑡 𝑆𝑥 − E𝑆𝑥 𝑑 = ∫ exp {− }𝑑𝑢=exp {− }. 󳨀→ N (0, 1) . (19) 0 2 2 √Var 𝑆𝑥 4 Mathematical Problems in Engineering

Proof. Denote Suppose that 𝑥≥4;by(A.1)and(21),wehave

∗ 𝑆𝑥 − (4𝑥 − 1) 󵄨 𝑆 := ,𝑥>0, ∗󵄨 𝑆 − (4𝑥 − 1) 󵄨 𝑥 𝑆 󵄨 = 𝑥 󵄨 √ 𝑥󵄨𝑈 =𝑡 󵄨 (32 ln 2−20) 𝑥 𝑥 √(32 2−20) 𝑥󵄨 ln 𝑈𝑥=𝑡 (20) 2 32𝑥 ln 𝑥−16𝑥 +8𝑥+8 𝑆(1) − E [𝑆(1)] 𝑆(2) − E [𝑆(2) ] ℎ (𝑥) := √ , 𝑥>0. { 𝑡 𝑡 𝑥−𝑡 𝑥−𝑡 { + , 2≤𝑡≤𝑥−2, (32 ln 2−20) 𝑥 {√ √ { (32 ln 2−20) 𝑥 (32 ln 2−20) 𝑥 { { (1) (1) (2) (2) { 𝑆 − E [𝑆 ] 𝑆 − E [𝑆 ] { 𝑡 𝑡 𝑥−𝑡 𝑥−𝑡 ThenfromLemmas1and2,weknowthat { + , 1≤𝑡<2, {√(32 ln 2−20) 𝑥 √(32 ln 2−20) 𝑥 { { { (1) (1) (2) (2) 𝑑 𝑆𝑡 − E [𝑆𝑡 ] 𝑆𝑥−𝑡 − E [𝑆𝑥−𝑡] 𝑆𝑥 − E𝑆𝑥 = { + , 𝑥−2<𝑡≤𝑥−1, { , 𝑥≥2, {√(32 2−20) 𝑥 √(32 2−20) 𝑥 { { ln ln { √Var 𝑆𝑥 { { { (2) (2) { {𝑆 − E [𝑆 ]−(4𝑡 − 2) {𝑆 − E𝑆 { 𝑥−𝑡 𝑥−𝑡 ,0<𝑡<1 ∗ 𝑥 𝑥 { 𝑆 = ⋅ℎ(𝑥) , 1≤𝑥<2, (21) { √(32 ln 2−20) 𝑥 𝑥 { √Var 𝑆 { { 𝑥 { { {𝑆(1) − E [𝑆(1)]−[4 (𝑥−𝑡) −2] { { 𝑡 𝑡 (27) { 2−4𝑥 { , 𝑥−1<𝑡<𝑥 { , 0<𝑥<1. { √(32 ln 2−20) 𝑥 {√(32 ln 2−20) 𝑥 𝑡 𝑥−𝑡 { ∗√ ̃∗ √ {𝑆𝑡 + 𝑆𝑥−𝑡 ,2≤𝑡≤𝑥−2; ∗ ∗ ∗ ∗ { 𝑥 𝑥 L(𝑆 )∈D 𝑥≥2 L(𝑆 )∉D { So, we have 𝑥 for and 𝑥 for { 0<𝑥<2 { 𝑡 𝑥−𝑡 . { ∗ √ ̃∗ √ {𝑆𝑡 ℎ (𝑡) + 𝑆𝑥−𝑡 ,1≤𝑡<2, According to the correlative inequality in [21], for any { 𝑥 𝑥 ∗ ∗ { L(𝑋) ∈ D , L(𝑌) ∈ D { , { 𝑡 𝑥−𝑡 𝑑 ∗√ ̃∗ √ = {𝑆𝑡 + 𝑆𝑥−𝑡ℎ (𝑥−𝑡) , 𝑥−2<𝑡≤𝑥−1, { 𝑥 𝑥 { Γ (2) 3 { 𝜁 (𝑋, 𝑌) ≤ ∫ |𝑡| 𝑑 |𝑃 (𝑋<𝑡) −𝑃(𝑌<𝑡)| , { 𝑥−𝑡 4𝑡 − 2 3 (22) {𝑆̃∗ √ − ,0<𝑡<1; Γ (4) R { 𝑥−𝑡 { 𝑥 √(32 ln 2−20) 𝑥 { { { 𝑡 4 (𝑥−𝑡) −2 {𝑆∗√ − , 𝑥−1<𝑡<𝑥, where Γ is the gamma function. Assume that the distribution 𝑡 { 𝑥 √(32 ln 2−20) 𝑥 of random variable 𝑍 is N(0, 1).ItfollowsfromProposition3 that

where 𝑈𝑥 =𝑡is the first point chosen from interval (0, 𝑥) and ∗ 4 ̃∗ ∗ sup E (𝑆𝑥) <∞. {𝑆𝑥,𝑥>0}is an independent copy of {𝑆𝑥,𝑥>0}. 𝑥≥4 (23) If we denote 𝑈:=𝑈𝑥/𝑥,then𝑈∼𝑈(0,1)and we can rewritetheaboveformulaas Therefore, there exists a constant 𝐶>0such that 󵄨 𝑆 − (4𝑥 − 1) 󵄨 ∗󵄨 𝑥 󵄨 𝑆𝑥󵄨 = 󵄨 𝑈=𝑢 √ 32 2−20 𝑥󵄨 (28) ∗ 󵄨 ∗󵄨3 3 ( ln ) 󵄨𝑈=𝑢 sup 𝜁3 (𝑆𝑥,𝑍)≤𝐶(sup E 󵄨𝑆𝑥󵄨 + E |𝑍| )<∞. (24) 𝑥≥4 𝑥≥4 2 2 {𝑆∗ √𝑢+𝑆̃∗ √1−𝑢, ≤𝑢≤1− ; { 𝑢𝑥 (1−𝑢)𝑥 { 𝑥 𝑥 { Denote { { ∗ ∗ 1 2 {𝑆 ℎ (𝑢𝑥) √𝑢+𝑆̃ √1−𝑢, ≤𝑢< , { 𝑢𝑥 (1−𝑢)𝑥 𝑥 𝑥 { ∗ ∗ { 𝑎𝑥 := 𝜁3 (L (𝑆𝑥) ,Φ) =𝜁3 (𝑆𝑥,𝑍) , (25) { 2 1 𝑑 𝑆∗ √𝑢+𝑆̃∗ ℎ ((1−𝑢) 𝑥) √1−𝑢, 1− <𝑢≤1− , = { 𝑢𝑥 (1−𝑢)𝑥 (29) { 𝑥 𝑥 { Φ 𝑍 { where is standard normal distribution and is standard { ∗ 4𝑢𝑥 −2 1 {𝑆̃ √1−𝑢− ,0<𝑢<; { (1−𝑢)𝑥 √(32 2−20) 𝑥 𝑥 normal random variable; then we can see that { ln { { 4 (1−𝑢) 𝑥−2 1 { ∗ 𝑆𝑢𝑥√𝑢− ,1−<𝑢<1. 0≤𝑏:=lim sup 𝑎𝑥 <∞. { √(32 ln 2−20) 𝑥 𝑥 𝑥→∞ (26)

∗ Now, we just need to prove that 𝑏=0; then the theorem According to the definition of D and 𝜁3,itcouldbe ∗ ∗ ∗ ∗ follows. found that (𝑆𝑥 |𝑈𝑥 =𝑡)∈D for 2<𝑡<𝑥−2and 𝑆𝑥 ∈ D . Mathematical Problems in Engineering 5

󸀠 ∗ If we define 𝑆𝑥 := (𝑆𝑥 |𝑈𝑥 <2or 𝑈𝑥 >𝑥−2),thenwe Thus, 󸀠 ∗ 󸀠 4 can also see that 𝑆 ∈ D .Furthermore,E((𝑆 ) )≤𝐶1 for 𝛿 𝛿 𝛿 𝑥 𝑥 3/2 3/2 3/2 some positive constant 𝐶1 by conditioning on 𝑈𝑥 and using 2 ∫ 𝑢 𝑎𝑥𝑢𝑑𝑢 ≤ 2𝛽 ∫ 𝑢 𝑑𝑢 ≤ 2𝛽 ∫ 𝑢 𝑑𝑢 the similar calculation in the appendix. Hence, 2/𝑥 2/𝑥 0 4𝛽𝛿5/2 𝜀 = < ; 󸀠 5 10 𝜁3 (𝑆 ,𝑍)≤𝛽, (30) 𝑥 (33) 1−2/𝑥 1−2/𝑥 3/2 3/2 2 ∫ 𝑢 𝑎𝑥𝑢𝑑𝑢 ≤2 (𝑏+𝜀) ∫ 𝑢 𝑑𝑢 forsomepositiveconstant𝛽. 𝛿 𝛿 As we had pointed out before, the standard normal 1 4 (𝑏+𝜀) distribution is the only distribution satisfying (10) in the set ≤2(𝑏+𝜀) ∫ 𝑢3/2𝑑𝑢 < , ∗ 5 D . From (25), (14), and Lemma 4, for 𝑥>4,wehave 0 where 𝛽 is the constant as before and 𝑥 is sufficiently large. It ∗ 󸀠 4 implies that 𝑎𝑥 =𝜁3 (𝑆 ,𝑍)≤𝜁3 (𝑆 ,𝑍)⋅ 𝑥 𝑥 𝑥 4𝛽 𝛿 1−2/𝑥 𝑎 ≤ +2∫ 𝑢3/2𝑎 𝑑𝑢 +2 ∫ 𝑢3/2𝑎 𝑑𝑢 1−2/𝑥 𝑥 𝑥 𝑥𝑢 𝑥𝑢 ∗ 4𝛽 2/𝑥 𝛿 + ∫ 𝜁3 ((𝑆𝑥 | 𝑈 = 𝑢) , 𝑍) 𝑑𝑢≤ (34) 2/𝑥 𝑥 𝜀 𝜀 4 (𝑏+𝜀) 4𝑏 < + + < +𝜀, 1−2/𝑥 10 10 5 5 ∗ √ ∗ √ √ + ∫ 𝜁3 (𝑆𝑥𝑢 𝑢+𝑆𝑥(1−𝑢) 1−𝑢,𝑍 𝑢 2/𝑥 when 𝑥 is sufficiently large. Therefore, √ 4𝑏 + 𝑍 1−𝑢)𝑑𝑢By ( (15) and (29)) 𝑏:=lim sup 𝑎𝑥 ≤ +𝜀. (35) 𝑥→∞ 5 4𝛽 1−2/𝑥 ∗ √ ∗ √ √ From this equation and the arbitrariness of 𝜀>0,wecan ≤ + ∫ 𝜁3 (𝑆𝑥𝑢 𝑢+𝑆𝑥(1−𝑢) 1−𝑢,𝑍 𝑢 𝑥 2/𝑥 conclude 𝑏=0and ∗ 1−2/𝑥 𝜁 (𝑆 ,𝑍)= 𝑎 =0 ∗ lim 3 𝑥 lim 𝑥 (36) √ √ 𝑥→∞ 𝑥→∞ + 𝑆𝑥(1−𝑢) 1−𝑢)𝑑𝑢+∫ 𝜁3 (𝑍 𝑢 2/𝑥 immediately. By (18), the theorem holds. (31) ∗ √ √ √ + 𝑆𝑥(1−𝑢) 1−𝑢,𝑍 𝑢+𝑍 1−𝑢)𝑑𝑢 4𝛽 1−2/𝑥 Appendix ∗ √ √ ≤ + ∫ 𝜁3 (𝑆𝑥𝑢 𝑢, 𝑍 𝑢) 𝑑𝑢 𝑥 2/𝑥 Proof of Proposition 3 1−2/𝑥 ∗ √ √ From the process of generating the binary interval trees, it is + ∫ 𝜁3 (𝑆𝑥(1−𝑢) 1−𝑢,𝑍 1−𝑢)𝑑𝑢 2/𝑥 obvious that, for given 𝑈𝑥 =𝑡, 𝑡 ∈ (0, 𝑥), 1−2/𝑥 󵄨 4𝛽 (𝑆𝑥 − E [𝑆𝑥])󵄨𝑈 =𝑡 ∗ √ √ 𝑥 = +2∫ 𝜁3 (𝑆𝑥𝑢 𝑢, 𝑍 𝑢) 𝑑𝑢 𝑥 2/𝑥 (1) (1) (2) (2) {(𝑆𝑡 − E [𝑆𝑡 ]) + (𝑆𝑥−𝑡 − E [𝑆𝑥−𝑡]) , 1 ≤ 𝑡 ≤ 𝑥 − 1; { 4𝛽 1−2/𝑥 { (A.1) = +2∫ 𝑢3/2𝜁 (𝑆∗ ,𝑍)𝑑𝑢 𝑑 (2) (2) 3 𝑥𝑢 = {𝑆𝑥−𝑡 − E [𝑆𝑥−𝑡]−(4𝑡 − 2) ,0<𝑡<1; 𝑥 2/𝑥 { { (1) (1) 1−2/𝑥 {𝑆 − E [𝑆 ]−[4 (𝑥−𝑡) −2] , 𝑥−1<𝑡<𝑥, 4𝛽 3/2 𝑡 𝑡 = +2∫ 𝑢 𝑎𝑥𝑢𝑑𝑢. 𝑥 2/𝑥 where 𝑈𝑥 =𝑡is the first point chosen from interval (0, 𝑥).For 𝑥≥1, if we denote 5/2 (1) (1) Given 𝜀>0,let𝛿>0be small enough such that 𝛽𝛿 <𝜀/8. 𝑇𝑥 := 𝑆𝑥 − E [𝑆𝑥 ], For any fixed 𝛿>0,when𝑥 is sufficiently large, then (A.2) ∗ (2) (2) 𝑇𝑥 := 𝑆𝑥 − E [𝑆𝑥 ], 4𝛽 𝜀 < , then we have 𝑥 10 ∗ {𝑇𝑡 +𝑇𝑥−𝑡,1≤𝑡≤𝑥−1; 2 { <𝛿, { (32) 󵄨 𝑑 ∗ 𝑥 𝑇𝑥󵄨𝑈 =𝑡 = {𝑇𝑥−𝑡 − (4𝑡 − 2) ,0<𝑡<1; (A.3) 𝑥 { { sup 𝑎𝑥𝑢 <𝑏+𝜀. 𝛿≤𝑢≤1 {𝑇𝑡 − [4 (𝑥−𝑡) −2] , 𝑥−1<𝑡<𝑥. 6 Mathematical Problems in Engineering

3 4 We need to calculate E𝑇𝑥 first before we get E𝑇𝑥.For𝑥>3, and when 𝑥>3,forthepart𝑀2,wehave we have 𝑥 1 6 3 3 1 𝑀2 =− ∫ [4 (𝑥−𝑡) −2] Var [𝑆𝑡]𝑑𝑡 E [𝑇𝑥] = E [E (𝑇𝑥 |𝑈𝑥)] = ∫ E [𝑇𝑥−𝑡 𝑥 𝑥−1 𝑥 0 𝑥 𝑥 6 3 1 =− ∫ [4 (𝑥−𝑡) −2][(32 ln 2−20) 𝑡] 𝑑𝑡 − (4𝑡 − 2)] 𝑑𝑡 + ∫ E {𝑇𝑡 𝑥 𝑥−1 𝑥 𝑥−1 (A.7) 1 𝑥−1 6 (32 ln 2−20) 3 1 ∗ 3 =− ∫ (4𝑡 − 2)(𝑥−𝑡) 𝑑𝑡 − [4 (𝑥−𝑡) −2]} 𝑑𝑡 + ∫ E [𝑇𝑡 +𝑇𝑥−𝑡] 𝑑𝑡 𝑥 0 𝑥 1 𝑥 2 (32 ln 2−20) 2 3 1 = . = ∫ E {𝑇𝑡 − [4 (𝑥−𝑡) −2]} 𝑑𝑡 + 𝑥 𝑥 𝑥−1 𝑥 𝑥−1 ∗ 3 2 Therefore, ⋅ ∫ E [𝑇𝑡 +𝑇𝑥−𝑡] 𝑑𝑡 = (− 1 𝑥 𝑥 𝑥 3 2 3 2 (32 ln 2−20) 3 6 (A.4) E [𝑇𝑥] = ∫ E [𝑇𝑡 ] 𝑑𝑡 + , 𝑥>3. (A.8) ⋅ ∫ [4 (𝑥−𝑡) −2] 𝑑𝑡 + 𝑥 1 𝑥 𝑥−1 𝑥 𝑥 2 6 That is, ⋅ ∫ [4 (𝑥−𝑡) −2] E [𝑇𝑡]𝑑𝑡 − 𝑥−1 𝑥 𝑥 𝑥 3 3 2 2 𝑥E [𝑇 ]=2∫ E [𝑇 ]𝑑𝑡+2 32 2−20 , ⋅ ∫ [4 (𝑥−𝑡) −2] E [𝑇 ]𝑑𝑡+ 𝑥 𝑡 ( ln ) 𝑡 1 𝑥−1 𝑥 (A.9) 𝑥 𝑥−1 𝑥>3. 3 1 3 ⋅ ∫ E [𝑇𝑡 ]𝑑𝑡)+ ∫ (E [𝑇𝑡] 𝑥−1 𝑥 1 Via differentiation with respect to 𝑥, we get the differential 2 ∗ ∗ 2 +3E [(𝑇𝑡) 𝑇𝑥−𝑡]+3E [𝑇𝑡 (𝑇𝑥−𝑡) ] equation:

∗ 3 + E [𝑇 ] )𝑑𝑡. 󸀠 1 𝑥−𝑡 (E [𝑇3]) − E [𝑇3]=0, 𝑥>3. 𝑥 𝑥 𝑥 (A.10) ∗ In view of the independence between 𝑇𝑡 and 𝑇𝑥−𝑡 and that E[𝑇 ]=E[𝑇∗]=0 1≤𝑡≤𝑥−1 𝑡 𝑡 holds for any ,wehave The solution to this differential equation is 𝑥 3 2 3 3 E [𝑇𝑥] =− ∫ [4 (𝑥−𝑡) −2] 𝑑𝑡 E [𝑇 ]=𝑘𝑥, 𝑥 > 3, 𝑥 𝑥−1 𝑥 0 (A.11) 6 𝑥 − ∫ 4 𝑥−𝑡 −2 E [𝑇2]𝑑𝑡 [ ( ) ] 𝑡 where 𝑘0 is a constant real number. 𝑥 𝑥−1 4 Similarly, for E[𝑇𝑥] ,when𝑥>4,wehave 2 𝑥 2 𝑥−1 + ∫ E [𝑇3]𝑑𝑡+ ∫ E [𝑇3]𝑑𝑡 𝑡 𝑡 𝑥 𝑥 𝑥−1 𝑥 1 2 E [𝑇 ]4 = ∫ E {𝑇 − [4 (𝑥−𝑡) −2]}4 𝑑𝑡 𝑥 𝑥 𝑥 𝑡 2 3 𝑥−1 =− ∫ [4 (𝑥−𝑡) −2] 𝑑𝑡 (A.5) (A.12) 𝑥 𝑥−1 1 𝑥−1 + ∫ E [𝑇 +𝑇∗ ]4 𝑑𝑡. 𝑥 𝑡 𝑥−𝑡 6 2 𝑥 1 − ∫ [4 (𝑥−𝑡) −2] E [𝑇𝑡 ]𝑑𝑡 𝑥 𝑥−1 𝑇 𝑇∗ E𝑇 =0 2 𝑥 Because 𝑡 is independent of 𝑥−𝑡,and 𝑡 holds for any 3 1≤𝑡≤𝑥−1 + ∫ E [𝑇𝑡 ]𝑑𝑡 ,weget 𝑥 1 2 𝑥 2 𝑥 := 𝑀 +𝑀 + ∫ E [𝑇3]𝑑𝑡. 4 4 1 2 𝑡 E [𝑇𝑥] = ∫ [4 (𝑥−𝑡) −2] 𝑑𝑡 𝑥 1 𝑥 𝑥−1 12 𝑥 It is easy to see that 2 2 + ∫ [4 (𝑥−𝑡) −2] E [𝑇𝑡 ]𝑑𝑡 𝑥 𝑥−1 2 1 3 8 𝑥 𝑀1 =− ∫ 𝑢 𝑑𝑢 = 0, (A.6) − ∫ [4 (𝑥−𝑡) −2] E [𝑇3]𝑑𝑡 2𝑥 −2 𝑡 𝑥 𝑥−1 Mathematical Problems in Engineering 7

𝑥 2 4 When 𝑥>4,forthepart𝐼5,wehave + ∫ E [𝑇𝑡 ]𝑑𝑡 𝑥 𝑥−1 𝑥−1 𝑥−1 6 ∗ 2 1 4 ∗ 4 ∗ 2 𝐼 = ∫ E [𝑇 𝑇 ] 𝑑𝑡 + ∫ (E [𝑇 ]+E [𝑇 ] +6E [𝑇 𝑇 ] )𝑑𝑡 5 𝑡 𝑥−𝑡 𝑡 𝑥−𝑡 𝑡 𝑥−𝑡 𝑥 1 𝑥 1 2 𝑥 2 = ∫ [4 (𝑥−𝑡) −2]4 𝑑𝑡 6 2 ∗ 2 = ∫ E [𝑇𝑡 ] E [𝑇𝑥−𝑡] 𝑑𝑡 𝑥 𝑥−1 𝑥 1 12 𝑥 + ∫ [4 (𝑥−𝑡) −2]2 E [𝑇2]𝑑𝑡 6 𝑥−1 𝑥 𝑡 2 ∗ 2 𝑥−1 + ∫ E [𝑇𝑡 ] E [𝑇𝑥−𝑡] 𝑑𝑡 𝑥 𝑥−2 8 𝑥 − ∫ [4 (𝑥−𝑡) −2] E [𝑇3]𝑑𝑡 (A.17) 𝑡 𝑥−2 𝑥 𝑥−1 6 2 ∗ 2 + ∫ E [𝑇𝑡 ] E [𝑇𝑥−𝑡] 𝑑𝑡 𝑥 𝑥−1 𝑥 2 2 4 6 ∗ 2 + ∫ E [𝑇𝑡] 𝑑𝑡 + ∫ E [𝑇𝑡𝑇𝑥−𝑡] 𝑑𝑡 :=1 𝐼 𝑥 1 𝑥 1 2 12 2 ∗ 2 = ∫ E [𝑇𝑡 ] E [𝑇𝑥−𝑡] 𝑑𝑡 +𝐼2 +𝐼3 +𝐼4 +𝐼5. 𝑥 1

(A.13) 𝑥−2 6 2 ∗ 2 + ∫ E [𝑇𝑡 ] E [𝑇𝑥−𝑡] 𝑑𝑡. 𝑥 2 In particular, for the part 𝐼1,wehave 2 Noting that E[𝑇𝑡 ]=Var[𝑆𝑡] and (6), we can see that 32 𝐼 = . 1 (A.14) 𝑥 5𝑥 2 ∗ 2 ∫ E [𝑇𝑡 ] E [𝑇𝑥−𝑡] 𝑑𝑡 1 When 𝑥>3,forthepart𝐼2,wehave 2 2 =2∫ (32𝑡 ln 𝑡−16𝑡 +8𝑡+8) 𝑥 1 12 2 2 𝐼2 = ∫ [4 (𝑥−𝑡) −2] E [𝑇𝑡 ]𝑑𝑡 𝑥 𝑥−1 ⋅ ((32 ln 2−20)(𝑥−𝑡)) 𝑑𝑡 12 𝑥 = ∫ [4 (𝑥−𝑡) −2]2 Var [𝑆 ]𝑑𝑡 𝑡 𝑥−2 𝑥 𝑥−1 + ∫ ((32 ln 2−20) 𝑡) 12 𝑥 2 = ∫ [4 (𝑥−𝑡) −2]2 [(32 2−20) 𝑡] 𝑑𝑡 (A.18) 𝑥 ln 𝑥−1 1 (A.15) ⋅ ((32 2−20)(𝑥−𝑡)) 𝑑𝑡 = (32 2−20)2 12 (32 2−20) 1 ln 6 ln = ln ∫ (4𝑡 − 2)2 (𝑥−𝑡) 𝑑𝑡 𝑥 0 3 4 2 2 1 48 24 (32 2−20) ⋅𝑥 − (32 ln 2−20) 𝑥 + (32 ln 2−20) = (32 2−20) − ln 3 3 3 ln 3𝑥 16 6𝑘 ⋅ (256 2 − 168) 𝑥+ (32 2−20) := 𝑎 𝑥3 := 12𝑘 − 1 , ln 9 ln 3 1 𝑥 2 +𝑎2𝑥 +𝑎1𝑥+𝑎0, where 𝑘1 := 4(32 ln 2 − 20)/3 is a constant. 𝑥>4 𝐼 When ,forthepart 3,wehave where

𝑥 16 8 3 𝑎0 = (32 ln 2−20) ; 𝐼3 =− ∫ [4 (𝑥−𝑡) −2] E [𝑇𝑡 ]𝑑𝑡 9 𝑥 𝑥−1 8 𝑥 1 =− ∫ [4 (𝑥−𝑡) −2] 𝑘0𝑡𝑑𝑡 (A.16) 𝑎1 = (32 ln 2−20)(256 ln 2 − 168) ; 𝑥 𝑥−1 3 (A.19) 8𝑘 1 8𝑘 =− 0 ∫ (4𝑡 − 2)(𝑥−𝑡) 𝑑𝑡 = 0 , 4 2 𝑎2 =− (32 ln 2−20) ; 𝑥 0 3𝑥 3

1 2 𝑎3 = (32 ln 2−20) . where 𝑘0 isthesameasthatin(A.11). 6 8 Mathematical Problems in Engineering

Therefore, Education Institutions of China (no. 15KJA110001), Qing Lan 2 𝑥 Project, PAPD, Program of Excellent Science and Technology E [𝑇 ]4 = ∫ E [𝑇 ]4 𝑑𝑡 +𝐼 +𝐼 +𝐼 +𝐼, Innovation Team of the Jiangsu Higher Education Institu- 𝑥 𝑥 𝑡 1 2 3 5 1 (A.20) tions of China, Project of Construction for Superior Subjects 𝑥>4. of Statistics of Jiangsu Higher Education Institutions, and ProjectoftheKeyLabofFinancialEngineeringofJiangsu That is, Province. 𝑥 4 4 𝑥E [𝑇𝑥]=2∫ E [𝑇𝑡] 𝑑𝑡 +1 (𝐼 +𝐼2 +𝐼3 +𝐼5)𝑥, References 1 (A.21) 𝑥>4. [1] M. Drmota, Random Trees, Springer, New York, NY,USA, 2009. [2] L. Devroye and S. Janson, “Protected nodes and fringe subtrees Via differentiation with respect to 𝑥, we get the differential in some random trees,” Electronic Communications in Probabil- equation: ity,vol.19,article6,2014. [3] Q. Feng and Z. Hu, “Phase changes in the topological indices of 󸀠 1 𝑘 1 (E𝑇4) − E𝑇4 =12 1 +6(3𝑎𝑥+2𝑎 +𝑎 ), scale-free trees,” Journal of Applied Probability,vol.50,no.2,pp. 𝑥 𝑥 𝑥 𝑥 3 2 1 𝑥 516–532, 2013. (A.22) 𝑥>4. [4] N. Broutin and P.Flajolet, “The distribution of height and diam- eter in random non-plane binary trees,” Random Structures and Algorithms,vol.41,no.2,pp.215–252,2012. The solution to this differential equation is [5] F. Dennert and R. Grubel,¨ “On the subtree size profile of binary 4 2 search trees,” Combinatorics, Probability and Computing,vol.19, E𝑇𝑥 =18𝑎3𝑥 +12𝑎2𝑥 ln 𝑥+𝑐𝑥−6𝑎1 − 12𝑘1, (A.23) no. 4, pp. 561–578, 2010. 𝑥>4, [6] H. M. Mahmoud, “One-sided variations on binary search trees,” Annals of the Institute of Statistical Mathematics,vol.55,no.4, where 𝑐 is a constant and the constants 𝑘1,𝑎1,𝑎2,𝑎3 are real pp. 885–900, 2003. numbers as defined before. From this equation, Proposition 3 [7]Q.FengandZ.Hu,“OntheZagrebindexofrandomrecursive follows. trees,” Journal of Applied Probability,vol.48,no.4,pp.1189–1196, 2011. Consent [8] R. Grubel¨ and I. Michailow, “Random recursive trees: a bound- ary theory approach,” Electronic Journal of Probability,vol.20, Informed consent was obtained from all individual partici- article 37, 2015. pants included in the study. [9] H. M. Mahmoud and M. D. Ward, “Asymptotic properties of protected nodes in random recursive trees,” JournalofApplied Conflict of Interests Probability, vol. 52, no. 1, pp. 290–297, 2015. [10] M. Sibuya and Y. Itoh, “Random sequential bisection and its The authors declare that they have no conflict of interests. associated binary tree,” Annals of the Institute of Statistical Mathematics,vol.39,no.1,pp.69–84,1987. Acknowledgments [11] H. Prodinger, “How to select a loser,” Discrete Mathematics,vol. 120, no. 1–3, pp. 149–159, 1993. The authors are most grateful to the referee and the editor [12] J. A. Fill, H. M. Mahmoud, and W. Szpankowski, “On the for their very thorough reading of the paper and valuable distribution for the duration of a randomized leader election suggestions, which greatly improve the original results and algorithm,” The Annals of Applied Probability,vol.6,no.4,pp. presentation of this paper. Jie Liu’s work was supported by 1260–1283, 1996. the National Natural Science Foundation of China (nos. [13] Y. Itoh and H. M. Mahmoud, “One-sided variations on interval 11101394, 71471168, and 71520107002), China Postdoctoral trees,” Journal of Applied Probability, vol. 40, no. 3, pp. 654–670, Science Foundation Funded Project (nos. 201104312 and 2003. 20100480688), Fund for the Doctoral Program of Higher [14] S. Janson, “One-sided interval trees,” Journal of the Iranian Education Foundation (no. 20113402120005), and the Fun- Statistical Society,vol.3,pp.149–164,2004. damental Research Funds for the Central Universities of [15] S. Janson and R. Neininger, “The size of random fragmentation China(no.WK2040160008).YangYang’sworkwassupported trees,” Probability Theory and Related Fields,vol.142,no.3-4,pp. by the National Natural Science Foundation of China (no. 399–442, 2008. 71471090), the Humanities and Social Sciences Foundation [16] M. Javanian, H. M. Mahmoud, and M. Vahidi-Asl, “Paths in m- of the Ministry of Education of China (no. 14YJCZH182), ary interval trees,” Discrete Mathematics,vol.287,no.1–3,pp. China Postdoctoral Science Foundation (nos. 2014T70449 45–53, 2004. and 2012M520964), Natural Science Foundation of Jiangsu [17]C.Su,J.Liu,andZ.S.Hu,“TheLawsoflargenumbersforthe Province of China (no. BK20131339), the Major Research size of complete interval trees,” Advances in Mathematics,vol. Plan of Natural Science Foundation of the Jiangsu Higher 36, no. 2, pp. 181–188, 2007. Mathematical Problems in Engineering 9

[18] W. Pan, S.-K. Li, X. Cai, and L. Zeng, “The construction algo- rithm of binary interval tree nodes based on lattice partition,” JournalofComputer-AidedDesign&ComputerGraphics,vol. 23, pp. 1115–1122, 2011. [19] V. M. Zolotarev, Modern Theory of Summation of Random Variables,VSP,Utrecht,TheNetherlands,1997. [20] V. M. Zolotarev, “Approximation of distributions of sums of independent random variables with values in infinite- dimensional spaces,” Theory of Probability and Its Applications, vol. 21, no. 4, pp. 721–737, 1976. [21] S. T. Rachev and L. Ruschendorf,¨ “Probability metrics and recursive algorithms,” Advances in Applied Probability,vol.27, no. 3, pp. 770–799, 1995. Advances in Advances in Journal of Journal of Operations Research Decision Sciences Applied Mathematics Algebra Probability and Statistics Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014

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