Horizontal visibility graph of a random restricted growth sequence

Abstract

We study the distributional properties of horizontal visibility graphs associated with random restrictive growth sequences and random set partitions of size n. Our main results are formulas expressing the expected degree of graph nodes in terms of simple explicit functions of a finite collection of Stirling and Bernoulli numbers.

1. Introduction and statement of results

We study here horizontal visibility graphs of restricted growth sequences. The latter class of sequences is of interest both independently and in connection with set partitions [19], q-analogues [2], certain combinatorial matrices [7], bargraphs [20], and Gray codes [5].

A horizontal visibility graph (HVG) [17] constitutes a paradigmatic complex network representation of sequential data, typically used to reveal order structures within the data set [8, 35]. HVG-based algorithms have been employed to characterize fractal behavior of dynamical systems [21, 31], study canonical routes to chaos (see [24] and references therein), discriminate between chaotic and stochastic time series [26], and test time series irreversibility [33]. There is a growing body of literature using these combinatorial data analysis techniques in applied fields such as optics [1], fluid dynamics [22], geophysics [30], physiology and neuroscience [18, 27], finance [25], image processing [13], and more [8, 35]. For other graph theoretic methods of applied time series analysis as well as many fruitful extensions of the horizontal visibility algorithm, we refer to recent surveys [8, 35].

From a combinatoric point of view, HVGs are outerplanar graphs with a Hamiltonian path, an important subclass of so-called non-crossing graphs of algebraic combinatorics [10]. An illuminating characterization of HVGs using “one-point compactified” times series and tools of algebraic topology is obtained in a recent work [29]. Theoretical body of work on the HVGs includes studies of their degree distributions [14, 16], information-theoretic [9, 15] and other [11] topological characteristics, motifs [12, 32], spectral properties [6, 17], and dependence of graph features on the parameter for a specific parametric family of chaotic [4] or stochastic processes [31, 34]. For more, see a recent comprehensive survey [35] and an extensive review of earlier results [23].

In this paper, our main focus is on the degree properties of the horizontal visibility graph associated with a random restricted growth sequence. Let π = π₁ ⋯ π_n be a sequence of elements of a totally ordered set. We say that (π_i, π_j) is a strong visible pair if

$$\underset{i<\mathcal{l}<j}{\text{max}}{\pi }_{\mathcal{l}}<\text{min}\left\{{\pi }_i,{\pi }_j\right\},$$

where we use the usual convention that max ∅ = −∞. Similarly, we refer to (π_i, π_j) as a weak visible pair if

$$\underset{i<\mathcal{l}<j}{\text{max}}{\pi }_{\mathcal{l}}\le \text{min}\left\{{\pi }_i,{\pi }_j\right\}.$$

We denote by ${\mathcal{V}}_{\pi }$ the set of all strong visible pairs of π, and let $V_{\pi }=\text{Card}\left({\mathcal{V}}_{\pi }\right)$ be the number of strong visible pairs in the sequence π. For example,

$${\mathcal{V}}_{12122}=\left\{(1,2),(2,3),(3,4),(4,5),(2,4)\right\},V_{12122}=5.$$

We use the above notation with addition of the superscript w to denote the corresponding weak visibility pairs statistics. For example,

$${\mathcal{V}}_{12122}^w=\left\{(1,2),(2,3),(3,4),(4,5),(2,4),(2,5)\right\},V_{12122}^w=6.$$

The graph ${\mathcal{G}}_{\pi }:=\left([n],{\mathcal{V}}_{\pi }\right)$ with the set of nodes [n] := {1, 2, … , n}is the horizontal visibility graph associated with π [17]. For i ∈ [n], we denote by d_π(i) the degree of the node i in the visibility graph ${\mathcal{G}}_{\pi }$. We set e_π(i, j) = 1 when $e_{\pi }(i,j)\in {\mathcal{V}}_{\pi }$ and e_π(i, j) = 0 otherwise. Thus,

$$d_{\pi }(i)=\sum _{j\in [n]\backslash \left\{i\right\}}{e}_{\pi }(i,j).$$ (1)

We now turn to the definition of a restricted growth sequence. A sequence of positive integers $\pi ={\pi }_1{\pi }_2\cdots {\pi }_n\in {\mathbb{N}}^n$ is called a restricted growth sequence if

$${\pi }_1=1\text{and}{\pi }_{j+1}\le 1+\text{max}\left\{{\pi }_1,\cdots ,{\pi }_j\right\}\text{for all }1\le j<n.$$

There is a bijective connection between these sequences and canonical set partitions. A partition of a set A is a collection of non-empty, mutually disjoint subsets, called blocks, whose union is the set A. A partition II with k blocks is called a k-partition and denoted by II = A₁|A₂|⋯|A_k. A k-partition A₁|A₂|⋯|A_k is said to be in the standard form if the blocks A_i are labeled in such a way that

$$\text{min}{A}_1<\text{min}{A}_2<\dots <\text{min}{A}_k.$$

The partition can be represented equivalently by the canonical sequential form π₁π₂ … π_n, where π_i ∈ [n] and $i\in A_{{\pi }_i}$ for all i [19]. In words, π_i is the label of the partition block that contains i. It is easy to verify that a word π ∈ [k]ⁿ is a canonical representation of a k-partition of [n] in the standard form if and only if it is a restricted growth sequence [19].

Example 1.1.

For instance, canonical partition {1, 4, 7} | {2, 3, 6, 9} | {5, 8} in the canonical sequential form is π = 122132132, where π₃ = 2 indicates that 3 belongs to the second block {2, 3, 6, 9}, etc. The (weak and strong) visibility graphs of π are given in Fig. 1 below.

Figure 1:

On he left is a picture of the strong visibility graph of the sequence 12132132231. On the right, is the weak visibility graph associated with same sequence.

We denote by ${\mathcal{R}}_n$ the set of all restricted growth sequences of length n. For a given $\pi \in {\mathcal{R}}_n$, we let $\mathcal{O}(\pi ):=\text{Card}\left\{{\pi }_i:i\in [n]\right\}$, the number of different letters in the word π. For example, $\mathcal{O}(1231)=3$. We denote by ${\mathcal{R}}_{n,k}$ the set of all restricted growth sequences π with $\mathcal{O}(\pi )=k$. Clearly, ${\mathcal{R}}_n:={\cup }_{k\in \left[n\right]}{\mathcal{R}}_{n,k}$.

It is well-known that $\text{Card}\left({\mathcal{R}}_{n,k}\right)=S_{n,k}$ and $\text{Card}\left({\mathcal{R}}_n\right)=B_n$ where S_n,k is a Stirling number of second kind and B_n is the n-th Bell number [19]. The Stirling numbers can be introduced algebraically in several different ways. For instance,

$$\frac{x^k}{{\prod }_{j=1}^k(1-jx)}=\sum _{n\ge 0}{S}_{n,k}{x}^n,\forall k\in \mathbb{N}.$$ (2)

Alternatively, one can define the sequence of Stirling numbers of the second kind as the solution to the recursion

$$S_{n,k}=k{S}_{n-1,k}+S_{n-1,k-1},n,k\in \mathbb{N},k\le n,$$ (3)

with S_0,0 = 1 and S_0,n = 0. The sequence of Bell numbers (B_n)_n≥0 can be then defined for instance, through the formula $B_n={\sum }_{k=0}^n{S}_{n,k}$, or, recursively via the formula $B_{n+1}={\sum }_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)B_k$ with B₀ = 1, or through Dobinski’s formula [3]

$$B_n=\frac{1}{e}\sum _{m=0}^{\infty }\frac{m^n}{m!},n\ge 0.$$ (4)

In what follows, we denote a random restricted growth sequence, sampled uniformly from ${\mathcal{R}}_{n,k}\left(\text{resp}.{\mathcal{R}}_n\right)$ by π⁽ⁿ⁾ (resp. π^(n,k)). That is,

$$P\left({\pi }^{(n,k)}=\pi \right)=\frac{1}{S_{n,k}}\text{for all}\pi \in {\mathcal{R}}_{n,k},$$

and

$$P\left({\pi }^{(n)}=\pi \right)=\frac{1}{B_n}\text{for all}\pi \in {\mathcal{R}}_n.$$

We denote by ${\mathcal{G}}_{n,k}:={\mathcal{G}}_{{\pi }^{(n,k)}}(\text{resp}.{\mathcal{G}}_n:={\mathcal{G}}_{{\pi }^{(n)}})$ the HVG of the random restrictive growth sequence π^(n,k) (resp. π⁽ⁿ⁾). Furthermore, we use the notations e_n(., .), d_n(.), and V_n to denote, respectively, $e_{{\pi }^{(n)}}(.,.)$, $d_{{\pi }^{(n)}}(.)$, and $V_{{\pi }^{(n)}}$. See Fig. 2 below for two instances of visibility graphs of uniformly sampled restrictive growth sequences of length n = 200.

Figure 2:

An instance of ${\mathcal{G}}_{200}$ (on the left) and the corresponding ${\mathcal{G}}_{200}^w$ (on the right).

For any $k\in \mathbb{N}$, we define the generating function

$$P_k(x,q):=\sum _{n=k}^{\infty }{x}^n{S}_{n,k}E\left(q^{V_{\pi }}\mid \pi \in {\mathcal{R}}_{n,k}\right)=\sum _{n=k}^{\infty }\sum _{\pi \in {\mathcal{R}}_{n,k}}{x}^n{q}^{V_{\pi }},x,q\in ℂ.$$ (5)

Knowing an explicit form of (5), would in principle give us the distribution of V_n in full details for all $n\in \mathbb{N}$. Unfortunately, so far we were unable to find an explicit form of P_k(x, q). In this paper, we calculate instead the following generating function:

$$\underset{\_}{Q}(x,y):=\sum _{k\ge 1}{y}^k\sum _{n\ge 0}\frac{x^n}{n!}{B}_{n}E\left(V_{\pi }\mid \pi \in {\mathcal{R}}_{n,k}\right).$$

Theorem 1.2.

We have:

$$\underset{\_}{Q}(x,y)=\frac{1}{y}{\int }_0^x{e}^{-y{e}^{x-t}-t}{\int }_0^t{e}^{y{e}^{x-r}+r}\left(e^{r-x}+y\right)\underset{\_}{T}\left(r,y{e}^{x-r}\right)\mathit{\text{drdt}},$$

where

$$\begin{gathered} \underset{\_}{T}(x,y)=y^3{\int }_0^x(x-t)e^{y{e}^t-y}{\int }_0^{t}Ei\left(1,y{e}^r\right)e^{y{e}^r+2r}\mathit{\text{drdt}} \\ +y{\int }_0^x(t-x)e^{y{e}^t-y}\left(Ei\left(1,y{e}^t\right)e^{y{e}^t}\left(y{e}^t-1\right)-y{e}^t\right)dt \\ +y(1-y){\int }_0^x(t-x)e^{y{e}^t}Ei\left(1,y{e}^t\right)dt, \end{gathered}$$

and $Ei(1,z)={\int }_1^{\infty }\frac{e^{-zt}}{t}dt$ is the exponential integral.

Example 1.3.

First several terms of the generating function Q(x, 1) are given by

$$x^2+\frac{5}{3}{x}^3+\frac{47}{24}{x}^4+\frac{113}{60}{x}^5+\frac{19}{12}{x}^6+\frac{1013}{840}{x}^7+\frac{11429}{13440}{x}^8+\frac{204361}{362880}{x}^9+\cdots$$

The proof of Theorem 1.2 is given in Section 2. The solution is derived from a PDE for T which is obtained in Lemma 2.6. Our next result, Theorem 1.4, gives a an alternative, closed form expression for E(V_n) through a different, probabilistic approach.

We partition I_n = {(i, j) : 1 ≤ i < j ≤ n} into three subsets

$$\begin{gathered} I_n^{(1)}:=\left\{(1,j):3\le j\le n\right\}, \\ {I}_n^{(2)}:=\left\{(i,i+1):1\le i\le n-1\right\}, \\ {I}_n^{(3)}:=\left\{(i,j):2\le i<j\le n,j>i+1\right\}. \end{gathered}$$

Clearly, e_π (i, j) = 0 on $I_n^{(1)}$ and e_π (i, j) = 1 on $I_n^{(2)}$ for all $\pi \in {\mathcal{R}}_n$. Therefore,

$$V_{\pi }=n-1+\sum _{(i,j)\in I_n^{(3)}}{e}_{\pi }(i,j).$$ (6)

The following theorem evaluates the probability that $(i,j)\in {\mathcal{V}}_n$ for a given $(i,j)\in I_n^{(3)}$ in terms of explicit multi-linear polynomials of S_k,i, B_i, and Bernoulli numbers. By virtue of (1), the result immediately yields the average degree E(d_n(i)) of any given node i ∈ V_n and the average number of edges E(V_n).

We will use the following Bernoulli formula for Faulhaber polynomials [3]:

$${\Psi }_n(t):=\sum _{k=0}^t{k}^{n-1}=\frac{1}{n}\sum _{\mathcal{l}=0}^{n-1}\left(\begin{array}{l}n\hfill \\ \mathcal{l}\hfill \end{array}\right)t^{n-\mathcal{l}}{\mathcal{B}}_{\mathcal{l}},n\in \mathbb{N},t\ge 0.$$ (7)

where ${\mathcal{B}}_{\mathcal{l}}$ are Bernoulli numbers. The latter can be calculated, for example, using the recursion

$$\sum _{\mathcal{l}=0}^{n-1}\left(\begin{array}{l}n\hfill \\ \mathcal{l}\hfill \end{array}\right){\mathcal{B}}_{\mathcal{l}}=0$$

with ${\mathcal{B}}_0=1$. See, for instance, [3] for alternative definitions of Bernoulli numbers.

We will also need the following well-known extension of Dobinski’s identity (4). For any integers n, t ≥ 0 we have:

$$\begin{gathered} {\Theta }_n(t):=\frac{1}{e}\sum _{m=t}^{\infty }\frac{m^n}{(m-t)!}=\frac{1}{e}\sum _{k=0}^{\infty }\frac{{(k+t)}^n}{k!}=\frac{1}{e}\sum _{\mathcal{l}=0}^n\left(\begin{array}{l}n\hfill \\ \mathcal{l}\hfill \end{array}\right)t^{n-\mathcal{l}}\sum _{k=0}^{\infty }\frac{k^{\mathcal{l}}}{k!} \\ =\sum _{\mathcal{l}=0}^n\left(\begin{array}{l}n\hfill \\ \mathcal{l}\hfill \end{array}\right)t^{n-\mathcal{l}}{B}_{\mathcal{l}}, \end{gathered}$$ (8)

were in the last step we applied the original formula (4).

Theorem 1.4.

For all n ≥ 3 and $(i,j)\in I_n^{(3)}$, we have

$$\begin{gathered} B_{n}P\left((i,j)\in {\mathcal{V}}_n\right)=\sum _{t=1}^{i-1}{S}_{i-1,t}{\Theta }_{n-j+1}(t){\Psi }_{j-i}(t-1) \\ +\sum _{t=1}^{i-1}{S}_{i-1,t}{\Theta }_{n-j}(t)\sum _{a=1}^t\left\{-a{(a-1)}^{j-i-1}+{\Psi }_{j-i}(a-1)\right\} \\ +\sum _{t=1}^{i-1}{S}_{i-1,t}{\Theta }_{n-j+1}(t+1)t^{j-i-1}+\sum _{t=1}^{i-1}{S}_{i-1,t}{\Theta }_{n-j}(t+1)\left\{-(t+1)t^{j-i-1}+{\Psi }_{j-i}(t)\right\}. \end{gathered}$$

The proof of Theorem 1.4 is deferred to Section 3. We next evaluate the probability that for a given pair of nodes i, j ∈ [n], we have $(i,j)\in {\mathcal{V}}_n^w$ but $(i,j)\notin {\mathcal{V}}_n$.

Theorem 1.5.

The following holds true for n ≥ 3:

1. If $(i,j)\in I_n^{(3)}$, then

   $$\begin{gathered} B_{n}P\left((i,j)\in {\mathcal{V}}_n^w\backslash {\mathcal{V}}_n\right)= \\ \sum _{t=1}^{i-1}{S}_{i-1,t}{\Theta }_{n-j}(t)\left\{-t^{j-i}+t^{j-i-1}+2{\Psi }_{j-i}(t-1)\right\} \\ +\sum _{t=1}^{i-1}{S}_{i-1,t}{\Theta }_{j-i+1}(t)t^{j-i-1}-\sum _{t=1}^{i-1}{S}_{i-1,t}{\Theta }_{j-i}(t+1)\left\{(t-1)t^{j-i-1}+t{(t+1)}^{j-i-1}\right\} \\ +\sum _{t=1}^{i-1}{S}_{i-1,t}{\Theta }_{j-i+1}(t+1)\left\{{(t+1)}^{j-i-1}-t^{j-i-1}\right\}. \end{gathered}$$
2. If $(i,j)\in I_n^{(1)}$, then

   $$P\left((i,j)\in {\mathcal{V}}_n^w\backslash {\mathcal{V}}_n\right)=\frac{B_{n-j+i+1}}{B_n}.$$

Since ${\mathcal{V}}_n\subset {\mathcal{V}}_n^w$, we have

$$E\left(V_n^w\right)=E\left(V_n\right)+\sum _{(i,j)\in I_n^{(1)}}P\left((i,j)\in {\mathcal{V}}_n^w\backslash {\mathcal{V}}_n\right)+\sum _{(i,j)\in I_n^{(3)}}P\left((i,j)\in {\mathcal{V}}_n^w\backslash {\mathcal{V}}_n\right),$$

which yields $E\left(V_n^w\right)$. The proof of Theorem 1.5 is included in Section 4.

2. Proof of Theorem 1.2

Throughout this section, for any given ordinary generating function $A(x)={\sum }_{n\ge 0}{a}_n{x}^n$, $x\in ℂ$, we use A to denote the corresponding exponential generating function. That is,

$$\underset{\_}{A}(x):=\sum _{n=0}^{\infty }{a}_n\frac{x^n}{n!}=\sum _{n=0}^{\infty }\frac{x^n}{n!}\left[x^n\right]A(x),$$

where [xⁿ]A(x) stands for the coefficient of xⁿ in the generating function A(x).

Note that each restricted growth sequence in ${\mathcal{R}}_{n,k}$ can be represented as a word in the form 1π⁽¹⁾2π⁽²⁾ ⋯ kπ^(k), where π^(j) is an arbitrary subword over the alphabet [j]. Therefore, we can rewrite (5) as

$$P_k(x,q)=x^k{L}_k(x,q)\prod _{j=1}^{k-1}{M}_j(x,q),$$ (9)

where L_k(x, q) and M_k(x, q) are given by

$$\begin{gathered} L_k(x,q)={\sum }_{n\ge 0}{\sum }_{\pi \in {[k]}^n}{x}^n{q}^{V(k\pi )}, \\ {M}_k(x,q)={\sum }_{n\ge 0}{\sum }_{\pi \in {[k]}^n}{x}^n{q}^{V(k\pi (k+1))}. \end{gathered}$$ (10)

This representation is instrumental in our proof of the following result:

Proposition 2.1.

For k ≥ 1,

$$P_k(x,q)=\frac{x^k}{1-x{\tilde{M}}_k(x,q)}\prod _{j=1}^{k-1}\frac{{\tilde{M}}_j(x,q)}{{\left(1-x{\tilde{M}}_j(x,q)\right)}^2},$$

where ${\tilde{M}}_k(x,q)$ is defined recursively by the equation

$${\tilde{M}}_k(x,q)={\tilde{M}}_{k-1}(x,q)+\frac{xq{\left({\tilde{M}}_{k-1}(x,q)\right)}^2}{1-x{\tilde{M}}_{k-1}(x,q)}$$

with the initial condition ${\tilde{M}}_1(x,q)=q$.

Proof of Proposition 2.1.

In view of (9) and (10), in order to prove the proposition it suffices to evaluate L_k(x, q) and M_k(x, q). These calculations are the content of the next two lemmas.

Lemma 2.2.

For all k ≥ 1,

$$L_k(x,q)={\prod }_{j=1}^k\frac{1}{1-x{\tilde{M}}_j(x,q)},$$

where ${\tilde{M}}_k(x,q)$ satisfies the recurrence relation

$${\tilde{M}}_k(x,q)={\tilde{M}}_{k-1}(x,q)+\frac{xq{\left({\tilde{M}}_{k-1}(x,q)\right)}^2}{1-x{\tilde{M}}_{k-1}(x,q)},$$

with ${\tilde{M}}_1(x,q)=q$.

Proof of Lemma 2.2.

Any word kπ ∈ [k]ⁿ can be written as

$$k\pi =k{\pi }^{(1)}k{\pi }^{(2)}\cdots k{\pi }^{(s)}$$

for some s ≥ 1 and subwords π^(j) ∈ [k − 1]. Thus, the contribution for a fixed s is ${\left(x{\tilde{M}}_k(x,q)\right)}^{s-1}{\tilde{L}}_k(x,q)$, where

$$\begin{gathered} {\tilde{L}}_k(x,q)=\sum _{n\ge 0}\sum _{\pi \in {[k-1]}^n}{x}^n{q}^{V(k\pi )}, \\ {\tilde{M}}_k(x,q)=\sum _{n\ge 0}\sum _{\pi \in {[k-1]}^n}{x}^n{q}^{V(k\pi k)}. \end{gathered}$$

Hence,

$$L_k(x,q)=\sum _{s\ge 1}{\left(x{\tilde{M}}_k(x,q)\right)}^{s-1}{\tilde{L}}_k(x,q)=\frac{{\tilde{L}}_k(x,q)}{1-x{\tilde{M}}_k(x,q)}.$$ (11)

Note that any word π ∈ [k − 1]ⁿ can be written as π⁽⁰⁾(k − 1)π⁽¹⁾ ⋯ (k − 1)π^(s) with s ≥ 0 and π^(j) is a word over alphabet [k − 2] for all j. Thus,

$${\tilde{L}}_k(x,q)=\sum _{s\ge 1}{\left(x{\tilde{M}}_{k-1}(x,q)\right)}^{s-1}{\tilde{L}}_{k-1}(x,q)=\frac{{\tilde{L}}_{k-1}(x,q)}{1-x{\tilde{M}}_{k-1}(x,q)},$$ (12)

where we used the fact that V (kπ′k) = V ((k − 1)π′ (k − 1)) for all π′ [k − 2]ⁿ. Hence, by (11) and (12), we see that ${\tilde{L}}_k(x,q)=L_{k-1}(x,q)$, which leads to

$$L_k(x,q)=\frac{L_{k-1}(x,q)}{1-x{\tilde{M}}_k(x,q)}.$$

By induction on k, and using the fact that $L_1(x,q)=\frac{1}{1-xq}$, we complete the proof for the formula L_k(x, q).

Now let us write an equation for ${\tilde{M}}_k(x,q)$. Clearly, ${\tilde{M}}_1(x,q)=q$, which counts the only empty word according the the visible pairs in 11. Note that for any word π ∈ [k − 1]ⁿ, the word kπk can be decomposed as kπ⁽⁰⁾(k − 1)π⁽¹⁾ ⋯ (k − 1)π^(s)k with π^(j) is a word over alphabet [k − 2] for all j. Thus,

$${\tilde{M}}_k(x,q)={\tilde{M}}_{k-1}(x,q)+\sum _{s\ge 1}{x}^{s}q{\left({\tilde{M}}_k(x,q)\right)}^{s+1}={\tilde{M}}_{k-1}(x,q)+\frac{xq{\left({\tilde{M}}_{k-1}(x,q)\right)}^2}{1-x{\tilde{M}}_{k-1}(x,q)}.$$

where we used that fact V (kπ′(k − 1)) = V ((k − 1)π′(k − 1)) for all π′ ∈ [k − 2]ⁿ. □

Lemma 2.3.

For all k ≥ 1,

$$M_k(x,q)=\frac{{\tilde{M}}_k(x,q)}{1-x{\tilde{M}}_k(x,q)}.$$

Proof of Lemma 2.3.

For any word kπ ∈ [k]ⁿ, the word kπ(k + 1) can be decomposed as either kπ′(k + 1) or kπ′kπ″(k + 1), where π′ is a word over alphabet [k − 1] and π″ is a word over alphabet [k]. Since V (kπ′(k + 1)) = V (kπ′k), we have

$$M_k(x,q)={\tilde{M}}_k(x,q)+x{\tilde{M}}_k(x,q)M_k(x,q),$$

which, by solving for M_k(x, q), complete the proof of Lemma 2.3. □

By Lemmas 2.2 and 2.3 and (9), we have

$$P_k(x,q)=x^k\prod _{j=1}^{k-1}\frac{1}{1-x{\tilde{M}}_j(x,q)}\prod _{j=1}^{k-1}\frac{{\tilde{M}}_j(x,q)}{1-x{\tilde{M}}_j(x,q)}.$$

The proof of Proposition 2.1 is complete. □

Example 2.4.

The first coe cients of the generating function $1+{\sum }_{k\ge 1}{P}_k(x,q)$ are given by 1 + x + 2qx² + 5q²x³ + (2q⁴ + 13q³)x⁴ + (18q⁵ + 34q⁴)x⁵ + (11q⁷ + 103q⁶ + 89q⁵)x⁶ + (6q⁹ + 160q⁸ + 478q⁷ + 233q⁶)x⁷ + (2q¹¹ + 206q¹⁰ + 1359q⁹ + 1963q⁸ + 610q⁷)x⁸ + (230q¹² + 3066q¹¹ + 8813q¹⁰ + 7441q⁹ + 1597q⁸)x⁹.

With Proposition 2.1 at hand, we turn now to the study of the expected number of vertexes in ${\mathcal{G}}_n$. More precisely, we obtain:

Proposition 2.5.

For all k ≥ 1,

$${\frac{\partial }{\partial q}{P}_k(x,q)|}_{q=1}=\frac{x^k}{\prod _{j=1}^k(1-jx)}{H}_k(x),$$

where

$$H_k(x)=\sum _{i=1}^{k-1}{f}_i(x)(1-ix)+2x\sum _{i=1}^{k-1}{f}_i(x)+x{f}_k(x),$$

with

$$f_i(x):=\frac{1+x\sum _{j=1}^{i-1}\frac{1-jx}{1-(j-1)x}}{(1-(i-1)x)(1-ix)}.$$

We use here the usual convention that an empty sum is zero.

Proof of Proposition 2.5.

By Proposition 2.1, the generating function ${\tilde{M}}_k(x,q)$ satisfies

$${\tilde{M}}_k(x,q)={\tilde{M}}_{k-1}(x,q)+\frac{xq{\left({\tilde{M}}_{k-1}(x,q)\right)}^2}{1-x{\tilde{M}}_{k-1}(x,q)}$$

with ${\tilde{M}}_1(x,q)=q$. Thus,

$${\tilde{M}}_k(x,1)=\frac{{\tilde{M}}_{k-1}(x,1)}{1-x{\tilde{M}}_{k-1}(x,1)}$$

with ${\tilde{M}}_1(x,1)=1$. Hence, by induction on k, we have ${\tilde{M}}_k(x,1)=\frac{1}{1-(k-1)x}$.

Moreover, by differentiation the recurrence relation at q = 1, we obtain

$${\frac{\partial }{\partial q}{\tilde{M}}_k(x,q)|}_{q=1}={\frac{\partial }{\partial q}{\tilde{M}}_{k-1}(x,q)|}_{q=1}+\frac{x{\left({\tilde{M}}_{k-1}(x,1)\right)}^2+{x{\tilde{M}}_{k-1}(x,1)\frac{\partial }{\partial q}{\tilde{M}}_{k-1}(x,q)|}_{q=1}\left(2-x{\tilde{M}}_{k-1}(x,1)\right)}{{\left(1-x{\tilde{M}}_{k-1}(x,1)\right)}^2},$$

which, by ${\tilde{M}}_k(x,1)=\frac{1}{1-(k-1)x}$, implies

$${\frac{\partial }{\partial q}{\tilde{M}}_k(x,q)|}_{q=1}=\frac{x}{{(1-kx)}^2}+{\frac{{(1-(k-1)x)}^2}{{(1-kx)}^2}\frac{\partial }{\partial q}{\tilde{M}}_{k-1}(x,q)|}_{q=1}.$$

We can now complete the proof of the proposition by using induction on k and the initial condition $\frac{\partial }{\partial q}{\tilde{M}}_1(x,q){|}_{q=1}=1$. □

By Proposition 2.5, we have:

$$\begin{gathered} \frac{\partial }{\partial q}{P}_k(x,q){|}_{q=1}-\frac{x}{1-kx}\frac{\partial }{\partial q}{P}_{k-1}\left(x,q\right){|}_{q=1} \\ =\frac{x^k}{{\prod }_{j=1}^k(1-jx)}\left(\frac{1+x{\sum }_{j=1}^{k-1}\frac{1-jx}{1-(j-1)x}}{1-(k-1)x}+\frac{x+x^2{\sum }_{j=1}^{k-1}\frac{1-jx}{1-(j-1)x}}{(1-(k-1)x)(1-kx)}\right) \end{gathered}$$

with $\frac{\partial }{\partial q}{P}_1(x,q){|}_{q=1}=\frac{x^2}{1-x}$. For all k ≥ 2, define

$$T_k(x)=\frac{x^k}{{\prod }_{j=1}^k(1-jx)}\cdot \frac{1+x\sum _{j=1}^{k-2}\frac{1-jx}{1-(j-1)x}}{1-(k-1)x}.$$

Then,

$${(1-kx)\frac{\partial }{\partial q}{P}_k(x,q)|}_{q=1}-{x\frac{\partial }{\partial q}{P}_{k-1}(x,q)|}_{q=1}=\frac{T_k(x)+T_{k+1}(x)}{1-(k-1)x}$$ (13)

with $\frac{\partial }{\partial q}{P}_1(x,q){|}_{q=1}=\frac{x^2}{{(1-x)}^2}$.

In order to solve (13), we first study the corresponding exponential generating functions Q_k(x) and T_k(x) of the ordinary generating functions $Q_k(x)={\frac{\partial }{\partial q}{P}_k(x,q)|}_{q=1}$ and T_k(x), respectively. In other words,

$${\underset{\_}{Q}}_k(x)={\sum _{n\ge 0}\frac{x^n}{n!}\left[x^n\right]\frac{\partial }{\partial q}{P}_k(x,q)|}_{q=1}$$

and

$${\underset{\_}{T}}_k(x)=\sum _{n\ge 0}\frac{x^n}{n!}\left[x^n\right]T_k(x),\underset{\_}{T}(x,y)=\sum _{k\ge 2}{\underset{\_}{T}}_k(x)y^k.$$

Lemma 2.6.

The generating function $\underset{\_}{T}(x,y)={\sum }_{k\ge 2}{\underset{\_}{T}}_k(x)y^k$ is given by

$$\begin{gathered} \underset{\_}{T}(x,y)=y^3{\int }_0^x(x-t)e^{y{e}^t-y}{\int }_0^{t}Ei\left(1,y{e}^r\right)e^{y{e}^r+2r}\mathit{\text{drdt}} \\ +y{\int }_0^x(t-x)e^{y{e}^t-y}\left(Ei\left(1,y{e}^t\right)e^{y{e}^t}\left(y{e}^t-1\right)-y{e}^t\right)dt \\ +y(1-y){\int }_0^x(t-x)e^{y{e}^t}Ei\left(1,y{e}^t\right)dt, \end{gathered}$$

where $Ei(1,z)={\int }_1^{\infty }\frac{e^{-zt}}{t}dt$.

Proof of Lemma 2.6.

By the definition of T_k(x), we have:

$$(1-(k-3)x)(1-(k-1)x)T_k(x)-x(1-(k-3)x{T}_{k-1}(x)=\frac{x^{k+1}}{{\prod }_{j=1}^{k-3}(1-jx)}$$

with $L_2(x)=\frac{x^2}{1-x}$. Rewriting this equation in terms of exponential generating functions, we obtain:

$$\begin{gathered} \frac{d^4}{d{x}^4}{\underset{\_}{T}}_k(x)-(2k-4)\frac{d^3}{d{x}^3}{\underset{\_}{T}}_k(x)+(k-3)(k-1)\frac{d^2}{d{x}^2}{\underset{\_}{T}}_k(x) \\ -\frac{d^3}{d{x}^3}{\underset{\_}{T}}_{k-1}(x)+(k-3)\frac{d^2}{d{x}^2}{\underset{\_}{T}}_{k-1}(x)=\frac{{\left(e^x-1\right)}^{k-3}}{(k-3)!}, \end{gathered}$$

where we used (2) and the fact that ${\sum }_{n\ge k}{S}_{n,k}\frac{x^n}{n!}=\frac{{\left(e^x-1\right)}^k}{k!}$.

Multiplying both sides of the last recurrence by y^k and summing over k ≥3, we obtain:

$$\begin{gathered} \frac{{\partial }^4}{\partial x^4}\left(\underset{\_}{T}(x,y)-{\underset{\_}{T}}_2(x)y^2\right)-2y\frac{{\partial }^4}{\partial x^3\partial y}\left(\underset{\_}{T}(x,y)-{\underset{\_}{T}}_2(x)y^2\right)+4\frac{{\partial }^3}{\partial x^3}\left(\underset{\_}{T}(x,y)-{\underset{\_}{T}}_2(x)y^2\right) \\ +y\frac{\partial }{\partial y}\left(y\frac{{\partial }^3}{\partial x^2\partial y}\left(\underset{\_}{T}(x,y)-{\underset{\_}{T}}_2(x)y^2\right)\right)-4y\frac{{\partial }^3}{\partial x^2\partial y}\left(\underset{\_}{T}(x,y)-{\underset{\_}{T}}_2(x)y^2\right) \\ +3\frac{{\partial }^2}{\partial x^2}\left(\underset{\_}{T}(x,y)-{\underset{\_}{T}}_2(x)y^2\right)-y\frac{{\partial }^3}{\partial x^3}\underset{\_}{T}(x,y)+y\frac{{\partial }^3}{\partial x^2\partial y}(y\underset{\_}{T}(x,y)) \\ -3y\frac{{\partial }^2}{\partial x^2}\underset{\_}{T}(x,y)=y^3{e}^{y\left(e^x-1\right)}, \end{gathered}$$

where T₂(x) = e^x − 1 − x. Note that

$$\underset{\_}{T}(0,y)=\frac{\partial }{\partial x}\underset{\_}{T}(x,y){|}_{x=0}=0,\frac{{\partial }^2}{\partial x^2}\underset{\_}{T}(x,y){|}_{x=0}=y^2,\frac{{\partial }^3}{\partial x^3}\underset{\_}{T}(x,y){|}_{x=0}=y^2+y^3.$$

Solving the partial differential equation with these initial conditions, we obtain the result in Lemma 2.6. □

Finally,

$$\begin{gathered} \frac{d^2}{d{x}^2}{\underset{\_}{Q}}_k(x)-(2k-1)\frac{d}{dx}{\underset{\_}{Q}}_k(x)+k(k-1){\underset{\_}{Q}}_k(x) \\ -\frac{d}{dx}{\underset{\_}{Q}}_{k-1}(x)+(k-1){\underset{\_}{Q}}_{k-1}(x)=\frac{d^2}{d{x}^2}\left({\underset{\_}{T}}_k(x)+{\underset{\_}{T}}_{k+1}(x)\right) \end{gathered}$$

with Q₁(x) = 1 + (x − 1)e^x.

Recall $\underset{\_}{Q}(x,y)={\sum }_{k\ge 1}{\underset{\_}{Q}}_k(x)y^k$. Multiplying both sides of this recurrence equation by y^k and summing over k ≥ 2, we obtain

$$\begin{gathered} \frac{{\partial }^2}{\partial x^2}\left(\underset{\_}{Q}(x,y)-{\underset{\_}{Q}}_1(x)y\right)-2y\frac{{\partial }^2}{\partial x\partial y}\left(\underset{\_}{Q}(x,y)-{\underset{\_}{Q}}_1(x)y\right) \\ +\frac{\partial }{\partial x}\left(\underset{\_}{Q}(x,y)-{\underset{\_}{Q}}_1(x)y\right)+y\frac{\partial }{\partial y}\left(y\frac{\partial }{\partial y}\left(\underset{\_}{Q}(x,y)-{\underset{\_}{Q}}_1(x)y\right)\right) \\ -y\frac{\partial }{\partial y}\left(\underset{\_}{Q}(x,y)-{\underset{\_}{Q}}_1(x)y\right)-y\frac{\partial }{\partial x}\underset{\_}{Q}(x,y)+y\frac{\partial }{\partial y}(y\underset{\_}{Q}(x,y))-y\underset{\_}{Q}(x,y) \\ =\frac{{\partial }^2}{\partial x^2}\left(\underset{\_}{T}(x,y)+1/y\left(\underset{\_}{T}(x,y)-{\underset{\_}{T}}_2(x)y^2\right)\right) \end{gathered}$$

with Q(0, y) = 0 and ${\frac{\partial }{\partial x}\underset{\_}{Q}(x,y)|}_{x=0}=0$ This along with Lemma 2.6 and an aid of Maple, yields the explicit formula for the generating function Q(x, y) stated in Theorem 1.2. □

3. Proof of Theorem 1.4

The proof relies on the use of a generator of a uniformly random set partition of [n] proposed by Stam [28]. We next describe Stam’s algorithm for a given n.

1. For $m\in \mathbb{N}$, let ${\mu }_n(m)=\frac{m^n}{em!B_n}$. Dobinski’s formula (4) shows that μ_n(·) is a probability distribution on $\mathbb{N}$.

   At time zero, choose a random $M\in \mathbb{N}$ distributed according to μ_n, and arrange M empty and unlabeled boxes.

2. Arranges n balls labeled by integers from the set [n].

   At time i ∈ [n], place the ball ‘i’ into of one the M boxes, chosen uniformly at random. Repeat until there are no balls remaining.

3. Label the boxes in the order that they get occupied by the balls. Once a box is labeled, the label does not change anymore.

4. Form a set partition π of [n] with i in the k-th block if and only if ball ‘i” is in the k-th box.

Let N_i be the random number of nonempty boxes right after placing the i-th ball and X_i be the label of the box where the i-th ball was placed. Notice that if the i-th ball is dropped in an empty box, then X_i = N_i−1 + 1 and N_i = N_i−1 + 1. Otherwise, if the box was occupied previously, X_i = X_j where j < i is the first ball that was dropped in that box and N_i = N_i−1. Then, X := X₁ ⋯ X_n is the random set partition of [n] produced by the algorithm.

We denote by P_m(·) conditional probability distribution P (· | M = m). Clearly N₁ = 1, N_i ≤ i, and

$$P_m\left(N_{i+1}=t+1\mid N_i=t\right)=\frac{m-t}{m}\text{and}{P}_m\left(N_{i+1}=t\mid N_i=t\right)=\frac{t}{m}.$$

Let α_i,t(m) := P_m(N_i = t). Then, taking in account that

$$P_m\left(N_i=t\right)=P_m\left(N_i=t,N_{i-1}=t-1\right)+P_m\left(N_i=t,N_{i-1}=t\right),$$

we obtain:

$${\alpha }_{i,t}(m)=\{\begin{array}{ll}\frac{t}{m}{\alpha }_{i-1,t}(m)+\frac{m-t+1}{m}{\alpha }_{i-1,t-1}(m)\hfill & \text{ if }2\le t\le m\text{ and }t\le i\hfill \\ 0\hfill & \text{ if }t>i\text{ or }t>m\hfill \\ \frac{1}{m^{i-1}}\hfill & \text{ if }t=1\text{ and }1\le i.\hfill \end{array}$$

A comparison with (3) reveals that for t ≤ m,

$$P_m\left(N_i=t\right)=\frac{S_{i,t}}{m^i}\frac{m!}{(m-t)!}.$$ (14)

In addition,

$$P_m\left(X_{i+1}=\mathcal{l}\mid N_i=t\right)=\{\begin{array}{ll}\frac{1}{m}\hfill & \text{ if }\mathcal{l}\le t\hfill \\ \frac{m-t}{m}\hfill & \mathcal{l}=t+1\hfill \\ 0\hfill & \text{ otherwise}.\hfill \end{array}$$

Notice that some of the boxes may remain empty at the end of the algorithm’s run.

In view of (6), in order to calculate E(V_n), we need to evaluate

$$E\left(e_n(i,j)\right)=E\left[E_M\left(e_n(i,j)\right)\right]=E\left(P_M\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{X_i,X_j\right\}\right)\right)$$

for $(i,j)\in I_n^{(3)}$. For any constant $m\in \mathbb{N}$ we have:

$$\begin{gathered} P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{X_i,X_j\right\}\right) \\ =\sum _{t=1}^{(i-1)\wedge m}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{X_i,X_j\right\}|N_{i-1}=t\right)P_m\left(N_{i-1}=t\right) \\ =\sum _{t=1}^{(i-1)\wedge m}\sum _{k=1}^{m\wedge (t+1)}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{k,X_j\right\}|N_{i-1}=t,X_i=k\right) \\ \times P_m\left(X_i=k_i|N_{i-1}=t\right)P_m\left(N_{i-1}=t\right) \\ =\frac{m!}{m^i}\sum _{t=1}^{(i-1)\wedge m}\frac{S(i-1,t)}{(m-t)!}\sum _{k=1}^t{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{k,X_j\right\}|N_{i-1}=t\right) \\ +\frac{m!}{m^i}\sum _{t=1}^{(i-1)\wedge (m-1)}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{t+1,X_j\right\}|N_i=t+1\right)\frac{S(i-1,t)}{(m-t-1)!}. \end{gathered}$$ (15)

Furthermore, for any a ≤ t ≤ m we have:

$$\begin{gathered} P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{a,X_j\right\}|N_i=t\right) \\ =\sum _{b=1}^{a-1}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{a,X_j\right\},X_{i+1}=b|N_i=t\right) \\ =\sum _{b=1}^{a-1}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{a,X_j\right\}|N_{i+1}=t\right)P_m\left(X_{i+1}=b|N_i=t\right) \\ =\frac{1}{m}\sum _{b=1}^{a-1}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{a,X_j\right\}|N_{i+1}=t\right). \end{gathered}$$

Iterating, we obtain:

$$\begin{gathered} P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{a,X_j\right\}|N_i=t\right) \\ =\frac{1}{m^{j-i-1}}\sum _{b_{i+1}=1}^{a-1}\cdots \sum _{b_{j-1}=1}^{a-1}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{b}_{\mathcal{l}}<X_j|N_{j-1}=t\right). \end{gathered}$$ (16)

Denote p := max_i<ℓ<j b_ℓ and q := |{ℓ ∈ (i, j) : b_ℓ = p}|. In this terms, the last summation can be written as

$$\begin{gathered} \sum _{b_{i+1}=1}^{a-1}\cdots \sum _{b_{j-1}=1}^{a-1}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{b}_{\mathcal{l}}<X_j|N_{j-1}=t\right) \\ =\sum _{p=1}^{a-1}\sum _{q=1}^{j-i-1}\left(\begin{array}{c}j-i-1\\ q\end{array}\right){\left(p-1\right)}^{j-i-1-q}{P}_m\left(X_j>p|N_{j-1}=t\right) \\ =\sum _{p=1}^{a-1}\left(p^{j-i-1}-{(p-1)}^{j-i-1}\right)\left(1-\frac{p}{m}\right)=\left(1-\frac{a}{m}\right){\left(a-1\right)}^{j-i-1}+\frac{1}{m}{\Psi }_{i-j}(a-1), \end{gathered}$$

where Ψ_i−j is introduced in (7). Thus,

$$P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}<\text{min}\left\{a,X_j\right\}|N_i=t\right)=\frac{1}{m^{j-i}}\left((m-a){(a-1)}^{j-i-1}+{\Psi }_{i-j}(a-1)\right).$$ (17)

Inserting (14) and (17) into (15) and taking expectation with respect to μ_n(·), we obtain:

$$\begin{gathered} e{B}_{n}P\left(e_n(i,j)=1\right)=\sum _{t=1}^{i-1}{S}_{i-1,t}{\Psi }_{i-j}(t-1)\sum _{m=t}^{\infty }\frac{m^{n-j+1}}{(m-t)!} \\ +\sum _{t=1}^{i-1}{S}_{i-1,t}\sum _{a=1}^t\left(-a{(a-1)}^{j-i-1}+{\Psi }_{i-j}(a-1)\right)\sum _{m=t}^{\infty }\frac{m^{n-j}}{(m-t)!} \\ +\sum _{t=1}^{i-1}{S}_{i-1,t}{t}^{j-i-1}\sum _{m=t+1}^{\infty }\frac{m^{n-j+1}}{(m-t-1)!} \\ +\sum _{t=1}^{i-1}{S}_{i-1,t}\left(-(t+1)t^{j-i-1}+{\Psi }_{i-j}(t)\right)\sum _{m=t+1}^{\infty }\frac{m^{n-j}}{(m-t-1)!}, \end{gathered}$$

as desired. □

4. Proof of Theorem 1.5

Write:

$$P\left((i,j)\in {\mathcal{V}}_n^w\backslash {\mathcal{V}}_n\right)=E\left(P_M\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=\text{min}\left\{X_i,X_j\right\}\right)\right).$$

Case I) If $(i,j)\in I_n^{(3)}$, then similarly to the calculation in (15), for any $m\in \mathbb{N}$ we have:

$$\begin{gathered} P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=\text{min}\left\{X_i,X_j\right\}\right) \\ =\frac{m!}{m^i}\sum _{t=1}^{(i-1)\wedge m}\frac{S_{i-1,t}}{(m-t)!}\sum _{a=1}^t{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=\text{min}\left\{a,X_j\right\}|N_i=t\right) \\ +\frac{m!}{m^i}\sum _{t=1}^{(i-1)\wedge (m-1)}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=\text{min}\left\{t+1,X_j\right\}|N_i=t+1\right)\frac{S_{i-1,t}}{(m-t-1)!}. \end{gathered}$$ (18)

Similarly to (16), for a ≤ t ≤ m we have:

$$\begin{gathered} P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=\text{min}\left\{a,X_j\right\}|N_i=t\right)= \\ {P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=a,X_j\ge a|N_i=t\right)+P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=X_j,X_j<a|N_i=t\right). \end{gathered}$$ (19)

The first term on the right hand-side of (19) can be written as

$$\begin{gathered} P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=a|N_i=t\right)P_n\left(a\le X_j|N_{j-1}=t\right) \\ =\left\{P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}\le a|N_i=t\right)-P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}\le a-1|N_i=t\right)\right\}\frac{m-a+1}{m} \\ =\frac{1}{m^{j-i}}(m-a+1)\left(a^{j-i-1}-{(a-1)}^{j-i-1}\right). \end{gathered}$$ (20)

Similarly, the second term in right hand side of (19) contributes:

$$\begin{gathered} \sum _{b=1}^{a-1}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=b,X_j=b|N_i=t\right) \\ =\sum _{b=1}^{a-1}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=b|N_i=t\right)P_m\left(X_j=b|N_{j-1}=t\right) \\ =\frac{1}{m}\sum _{b=1}^{a-1}{P}_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=b|N_i=t\right) \\ =\frac{1}{m}\sum _{b=1}^{a-1}\left\{{\left(\frac{b}{m}\right)}^{j-i-1}-{\left(\frac{b-1}{m}\right)}^{j-i-1}\right\}=\frac{{(a-1)}^{j-i-1}}{m^{j-i}}. \end{gathered}$$ (21)

Inserting (20) and (21) back into (19), we obtain:

$$\begin{gathered} P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=\text{min}\left\{a,X_j\right\}|N_i=t\right) \\ =\frac{1}{m^{j-i}}\left((m-a+1)a^{j-i-1}-(m-a){(a-1)}^{j-i-1}\right). \end{gathered}$$

Plugging the result into (18) and taking expectation with respect to μ_n(·) gives:

$$\begin{gathered} e{B}_{n}P\left((i,j)\in {\mathcal{V}}_n^w\backslash {\mathcal{V}}_n\right) \\ =\sum _{m=1}^{\infty }\sum _{t=1}^{(i-1)\wedge m}\frac{m^{n-j}}{(m-t)!}{S}_{i-1,t}\left(m{t}^{j-i-1}-t^{j-i}+t^{j-i-1}+2\sum _{a=1}^{t-1}{a}^{j-i-1}\right)+\sum _{m=1}^{\infty }\sum _{t=1}^{(i-1)\wedge (m-1)}\frac{m^{n-j}}{(m-t-1)!}{S}_{i-1,t}\left((m-t)\left({(t+1)}^{j-i-1}-t^{j-i-1}\right)+t^{j-i-1}\right). \end{gathered}$$

The result in case (i) follows from this formula by changing the order of summation and applying (8).

Case (ii) If $(i,j)\in I_n^{(1)}$, then

$$P_m\left(\underset{i<\mathcal{l}<j}{\text{max}}{X}_{\mathcal{l}}=\text{min}\left\{1,X_j\right\}\right)=\frac{1}{m^{j-i-1}}.$$

Hence, an application of Dobinski’s identity (4) yields

$$P\left((i,j)\in {\mathcal{V}}_n^w\backslash {\mathcal{V}}_n\right)=\frac{1}{e{B}_n}\sum _{m=1}^{\infty }\frac{m^{n-j+i+1}}{m!}=\frac{B_{n-j+i+1}}{B_n},$$

as desired. □

Figure 3:

Empirical distributions of V₂₀₀ (left) and $V_{200}^w$ (right) based on 1500 samples.