向右之具長域Domany-Kinzel模型的漸進行為 - 政大學術集成

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(1)國立政治大學應用數學系碩士學位論文. 治向右之具長域政 Domany-Kinzel 模型的大. 漸進行為. 學. ‧ 國. 立. ‧. Asymptotic behavior for a long-range. n. direction Ch. engchi. er. io. al. sit. y. Nat. Domany-Kinzel model with right i n U. v. 研究生：林芳誼撰指導教授：陳隆奇博士中華民國 108 年 6 月 DOI:10.6814/NCCU201900928.

(2) 致謝在政大的這兩年碩士生涯裡，若以充實來總結似乎不足以形容，應該是幸福的我想，無論是扎實的課程內容、實際互動的助教經驗，亦或是生活. 政治大. 中師長與同學的支持與鼓勵，點點滴滴匯聚成如今充滿回憶與感謝的一段學生生涯。. 立. 在這篇論文的寫作過程中，首先要感謝的是我的指導教授陳隆奇老師，. ‧ 國. 學. 若非他的循循善誘、諄諄教誨，就沒有如今的這篇論文，老師從一開始指導王守朋與我時，便積極穩定的為我們安排論文研討與夏季課程，讓我們. ‧. 穩步的累積論文研究的先行知識，而後從論文主題的挑選、內容的研究，一直到結論的確立，全有賴於老師悉心的提攜指點，誠摯感謝老師這段時. y. Nat. sit. 間的指導與關照。. er. io. 接下來我要感謝這一路上提供我許多幫助的老師們，感謝教導我們實變. al. n. v i n Ch 最堅實的基礎；感謝宋傳欽老師、姜志銘老師、李明融老師、陸行老師以 engchi U 函數論的陳天進老師，風趣的風格、充實的講課為我們奠定碩士學術生涯. 及曾睿彬老師，老師們細心的教導與溫暖的關懷，扶持著我順利完成碩士學位的修習，在政大的每一天都是滿滿的感謝。. 最後要感謝系上助教們的傾力協助，沒有他們的幫忙，對許多規章程序懵懵懂懂的我難以完成課業的修習；感謝學長姐們和同學們時刻的關心與陪伴，因為有這份同儕之間的情誼，在政大的生活精彩許多；最後的最後，還要感謝我的父母對我無私的、全心全意的支持，他們的愛伴隨著我這一路的求學生涯使我成長，真心感謝他們。在此，再一次誠摯的對碩士生涯一路上幫助過我的老師們與同學們表示深深的感謝。. i. DOI:10.6814/NCCU201900928.

(3) 中文摘要在本篇文章中，我們介紹一種向右之具長域的 Domany-Kinzel 模型，其模型定義在二維方格座標上，假設 n 為一個非負整數，每個座標點 (a, b) 都擁有具機率一的向右有向鏈結，並擁有 n + 1 個分別具有 pk ∈ (0, 1) 機率的. 政治大. 從 (a, b) 到 (a + k, b + 1) 之有向鏈結，其中 a, b ∈ Z+ 且 k = 0, 1, · · · , n。假設. 立. τn (N, M ) 為從 (0, 0) 到 (N, M ) 至少有一個由被滲透的邊組成之連通的有向. ‧ 國. 學. 路徑之機率，定義長寬比以 α = N /M 表示，我們求得臨界值 αn,c ∈ R+ 使得當 α = αn,c 時在 M 趨近於無限下 τn (N, M ) 趨近於 1/2，並對其收斂速率. y. e−λ λm m!. 其中 p ∈ (0, 1)、s > 1，. p ms. 其中 λ > 0 這兩種假設情況進行討論，我們發現當 s 和 λ. sit. 以及 pm =. Nat. 且 pm ∈ [0, 1) 的前提下，特別聚焦於 pm ≈m→∞. ‧. 進行研討。進而我們研究對 n 趨近於無限時模型的表現，在 m 為非負整數. n. al. er. io. 的值符合前述情境時，limn→∞ τn (N, M ) 的極值表現與先前 n 為非負整數時. i n U. v. 的結果相似，並且在 n 趨近於無限的模型中，limn→∞ τn (N, M ) 的極值表現. Ch. engchi. 受 α 逼近 αn,c 的速度影響甚劇。. 關鍵字：Domany-Kinzel 模型、定向滲流、隨機漫步、漸進行為、臨界值行為、Berry-Esseen 定理、大離差定理. ii. DOI:10.6814/NCCU201900928.

(4) Abstract In this thesis, we introduce a certain type of Domany-Kinzel model which may be regarded as a long-range model with right direction in two-dimension. 政治大. rectangular lattices. For a fixed non-negative integer n, every site (a, b) possesses not only a directed bond from site (a, b) to (a + 1, b) with probability one but also. 立. n + 1 directed bonds from (a, b) to (a + k, b + 1) with respectively probabilities. ‧ 國. 學. pk ∈ (0, 1), ∀a, b ∈ Z+ , k = 0, 1 · · · n. Let τn (N, M ) be the probability that there is at least one connected-directed path of occupied edges from (0, 0) to (N, M ) and. ‧. let α be the aspect ratio which means α = N /M . We conclude that τn (N, M ). y. Nat. converges to 1, 0, and 1/2 as M → ∞ for α > αn,c , α < αn,c , and α = αn,c ,. sit. respectively, where αn,c ∈ R+ is the critical value. The rate of convergence is. n. al. er. io. discussed, too. Moreover, we study the cases that n tends to infinity. Specifically,. v. for pm ∈ [0, 1) with m ∈ Z+ , we discuss the two cases in detail which are pm ≈m→∞. p ms. Ch. engchi. with p ∈ (0, 1), s > 1 and pm. i n =Ue λ. −λ m. m!. with λ > 0. We discover. that the behavior of limn→∞ τn (N, M ) is similar to the case that n is a non-negative integer when s and λ fit the definition. Moreover, the speed of α approaching to the critical apect ratio highly influences the behavior of limn→∞ τn (N, M ).. Keywords: Domany-Kinzel model, directed percolation, random walk, asymptotic behavior, critical behavior, Berry-Esseen theorem, large deviation.. iii. DOI:10.6814/NCCU201900928.

(5) Contents 致謝. i. 政治大. 中文摘要. 立. Abstract. ‧ 國. sit. al. n 3 Random walk. 1 5. er. io. 2 Main results. v. y. Nat. 1 Introduction. iv. ‧. List of Figures. iii. 學. Contents. ii. Ch. . e. n . .g. c . .h. i .. i n U. v. 9. 3.1. Derivation of Dn . . . . .. . . . . . . . . . . . . . . . . . . .. 9. 3.2. Derivation of αn,c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 11. 3.3. Derivation of σn2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 15. 3.4. Behavior of αn,c and σn as n → ∞ . . . . . . . . . . . . . . . . . . . . . . . .. 17. 3.4.1 3.4.2. The case that pm ≈m→∞ The case that pm =. p ms. e−λ λm m!. . . . . . . . . . . . . . . . . . . . . . . .. 17. . . . . . . . . . . . . . . . . . . . . . . . .. 21. 4 The proof of main theorem. 23. 4.1. Proof of Theorem 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 23. 4.2. Proof of Theorem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 26. 4.3. Proof of Theorem 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 28. Bibliography. 31. DOI:10.6814/NCCU201900928.

(6) List of Figures 1.1. The long-range right directed bond Domany-Kinzel model. . . . . . . . . . . .. 2. 3.1. The illustration for D3 (1). . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 治 .................. The illustration for D (2). 政 . . . . . . . . . .大立. 10. 3. 11. 學 ‧. ‧ 國 io. sit. y. Nat. n. al. er. 3.2. Ch. engchi. i n U. v. DOI:10.6814/NCCU201900928.

(7) Chapter 1 Introduction 政治大. In 1957, percolation and directed percolation were first studied by Broadbent and Ham-. 立. mersley [2]. For the reseachers in probability and statistical mechanics, these topics are still one. ‧ 國. 學. of the most extraordinary fascinating problems in this day. With referring to [1, 11] and their references, we can find numerous properties, conjectures and results of percolation and directed. especially to the directed percolation problems.. Nat. y. ‧. percolation. But not much is known about the exact solutions to the percolation problem,. sit. In 1981, a doable version of directed percolation on the square lattice was defined by. er. io. Domany and Kinzel [8] as follows. Each vertical bond is directed upward with occupation. al. n. v i n C h 1. An essentialUpoint of this solvable version is that rightward with occupation probability engchi. probability p, where p is a fixed real number in (0, 1), and each horizontal bond is directed. the occupation probability of every single bond is independent with each other. Moreover, the boundary of the Domany-Kinzel model is known to have the same distribution as the onedimensional last passage percolation model (see [10]). Since the model mentioned above is highly specialized, it has been considered in more general cases recently. There are several directed percolation models that are considered on the square lattice. For instance, the model’s horizontal edges are occupied with probability 1 in the even rows and ph ∈ [0, 1] in the odd rows while the vertical edges are defined with occupation probability pv ∈ (0, 1) [7]; the model’s horizontal edges are defined with occupation probability 1 while the vertical edges in the n-th column are occupied with probabilities p1 ∈ [0, 1), p2 ∈ [0, 1) alternatively if n is even and probabilities p2 , p1 alternatively if n is odd [6]. Furthermore, there are some directed percolation models that are considered on more complicated lattices such. 1. DOI:10.6814/NCCU201900928.

(8) as triangular lattice and honeycomb lattice. For example, the model on the triangular lattice whose horizontal edges are directed rightward with occupation probabilities 1 and x ∈ [0, 1] alternatively, vertical edges are directed upward with occupation probability y ∈ (0, 1), and diagonal edges from lower-left to upper-right or from lower-right to upper-left with occupation probability d ∈ [0, 1) [3]; the model on the honeycomb lattice as bricks whose horizontal edges are directed rightward with occupation probabilities 1 and x ∈ [0, 1] in alternate rows while vertical edges are directed upward with occupation probability y ∈ (0, 1) [4]. In 1983 [13], Li and Zhang introduce the long-range Domany-Kinzel model with left direction on the two-dimensional lattice as follows. For every site (a, b), where a, b ∈ Z+ , there is a directed bond present from site (a, b) to (a + 1, b) with probability 1. There are also. 政治大. n + 1 directed bonds present from (a, b) to (a − k + 1, b + 1), k = 0, 1, 2, · · · , n with respective. 立. probabilities pk ∈ (0, 1) where n ∈ Z+ . They also obtained the limiting behavior of this model.. ‧ 國. 學. The model had been extended to a more general case by Chang and Chen in 2018 [5]. Moreover, they obtained the asymptotic behavior of it.. ‧. In this thesis, we introduce a certain type of Domany-Kinzel model on the two-dimensional lattice which may be regarded as a long-range model with right direction, instead of left. Nat. sit. y. direction, and we define the model as follows. For every site (a, b), where a, b ∈ Z+ , there. er. io. is a directed bond present from site (a, b) to (a + 1, b) with probability 1 and there are n + 1. al. directed bonds present from (a, b) to (a + k, b + 1), with respectively probabilities pk ∈ (0, 1),. n. v i n Ch k = 0, 1, · · · , n, where n is any non-negative (see Fig 1.1). e n ginteger chi U. Figure 1.1: The long-range right directed bond Domany-Kinzel model.. Throughout this thesis, we define some notations as follows. We denote p′ k = 1 − pk for Q k = 0, 1, · · · , n and p¯′ n = nj=0 p′ j . f (M ) ≈M →∞ g(M ) means that limM →∞ f (M )/g(M ) ∈ (0, ∞). Similarly, for a fixed α0 , f (α) ≈α→α0 g(α) means that limα→α0 f (α)/g(α) ∈ (0, ∞). 2. DOI:10.6814/NCCU201900928.

(9) The vertex (a, b) is said to be percolating if there is at least one connected-directed path of occupied edges from (0, 0) to (a, b). And the notation (0, 0) ; (a, b) is used to denote that (a, b) is percolating in this thesis. For any α ∈ R, denote Mα = ⌊αM ⌋ = sup{i ∈ Z : i ≤ αM } with non-negative integer M . Let P be the probability distribution of the bond variables, and define the two point correlation function, with respect to n, τn (Mα , M ) = P (0, 0) ; (Mα , M ) .. (1.1). For triangle lattices (in our model is n=1), it was shown in [15] by the method of steepest descent that there is. p p , 政治 1−p p 大 ′2 ′ 0 1 ′ ′ 0 1. α1,c = p′ 0 +. 立. lim τ1 (Mα , M ) =. 1, if α > α1,c ,. (1.2). 0, if α < α1,c ,.        . if α = α1,c .. io. p′ 0 . 1−p′ 0. er. Remark 1.1. For n = 0, it was shown that α0,c =. sit. y. Nat. 1 , 2. ‧. M →∞.         . 學. ‧ 國. such that. al. n. v i n C we can get α1,c and the result of (1.2)heasily i Uof large number rather than the method e n gbycthehlaw. In fact, probability theory is extremely effective to deal with this model. More specifically,. of steepest descent. Moreover, we can extend the result in (1.2) to all n ∈ Z+ in this thesis (see. Theorem 2.1 in the next chapter). Note that it is square lattices for n = 0 and triangle lattices for n = 1. The limiting behavior of two-point function in (1.2) is really interesting as the critical point is discontinuous. It is appropriate to define some of the standard critical exponents and to sketch the phenomenological scaling theory of τn (Mα , M ). For α near αn,c , the scaling theory of critical behavior now asserts that the singular part of τn (Mα , M ) varies asymptotically as (see [12]) Aα τn (Mα , M ) ∼ µ exp M. . −Bα M (αn,c − α)−ν. ,. (1.3). where f (M ) ∼ g(M ) means that limM →∞ f (M )/g(M ) = 1, the constants Aα and Bα depend on α, and µ, ν ∈ (0, ∞) are universal constants. Furthermore, µ is called the critical exponent 3. DOI:10.6814/NCCU201900928.

(10) and ν is called the critical exponent of the correlation length [13]. Note that there has been no general proof of the existence of critical exponents. This allows us analyze (1.2) in detail. In this thesis, we use the Berry-Esseen theorem and large deviation argument to investigate the asymptotic behavior of τn (Mα , M ) for any n ∈ Z+ and as n → ∞. The rest of this thesis is organized as follows. The main results are presented in chapter 2. In chapter 3, we describe how we derive αn,c and σn2 in section 3.2, and 3.3, respectively. Moreover, we describe the behavior of αn,c and σn2 as n → ∞ in 3.4. The proofs of the main results are presented in chapter 4.. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. 4. DOI:10.6814/NCCU201900928.

(11) Chapter 2 Main results 政治大. In this chapter, we present the main theorems of this thesis. As beginning, we study the. 立. model in the case that n is a finite non-negative integer. In this case, the critical aspect ratio,. ‧ 國. 學. αn,c , and the rate of convergence are derived. Then we study the case n → ∞. With different assumptions of pk , the limit behavior of τn (Mα , M ) are learned. Finally, we investigate the. ‧. asymptotic behavior of τn (Mα , M ) in the large M limit and n → ∞. Notice that the differences between [5] and this thesis are mentioned in the remarks after the theorems.. sit. y. Nat. io. al. n. aspect ratio. er. Theorem 2.1. Given a finite n ∈ Z+ and pk ∈ (0, 1) , k = 0, 1, 2, · · · , n, there is a critical. αn,c =. n X. Ch. en c h+i · · ·g p′ j−1. p′ 0 p ′ 1 j. j−1. j=1. iv n n U Y. 1 1 − p¯′ n. p′ l. n+1−l. ,. (2.1). l=0. such that, in the large M limit,. τn (Mαn,c , M ) =. 1 1 + O( √ ) 2 M. (2.2). and when α is close to αn,c but not equal to αn,c τn (Mα , M ) ≤ e−M I(α) for α < αn,c , 1 − τn (Mα , M ) ≤ e−M I(α) for α > αn,c , where I(α) ≈α→αn,c (α − αn,c )2 .. (2.3). 5. DOI:10.6814/NCCU201900928.

(12) Notice that the rate function I(α) in (2.3) is optimal [14], which gives the upper bound of τn (Mα , M ) or 1 − τn (Mα , M ) as e−M I(α) when α is smaller or larger than αn,c , respectively. Remark 2.1. For the long-range Domany-Kinzel model with left direction in [5], we have the critical aspect ratio αn,c = −. n X. (1 − p. ′. ′2 j p j+1. n−j+1 · · · p′ n ). j=2. n 1 Y ′l + p 1 − p¯′ n l=1 l. which has several differences to (2.1) and this leads to inconsistent results in the following theorems with those of [5].. 政治大 following passages of this chapter 立 is dedicated to it. For convenience, we define. Since researching into the limit behavior as n → ∞ brings quite a sense of interest, the. to represent τn (Mα , M ) in the case n → ∞.. j. y Pj−1. · · · p′ j−1 = e. n. al. j−1. Ch. sit. k=0. and p′ 0 p′ 1. log(1−pk ). io. Pn. k=0 (j−k) log(1−pk ). er. Nat. It’s essential that pk → 0 as k → ∞, we rewrite p¯′ n = e. (2.4). ‧. ‧ 國. 學. τ (Mα , M ) = P (0, 0) ; (Mα , M ). By Maclaurin series of log(1 − x) where x < 1,. engchi. i n U. .. (2.5). v. −pk (1 + pk ) ≤ log(1 − pk ) ≤ −pk for pk ∈ (0, 0.6838026238) , so we can get the approximations p¯′ n ≈n→∞ e−. Pn k=0. and for j ≥ 1 p′ 0 p′ 1 j. j−1. · · · p′ j−1 ≈n→∞ e−. pk. (2.6). Pj−1. k=0 (j−k)pk. (2.7). Hence, under the condition pk → 0 as k → ∞, we obtain from (2.1) that αc := lim αn,c ≈ lim e− n→∞. Pj−1. k=0 (j−k)pk. n→∞. Pn. e− k=0 (n+1−k)pk Pn + . 1 − e− k=0 pk. (2.8). 6. DOI:10.6814/NCCU201900928.

(13) As a result of convenience, the subscript n is omitted in the case n → ∞ henceforth. The following two cases for n → ∞ deserve further consideration. (i) pm ≈m→∞ (ii) pm =. p ms. e−λ λm m!. with p ∈ (0, 1), s > 0;. with λ > 0 (note that pm ∈ [0, 1) with m ∈ Z+ ).. The behaviors of these two cases are illustrate in Theorem 2.2 and Theorem 2.3, respectively. Theorem 2.2. Let pm ∈ [0, 1) with m ∈ Z+ and pm ≈m→∞. p ms. with p ∈ (0, 1), s > 0. For. s > 1, we have αc ∈ (0, ∞), and, in the large M limit,. 立. (2.9). 政治大 1 1 + O( √ ), 2 M. s > 1.. 學. τ (Mαc , M ) =. ‧ 國. σ ∈ (0, ∞),. (2.10). Furthermore, for s>1, when α is close to αc but not equal to αc , we have. for α > αc ,. sit. (2.11). er. n. al. y. 1 − τ (Mα , M ) ≤ e−M I(α). io. where. ‧. for α < αc ,. Nat. τ (Mα , M ) ≤ e−M I(α). C hI(α) ≈ (α − αc)U. n i engchi 2. v. (2.12). Remark 2.2. For the long-range Domany-Kinzel model with left direction in [5] with the same assumption in Theorem 2.2, αc ∈ (−∞, 1) for s > 3 and σ 2 ∈ (0, ∞) for s > 4 and these lead to different limit behavior of τ (Mαc , M ) (see [5] Theorem 2.4). Theorem 2.3. Let pm ∈ [0, 1) with m ∈ Z+ and pm =. e−λ λm m!. with λ > 0. For λ > 0, we have. αc ∈ (0, ∞),. σ ∈ (0, ∞),. (2.13). τ (Mαc , M ) =. 1 1 + O( √ ). 2 M. (2.14). and, in the large M limit,. 7. DOI:10.6814/NCCU201900928.

(14) Furthermore, when α is close to αc but not equal to αc , we have τ (Mα , M ) ≤ e−M I(α). for α < αc ,. 1 − τ (Mα , M ) ≤ e−M I(α). for α > αc ,. (2.15). where I(α) ≈ (α − αc )2 .. (2.16). The following theorem is dealing with the asymptotic behavior of τ (Mα , M ) in the large M limit. with p ∈ (0, 1) and s > 1, 治政 let α = α − M and α = α + M , where ρ大 ∈ (0, ∞) and {ℓ } is a positive 立 slowly varying sequence. Note that the positive slowly varying sequence is defined as ∀M ∈ N Theorem 2.4. Given pm ∈ [0, 1) with m ∈ Z+ and pm ≈m→∞ ρ ℓ 2 M. + M. ℓaN ℓM. ∞ M M =1. = 1 ∀a > 0. We obtain the following results in the large M limit.. 學. ℓM > 0 and limM →∞. ρ ℓ 2 M. c. ‧. τ (Mα−M , M ), 1 − τ (Mα+M , M ) =  M 1−ρ ℓ2 σ.  M  −  √1 2σ 2 O(1) max , e  1−ρ  M M 2 ℓM      .  L  √1 Ψ( ) + O(1) max |ℓ − L|, ,  M σ  M    σ − ℓ2M 1 O(1) max e 2σ2 , √M ℓM        ℓM 1  + O ,  ρ−1  2  M 2       1 + O √1 , 2 M. , if ρ = 1, and limM →∞ ℓM = L ∈ (0, ∞),. n. er. io. al. sit. Nat. , if ρ ∈ (0, 1),. Ch. y. c. ‧ 國. − M. p ms. i n U. v. , if ρ = 1, and limM →∞ ℓM = ∞,. engchi. , if ρ ∈ (1, 2], , if ρ > 2. (2.17). 8. DOI:10.6814/NCCU201900928.

(15) Chapter 3 Random walk 3.1 Derivation of 立 Dn. 政治大. ‧ 國. 學. For any M ∈ N, an occupied vertical edge in a bond configuration is said to be wet if it lies on a percolating path where (Mα , M ) is percolating. A certain occupied vertical edge ending. ‧. at (k, m) is said to be primary wet if it is the wet edge with smallest k value for that m. In. y. Nat. a percolating configuration where (Mα , M ) is percolating, there is only one primary wet edge. sit. for each m ∈ {1, 2, · · · , M }. Define Cn,M (k) as the probability that the primary wet edge for. al. n. delta, i.e.,. er. io. m = M ending at (k, M ), and let us formally define Cn,0 (k) = δ0,k where δ is the Kronecker. Ch. i n U.     . v. e n g1,cifhk i= 0 ,. δ0,k =.    . 0, if k ̸= 0 .. Since the primary wet edge can occur at any value of k ≤ Mα , we obtain X. τn (Mα , M ) =. Cn,M (k). (3.1.1). k≤Mα. for M ∈ N. By the definition of the model in this thesis and the properties of random walk, for any integer k and non-negative integer m we have. Cn,m+1 (k) =. k X. Cn,m (k − j)Dn (j),. (3.1.2). j=0. 9. DOI:10.6814/NCCU201900928.

(16) where               Dn (j) =.             . 0. , if j < 0 ,. 1 − p′ 0. , if j = 0 , (3.1.3). (p′ j0 p′ j−1 · · · p′ j−1 )(1 − p′ 0 p′ 1 · · · p′ j−1 p′ j ) 1. , if 1 ≤ j ≤ n ,. (p¯′ n )j−n (p′ n0 p′ n−1 · · · p′ n−1 )(1 − p¯′ n ) 1. , if j > n .. · · · p′ j−1 )(1 − p′ 0 p′ 1 · · · p′ j−1 p′ j ) , if 0 ≤ j ≤ n. For convenience, define Un (j) = (p′ j0 p′ j−1 1. 政治大. As an example, we illustrate D3 (1) in Figure 3.1 and D3 (2) in Figure 3.2 for n = 3. In the case of D3 (1), we mention two paths in Figure 3.1, that is (a) and (b). The connection. 立. with an arrow is occupied, while the others with double bar is unoccupied. The dotted lines can. ‧ 國. 學. be either occupied or unoccupied. The probabilities of the two paths, (a) P ({(0, 0) → (1, 1)}) = p′ 0 p1 and (b) P ({(1, 0) → (1, 1)}) = p′ 0 p′ 1 p0 , lead us to the operation of D3 (1) in the following.. ‧. D3 (1) = P ({(0, 0) → (1, 1)}) + P ({(1, 0) → (1, 1)}). y. Nat. n. al. = p′ 0 (1 − p′ 0 p′ 1 ) .. Ch. engchi. er. io. = p′ 0 (1 − p′ 1 + p′ 1 (1 − p′ 0 )). sit. = p′ 0 p1 + p′ 0 p′ 1 p0. i n U. v. Figure 3.1: The illustration for D3 (1).. In the case of D3 (2), we mention three paths in Figure 3.2, that is (a), (b), and (c). The connection with an arrow is occupied, while the others with double bar is unoccupied. The dotted lines can be either occupied or unoccupied. The probabilities of the three paths, (a). 10. DOI:10.6814/NCCU201900928.

(17) P ({(0, 0) → (2, 1)}) = p′ 20 p′ 1 p2 , (b) P ({(1, 0) → (2, 1)}) = p′ 20 p′ 1 p′ 2 p1 , and (c) P ({(2, 0) → (2, 1)}) = p′ 20 p′ 21 p′ 2 p0 , lead us to the operation of D3 (1) in the following. D3 (2) = P ({(0, 0) → (2, 1)}) + P ({(1, 0) → (2, 1)}) + P ({(2, 0) → (2, 1)}) = p′ 0 p′ 1 p2 + p′ 0 p′ 1 p′ 2 p1 + p′ 0 p′ 1 p′ 2 p0 2. 2. 2. 2. = p′ 0 p′ 1 [1 − p′ 2 + p′ 2 (1 − p′ 1 ) + p′ 1 p′ 2 (1 − p′ 0 )] 2. = p′ 0 p′ 1 (1 − p′ 0 p′ 1 p′ 2 ) 2. 政治大. 立. ‧. ‧ 國. 學. io. sit. y. Nat. n. al. er. Figure 3.2: The illustration for D3 (2).. Ch. 3.2 Derivation of αn,c. engchi. i n U. v. The generating function of a probability distribution f : Z+ → R+ can be defined as fˆ(t) =. X. f (j)tj , where |t| is less than the radius of convergence.. (3.2.1). j∈Z. 11. DOI:10.6814/NCCU201900928.

(18) Note that Cˆn,m (t) = =. ∞ X. Cn,m (k)tk. k=0 ∞ hX k X j=0. k=0. = =. X. i Cn,m−1 (k − j)Dn (j) tk−j tj j. Dn (j)t. ∞ X. j∈Z. k=j. X. ∞ X. Dn (j)tj. j∈Z. Cn,m−1 (k − j)tk−j. (3.2.2). by Fubini’s thm. Cn,m−1 (k)tk. k=0. ˆ n (t)Cˆn,m−1 (t) . =D. 政治大. By repeating (3.2.2) m times, it follows that. 立. ˆ n (t)m . Cˆn,m (t) = D. (3.2.3). ‧ 國. 學. By (3.1.3), we have. Dn (j)tj. y. al. 0. n X. sit. ′. Dn (j)t + (1 − p 0 )t + j. n. j∈Z,j<0. io. X. j. Un (j)t +. i n ∞ UX. j=1. C n−1 Y h ep′nn−lg)(1c−hp¯i′ )( n l. = 0 + (1 − p′ 0 ) + Uˆn (t) + ( = (1 − p′ 0 ) + Uˆn (t) + ( = (1 − p′ 0 ) + Uˆn (t) + (. l=0 n−1 Y. v. n−1 Y. (p¯′ n )j−n (. j∈Z,j>n. (p¯′ n )j−n tj −. j=1. l=0 n−1 Y. X. er. j∈Z. =. ‧. X. Nat. ˆ n (t) = D. p′ l. n−l. )(1 − p¯′ n )tj. l=0 n X. (p¯′ n )j−n tj ). j=1. p′ l. (p¯′ )1−n t (p¯′ )1−n t[1 − (p¯′ n t)n ] )(1 − p¯′ n )( n ¯′ − n ) 1 − p nt 1 − p¯′ n t. p′ l. p¯′ tn+1 )(1 − p¯′ n )( n ¯′ ) , 1 − p nt. n−l. n−l. l=0. (3.2.4) where Uˆn (t) =. n X. (p′ 0 p′ 1 j. j−1. · · · p′ j−1 )(1 − p′ 0 p′ 1 · · · p′ j−1 p′ j )tj .. j=1. Note that. 12. DOI:10.6814/NCCU201900928.

(19) 1 − p′ 0 + Uˆn (1) = 1 − p′ 0 +. n X. (p′ 0 p′ 1. · · · p′ j−1 )(1 − p′ 0 p′ 1 · · · p′ j−1 p′ j ). p′ 0 p ′ 1. · · · p′ j−1 −. j. j−1. j=1. = 1 − p′ 0 +. j. j−1. n X. j=1. j=1. n X. n+1 X. p′ 0 p ′ 1 j. j−1. · · · p′ j−1 −. j=1 ′. ′. p′ 0 p′ 1 · · · p′ j−1 p′ j j+1. p′ 0 p′ 1 j. n X. j j−1 p′ 0 p′ 1. ′. · · · p j−1 −. j=2 n−1 Y. p′ l. n−l. l=0. 立. j−1. n X. p′ 0 p ′ 1 j. j−1. · · · p′ j−1 − p′ 0. n+1 ′ n p1. · · · p′ n. j=2. 政治大. ,. ˆ n (1) = 1 − p′ 0 + Uˆn (1) + D. n−1 Y. p′ l. n−l. p′ l. n−l. + p¯′ n. n−1 Y. p′ l. n−l. l=0. sit. l=0. p¯′ n (1 − p¯′ n ) 1 − p¯′ n. io. =1. n. er. Nat. = 1 − p¯′ n. n−1 Y. (3.2.5). ‧. l=0. al. · · · p′ j−1. 學. ‧ 國. such that. 2. j=2. =1−p0+p0+ = 1 − p¯′ n. j. y. = 1 − p′ 0 +. n X. The average mean of 1-step walk is defined as. Ch. µn =. eXn g c h i. i n U. v. ˆ n′ (1) . jDn (j) = D. (3.2.6). j∈Z. We will show in next section that αn,c = µn .. (3.2.7). By (3.2.4), we have n−1 Y n−l d ˆ d ˆ (n + 1)tn − np¯′ n tn+1 Dn (t) = Un (t) + ( , p′ l )(1 − p¯′ n )p¯′ n ¯′ n t)2 dt dt (1 − p l=0. where. X j j−1 d ˆ (p′ 0 p′ 1 · · · p′ j−1 )(1 − p′ 0 p′ 1 · · · p′ j−1 p′ j )jtj−1 . Un (t) = dt j=1. (3.2.8). n. (3.2.9). 13. DOI:10.6814/NCCU201900928.

(20) Taking t = 1 in (3.2.9), we have X j j−1 d ˆ Un (1) = (p′ 0 p′ 1 · · · p′ j−1 )(1 − p′ 0 p′ 1 · · · p′ j−1 p′ j )j dt j=1 n. =. n X. j j−1 jp′ 0 p′ 1. ′. · · · p j−1 −. j=1. n X. jp′ 0 p′ 1 · · · p′ j−1 p′ j j+1. j. 2. j=1. ′. =p0+. n X. j j−1 jp′ 0 p′ 1. ′. · · · p j−1 −. n X. j=2. = p′ 0 +. n−1 X. = p′ 0 +. j+1. j. j=1. (j + 1)p′ 0 p′ 1 · · · p′ j − j+1. j. n−1 X. j=1. = p′ 0 +. jp′ 0 p′ 1 · · · p′ j. n−1 X j=1 n X. jp′ 0 p′ 1 · · · p′ j − np′ 0 j+1. ·p 政治p · ·大 ′. p′ 0 p′ 1 · · · p′ j − np′ 0. 立. p′ 0 p′ 1 j. n+1 ′ n 1. j. j+1. j−1. · · · p′ j−1 − np′ 0. n+1 ′ n p1. n. · · · p′ n. ‧ 國. p′ 0 p′ 1 j. j−1. · · · p′ j−1 − np′ 0. n+1 ′ n p1. 學. =. · · · p′ n. j=1. j=2. n X. n+1 ′ n p1. j. · · · p′ n .. ‧. j=1. (3.2.10). =. y. al. p′ 0 p′ 1 j. j−1. = =. j=1 n X j=1. · · · p n + p¯′ n. p′ 0 p′ 1 j. j−1. Cn+1 h ep′n1n· ·g· pc′n +h pi¯′n. · · · p′ j−1 − np¯′ n. n−1 Y. p′ l. n−l. + np¯′ n. l=0. p′ 0 p′ 1 j. j−1. · · · p′ j−1 +. n−1 Y. 1 1 − p¯′ n. n−1 Y. p′ l. n+1−l. pl. − np¯′ n + 1 1 − p¯′ n. n+. l=0. n−1 Y. p′ l. l=0 n Y. n−l n. p′ l. iv n U ′n−l l=0. · · · p′ j−1 − np′ 0. j=1 n X. ′. sit. · · · p j−1 −. n+1 n np′ 0 p′ 1. n. j=1 n X. ′. er. j j−1 p ′ 0 p′ 1. io. αn,c =. n X. Nat. Therefore from (3.2.6), (3.2.7), (3.2.8), and (3.2.10), we obtain. n−l. +. 1 1 − p¯′ n. 1 ¯′ p 1 − p¯′ n n. n−1 Y. (3.2.11) p′ l. n−l. l=0. .. l=0. Remark 3.1. From (3.2.11) we have α1,c = p′ 0 +. p′ 20 p′ 1 1−p′ 0 p′ 1. and α0,c =. p′ 0 1−p′ 0. which are equal to. α1,c and α0,c that are mentioned in chapter 1.. 14. DOI:10.6814/NCCU201900928.

(21) 3.3 Derivation of σn2 Define the variance of 1-step walk as σn2 ≡. X. (jDn (j) − µn )2. j∈Z. =. X. (3.3.1) j Dn (j) − 2. µ2n .. j∈Z. Since. X d2 ˆ D (t) = j(j − 1)Dn (j)tj−2 n dt2 j=1 ∞. 治 = 政 j D (j)t ∞ X. 立. and. 2. j−2. n. j=1. −. ∞ X. 大. jDn (j)tj−2. j=1. X d2 ˆ = j 2 Dn (j) − αn,c , D (1) n dt2 j=1. (3.3.2). er. n. 2. io. 2. al. d2 ˆ 2 Dn (1) + αn,c − αn,c . dt2. sit. Nat By (3.2.8), we have. σn2 =. ‧. a representation of the variance is given by. y. ‧ 國. 學. ∞. Ch. engchi. i n U. v. d ˆ d Dn (t) = 2 Uˆn (t) + (1 − p¯′ n )p¯′ n 2 dt dt n−1 Y n−l n(n + 1)tn−1 (1 − p¯′ t)2 + 2mp¯′ tn (1 − p¯′ t) + 2p¯′ tn n n n n × p′ l . ¯′ n t)3 (1 − p l=0. (3.3.3). 15. DOI:10.6814/NCCU201900928.

(22) Note that hX i d2 ˆ ′ j ′ j−1 ′ ′ ′ ′ ′ j−2 = (p p · · · p )(1 − p p · · · p p )j(j − 1)t U (1) n j−1 0 1 j−1 j 0 1 dt2 t=1 j=1 n. = =. n X. (j 2 − j)(p′ 0 p′ 1. j=2 n X. j. j 2 (p′ 0 p′ 1. j−1. j. j−1. · · · p′ j−1 − p′ 0 p′ 1 · · · p′ j−1 p′ j ) j+1. j. 2. · · · p′ j−1 − p′ 0 p′ 1 · · · p′ j ) − j+1. j. n X. = 4p′ 0 p′ 1 + 2. n X. j 2 (p′ 0 p′ 1 j. j−1. n X. · · · p′ j−1 ) −. j=3. h. 2. j(p′ 0 p′ 1 j. j−1. 2 2p′ 0 p′ 1. +. n−1 X. · · · p′ j−1 ) −. j. (j 立 + 1) (p. ‧ 國. i j+1 j j(p′ 0 p′ 1 · · · p′ j ). n−1. ′ j+1 ′ j 0 p1. 2. ′. j 2 (p′ 0 p′ 1 · · · p′ j ) − n2 (p′ 0. · · · p j) −. j+1. j=2. (j + 1)(p′ 0 p′ 1 · · · p′ j ) − j+1. j. n−1 X. j(p′ 0 p′ 1 · · · p′ j ) − n(p′ 0 j+1. j. n+1 ′ n p1. · · · p′ n ). i · · · p′ n ). j=2. n−1 X. 2j(p′ 0 p′ 1 · · · p′ j ) − (n2 − n)(p′ 0 j+1. n+1 ′ n p1. j. Nat. n+1 ′ n p1. io. n. al. · · · p′ n ) .. er. j. sit. 2j(p′ 0 p′ 1 · · · p′ j ) − (n2 − n)(p′ 0 j+1. · · · p′ n ). y. j=2. j=1. n+1 ′ n p1. j. ‧. = 2p′ 0 p′ 1 +. j. j=2. j=2. n−1 X. j+1. 學. n−1 hX. j+1. n X. j=2. =. · · · p′ j−1 − p′ 0 p′ 1 · · · p′ j ). j 2 (p′ 0 p′ 1 · · · p′ j ). 政治大 X. j=3. 2. j−1. j=2 n X. − 2p′ 0 p′ 1 +. −. j. j=2. j=2. =. j(p′ 0 p′ 1. Ch. Then we take t = 1 into (3.3.3) and obtain. engchi. i n U. v. X d2 ˆ j+1 j n+1 n D (1) = 2j(p′ 0 p′ 1 · · · p′ j ) − (n2 − n)(p′ 0 p′ 1 · · · p′ n ) n 2 dt j=1 n−1. n(n + 1)(1 − p¯′ n )2 + 2mp¯′ n (1 − p¯′ n ) + 2p¯′ n (1 − p¯′ n )2 n−1 h X 1 2p¯′ n i ′ ′ ′ n+1 ′ n ′ j+1 ′ j − 1) + = 2j(p 0 p 1 · · · p j ) + (p 0 p 1 · · · p n ) 2m + 2m( 1 − p¯′ n (1 − p¯′ n )2 j=1 + (p′ 0. n+1 ′ n p1. =2. n−1 X j=1. · · · p′ n ). j(p′ 0 p′ 1 · · · p′ j ) + j+1. j. 2 p¯′ n ′ n+1 ′ n ′ (p p · . · · p ) n + n 1 1 − p¯′ n 0 1 − p¯′ n (3.3.4). Therefore, we find. 16. DOI:10.6814/NCCU201900928.

(23) σn2. =2. n−1 X. j+1 j j(p′ 0 p′ 1. j=1. 2 p¯′ n ′ n+1 ′ n ′ 2 · · · p j )+ (p 0 p 1 · · · p n ) n+ +αn,c −αn,c ′ ′ ¯ ¯ 1−pn 1−pn ′. . (3.3.5). Remark 3.2. For the long-range Domany-Kinzel model with left direction in [5], Qn. n p′ jj X 2 n−k+1 − αn,c = (1 − p′ k p′ k+1 · · · p′ n ) , and ′ ¯ 1−pn k=2 Q n−1 X 2p¯′ n ( nj=1 p′ jj ) 2 ′ ′ n−k 2 σn = n(n − 1) − 2 kp k+1 · · · p n + ¯′ n )2 + αn,c − (αn,c ) . (1 − p k=1 j=1. n→∞ 3.4 Behavior of αn,c and 政σn as治. 大. 2. = limn→∞ σn2 . p ms. In this subsection we let pm ≈m→∞. p ms. ‧. 3.4.1 The case that pm ≈m→∞. 學. ‧ 國. Let αc = limn→∞ αn,c. 立 and σ. with p ∈ (0, 1) and s > 1. We describe the behavior. n. aj=1l. Proving limn→∞. Pn. ′ j ′ j−1 j=1 p 0 p 1. y. sit. io. αn,c =. n X. j j−1 p′ 0 p ′ 1. Ch ′. n 1 Y ′ n+1−l p . · · · p j−1 + 1 − p¯′ n l=0 l. er. Nat. of αn,c as n → ∞ in the following. Recall that ′. engchi. i n U. . · · · p j−1 < ∞ and limn→∞. us to αc ∈ (0, ∞). Then, first, since p′ j ≤ 1 ∀j ∈ Z+ ,. lim. n→∞. n X. p′ 0 p′ 1 j. j−1. · · · p′ j−1 ≤ lim. n→∞. j=1. v. 1 1−p¯′. . Qn n. ′ n+1−l l=0 p l. = 0 will lead. p′ j0 p′ j−1 · · · p′ j−1 ≤ p′ j0 . We have 1 n X. p′ 0 < ∞. j. (3.4.1.1). j=1. Secondly, we need to prove that limn→∞ p¯′ n ∈ (0, 1) and limn→∞. Qn l=0. = 0. By p′ n+1−l l. definition, Pn. ′ ′ ′ p¯′ n = elog(p 0 p 1 ···p n ) = e. k=0. log(1−pk ). ≈ e−. Pn k=1. pk. ≈ e−. Pn. p k=1 ks. .. (3.4.1.2). 17. DOI:10.6814/NCCU201900928.

(24)     . Note that. ∞, if s ≤ 1 ,. ∞ X p = ks   k=1  . Hence, we have, lim p¯′ n =. n→∞. Next, we consider. n Y l=0.    . n+1−l. 0, if s ≤ 1 , (3.4.1.3) ∈ (0, 1), if s > 1 .. ′ n+1 p′ n ···p′ ) 1 n. = elog(p 0. 政治大 =e Pn. k=0 (n+1−k) log(1−pk ). ≈ e− ≈ e−. Qn l=0. (3.4.1.4). Pn. k=1 (n+1−k)pk. Pn. p k=1 (n+1−k) ks. .. = 0 , it is sufficient to show that limn→∞ p′ n+1−l l. ‧. To show that limn→∞.     . 學. ‧ 國. 立. p′ l. ∈ (0, ∞), if s > 1 .. Pn. k=1 (n. +1−. y. al. n. k=1. n h X p p p i 1) = lim (n + − k s n→∞ k=1 k s k s−1 n n X X 1 1 −p = lim (n + 1)p s n→∞ k k s−1 k=1 k=1. sit. (n + 1 − k). er. n→∞. n X. io. lim. Nat. k) kps = ∞. We seperate it into two cases. In the first case that s ≥ 2, we derive. Ch. engchi. i n U. v. (3.4.1.5). = ∞. In the second case that s ∈ (1, 2), first we consider n X. X X p 1 j (n + 1 − k) s ≈ (n + 1 − k) s = . k k (n + 1 − j)s j=1 k=1 k=1 n. n. 18. DOI:10.6814/NCCU201900928.

(25) Note that n+1 X j=1. j = (n + 1 − j)s. ≥. n+1 2 X j=1. n+1 2 X j=1. n+1 X. j + (n + 1 − j)s j=. . n+1 2. j (n + 1 − j)s. . +1. j (n + 1 − j)s. n+1 2 X j ≥ ns j=1 n+1 n+1 + 1 2 1 2 = s n 2. ‧ 國. n→∞. p′ l. n+1−l. = 0 , if s > 1.. l=0. Nat. io. n. al. er. Finally, by (3.4.1.1), (3.4.1.3), and(3.4.1.7), we conclude that. (3.4.1.7). ‧. lim. n Y. 學. Hence, by (3.4.1.5) and (3.4.1.6), we have,. y. 立. 政治大 → ∞ as n → ∞.. sit. ≈ n2−s. (3.4.1.6). αc ∈ (0, ∞) , when s > 1.. Ch. engchi U. (3.4.1.8). v ni. Next, we describe the behavior of σn as n → ∞ in the following. Recall that σn2. =2. n−1 X. j(p′ 0 p′ 1 · · · p′ j ) + j+1. j. j=1. 2 p¯′ n ′ n+1 ′ n ′ 2 · · p ) n + (p p · + αn,c − αn,c . n 1 1 − p¯′ n 0 1 − p¯′ n. To say that limn→∞ σn2 ∈ (0, ∞), it is sufficient to show that limn→∞. Pn−1 j=1. ′j ′ j(p′ j+1 0 p 1 · · · p j) ∈. (0, ∞) and limn→∞ n(p′ n+1 p′ n1 · · · p′ n ) = 0. First, we rewrite 0 n−1 X. j+1 j j(p′ 0 p′ 1. n−1 X. ′. · · · p j) =. je. Pj. k=0 (j+1−k) log(1−pk ). j=1. j=1. ≈. n−1 X. je−. Pj. 1 k=1 (j+1−k) ks. (3.4.1.9) .. j=1. 19. DOI:10.6814/NCCU201900928.

(26) Similar to the proof of αc , we seperate it into three cases. In the case that s > 2, j X. j j X X 1 1 1 (j + 1 − k) s = (j + 1) − k k s k=1 k s−1 k=1 k=1. (3.4.1.10). ≈ j + 1. In the case that s = 2, j j X X 1 1 1 (j + 1 − k) s = (j + 1) − 2 k k k k=1 k=1 k=1. j X. (3.4.1.11). ≈ j + 1 − log j.. In the case that s ∈ (1, 2),. 立. ‧ 國. sit. y. ≈ j 2−s .. + 1 − k) k1s back in (3.4.1.9), we get. al. er. k=1 (j. (3.4.1.12). ‧. j X l ≈ js l=1. n. n−1 X. k=1. io. Pj. X 1 l = ks (j + 1 − l)s l=1 j. (j + 1 − k). 學. j X. Nat. Put the results of. 政治大. Ch.         . j e n g c hPin−1 j=1 e. j(p′ 0 p′ 1 · · · p′ j ) ≈n→∞ j+1. i n U. j.        . j=1. j+1. v. , if s > 2 ,. Pn−1. jelog j j=1 ej+1 ,. Pn−1. j j=1 ej 2−s ,. if s = 2 , if s ∈ (1, 2) .. Hence, we conclude that, as n → ∞, lim. n→∞. In the case that s > 1, since. n−1 X. j(p′ 0 p′ 1 · · · p′ j ) < ∞ , if s > 1. j+1. j. (3.4.1.13). j=1. Pn−1 j=1. ′j ′ j(p′ j+1 0 p 1 · · · p j ) < ∞ by (3.4.1.13), using the fact that if. 20. DOI:10.6814/NCCU201900928.

(27) P∞ n=0. an converges, then limn→∞ an = 0, lim np′ 0. n+1 ′ n p1. n→∞. · · · p′ n = 0.. (3.4.1.14). Thus, by (3.4.1.3), (3.4.1.13), and (3.4.1.14), we conclude that σ 2 ∈ (0, ∞) , when s > 1.. (3.4.1.15). Remark 3.3. For the long-range Domany-Kinzel model with left direction in the case that pm ≈m→∞. p ms. in [5], αc ∈ (−∞, 1) for s > 3 and σ 2 ∈ (0, ∞) for s > 4.. 3.4.2 The case that pm =. 立. 政治大. e−λ λm m!. ‧ 國. with λ > 0. We use the same argument as 3.4.1 as. 學. e−λ λm m!. In this subsection we let pm = following. Recall that. Nat. ′. al. n. us to αc ∈ (0, ∞). Then, first, since p′ j ≤ 1 ∀j ∈ Z+ , We have lim. n X. n→∞. Ch. engchi. j j−1 p′ 0 p′ 1. i n U. · · · p j−1 ≤ lim. . Qn. n. ′ n+1−l l=0 p l. = 0 will lead. v. p′ j0 p′ j−1 · · · p′ j−1 ≤ p′ j0 . 1. n X. ′. n→∞. j=1. 1 1−p¯′. er. · · · p j−1 < ∞ and limn→∞. sit. . ′ j ′ j−1 j=1 p 0 p 1. io. Proving limn→∞. Pn. j=1. n 1 Y ′ n+1−l · · · p j−1 + p . 1 − p¯′ n l=0 l ′. y. αn,c =. j j−1 p′ 0 p ′ 1. ‧. n X. p′ 0 < ∞.. (3.4.2.1). = lim e−1 .. (3.4.2.2). j. j=1. Secondly, we have lim p¯′ n ≈ lim e−. n→∞. Pn k=0. n→∞. pk. ≈ lim e−. Pn k=0. λk e−λ k!. n→∞. n→∞. Thirdly, we consider n Y. p′ l. n+1−l. ≈ e−. Pn. k=0 (n+1−k)pk. ≈ e−. Pn. k=0 (n+1−k). λk e−λ k!. .. (3.4.2.3). l=0. 21. DOI:10.6814/NCCU201900928.

(28) Note that. P∞. λk k=0 k!. n X. conveges for all λ > 0. Then, we know that. (n + 1 − k). k=0. ∞ ∞ h X λk e−λ λk X λk i = e−λ (n + 1) − k! k! k=0 (k − 1)! k=0. (3.4.2.4). ≈ n + 1 → ∞ as n → ∞. Hence, we have. n Y. lim. n→∞. p′ l. n+1−l. = 0.. (3.4.2.5). l=0. Finally, by (3.4.2.1), (3.4.2.2), and(3.4.2.5), we conclude that e−λ λm , m!. 政治大. αc ∈ (0, ∞) , when pm =. 立. ∀λ > 0.. (3.4.2.6). j+1 j j(p′ 0 p′ 1. n−1 X. ′. · · · p j) =. j=1. j=1. λk e−λ k=1 (j+1−k) k!. al. n−1 X. Cj+1 h. k e−λ. k!. ≈ j + 1.. ej n g c h i. i n n−1 UX. j(p′ 0 p′ 1 · · · p′ j ) ≈ lim. n→∞. j=1. (3.4.2.7) .. y. je−. Pj. sit. + 1 − k) λ. n. lim. n−1 X j=1. k=1 (j. n→∞. Pn−1. k=0 (j+1−k) log(1−pk ). ‧. ≈. io. Pj. Hence, we have,. Since limn→∞. Pj. j=1. Nat. Similar to (3.4.2.4),. je. er. n−1 X. 學. ‧ 國. Next, we describe the behavior of σn as n → ∞ in the following. We rewrite. j=1. v. j ej+1. < ∞.. (3.4.2.8). ′j ′ ′ n+1 ′ n j(p′ j+1 p 1 · · · p′ n = 0 . Thus, by (3.4.2.2) 0 p 1 · · · p j ) < ∞ , limn→∞ np 0. and (3.4.2.8), we conclude that σ 2 ∈ (0, ∞) , when pm =. e−λ λm , m!. ∀λ > 0.. (3.4.2.9). 22. DOI:10.6814/NCCU201900928.

(29) Chapter 4 The proof of main theorem 4.1. 政治大 Proof of Theorem 立 2.1. ‧ 國. 學. Let n ∈ N be fixed in this section. The following proof is based on section 4 in [5]. Since the expression of αn,c has been obtained in section 3.2, we shall go into the behavior of. ‧. τn (Mα , M ) when α is close to αn,c . Define a M -step random walk Sn,M with the distribution. y. Nat. Dn for each step. To avoid confusion with the probability P defined in chapter 1, we define the. sit. probability Probn such that Probn (Sn,M = j) = Cn,M (j) with j ∈ Z and Probn (Sn,0 = j) =. n. al. er. io. δ0,j , the Kronecker delta. The expectation for Probn is denoted by Expn . Let X1 , X2 , · · · , XM. i n 2U. v. be i.i.d. random variables with distribution Dn . Then, by the result in section 3.2 and 3.3, the. Ch. engchi. random variables have finite mean (αn,c ), variance (σn ), and third moment. Since each step of Sn,M is independent, we obtain that Sn,M = X1 + X2 + · · · + XM . By the law of large number, we have Sn,M → αn,c M. a.s. when M → ∞.. (4.1.1). Notice that, since Sn,M = X1 + X2 + · · · + XM , the mean and variance of Sn,M are given by αn,c M and M σn2 , respectively. Berry-Esseen theorem (c.f. [9]) asserts that (α−αn,c )

(30)

(31) S Z M√ P |j|3 Dn (j)

(32)

(33) u2 − α M M (α − α ) 1 2 n,M n,c n,c j∈Z M σn

(34) Probn p √ e− 2 du

(35)

(36) ≤ O √ ≤ p − .

(37) 2 2 2π M σn M σn M σn3 −∞ (4.1.2). 23. DOI:10.6814/NCCU201900928.

(38) With the definition of Mα given in the introduction, we have Probn (Sn,M ≤ αM − 1) ≤ τn (Mα , M ) = Probn (Sn,M ≤ Mα ) ≤ Probn (Sn,M ≤ αM ).. Setting α = αn,c and using. (4.1.3). P j∈Z. |j|3 Dn (j) < ∞, we obtain. S n,M − αn,c M √ τn (Mαn,c ,M ) = Probn (Sn,M ≤ αn,c M ) = Probn ≤0 M σn Z 0 2 u 1 1 1 1 √ e− 2 du + O( √ ) = + O( √ ), = 2 2π M M −∞ which gives (2.2).. (4.1.4). 政治大. Note that the Chernoff inequality states that. E(erX ) , r>0 era. P (X ≥ a) = P (erX ≥ era ) ≤ inf. 學. ‧ 國. 立. ‧. where X is the sum of n independent random variables and a ∈ R.. We consider a general α ̸= αn,c in the rest part of this section. When α < αn,c , we let η =. sit. y. Nat. − log r > 0 and deal P robn (Sn,M ≤ Mα ) with Chernoff inequality to obtain. n. al. er. io. Probn (Sn,M ≤ Mα ) ≤ Probn (Sn,M ≤ αM ) = Probn (e−ηSn,M ≥ e−ηαM ). i n U. v. Expn (e−ηSn,M ) Sˆn,M (r) = inf η>0 r∈(0,1) e−ηαM rαM ˆ n (r)M D = inf ≤ e−M In (α) , r∈(0,1) rαM. Ch. ≤ inf. where. engchi. (4.1.5). . ˆ n (r) := α log rα − log D ˆ n (rα ). In (α) = sup α log r − log D. (4.1.6). r>0. With similar argument, when α > αn,c , we let η = log r > 0 to obtain Probn (Sn,M > Mα ) = inf Probn (eηSn,M > eηαM −ηc ) , where c ∈ [0, 1) η>0. ≈M →∞ inf Probn (eηSn,M > eηαM ) η>0. (4.1.7). Expn (eηSn,M ) ≤ inf ≤ e−M In (α) . ηαM η>0 e. 24. DOI:10.6814/NCCU201900928.

(39) By (4.1.6), we have ˆ ′ (rα ) α D , = n ˆ n (rα ) rα D such that. (4.1.8). α ′ r = log rα . = log rα − − ˆ n (rα ) rα α D D ˆ ′ (rα ). In′ (α). n. (4.1.9). ˆ n (1) = 1 and D ˆ n′ (1) = αn,c , replacing rα = 1 in (4.1.8) we have rαn,c = D ˆ n (rαn,c ) = 1 As D and hence In (αn,c ) = In′ (αn,c ) = 0. Moreover, using the generating function of D and (4.1.8), we get. P ˆ n′ (rα ) mDn (m)rαm rα D α= = Pm∈Z . m ˆ n (rα ) D m∈Z Dn (m)rα. (4.1.10). 政治大. Let us take derivative on both sides of (4.1.8) with respect to α. Then we obtain. Again, taking derivative on (4.1.9), we obtain, for all α ∈ R, P. m2 Dn (m)rαm − α2 ∈ (0, ∞). ˆ n (rα ) D. m∈Z. (4.1.11). sit. y. where Vn (α) =. ‧. Nat. r′ 1 In′′ (α) = α = , rα Vn (α). 學. ‧ 國. 立. D ˆ n′′ (rα ) ˆ n′ (rα ) 2 1 αr′ D − 2α = − rα′ . ˆ ˆ rα rα D(rα ) Dn (rα ). er. io. Hence, we can conclude that In (α) is strictly convex ∀α ∈ R and it has local minimum at αn,c. al. n. v i n C h with respect to α,Uwe acquire |rα − rα Since rα is a continuous function engchi. with In (αn,c ) = 0.. n,c. | ∈ (0, 1) as α is. near αn,c . From (4.1.9) we have. In′ (α) = log rα = log rαn,c + (rα − rαn,c ) = (rα − rαn,c ) + O(1)(rα − rαn,c )2 . Applying mean value theorem and the fact that In (α) is strictly convex, it follows that In′ (α) ≈ In′′ (αn,c )(α − αn,c ). Therefore, rα − rαn,c ≈α→αn,c α − αn,c , so that Z. α. In (α) = αn,c. In′ (u)du. Z. Z. α. α. (u − αn,c )du =. (ru − rαn,c )du ≈α→αn,c. ≈α→αn,c. αn,c. αn,c. (α − αn,c )2 . 2 (4.1.12). Thus, the proof of Theorem 2.1 is completed.. 25. DOI:10.6814/NCCU201900928.

(40) 4.2 Proof of Theorem 2.2 In the case of n → ∞, we consider pm ∈ [0, 1) for m ∈ Z+ and pm ≈m→∞. p ms. with. p ∈ (0, 1) and s > 1. By the result of section 3.4.1, αc ∈ (0, ∞) for s > 1 and σ 2 ∈ (0, ∞) for s > 1. Denote the probability measure Prob such that Prob(SM = j) = CM (j) with j ∈ Z and P Prob(S0 = j) = C0 (j) = δ0,j where CM (k) = j∈Z CM −1 (k − j)D(j) for M ≥ 1. The expectation with respect to Prob is denoted by Exp. To use Berry-Esseen theorem, we P 3 need to show that ∞ k=1 k D(k) < ∞. Recall that, in (3.1.3) with n → ∞, D(j) = (p′ j0 p′ j−1 · · · p′ j−1 )(1 − p′ 0 p′ 1 · · · p′ j ) for j ≥ 1. 1. j 3 p′ 0 p′ 1 j. j−1. ′. ′. j−1. ′. 0. ′. 1. · · · p′ j−1 − j 3 p′ 0 p′ 1 · · · p′ j j+1. j. j=1. Nat. j=2 ∞ X. = p′ 0 +. io. al. n. = p′ 0 +. j=1 ∞ X. ≤ p′ 0 +. ′. · · · p j−1 −. ∞ X. j 3 p′ 0 p′ 1 · · · p′ j j+1. j=1. (j +. j+1 j 1)3 p′ 0 p′ 1. · · · p′ j −. j. ∞ X j=1. j+1 j j 3 p′ 0 p′ 1. (4.2.1) · · · p′ j. er. =p0+. j j−1 j 3 p′ 0 p ′ 1. . ‧. ∞ X. ′. j. 學. =. 立. j=1 ∞ X. ‧ 國. j=1. 政治大 )(1 − p p · · · p ) ···p. j j−1 j 3 (p′ 0 p′ 1. y. j 3 D(j) =. ∞ X. sit. ∞ X. (3j 2 + 3j + 1)(p′ 0 p′ 1 · · · p′ j ). j=1 ∞ X. j+1. Ch. j. i n U. v. e n g c hj+1i < ∞. (3j 2 + 3j + 1)p′ 0. j=1. Hence, we can use Berry-Esseen theorem and, with similar argument as (4.1.2) to (4.1.4), we have τ (Mαc , M ) =. 1 1 + O( √ ), 2 M. s > 1,. (4.2.2). which gives (2.10). We consider a general α ̸= αc in the rest part of this section. To use similar method in ˆ section 4.1 to generate the rate function, we need to check the range of r such that D(r) is convergent. Note that, with similar argument as (3.4.1.2), p′ 0 p′ 1 · · · p′ j ≈ e −. Pj. p k=1 ks. ≈ e−. Rj. p 1 ks dk. 1. ≈ e js−1. (4.2.3). 26. DOI:10.6814/NCCU201900928.

(41) and, with similar arguments as (3.4.1.5) and (3.4.1.6),. j j−1 p′ 0 p′ 1. ′. · · · p j−1 =.         . Hence, ˆ D(r) =. ∞ X. D(j)r ≈ j. j=0. e−cj , c ∈ R,. if s ≥ 2, (4.2.4). −j 2−s. e. if s ∈ (1, 2).. ,.     . P∞.    . P∞. rj j=1 ecj ,. if s ≥ 2, (4.2.5). rj. if s ∈ (1, 2). j=1 ej 2−s ,. 政治大. is convergent when r ∈ (0, 1 + ϵ) for some ϵ > 0.. Thus, we can use similar method in section 4.1. First, we consider the case that α < αc . By. 立. 學. ‧ 國. using the same argument as (4.1.5), we obtain Exp(e−ηSM ) ≤ e−M I(α) , −ηαM r∈(0,1) e. Prob(SM ≤ Mα ) ≤ inf. y. ˆ ˆ α ). α log r − log D(r) := α log rα − log D(r. (4.2.7). sit. Nat. . I(α) = sup. ‧. where. (4.2.6). r∈(0,1). io. er. Secondly, we consider the case that α > αc . By using the same argument as (4.1.7), we obtain. al. n. v i n C ) ≤ inf Exp(eU ) ≤ e−M I(α), Prob(SM > Mh α en g c h i eηαM r∈(1,1+ϵ) ηSM. (4.2.8). According to (4.2.7), we have ˆ ′ (rα ) α D , and I ′ (α) = log rα . = ˆ rα D(rα ). (4.2.9). ˆ ˆ ′ (1) = αc , setting rα = 1 in (4.2.5) leads to rαc = D(r ˆ αc ) = 1 and As D(1) = 1 and D I(αc ) = I ′ (αc ) = 0. Furthermore, using similar argument from (4.1.10) to (4.1.12), we get I(α) ≈α→αc (α − αc )2. for s > 1,. which completes the proof of Theorem 2.2. Remark 4.1. To prove Theorem 2.3, we use similar method as the proof of Theorem 2.2. Here 27. DOI:10.6814/NCCU201900928.

(42) we omit the proof.. 4.3 Proof of Theorem 2.4 − + = αc − M − 2 ℓM , αM = We consider the same definition of pm in section 4.2 and let αM ρ. ρ. αc + M − 2 ℓM , where ρ ∈ (0, ∞) and {ℓM }∞ M =1 is a positive slowly varying sequence. Note that, by (4.2.1), we can use Berry-Esseen theorem to analysis τ (Mα−M , M ) and τ (Mα+M , M ). Hence, we have. Z τ (Mα−M , M ) =. −∞. 立. u2 1 1 √ e− 2 du + O √ , 2π M. 政治 √1 大 e 2π Z. 1 − τ (Mα+M , M ) =. ρ − M (−M 2 ℓM ) √ σ M. ρ − M (−M 2 ℓM ) √ σ M. 2. − u2. −∞. (4.3.1). 1 du + O √ . M. ‧ 國. following.. 學. Here we analyse τ (Mα−M , M ) and 1 − τ (Mα+M , M ) in large M limit for several cases in the First, we consider ρ > 1. Recall that, by Tylor’s expansion, b. io. sit. y. Nat. a. Z. 1 u2 √ + o(u4 ) du 1− 2 2π a 3

(43) b u

(44) ≃ (u − )

(45) 6 a ≈ (b − a) 1 + O (b − a)2 , where a, b ∈ R, a < b, and b − a is small. (4.3.2). u2 1 √ e− 2 du ≃ 2π. n. al. er. b. ‧. Z. Ch. We simplify the upper limit of (4.3.1) ρ M (−M − 2 ℓM ). √ σ M. eMn g cℓ h) i (−M ρ −2. √. σ M. M. i n U. = −. v. ℓM σM. ρ−1 2. . Since. ρ−1 2. > 0 in this case,. → 0 as M → ∞. Hence, we obtain Z. ℓM ρ−1 2. u2 1 1 √ e− 2 du + O √ 2π M −∞ ℓM Z ∞ Z ρ−1 u2 u2 1 1 1 √ e− 2 du − σM 2 √ e− 2 du + O √ = 2π 2π M 0 0 1 ℓM 1 ≈ −O +O √ ρ−1 2 2 M σM      12 + O √1 if ρ > 2, M =     1 + O ℓM if ρ ∈ (1, 2].. τ (Mα−M , M ) =. −. σM. 2. M. (4.3.3). ρ−1 2. 28. DOI:10.6814/NCCU201900928.

(46) Secondly, we consider ρ = 1. Recall that Z. 1 Ψ(x) = √ 2π. ∞. 2. e. − u2. x. x2. e− 2 √ 1 + O(x−2 ) for large x. du = 2πx ρ. We simplify the upper limit of (4.3.1). M (−M − 2 ℓM ) √ σ M. (4.3.4). = − ℓσM in this case. Consider that. limM →∞ ℓM = L, L ∈ [0, ∞). Then we obtain Z. u2 1 1 √ e− 2 du + O √ ℓM 2π M σ L Z Z ∞ σ u2 u2 1 1 1 √ e− 2 du + √ e− 2 du + O √ = ℓM L 2π 2π M σ σ L 1 . = Ψ( ) + O(1) max |ℓM − L|, √ σ M. ∞. τ (Mα−M , M ) =. (4.3.5). 政治大. 立. ‧ 國. Z. 學. Next we consider that limM →∞ ℓM = ∞. Then we obtain, by using (4.3.4), u2 1 1 √ e− 2 du + O √ ℓM 2π M σ σ − ℓ2M 1 = O(1) max e 2σ2 , √ . ℓM M. ∞. ‧. τ (Mα−M , M ) =. Nat. sit. y. (4.3.6). al. er. io. In this case, we need to consider another situation which is the result by the large deviation. v. n. argument. Recall that τ (Mα , M ) ≤ e−M I(α) and I(α) ≈α→αc (α − αc )2 . We simplify that − αM − αc = −M. − 21. Ch. i n U. engchi. ℓM in this case. Then we obtain −. τ (Mα−M , M ) ≤ e−M I(αM ) −. ≈ O(1)e−tM (αM −αc ) , where t ∈ (0, ∞) 2. (4.3.7). = O(1)e−tℓM . 2. At last, we consider ρ ∈ (0, 1). We simplify that. ρ. M (−M − 2 ℓM ) √ σ M. = −M. 1−ρ 2 ℓM. σ. → −∞ as. M → ∞ in this case. Then we have, by (4.3.1) and (4.3.4), Z. u2 1 1 √ e− 2 du + O √ 2π M 1−ρ 2 M ℓM σ 1 = O(1) max e− 2σ2 , √ 1−ρ M M 2 ℓM. τ (Mα−M , M ) =. ∞. 1−ρ M 2 ℓM σ. (4.3.8). 29. DOI:10.6814/NCCU201900928.

(47) and by using large deviation argument we obtain −. τ (Mα−M , M ) ≤ e−M I(αM ) −. ≈ O(1)e−tM (αM −αc ) where t ∈ (0, ∞) = O(1)e−tM. 1−ρ ℓ2 M. 2. (4.3.9). .. + By the same method, we can get the corresponding result for α = αM . Thus, the proof of. Theorem 2.4 is completed.. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. 30. DOI:10.6814/NCCU201900928.

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