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Connectivity of a Family of Bilateral Agreement Random Graphs

Hossein Dabirian [email protected] EECS Department, University of Michigan Vijay Subramanian [email protected] EECS Department, University of Michigan
Abstract

Bilateral agreement based random undirected graphs were introduced and analyzed by La and Kabkab [9]. The construction of the graph with nn vertices in this model uses a (random) preference order on other n1n-1 vertices and each vertex only prefers the top kk other vertices using its own preference order; in general, kk can be a function of nn. An edge is constructed in the ensuing graph if and only if both vertices of a potential edge prefer each other. This random graph is a generalization of the random kthk^{th}-nearest neighbor graphs of Cooper and Frieze [3] that only consider unilateral preferences of the vertices. The authors in [10] studied the emergence of a giant component and its size in this new random graph family in the limit of nn going to infinity when kk is finite. Connectivity properties of this random graph family have not yet been completely analyzed: for example, in their original paper [9], La and Kabkab conjectured that for k(t)=tlognk(t)=t\log n, with high probability connectivity happens when t>1t>1 and the graph is disconnected for t<1t<1. We prove this conjecture. Further, we also prove an asymptote for the average degree of this graph for large nn and kk.

1 Introduction

Graphs are powerful mathematical tools used for modeling objects in a wide variety of topics [5, 8, 11]: from the Internet and chemistry to social science and economics. These mathematical objects have been studied using algebraic, combinatorial, and stochastic methods. Studying graphs using a stochastic framework [2, 4, 12] has a long history starting from when Gilbert introduced his random graph model in [7] in 19591959, and Erdős and Rényi introduced the well-known Erdős-Rényi model in their celebrated paper [6] in 19601960. These fundamental random graph models have been extensively analyzed over the last 5050 years.

An Erdős-Rényi random graph with nn vertices is characterized by a parameter 0p(n)10\leq p(n)\leq 1 indicating the probability of existence of each potential undirected edge between two vertices. Edges then appear independently based on Bernoulli coin tosses with no intrinsic preference between the edges on the part of the vertices. One question that immediately arises by specifying the parameter p(n)p(n) is whether the Erdős-Rényi graphs are connected when nn\rightarrow\infty. In particular, Erdős and Rényi [6, 2, 4, 12] prove that if p(n)=tlog(n)/np(n)=t\log(n)/n, with high probability, the graph is connected for t>1t>1 and disconnected for t<1t<1; note that the mean degree in this setting is tlog(n)t\log(n). In fact, the refined result is that if (for large enough nn) p(n)=(log(n)+c)/np(n)=(\log(n)+c)/n, then the graph is connected with probability eece^{-e^{-c}}.

Cooper and Frieze in [3] described a new family of random graphs based on preferences constructed using a distance measure. They considered the complete graph KnK_{n} and assigned independent uniform [0,1][0,1] random variables as distances to all edges. They then keep the k(n)k(n) shortest edges incident to each vertex. In this model, the concept of distance induces a preference order on all edges. Indeed, the shorter an edge is the higher its preference is. Since the distances are i.i.d., the preference order on all the edges will be chosen uniformly over the set of all permutations. We say a vertex proposes an edge if the edge belongs to the set of the k(n)k(n) most-preferred edges incident to it. Note that in this model, an edge is kept if at least one of its endpoints proposes it, which is a unilateral perspective.

Preference relations of the sort used by Cooper and Frieze create a complicated structure on any resulting graph. For example, keeping edges based on their ranking even in the unilateral mode as done by Cooper and Frieze, induces an edge dependency which is not the case for the Erdős-Rényi model. Cooper and Frieze [3] studied the connectivity of their model and proved that the graph is connected with high probability for k3k\geq 3. They also provided upper and lower bounds for the probability of connectivity as nn\rightarrow\infty when k=2k=2.

Just like the Cooper-Frieze unilaterally proposed random graphs, bilateral agreement random graphs introduced by La and Kabkab [9], are constructed based on the same parameter k(n)k(n), which is the maximum number of other vertices each vertex wants as its neighbors. Considering vertices as agents, once again, all agents have their own preferences on the potential edges with others via a priority or preference order over other agents. Then, in contrast to both the Erdos-Renyi and the Cooper-Frieze models, an edge is drawn if and only if each end vertex has the other vertex in its k(n)k(n) preferred vertices, i.e., if and only if both end vertices propose an edge, which then leads to bilateral agreement.

One can interpret the bilateral agreement model as a network formation process conducted via a game (selfish objective maximization) among agents. These kinds of network formation processes [5, 8, 11] have been studied in social science and economics. Assume there are nn agents that are aware of their own benefits from the potential pairwise contracts with others where the benefit is assessed via an appropriate reward, cost or even distance measure. Then each agent agrees on its own k(n)k(n) most profitable contracts. Finally, each contract will be concluded if and only if both parties mutually agree to it, and the graph of pairwise connections will be generated. The interpretation above highlights the importance of bilateral agreement random graphs from a game-theoretic (i.e., individual agent oriented) point of view.

A core question studied in the La and Kabkab paper [9] is the determination of the choice of k(n)k(n) that results in the graphs produced being connected. The authors showed the following results hold with high probability: 1) if k(n)>Clog(n)k(n)>C\log(n) for C=2.4625C=2.4625, then the graph is connected; and 2) k(n)<clog(n)k(n)<c\log(n) for c=0.5c=0.5, then the graph has isolated vertices and so is not connected. Furthermore, using extensive simulation-based experimentation, they also conjectured that the connectivity threshold was exactly k(n)=log(n)k(n)=\log(n), surprisingly, which is the same threshold for connectivity of the Erdős-Rényi random graphs family in terms of the mean degree; note, however, that k(n)k(n) is the maximum degree in the La-Kabkab random graph family. In this paper, we show the following:

  1. 1.

    In Lemma 4.6 we show that the expected number of isolated vertices increases without bound if klog(n)3log(n)loglog(n)k\leq\log(n)-3\sqrt{\log(n)\log\log(n)}; see Section 4.

  2. 2.

    Using Lemma 4.6 and additional results, in Theorem 4.1 we prove that graphs from the La-Kabkab random graph family are not connected with high probability if klog(n)+t(loglog(n))log(n)k\leq\log(n)+t^{\prime}(\log\log(n))\sqrt{\log(n)} for t<1/2t^{\prime}<-1/2; see Section 4.

  3. 3.

    Theorem 5.3 then establishes that graphs from the La-Kabkab random graph family are connected with high probability if klog(n)+t(loglog(n))log(n)k\geq\log(n)+t^{\prime}(\log\log(n))\sqrt{\log(n)} for t>1/2t^{\prime}>1/2; see Section 5.

  4. 4.

    Finally, in Theorem 6.2 assuming that kk grows to infinity with k=o(n1/3)k=o(n^{1/3}), we show that the mean degree for graphs from the La-Kabkab random graph family asymptotically behaves like kkπ+18πk+o(1k)k-\sqrt{\tfrac{k}{\pi}}+\tfrac{1}{8\sqrt{\pi k}}+o(\tfrac{1}{\sqrt{k}}), i.e., the mean degree is asymptotically close to the maximum possible degree; see Section 6.

Whereas Theorem 4.1 and Theorem 5.3 establish the conjecture by La and Kabkab, they still do not precisely identify the connectivity threshold in comparison to existing results on the Erdős-Rényi random graph family [4, 12]. We believe this arises due to our use of analytically simpler sufficient conditions to prove both results, and hence, a different analysis could overcome this lacuna.

A different but related line of work from the connectivity question is that of Moharrami et al. [10], where they introduced a new branching process called the Erlang Weighted Tree (EWT) as the local weak limit of the bilateral agreement random graphs of La and Kabkab in the sparse regime. In [10] the parameter k(n)k(n) is a (finite) random parameter kk for each vertex that is independently chosen and identically distributed with a distribution on \mathbb{N} with finite mean. The authors then studied the degree distribution of the root vertex and its average. We will return to this specific result of [10] in Section 6 when we discuss the asymptotics of the mean degree. The authors of [10] also discuss the probability of extinction of an EWT, and conjecture its relevance to the size of the giant component [6, 4, 12] of the bilateral agreement random graphs of La and Kabkab, when there is one.

2 Mathematical Model

Consider the complete undirected graph KnK_{n} with vertices [n]={1,2,,n}[n]=\{1,2,\cdots,n\} and edges {(i,j):1i,jn,ij}\{(i,j):1\leq i,j\leq n,i\neq j\} where we assume (i,j)=(j,i)(i,j)=(j,i). Let k(n)k(n) be an integer such that 1k(n)n1\leq k(n)\leq n; henceforth, to avoid cumbersome notation, we will use kk instead of k(n)k(n). We assign independently and identically distributed random variables called priority scores to all edges of KnK_{n}. For any explicit calculations, one can assume they are uniformly distributed in [0,1][0,1] or exponentially distributed with parameter 11; this holds because we will only be interested in the order statistics. We denote the score of edge (i,j)(i,j) by V(i,j)=V(j,i)V(i,j)=V(j,i). The set of all scores of the edges of vertex i[n]i\in[n] is denoted by 𝒱i={V(i,1),,V(i,i1),V(i,i+1),,V(i,n)}\mathcal{V}_{i}=\{V(i,1),\cdots,V(i,i-1),V(i,i+1),\cdots,V(i,n)\}, and all the associated edges by i={(i,1),,(i,i1),(i,i+1),,(i,n)}\mathcal{E}_{i}=\{(i,1),\cdots,(i,i-1),(i,i+1),\cdots,(i,n)\}. Without loss of generality, we can also assume that the scores are distinct, then (Rij)1jn1(R_{i}^{j})_{1\leq j\leq n-1} represents an order on [n]{i}={1,2,,i1,i+1,,n}[n]\setminus\{i\}=\{1,2,\cdots,i-1,i+1,\cdots,n\} based on V(i,j)V(i,j) values. In other words, for each 1in1\leq i\leq n, the random vector i=(Ri1,Ri2,,Rin1)\mathcal{R}_{i}=(R_{i}^{1},R_{i}^{2},\cdots,R_{i}^{n-1}) is a permutation of [n]{i}[n]\setminus\{i\} in which

V(i,Ri1)>V(i,Ri2)>>V(i,Rin1).V(i,R_{i}^{1})>V(i,R_{i}^{2})>\cdots>V(i,R_{i}^{n-1}).

As the scores are chosen i.i.d., the distribution of the random vector i\mathcal{R}_{i} is uniform among all permutations of [n]{i}[n]\setminus\{i\} as it only depends on the order-statistics. In general, the scores also impose a permutation over all the edges. This plays an important role in defining the bilateral agreement random graphs.

Let V(i,j)V(i,j) be realized for all edges (i,j)(i,j) and parameter kk be fixed. Then we can determine two different classes of random graphs on vertices [n][n]. We first define the notion of agreement. If V(i,j)V(i,j) is among the kk largest scores in 𝒱i\mathcal{V}_{i}, i.e., jik:={Ri1,Ri2,,Rik}j\in\mathcal{R}_{i}^{\leq k}:=\{R_{i}^{1},R_{i}^{2},\cdots,R_{i}^{k}\} or (i,j)ik:={(i,Ri1),,(i,Rik)}(i,j)\in\mathcal{E}_{i}^{\leq k}:=\{(i,R_{i}^{1}),\cdots,(i,R_{i}^{k})\}, we say that vertex ii proposes edge (i,j)(i,j). The first model introduced by Cooper and Frieze [3] let the edge (i,j)(i,j) be present if at least one of ii or jj proposes (i,j)(i,j). We denote this class of random graphs by 𝔽(n,k)\mathbb{F}(n,k), which we call the class of unilaterally proposed graphs. The second model described by La and Kabkab [9] requires the proposal by both vertices for an edge to appear, which we deem as a bilateral agreement. We denote this class by 𝔾(n,k)\mathbb{G}(n,k) and we call it the class of bilateral agreement random graphs.

Both 𝔽(n,k),𝔾(n,k)\mathbb{F}(n,k),\mathbb{G}(n,k) only depend on the order of V(i,j)V(i,j) not the values. This is the consequence of the scores being i.i.d., which results in a uniformly drawn permutation among all permutations of edges of KnK_{n}. Assigning a random variable to each edge only helps us to understand the extent of independence in this problem. However, in the analysis of any underlying probability, we will always refer to the permutation viewpoint of the construction of these random graphs.

In the following sections, we will analyze the connectivity of 𝔾(n,k)\mathbb{G}(n,k) when kk is around log(n)\log(n) as nn\rightarrow\infty. More precisely, we will prove that when k=tlog(n)k=t\log(n) with high probability this graph will be connected for t>1t>1 and disconnected for t<1t<1, and also attempt to determine the connectivity threshold with greater precision.

3 Asymptotically Equivalent Distribution

In our analysis we will use the abstraction of an infinite urn model. The next two lemmas allow us to link the probability of an event in finite permutations space to an event in the infinite urn model. In essence, we will show that the appropriate probabilities converge to a negative multinomial distribution. Before delving into details, we will consider the set of all permutations with order restrictions over some elements. The notation xyx\succ y states that xx appears earlier than yy in the permutation. We need to work on the set of all permutations with some order restrictions. For instance, whenever we say the set of permutations of {1,2,3}\{1,2,3\} with order restrictions {12,13}\{1\succ 2,1\succ 3\}, this narrows down the the set of all 3!=63!=6 permutations to {(1,2,3),(1,3,2)}\{(1,2,3),(1,3,2)\}.

Lemma 3.1.

Let nn\in\mathbb{Z} grow to infinity. Assume s=o(n1/4)s=o(n^{1/4}) is a positive integer depending on nn. Moreover, m,m0,m1,,msm,m_{0},m_{1},\cdots,m_{s} are positive integer variables depending on nn and M=m0+m1++msM=m_{0}+m_{1}+\cdots+m_{s}. Suppose there are mim_{i} distinct objects of type ii for 0is0\leq i\leq s. Furthermore, m=o(n1/4)m=o(n^{1/4}) and for each ii, mi=hin+o(n1/4)m_{i}=h_{i}n+o(n^{1/4}) where hi>0h_{i}>0.

We consider a uniformly random permutation of all MM objects with some order restrictions which are only within type 0. Let XX denote the number of objects of type 0 that lie at the first mm places of this permutation. The law of XX for j<mj<m is given by the following formula:

pj:={X=j}=(mj)(l1=0j1(m0l1))(l2=0mj1(Mm0l2))(Mm)!M!\begin{split}p_{j}:=\mathbb{P}\{X=j\}=&\binom{m}{j}\left(\prod_{l_{1}=0}^{j-1}(m_{0}-l_{1})\right)\left(\prod_{l_{2}=0}^{m-j-1}(M-m_{0}-l_{2})\right)\frac{(M-m)!}{M!}\end{split} (1)

which is asymptotically given by

(mj)h0j(h1++hs)mj(h0+h1++hs)m(1+o(n1/2))\binom{m}{j}\frac{h_{0}^{j}(h_{1}+\cdots+h_{s})^{m-j}}{(h_{0}+h_{1}+\cdots+h_{s})^{m}}(1+o(n^{-1/2}))

as nn\rightarrow\infty.

Proof.

To compute pjp_{j} we will count the number of permutations with jj elements of type 0 at the first mm place then divide it by the number all permutations. Any given order restrictions on objects of type 0 make both numerator and denominator of this fraction divided by the number of symmetries. As a result, we can assume no restriction on objects of type 0.

First part is straightforward by choosing those jj places at the first mm observations for objects of type 0 and m0jm_{0}-j places in the remaining part. Next, counting the number of desired arrangements for objects of type 0 and other types we arrive at the following formula for the probability

pj=(mj)(Mmm0j)m0!(Mm0)!M!.p_{j}=\frac{\binom{m}{j}\binom{M-m}{m_{0}-j}m_{0}!(M-m_{0})!}{M!}.

This yields the first part by a few lines of algebra.

For the second part, we can write

pj\displaystyle p_{j} =l1=0j1(m0l1)l2=0mj1(Mm0l2)×(Mm)!M!\displaystyle=\frac{\prod_{l_{1}=0}^{j-1}(m_{0}-l_{1})\prod_{l_{2}=0}^{m-j-1}(M-m_{0}-l_{2})\times(M-m)!}{M!}
=(i=1jm0j+iMm+i)(i=1mjMm0m+j+iMm+j+i)\displaystyle=\left(\prod_{i=1}^{j}\frac{m_{0}-j+i}{M-m+i}\right)\left(\prod_{i=1}^{m-j}\frac{M-m_{0}-m+j+i}{M-m+j+i}\right)
=(h0n+o(n1/4))j((h1++hs)n+o(n1/4))mj((h0+h1++hs)n+o(n1/4))m\displaystyle=\frac{\left(h_{0}n+o(n^{1/4})\right)^{j}\left((h_{1}+\cdots+h_{s})n+o(n^{1/4})\right)^{m-j}}{\left((h_{0}+h_{1}+\cdots+h_{s})n+o(n^{1/4})\right)^{m}}
=(h0j(h1++hs)mj(h0+h1++hs)m)\displaystyle=\left(\frac{h_{0}^{j}(h_{1}+\cdots+h_{s})^{m-j}}{(h_{0}+h_{1}+\cdots+h_{s})^{m}}\right)
×(1+(1/h0)o(n3/4))j(1+(1/(h1++hs))o(n3/4))mj(1+(1/(h0+h1++hs))o(n3/4))m\displaystyle\qquad\times\frac{\left(1+(1/h_{0})o(n^{-3/4})\right)^{j}\left(1+(1/(h_{1}+\cdots+h_{s}))o(n^{-3/4})\right)^{m-j}}{\left(1+(1/(h_{0}+h_{1}+\cdots+h_{s}))o(n^{-3/4})\right)^{m}}
=(h0j(h1++hs)mj(h0+h1++hs)m)(1+o(n3/4))m\displaystyle=\left(\frac{h_{0}^{j}(h_{1}+\cdots+h_{s})^{m-j}}{(h_{0}+h_{1}+\cdots+h_{s})^{m}}\right)(1+o(n^{-3/4}))^{m}
=(h0j(h1++hs)mj(h0+h1++hs)m)(1+o(n3/4))o(n1/4)\displaystyle=\left(\frac{h_{0}^{j}(h_{1}+\cdots+h_{s})^{m-j}}{(h_{0}+h_{1}+\cdots+h_{s})^{m}}\right)(1+o(n^{-3/4}))^{o(n^{1/4})}
=(h0j(h1++hs)mj(h0+h1++hs)m)(1+o(n1/2)).\displaystyle=\left(\frac{h_{0}^{j}(h_{1}+\cdots+h_{s})^{m-j}}{(h_{0}+h_{1}+\cdots+h_{s})^{m}}\right)(1+o(n^{-1/2})).

This completes the proof. ∎

Lemma 3.2.

Under the same assumption of Lemma 3.1 let ij=o(n1/4)<mji_{j}=o(n^{1/4})<m_{j} be positive integers for each jj. Suppose vector XX represents the number of appearance of each type before the first observation of type 0 in a uniformly chosen permutation. Then

{X=(i1,,is)}\displaystyle\mathbb{P}\{X=(i_{1},\cdots,i_{s})\} =m0(m1i1)(msis)(i1++is)!(Mi1is1)!M!\displaystyle=\frac{m_{0}\binom{m_{1}}{i_{1}}\cdots\binom{m_{s}}{i_{s}}(i_{1}+\cdots+i_{s})!(M-i_{1}-\cdots-i_{s}-1)!}{M!}

which asymptotically is

h0h1i1hsis(h0+h1++hs)i1++is+1(i1+i2++is)!i1!i2!is!(1+o(n1/4)).\frac{h_{0}h_{1}^{i_{1}}\cdots h_{s}^{i_{s}}}{(h_{0}+h_{1}+\cdots+h_{s})^{i_{1}+\cdots+i_{s}+1}}\frac{(i_{1}+i_{2}+\cdots+i_{s})!}{i_{1}!i_{2}!\cdots i_{s}!}(1+o(n^{-1/4})).
Proof.

It is easy to check that the number of permutations of MM objects where at the first i1+i2++isi_{1}+i_{2}+\cdots+i_{s} places there are exactly iji_{j} elements of type jj for each 1jk1\leq j\leq k, and the first observation of type 0 happens at the (i1++is+1)(i_{1}+\cdots+i_{s}+1)-th position is as follows

m0(m1i1)(msis)(i1++is)!(Mi1is1m01)(Mm0i1ik)!m_{0}\binom{m_{1}}{i_{1}}\cdots\binom{m_{s}}{i_{s}}(i_{1}+\cdots+i_{s})!\binom{M-i_{1}-\cdots-i_{s}-1}{m_{0}-1}(M-m_{0}-i_{1}-\cdots-i_{k})!

Therefore, using hi1,h0+h1++hssh_{i}\geq 1,h_{0}+h_{1}+\cdots+h_{s}\geq s the probability of this event is as follows:

pi1,,is\displaystyle p_{i_{1},\cdots,i_{s}} =m0(m1i1)(msis)(i1++is)!(Mi1is1m01)(Mm0i1is)!M!\displaystyle=\frac{m_{0}\binom{m_{1}}{i_{1}}\cdots\binom{m_{s}}{i_{s}}(i_{1}+\cdots+i_{s})!\binom{M-i_{1}-\cdots-i_{s}-1}{m_{0}-1}(M-m_{0}-i_{1}-\cdots-i_{s})!}{M!}
=m0(m1i1)(msis)(i1++is)!(Mi1is1)!M!\displaystyle=\frac{m_{0}\binom{m_{1}}{i_{1}}\cdots\binom{m_{s}}{i_{s}}(i_{1}+\cdots+i_{s})!(M-i_{1}-\cdots-i_{s}-1)!}{M!}
=(h0n+o(n1/4))(h1n+o(n1/4))i1(hsn+o(n1/4))is(i1++is)!((h0+h1++hs)n+so(n1/4))i1++is+1i1!is!\displaystyle=\frac{(h_{0}n+o(n^{1/4}))(h_{1}n+o(n^{1/4}))^{i_{1}}\cdots(h_{s}n+o(n^{1/4}))^{i_{s}}(i_{1}+\cdots+i_{s})!}{((h_{0}+h_{1}+\cdots+h_{s})n+so(n^{1/4}))^{i_{1}+\cdots+i_{s}+1}i_{1}!\cdots i_{s}!}
=(h0h1i1hsis(i1++is)!(h0+h1++hs)i1++is+1i1!is!)(1+o(n3/4)1+(sh1++hs)o(n3/4))i1++is+1\displaystyle=\left(\frac{h_{0}h_{1}^{i_{1}}\cdots h_{s}^{i_{s}}(i_{1}+\cdots+i_{s})!}{(h_{0}+h_{1}+\cdots+h_{s})^{i_{1}+\cdots+i_{s}+1}i_{1}!\cdots i_{s}!}\right)\left(\frac{1+o(n^{-3/4})}{1+(\frac{s}{h_{1}+\cdots+h_{s}})o(n^{-3/4})}\right)^{i_{1}+\cdots+i_{s}+1}
=(h0h1i1hsis(i1++is)!(h0+h1++hs)i1++is+1i1!is!)(1+o(n3/4))i1++is+1\displaystyle=\left(\frac{h_{0}h_{1}^{i_{1}}\cdots h_{s}^{i_{s}}(i_{1}+\cdots+i_{s})!}{(h_{0}+h_{1}+\cdots+h_{s})^{i_{1}+\cdots+i_{s}+1}i_{1}!\cdots i_{s}!}\right)\left(1+o(n^{-3/4})\right)^{i_{1}+\cdots+i_{s}+1}
=(h0h1i1hsis(i1++is)!(h0+h1++hs)i1++is+1i1!is!)(1+o(n3/4))o(n1/2)\displaystyle=\left(\frac{h_{0}h_{1}^{i_{1}}\cdots h_{s}^{i_{s}}(i_{1}+\cdots+i_{s})!}{(h_{0}+h_{1}+\cdots+h_{s})^{i_{1}+\cdots+i_{s}+1}i_{1}!\cdots i_{s}!}\right)\left(1+o(n^{-3/4})\right)^{o(n^{1/2})}
=(h0h1i1hsis(i1++is)!(h0+h1++hs)i1++is+1i1!is!)(1+o(n1/4)).\displaystyle=\left(\frac{h_{0}h_{1}^{i_{1}}\cdots h_{s}^{i_{s}}(i_{1}+\cdots+i_{s})!}{(h_{0}+h_{1}+\cdots+h_{s})^{i_{1}+\cdots+i_{s}+1}i_{1}!\cdots i_{s}!}\right)\left(1+o(n^{-1/4})\right).

This finishes the proof. ∎

Lemmas 3.1 and 3.2 demonstrate that a negative multinomial distribution is the limit for the distribution of number of other types before the first appearance of a given type. As mentioned earlier, this distributional convergence will help us greatly in our analysis. The special case required in this paper fixes all hi=1h_{i}=1. Therefore, we state the following lemma.

Lemma 3.3.

Same as Lemma 3.1, there are mim_{i} objects of type ii where 0is0\leq i\leq s and s=o(n1/4)s=o(n^{1/4}). Let mi=n+o(n1/4)m_{i}=n+o(n^{1/4}) and m=o(n1/4)m=o(n^{1/4}) grow to infinity with an extra condition that s=o(m)s=o(m). Consider a uniformly random permutation of all objects with some order restrictions only among objects of type 0. We denote the number of objects of type 0 that within the first mm places of this permutation by XX. We have the following bound for the lower tail of XX

{Xt}exp(ms+1+tlog(mts)+t+log(t+1))(1+o(n1/2))\mathbb{P}\{X\leq t\}\leq\exp\left(\frac{-m}{s+1}+t\log\left(\frac{m}{ts}\right)+t+\log(t+1)\right)(1+o(n^{-1/2}))

where tm/st\leq m/s. In particular, if s,t=k+O(1)s,t=\sqrt{k}+O(1) and m=(sb)(kls1)m=(s-b)(k-ls-1) where k=o(n1/6)k=o(n^{1/6}) goes to infinity and b,l=O(1)b,l=O(1), for sufficiently large nn we have

{Xt}exp(k+12klog(k)+O(k))\mathbb{P}\{X\leq t\}\leq\exp\left(-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right)
Proof.

Let k>sk>s (ss constant or non-constant). It follows from tm/st\leq m/s that (mj)smj(s+1)m\binom{m}{j}\frac{s^{m-j}}{(s+1)^{m}} is increasing for 0jt0\leq j\leq t. Using this fact and Lemma 3.1 with hi=1h_{i}=1 we have

{Xt}=\displaystyle\mathbb{P}\{X\leq t\}= j=0t(mj)smj(s+1)m(1+o(n1/2))\displaystyle\sum_{j=0}^{t}\binom{m}{j}\frac{s^{m-j}}{(s+1)^{m}}(1+o(n^{-1/2}))
\displaystyle\leq (t+1)(mt)smt(s+1)m(1+o(n1/2))\displaystyle(t+1)\binom{m}{t}\frac{s^{m-t}}{(s+1)^{m}}(1+o(n^{-1/2}))
=\displaystyle= t+1st(mt)(11s+1)m(1+o(n1/2))\displaystyle\frac{t+1}{s^{t}}\binom{m}{t}\left(1-\frac{1}{s+1}\right)^{m}(1+o(n^{-1/2}))
=\displaystyle= t+1st(mt)((11s+1)s+1)m/(s+1)(1+o(n1/2))\displaystyle\frac{t+1}{s^{t}}\binom{m}{t}\left(\Big{(}1-\frac{1}{s+1}\Big{)}^{s+1}\right)^{m/(s+1)}(1+o(n^{-1/2}))
\displaystyle\leq t+1st(mt)exp(ms+1)(1+o(n1/2))\displaystyle\frac{t+1}{s^{t}}\binom{m}{t}\exp\left(\frac{-m}{s+1}\right)(1+o(n^{-1/2}))
\displaystyle\leq t+1stmtt!exp(ms+1)(1+o(n1/2)).\displaystyle\frac{t+1}{s^{t}}\frac{m^{t}}{t!}\exp\left(\frac{-m}{s+1}\right)(1+o(n^{-1/2})).

One can use Striling’s approximation t!>(t/e)tt!>(t/e)^{t} to get

{Xt}exp(ms+1+tlog(mts)+t+log(t+1))(1+o(n1/2))\mathbb{P}\{X\leq t\}\leq\exp\left(\frac{-m}{s+1}+t\log\left(\frac{m}{ts}\right)+t+\log(t+1)\right)(1+o(n^{-1/2}))

Setting s,t=k+O(1)s,t=\sqrt{k}+O(1) and m=(sb)(kls1)m=(s-b)(k-ls-1), it follows that

ms+1+t+log(t+1)\displaystyle\frac{-m}{s+1}+t+\log(t+1) =k+(bl+1)ks+1+s+t+log(t+1)+O(1)\displaystyle=-k+\frac{(bl+1)k}{s+1}+s+t+\log(t+1)+O(1)
=k+O(k)\displaystyle=-k+O(\sqrt{k})

Additionally,

tlog(mts)\displaystyle t\log\left(\frac{m}{ts}\right) =(k+O(1))log(k3/2(b+l)k+O(k)k+O(k))\displaystyle=\left(\sqrt{k}+O(1)\right)\log\left(\frac{k^{3/2}-(b+l)k+O(\sqrt{k})}{k+O(\sqrt{k})}\right)
=(k+O(1))log(k+O(1))\displaystyle=\left(\sqrt{k}+O(1)\right)\log\left(\sqrt{k}+O(1)\right)
=12klog(k)+O(k).\displaystyle=\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k}).

This completes the proof. ∎

4 Disconnectivity of graph for t<1t<1

This section begins by recalling the negative multinomial distribution. In the special case that we need here, we present an urn model to interpret the distribution. Suppose we have a sequence of i.i.d. random variables with k+1k+1 possible outcomes chosen uniformly at random. We describe these outcomes as the type of objects. Therefore, there are k+1k+1 types. This sequence stops as soon as the first object of type k+1k+1 occurs in the sequence. The probability that there are i1i_{1} objects of type 1, i2i_{2} objects of type 2, … , iki_{k} objects of type kk, and finally the nn-th observation which is of type k+1k+1, follows the negative multinomial distribution below

f(i1,i2,,ik)=1(k+1)i1+i2++ik+1(i1+i2++ik)!i1!i2!ik!,f(i_{1},i_{2},\cdots,i_{k})=\frac{1}{(k+1)^{i_{1}+i_{2}+\cdots+i_{k}+1}}\frac{(i_{1}+i_{2}+\cdots+i_{k})!}{i_{1}!i_{2}!\cdots i_{k}!}, (2)

in which it0i_{t}\geq 0 for 1tk1\leq t\leq k and i1+i2++ik=n1i_{1}+i_{2}+\cdots+i_{k}=n-1.

We now introduce some lemmas that help with any required computation.

Lemma 4.1.

With the definition (2)

i1+i2++ik=n1f(i1,i2,,ik)=1k+1(kk+1)n1\sum_{i_{1}+i_{2}+\cdots+i_{k}=n-1}f(i_{1},i_{2},\cdots,i_{k})=\frac{1}{k+1}\left(\frac{k}{k+1}\right)^{n-1}
Proof.

There is an algebraic proof to this lemma using the multinomial expansion formula. However, we give a simpler probabilistic way. This sum is over the probability of all possible arrangements of nn items in which all the first n1n-1 items are of type 1,2,,k1,2,\cdots,k and the last one is of type k+1k+1. Hence, the probability of this event is (kk+1)n1(1k+1)(\frac{k}{k+1})^{n-1}(\frac{1}{k+1}) due to the independent choice made in each position. ∎

Lemma 4.2.

Let 0<δ<10<\delta<1. Then:

k2(1+δ)<i1+i2++ik<4(k+1)2f(i1,i2,,ik)k2(1+δ)<i1+i2++ikf(i1,i2,,ik)e4(k+1)\sum_{k^{2}(1+\delta)<i_{1}+i_{2}+\cdots+i_{k}<4(k+1)^{2}}f(i_{1},i_{2},\cdots,i_{k})\geq\sum_{k^{2}(1+\delta)<i_{1}+i_{2}+\cdots+i_{k}}f(i_{1},i_{2},\cdots,i_{k})-e^{-4(k+1)}
Proof.
r\displaystyle r =k2(1+δ)<i1+i2++ikf(i1,i2,,ik)k2(1+δ)<i1+i2++ik<4(k+1)2f(i1,i2,,ik)\displaystyle=\sum_{k^{2}(1+\delta)<i_{1}+i_{2}+\cdots+i_{k}}f(i_{1},i_{2},\cdots,i_{k})-\sum_{k^{2}(1+\delta)<i_{1}+i_{2}+\cdots+i_{k}<4(k+1)^{2}}f(i_{1},i_{2},\cdots,i_{k})
=4(k+1)2i1+i2++ikf(i1,i2,,ik)\displaystyle=\sum_{4(k+1)^{2}\leq i_{1}+i_{2}+\cdots+i_{k}}f(i_{1},i_{2},\cdots,i_{k})
=n=4(k+1)21k+1(kk+1)n\displaystyle=\sum_{n=4(k+1)^{2}}^{\infty}\frac{1}{k+1}(\frac{k}{k+1})^{n}
=(kk+1)4(k+1)2\displaystyle=(\frac{k}{k+1})^{4(k+1)^{2}}
=((11k+1)(k+1))4(k+1)\displaystyle=\left(\left(1-\frac{1}{k+1}\right)^{(k+1)}\right)^{4(k+1)}
e4(k+1),\displaystyle\leq e^{-4(k+1)},

and the result follows. ∎

Lemma 4.3.

Let 0<δ<10<\delta<1. Then:

k2(1+δ)<i1+i2++ik<4(k+1)2,j,ijkf(i1,i2,,ik)\sum_{k^{2}(1+\delta)<i_{1}+i_{2}+\cdots+i_{k}<4(k+1)^{2},\;\forall j,\,i_{j}\geq k}f(i_{1},i_{2},\cdots,i_{k})\geq
(1kekδ27)k2(1+δ)<i1+i2++ik<4(k+1)2f(i1,i2,,ik)(1-ke^{\frac{-k\delta^{2}}{7}})\sum_{k^{2}(1+\delta)<i_{1}+i_{2}+\cdots+i_{k}<4(k+1)^{2}}f(i_{1},i_{2},\cdots,i_{k})
Proof.

Assume n=i1++ik+1>k2(1+δ)n=i_{1}+\cdots+i_{k}+1>k^{2}(1+\delta) is a fixed number. Then, at each step prior to the nn-th observation, an outcome of type mm has a Bernoulli distribution with parameter 1k\frac{1}{k}. Denote this random variable by Xm(j)X_{m}^{(j)} where jj refers to the step number. For an arbitrary 1mk1\leq m\leq k, we have 𝔼(Xm(j))=1k\mathbb{E}(X_{m}^{(j)})=\frac{1}{k} and Var(Xm(j))=k1k2\mathrm{Var}(X_{m}^{(j)})=\frac{k-1}{k^{2}}. If im<ki_{m}<k, then

Xm(1)++Xm(n1)n1<kk2(1+δ)=1k(1+δ).\frac{X_{m}^{(1)}+\cdots+X_{m}^{(n-1)}}{n-1}<\frac{k}{k^{2}(1+\delta)}=\frac{1}{k(1+\delta)}.

However, the expected value of the time average above is 1k\frac{1}{k}, and 1k(1+δ)1k=δk(1+δ)\frac{1}{k(1+\delta)}-\frac{1}{k}=-\frac{\delta}{k(1+\delta)}, so this is a (lower) tail event. Now, we apply the Bernstein inequality:

{im<k|i1++ik=n1}\displaystyle\mathbb{P}\{i_{m}<k|i_{1}+\cdots+i_{k}=n-1\} {Xm(1)++Xm(n1)n1<1k(1+δ)}\displaystyle\leq\mathbb{P}\left\{\frac{X_{m}^{(1)}+\cdots+X_{m}^{(n-1)}}{n-1}<\frac{1}{k(1+\delta)}\right\}
exp((n1)×δ2k2(1+δ)22(k1)k2+2δ3k(δ+1))\displaystyle\leq\exp\left(\frac{-(n-1)\times\frac{\delta^{2}}{k^{2}(1+\delta)^{2}}}{\frac{2(k-1)}{k^{2}}+\frac{2\delta}{3k(\delta+1)}}\right)
exp(k2(1+δ)×δ2k2(1+δ)22(k1)k2+2δ3k(δ+1))\displaystyle\leq\exp\left(\frac{-k^{2}(1+\delta)\times\frac{\delta^{2}}{k^{2}(1+\delta)^{2}}}{\frac{2(k-1)}{k^{2}}+\frac{2\delta}{3k(\delta+1)}}\right)
=exp(3k2δ26(k1)(δ+1)+2δk)\displaystyle=\exp\left(\frac{-3k^{2}\delta^{2}}{6(k-1)(\delta+1)+2\delta k}\right)
exp(kδ25).\displaystyle\leq\exp\left(\frac{-k\delta^{2}}{5}\right).

Using the union bound yields

{j:ijk|m=1kim=n1}1kexp(kδ25).\mathbb{P}\left\{\forall j:i_{j}\geq k\bigg{|}\sum_{m=1}^{k}i_{m}=n-1\right\}\geq 1-k\exp\left(\frac{-k\delta^{2}}{5}\right).

Therefore,

k2(1+δ)<im<4(k+1)2,j,ijkf(i1,i2,,ik)\displaystyle\sum_{k^{2}(1+\delta)<\sum i_{m}<4(k+1)^{2},\;\forall j,\,i_{j}\geq k}f(i_{1},i_{2},\cdots,i_{k})
=k2(1+δ)<n<4(k+1)2im=n1,j,ijkf(i1,i2,,ik)\displaystyle=\sum_{k^{2}(1+\delta)<n<4(k+1)^{2}}\sum_{\sum i_{m}=n-1,\;\forall j,\,i_{j}\geq k}f(i_{1},i_{2},\cdots,i_{k})
=k2(1+δ)<n<4(k+1)2(k+1k)n{j:ijk,im=n1}\displaystyle=\sum_{k^{2}(1+\delta)<n<4(k+1)^{2}}\left(\frac{k+1}{k}\right)^{n}\mathbb{P}\Big{\{}\forall j:i_{j}\geq k,\sum i_{m}=n-1\Big{\}}
=k2(1+δ)<n<4(k+1)2(k+1k)n{im=n1}{j:ijk|im=n1}\displaystyle=\sum_{k^{2}(1+\delta)<n<4(k+1)^{2}}\left(\frac{k+1}{k}\right)^{n}\mathbb{P}\Big{\{}\sum i_{m}=n-1\Big{\}}\mathbb{P}\Big{\{}\forall j:i_{j}\geq k\bigg{|}\sum i_{m}=n-1\Big{\}}
k2(1+δ)<n<4(k+1)2(k+1k)n{im=n1}(1kexp(kδ25))\displaystyle\geq\sum_{k^{2}(1+\delta)<n<4(k+1)^{2}}\left(\frac{k+1}{k}\right)^{n}\mathbb{P}\Big{\{}\sum i_{m}=n-1\Big{\}}\left(1-k\exp\left(\frac{-k\delta^{2}}{5}\right)\right)
=(1kexp(kδ25))k2(1+δ)<im<4(k+1)2f(i1,i2,,ik).\displaystyle=\left(1-k\exp\left(\frac{-k\delta^{2}}{5}\right)\right)\sum_{k^{2}(1+\delta)<\sum i_{m}<4(k+1)^{2}}f(i_{1},i_{2},\cdots,i_{k}).

This completes the proof. ∎

Lemma 4.4.

Let kk be a positive integer and 0δ10\leq\delta\leq 1 depends on kk such that for sufficiently large kk we have δ6log(k)k\delta\geq\sqrt{\frac{6\log(k)}{k}}. The following lower bound holds

(k2(1+δ)<i1+i2++ik<4(k+1)2,j,ijkf(i1,i2,,ik))O(ek(δ+1))\left(\sum_{k^{2}(1+\delta)<i_{1}+i_{2}+\cdots+i_{k}<4(k+1)^{2},\;\forall j,\,i_{j}\geq k}f(i_{1},i_{2},\cdots,i_{k})\right)\geq O(e^{-k(\delta+1)})

as kk\rightarrow\infty.

Proof.

Using Lemmas 4.3, 4.2, 4.1:

k2(1+δ)<i1+i2++ik<4(k+1)2,j,ijkf(i1,i2,,ik)\displaystyle\sum_{k^{2}(1+\delta)<i_{1}+i_{2}+\cdots+i_{k}<4(k+1)^{2},\;\forall j,\,i_{j}\geq k}f(i_{1},i_{2},\cdots,i_{k})
(1kekδ25)k2(1+δ)<i1+i2++ik<4(k+1)2f(i1,i2,,ik)\displaystyle\geq\Big{(}1-ke^{\frac{-k\delta^{2}}{5}}\Big{)}\sum_{k^{2}(1+\delta)<i_{1}+i_{2}+\cdots+i_{k}<4(k+1)^{2}}f(i_{1},i_{2},\cdots,i_{k})
(1kekδ25)(k2(1+δ)<i1+i2++ikf(i1,i2,,ik)e4(k+1))\displaystyle\geq\Big{(}1-ke^{\frac{-k\delta^{2}}{5}}\Big{)}\left(\sum_{k^{2}(1+\delta)<i_{1}+i_{2}+\cdots+i_{k}}f(i_{1},i_{2},\cdots,i_{k})-e^{-4(k+1)}\right)
=(1kekδ25)(N=k2(1+δ)+11k+1(kk+1)N1e4(k+1))\displaystyle=\Big{(}1-ke^{\frac{-k\delta^{2}}{5}}\Big{)}\left(\sum_{N=k^{2}(1+\delta)+1}^{\infty}\frac{1}{k+1}(\frac{k}{k+1})^{N-1}-e^{-4(k+1)}\right)
=(1kekδ25)((11k+1)k2(δ+1)e4(k+1)).\displaystyle=\Big{(}1-ke^{\frac{-k\delta^{2}}{5}}\Big{)}\left((1-\frac{1}{k+1})^{k^{2}(\delta+1)}-e^{-4(k+1)}\right).

We know that (11k+1)ke1(1-\frac{1}{k+1})^{k}\approx e^{-1} (more precisely (11k+1)k>e1(1-\frac{1}{k+1})^{k}>e^{-1}) and 1kekδ2511k611-ke^{\frac{-k\delta^{2}}{5}}\geq 1-\frac{1}{\sqrt[6]{k}}\rightarrow 1 as kk\rightarrow\infty. Hence, the mentioned sum is asymptotically greater than:

ek(δ+1)e4(k+1)e^{-k(\delta+1)}-e^{-4(k+1)}

Furthermore, 0<δ<10<\delta<1 causes e4(k+1)e^{-4(k+1)} to vanish faster than ek(δ+1)e^{-k(\delta+1)}. ∎

We are now in a position to relate the lemmas above to the disconnectivity result. Assume IjI_{j} represents the event of vertex jj being isolated. We want to show that with high probability there exists a jj such that Ij=1I_{j}=1.

Lemma 4.5.

For the random graph model 𝔾(n,k)\mathbb{G}(n,k) when k=o(n8)k=o(\sqrt[8]{n}) and kk\rightarrow\infty we have

{I1=1}O(ek(1+δ))\mathbb{P}\{I_{1}=1\}\geq O(e^{-k(1+\delta)})

for any δ=δ(k)\delta=\delta(k) with 6log(k)kδ(k)<1\sqrt{\frac{6\log(k)}{k}}\leq\delta(k)<1.

Proof.

As mentioned earlier, we can look at each configuration of independent random variables {V(i,j)}i,j\{V(i,j)\}_{i,j} as a permutation over all edges \mathcal{E}. Without loss of generality, one can assume vertex labels are sorted according to the scores of edges connecting to vertex 1. In other words, V(1,2)>V(1,3)>>V(1,n)V(1,2)>V(1,3)>\cdots>V(1,n), or equivalently R1j=j+1R_{1}^{j}=j+1 for 1jn11\leq j\leq n-1. Then, for vertex 1 to be isolated, it is sufficient that for each j{2,3,,k+1}j\in\{2,3,\cdots,k+1\} there exist at least kk elements of j\mathcal{E}_{j} (except (1,j)(1,j)) appearing before all elements of 1\mathcal{E}_{1} in the permutation. Strictly speaking, if kk elements of j\mathcal{E}_{j} appear before (1,j+1)(1,j+1), then that would be necessary and sufficient, but we consider a stricter condition which insists that kk elements of j\mathcal{E}_{j} appear before all elements of 1\mathcal{E}_{1}. This way, vertex jj does not agree on edges {1,j}\{1,j\} for 2jk+12\leq j\leq k+1 and vertex 11 also does not agree upon edges (1,j)(1,j) for k+2jnk+2\leq j\leq n.

To make this condition even stronger, we insist on no intersections. Define for each 2ik+12\leq i\leq k+1: i=i{(i,1),,(i,k+1)}\mathcal{E^{\prime}}_{i}=\mathcal{E}_{i}\setminus\{(i,1),\cdots,(i,k+1)\}. As a result, {1,2,3,,k+1}\{\mathcal{E}_{1},\mathcal{E}^{\prime}_{2},\mathcal{E}^{\prime}_{3},\cdots,\mathcal{E}^{\prime}_{k+1}\} is a pairwise disjoint collection with |1|=n1,|i|=nk1|\mathcal{E}_{1}|=n-1,|\mathcal{E}^{\prime}_{i}|=n-k-1. Note that we only determined the order over 1\mathcal{E}_{1}. Thus, any configuration of edge scores induces a uniformly random permutation on 12k+2\mathcal{E}_{1}\cup\mathcal{E^{\prime}}_{2}\cup\cdots\cup\mathcal{E^{\prime}}_{k+2} with the order restriction {1,2}{1,3}{1,n}\{1,2\}\succ\{1,3\}\succ\cdots\succ\{1,n\} on 1\mathcal{E}_{1}.

Assume 1\mathcal{E}_{1} are objects of type 0 and i\mathcal{E^{\prime}}_{i} are objects of type i1i-1 for 2ik+22\leq i\leq k+2. This satisfies all the conditions of Lemma 3.2. Therefore, the probability of having iji_{j} elements of type jj before the first observation of type 0 when k2(δ+1)<i1++ik<4(k+1)2=o(n1/4)k^{2}(\delta+1)<i_{1}+\cdots+i_{k}<4(k+1)^{2}=o(n^{1/4}) can be approximated by the distribution f()f(\cdot) from (2). Hence, for large enough nn we have:

{X=(i1,,ik)}12f(i1,,ik)\mathbb{P}\{X=(i_{1},\cdots,i_{k})\}\geq\frac{1}{2}f(i_{1},\cdots,i_{k})

From what we explained earlier, the extra condition ijki_{j}\geq k guarantees that vertex 1 is isolated. Applying Lemma 4.4 then allows us to conclude that

{I1=1}O(ek(1+δ)),\mathbb{P}\{I_{1}=1\}\geq O(e^{-k(1+\delta)}),

which completes the proof. ∎

Lemma 4.6.

For the random graph model 𝔾(n,k)\mathbb{G}(n,k) where klog(n)3log(n)loglog(n)k\leq\log(n)-3\sqrt{\log(n)\log\log(n)}, we have

𝔼{number of isolated vertices}\mathbb{E}\{\text{number of isolated vertices}\}\rightarrow\infty

In particular, if k=tlog(n)k=\lfloor t\log(n)\rfloor for t<1t<1, the statement is true.

Proof.

Let δ=6log(k)k\delta=\sqrt{\frac{6\log(k)}{k}}, according to Lemma 4.5 for sufficiently large nn and kk, we have

𝔼{number of isolated vertices}\displaystyle\mathbb{E}\{\textrm{number of isolated vertices}\} =𝔼{i=1nIi}\displaystyle=\mathbb{E}\left\{\sum_{i=1}^{n}I_{i}\right\}
=n𝔼{I1}\displaystyle=n\mathbb{E}\{I_{1}\}
=n{I1=1}\displaystyle=n\mathbb{P}\{I_{1}=1\}
nexp(k(1+δ))\displaystyle\geq n\exp(-k(1+\delta))
=exp(log(n)k6klog(k))\displaystyle=\exp(\log(n)-k-\sqrt{6k\log(k)})
exp(3log(n)loglog(n)6klog(k))\displaystyle\geq\exp(3\sqrt{\log(n)\log\log(n)}-\sqrt{6k\log(k)})
exp((36)(log(n)loglog(n))),\displaystyle\geq\exp((3-\sqrt{6})(\sqrt{\log(n)\log\log(n)})),

wherein the right hand side goes to infinity as nn\rightarrow\infty. ∎

There is one more step needed to prove the desired result of there being at least one isolated vertex (with high probability) as the mean of a non-negative random variable being unbounded doesn’t necessarily imply that the probability of it being 0 is 0. For this we will employ the second moment method [1]. To apply the second moment method we need to study the correlation between pairs of indicator random variables corresponding to the isolation of a vertex. Consider two arbitrary vertices, which could be (without loss of generality) chosen to be vertices 1 and 2. We want to study {I1=1,I2=1}={I1I2=1}=𝔼{I1I2}\mathbb{P}\{I_{1}=1,I_{2}=1\}=\mathbb{P}\{I_{1}I_{2}=1\}=\mathbb{E}\{I_{1}I_{2}\}. Define the following events:

  • B1;2:={21k}B_{1;2}:=\{2\in\mathcal{R}^{\leq k}_{1}\}, representing the event in which vertex 2 belongs to the kk-most-preferred set of vertex 1.

  • B2;1:={12k}B_{2;1}:=\{1\in\mathcal{R}^{\leq k}_{2}\}, representing the event in which vertex 1 belongs to the kk-most-preferred set of vertex 2.

  • B1 2:={1k2kϕ}B_{1\cup\,2}:=\{\mathcal{R}^{\leq k}_{1}\cap\mathcal{R}^{\leq k}_{2}\neq\phi\}, identifying the event where kk-most-preferred elements of 1 and 2 have intersections.

  • B12:={i1k such that (2k{2})ikϕ}(B1;2B2;1B1 2)B_{1\rightarrow 2}:=\{\exists i\in\mathcal{R}^{\leq k}_{1}\textrm{ such that }(\mathcal{R}^{\leq k}_{2}\cup\{2\})\cap\mathcal{R}^{\leq k}_{i}\neq\phi\}\setminus(B_{1;2}\cup B_{2;1}\cup B_{1\cup\,2}). This represents the event where there is a vertex ii within the kk-most-preferred of vertex 1 such that 2 or one of kk-most-preferred vertices of it belongs to the kk-most-preferred vertices of ii. We also assume that the kk-most-preferred of vertices 1 and 2 are disjoint and none of them is among the other’s kk-most-preferred vertices. Note that this event implies that the hop distance between nodes 11 and 22 is in {2,3}\{2,3\}.

  • B21:={i2k such that (1k{1})ikϕ}(B1;2B2;1B1 2)B_{2\rightarrow 1}:=\{\exists i\in\mathcal{R}^{\leq k}_{2}\textrm{ such that }(\mathcal{R}^{\leq k}_{1}\cup\{1\})\cap\mathcal{R}^{\leq k}_{i}\neq\phi\}\setminus(B_{1;2}\cup B_{2;1}\cup B_{1\cup\,2}). This represents the event where there is a vertex ii within the kk-most-preferred of vertex 2 such that 1 or one of kk-most-preferred vertices of it belongs to the kk-most-preferred vertices of ii. We also assume that the kk-most-preferred of vertices 1 and 2 are disjoint and none of them is among the other’s kk-most-preferred vertices. Again, this event implies that the hop distance between nodes 11 and 22 is in {2,3}\{2,3\}.

  • B12:={i1k such that |(2k{2})ik|2}(B1;2B2;1B1 2)B_{1\rightsquigarrow 2}:=\{\exists i\in\mathcal{R}^{\leq k}_{1}\textrm{ such that }|(\mathcal{R}^{\leq k}_{2}\cup\{2\})\cap\mathcal{R}^{\leq k}_{i}|\geq 2\}\setminus(B_{1;2}\cup B_{2;1}\cup B_{1\cup\,2}). This is very similar to B12B_{1\rightarrow 2}, but we assume that at least one vertex in 1k\mathcal{R}^{\leq k}_{1} has the intersection of its preferred set with 22 and its preferred set be of size at least 22; this implies at least two paths from 11 to 22 while retaining the distance to be in {2,3}\{2,3\}. Obviously B12B12B_{1\rightsquigarrow 2}\subset B_{1\rightarrow 2}.

  • B21:={i2k such that |(1k{1})ik|2}(B1;2B2;1B1 2)B_{2\rightsquigarrow 1}:=\{\exists i\in\mathcal{R}^{\leq k}_{2}\textrm{ such that }|(\mathcal{R}^{\leq k}_{1}\cup\{1\})\cap\mathcal{R}^{\leq k}_{i}|\geq 2\}\setminus(B_{1;2}\cup B_{2;1}\cup B_{1\cup\,2}). This is very similar to B21B_{2\rightarrow 1}, but we assume that at least one vertex in 2k\mathcal{R}^{\leq k}_{2} has the intersection of its preferred set with 11 and its preferred set be of size at least 22; this implies at least two paths from 22 to 11 while retaining the distance to be in {2,3}\{2,3\}. Obviously B^21B^21\hat{B}_{2\rightsquigarrow 1}\subset\hat{B}_{2\rightarrow 1}.

  • B:=B1;2B2;1B1 2B12B21.B:=B_{1;2}\cup B_{2;1}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2}\cup B_{2\rightarrow 1}.

For ease of presentation, we omit the value 11 from the events {I1=1}\{I_{1}=1\}, {I2=1}\{I_{2}=1\} or {I1I2=1}\{I_{1}I_{2}=1\}: for example, we will use the shorthand {I1I2}:={I1I2=1}\mathbb{P}\{I_{1}I_{2}\}:=\mathbb{P}\{I_{1}I_{2}=1\}. Furthermore, we assume k<log(n)k<\log(n) for the remainder of this section. We start with bounding the event BB.

Lemma 4.7.

We have the following bound for BB

{B}O(k3/n).\mathbb{P}\{B\}\leq O(k^{3}/n).
Proof.

First, note that {B1;2}={B2;1}=kn1\mathbb{P}\{B_{1;2}\}=\mathbb{P}\{B_{2;1}\}=\frac{k}{n-1}. Now, the event B1 2(B1;2B2;1)B_{1\cup\,2}\setminus(B_{1;2}\cup B_{2;1}) can be related to the problem of finding the probability of having an intersection for two completely random subsets of {3,,n}\{3,\cdots,n\} of size kk, which is a much simpler problem. We will now lower bound the probability of (B1 2(B1;2B2;1))c(B_{1\cup\,2}\setminus(B_{1;2}\cup B_{2;1}))^{c} using the fact that for each ii, ik\mathcal{R}_{i}^{\leq k} is uniform among all subsets of size kk.

Counting the number of ways to choose a pair of two disjoint subsets of size kk yields:

12(n2k)(nk2k),\frac{1}{2}\binom{n-2}{k}\binom{n-k-2}{k},

and for choosing two arbitrary subsets the number is:

12(n1k)2.\frac{1}{2}\binom{n-1}{k}^{2}.

Therefore, the probability of not having intersection is

{(B1 2(B1;2B2;1))c}\displaystyle\mathbb{P}\{(B_{1\cup\,2}\setminus(B_{1;2}\cup B_{2;1}))^{c}\} =(n2k)(nk2k)(n1k)2\displaystyle=\frac{\binom{n-2}{k}\binom{n-k-2}{k}}{\binom{n-1}{k}^{2}} (3)
=(nk2)2(nk3)(nk4)(n2k2)(n1)2(n2)(n3)(nk1)\displaystyle=\frac{(n-k-2)^{2}(n-k-3)(n-k-4)\cdots(n-2k-2)}{(n-1)^{2}(n-2)(n-3)\cdots(n-k-1)}
(12k+1n1)k\displaystyle\geq\left(1-\frac{2k+1}{n-1}\right)^{k}
=1O(k2/n).\displaystyle=1-O(k^{2}/n).

This implies {(B1 2(B1;2B2;1))}O(k2/n)\mathbb{P}\{(B_{1\cup\,2}\setminus(B_{1;2}\cup B_{2;1}))\}\leq O(k^{2}/n); therefore,

{B1 2B1;2B2;1}={(B1 2(B1;2B2;1))}+{B1;2B2;1}O(k2/n).\mathbb{P}\{B_{1\cup\,2}\cup B_{1;2}\cup B_{2;1}\}=\mathbb{P}\{(B_{1\cup\,2}\setminus(B_{1;2}\cup B_{2;1}))\}+\mathbb{P}\{B_{1;2}\cup B_{2;1}\}\leq O(k^{2}/n). (4)

To estimate the probability B12B_{1\rightarrow 2} we look at {i1k such that (2k{2})ikϕ}\{\exists i\in\mathcal{R}^{\leq k}_{1}\textrm{ such that }(\mathcal{R}^{\leq k}_{2}\cup\{2\})\cap\mathcal{R}^{\leq k}_{i}\neq\phi\} given (B1 2B1;2B2;1)c(B_{1\cup\,2}\cup B_{1;2}\cup B_{2;1})^{c} instead; note that the probability estimate increases with this conditioning. Let i1ki\in\mathcal{R}_{1}^{\leq k} be a vertex. A similar argument to (3) follows that the probability where for all zikz\in\mathcal{R}_{i}^{\leq k}: z{2}2kz\notin\{2\}\cup\mathcal{R}_{2}^{\leq k} is bounded below by (1k+1n1)k(1-\tfrac{k+1}{n-1})^{k}. As a result, using union bound, with probability at most k(1(1k+1n1)k)k\big{(}1-(1-\tfrac{k+1}{n-1})^{k}\big{)} there exists i1ki\in\mathcal{R}_{1}^{\leq k} without this property. Hence

{B12}k(1(1k+1n1)k)=O(k3/n),\mathbb{P}\{B_{1\rightarrow 2}\}\leq k\left(1-\left(1-\frac{k+1}{n-1}\right)^{k}\right)=O(k^{3}/n), (5)

The same is true for B21B_{2\rightarrow 1}. From (4), (5) we establish the desired bound. ∎

The following Lemma shows that B12B21B_{1\rightsquigarrow 2}\cup B_{2\rightsquigarrow 1} is very small; consequently, one can suppose (B12B21)c(B_{1\rightsquigarrow 2}\cup B_{2\rightsquigarrow 1})^{c} holds. This way we can control the size of both intersections (2k{2})ik(\mathcal{R}^{\leq k}_{2}\cup\{2\})\cap\mathcal{R}^{\leq k}_{i}, and (1k{1})ik(\mathcal{R}^{\leq k}_{1}\cup\{1\})\cap\mathcal{R}^{\leq k}_{i}.

Lemma 4.8.

The following bound for the probability of B12B21B_{1\rightsquigarrow 2}\cup B_{2\rightsquigarrow 1} holds.

{B12B21}O(k5/n2).\mathbb{P}\{B_{1\rightsquigarrow 2}\cup B_{2\rightsquigarrow 1}\}\leq O(k^{5}/n^{2}).
Proof.

We again look at {i1k such that |(2k{2})ik|2}\{\exists i\in\mathcal{R}^{\leq k}_{1}\textrm{ such that }|(\mathcal{R}^{\leq k}_{2}\cup\{2\})\cap\mathcal{R}^{\leq k}_{i}|\geq 2\} given (B1 2B1;2B2;1)c(B_{1\cup\,2}\cup B_{1;2}\cup B_{2;1})^{c} instead. For an arbitrary i1ki\in\mathcal{R}_{1}^{\leq k}, we already saw in the proof of Lemma 4.7 that the probability of |(2k{2})ik|=0|(\mathcal{R}^{\leq k}_{2}\cup\{2\})\cap\mathcal{R}^{\leq k}_{i}|=0 is bounded below by (1k+1n1)k(1-\tfrac{k+1}{n-1})^{k}. We now lower bound the probability of |(2k{2})ik|=1|(\mathcal{R}^{\leq k}_{2}\cup\{2\})\cap\mathcal{R}^{\leq k}_{i}|=1. The number of ways to determine ik\mathcal{R}^{\leq k}_{i} in order to having only one vertex in common with (2k{2})(\mathcal{R}^{\leq k}_{2}\cup\{2\}) is (k+1)(nk2k1)(k+1)\binom{n-k-2}{k-1}; therefore, the probability becomes

(k+1)(nk2k1)(n1k)k(k+1)n(12k1n1)k.\frac{(k+1)\binom{n-k-2}{k-1}}{\binom{n-1}{k}}\geq\frac{k(k+1)}{n}\left(1-\frac{2k-1}{n-1}\right)^{k}.

We now apply the union bound to obtain

{B12}k(1(1k+1n1)kk(k+1)n(12k1n1)k)O(k5/n2).\mathbb{P}\{B_{1\rightsquigarrow 2}\}\leq k\left(1-\left(1-\frac{k+1}{n-1}\right)^{k}-\frac{k(k+1)}{n}\left(1-\frac{2k-1}{n-1}\right)^{k}\right)\leq O(k^{5}/n^{2}).

A similar argument for B21B_{2\rightsquigarrow 1} finishes the proof. ∎

Lemma 4.9.

The following two probability bounds hold:

{I1(B1;2B1 2B12)|B12c}\displaystyle\mathbb{P}\{I_{1}\cap(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})|B_{1\rightsquigarrow 2}^{c}\} exp(log(n)k+12klog(k)+O(k)),\displaystyle\leq\exp\left(-\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right),
{I2(B2;1B1 2B21)|B21c}\displaystyle\mathbb{P}\{I_{2}\cap(B_{2;1}\cup B_{1\cup\,2}\cup B_{2\rightarrow 1})|B_{2\rightsquigarrow 1}^{c}\} exp(log(n)k+12klog(k)+O(k)).\displaystyle\leq\exp\left(-\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right).
Proof.

We prove the first one, and then the second one will follow by symmetry. Let s=kks=\lfloor\sqrt{k}\rfloor\leq k. Since B12cB_{1\rightsquigarrow 2}^{c} holds, for any i1si\in\mathcal{R}^{\leq s}_{1} there is at most one element ti2kt_{i}\in\mathcal{R}^{\leq k}_{2} such that tiikt_{i}\in\mathcal{R}^{\leq k}_{i} too. Define i=i{(i,j):j1s2k{1,2}}\mathcal{E^{\prime}}_{i}=\mathcal{E}_{i}\setminus\{(i,j):j\in\mathcal{R}^{\leq s}_{1}\cup\mathcal{R}^{\leq k}_{2}\cup\{1,2\}\}. Therefore, if we consider the following union

=1i1si,\mathcal{E}^{\prime}=\mathcal{E}_{1}\bigcup_{i\in\mathcal{R}^{\leq s}_{1}}\mathcal{E^{\prime}}_{i},

then it will be a partition in which |1|=n1|\mathcal{E}_{1}|=n-1, and |i|=nks3|\mathcal{E^{\prime}}_{i}|=n-k-s-3 for all i1si\in\mathcal{R}^{\leq s}_{1}. Define 1\mathcal{E}_{1} objects to be of type 0 and i\mathcal{E^{\prime}}_{i} objects to be of type ii for i1si\in\mathcal{R}_{1}^{\leq s}. One can easily see that each realization of 𝔾(n,k)\mathbb{G}(n,k) induces a uniformly random permutation over \mathcal{E}^{\prime} which has only a fixed order on the first ss or s+1s+1 elements of 1\mathcal{E}_{1}, i.e., type 0.

A necessary condition for vertex 1 to be isolated given that B1;2B1 2B12B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2} and B12cB_{1\rightsquigarrow 2}^{c} hold, is that for all i(2)1si(\neq 2)\in\mathcal{R}^{\leq s}_{1}, there should be at least kk edges of i\mathcal{E}_{i} appearing before the sths^{th} element of 1\mathcal{E}_{1}. Excluding possible edges to 1s\mathcal{R}^{\leq s}_{1}, ti2kt_{i}\in\mathcal{R}^{\leq k}_{2}, and {1,2,i}\{1,2,i\} implies that there must be at least ks4k-s-4 edges of i\mathcal{E^{\prime}}_{i} before the sths^{th} element of 1\mathcal{E}_{1}. As a result, in the first (s1)(ks4)(s-1)(k-s-4) elements of this permutation there are at most s1s-1 elements of type 0. Lemma 3.3 bounds the probability of the event above by

{I1|(B1;2B1 2B12)B12c}exp(k+12klog(k)+O(k)).\mathbb{P}\{I_{1}|(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})\cap B_{1\rightsquigarrow 2}^{c}\}\leq\exp\left(-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right).

We also have

{(B1;2B1 2B12)|B12c}\displaystyle\mathbb{P}\{(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})|B_{1\rightsquigarrow 2}^{c}\}\leq {B1;2B1 2B12}{B12c}\displaystyle\frac{\mathbb{P}\{B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2}\}}{\mathbb{P}\{B_{1\rightsquigarrow 2}^{c}\}}
\displaystyle\leq O(k3/n).\displaystyle O(k^{3}/n).

And finally,

{I1(B1;2B1 2B12)|B12c}=\displaystyle\mathbb{P}\{I_{1}\cap(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})|B_{1\rightsquigarrow 2}^{c}\}= {I1|(B1;2B1 2B12)B12c}\displaystyle\mathbb{P}\{I_{1}|(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})\cap B_{1\rightsquigarrow 2}^{c}\}
{(B1;2B1 2B12)|B12c}\displaystyle\mathbb{P}\{(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})|B_{1\rightsquigarrow 2}^{c}\}
\displaystyle\leq exp(log(n)k+12klog(k)+O(k)),\displaystyle\exp\left(-\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right),

which concludes the proof. ∎

Lemma 4.10.

The following holds

{I1I2B}exp(log(n)k+12klog(k)+O(k))\mathbb{P}\{I_{1}I_{2}\cap B\}\leq\exp\left(-\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right)
Proof.
{I1I2B}\displaystyle\mathbb{P}\{I_{1}I_{2}\cap B\} {I1I2(B1;2B1 2B12)}+{I1I2(B2;1B1 2B21)}\displaystyle\leq\mathbb{P}\{I_{1}I_{2}\cap(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})\}+\mathbb{P}\{I_{1}I_{2}\cap(B_{2;1}\cup B_{1\cup\,2}\cup B_{2\rightarrow 1})\}
{I1(B1;2B1 2B12)}+{I2(B2;1B1 2B21)}\displaystyle\leq\mathbb{P}\{I_{1}\cap(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})\}+\mathbb{P}\{I_{2}\cap(B_{2;1}\cup B_{1\cup\,2}\cup B_{2\rightarrow 1})\}
{I1(B1;2B1 2B12)B12}\displaystyle\leq\mathbb{P}\{I_{1}\cap(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})\cap B_{1\rightsquigarrow 2}\}
+{I1(B1;2B1 2B12)B12c}\displaystyle\quad+\mathbb{P}\{I_{1}\cap(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})\cap B_{1\rightsquigarrow 2}^{c}\}
+{I2(B2;1B1 2B21)B21}\displaystyle\quad+\mathbb{P}\{I_{2}\cap(B_{2;1}\cup B_{1\cup\,2}\cup B_{2\rightarrow 1})\cap B_{2\rightsquigarrow 1}\}
+{I2(B2;1B1 2B21)B21c}\displaystyle\quad+\mathbb{P}\{I_{2}\cap(B_{2;1}\cup B_{1\cup\,2}\cup B_{2\rightarrow 1})\cap B_{2\rightsquigarrow 1}^{c}\}
(B12B21)++{I1(B1;2B1 2B12)|B12c}\displaystyle\leq\mathbb{P}(B_{1\rightsquigarrow 2}\cup B_{2\rightsquigarrow 1})++\mathbb{P}\{I_{1}\cap(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})|B_{1\rightsquigarrow 2}^{c}\}
+{I2(B2;1B1 2B21)|B21c}\displaystyle\quad+\mathbb{P}\{I_{2}\cap(B_{2;1}\cup B_{1\cup\,2}\cup B_{2\rightarrow 1})|B_{2\rightsquigarrow 1}^{c}\}
O(k5/n2)+{I1(B1;2B1 2B12)|B12c}(Lemma4.8)\displaystyle\leq O(k^{5}/n^{2})+\mathbb{P}\{I_{1}\cap(B_{1;2}\cup B_{1\cup\,2}\cup B_{1\rightarrow 2})|B_{1\rightsquigarrow 2}^{c}\}\quad(\text{Lemma}\ref{bound-B4})
+{I2(B2;1B1 2B21)|B21c}\displaystyle\quad+\mathbb{P}\{I_{2}\cap(B_{2;1}\cup B_{1\cup\,2}\cup B_{2\rightarrow 1})|B_{2\rightsquigarrow 1}^{c}\}
O(k5/n2)+2exp(log(n)k+12klog(k)+O(k))(Lemma 4.9)\displaystyle\leq O(k^{5}/n^{2})+2\exp\left(-\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right)\quad(\text{Lemma \ref{bound-IB123}})
=exp(log(n)k+12klog(k)+O(k)),\displaystyle=\exp\left(-\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right),

which establishes the bound. ∎

Lemma 4.11.

We have the following conditional independence

{I1I2|Bc}={I1|Bc}{I2|Bc}.\mathbb{P}\{I_{1}I_{2}|B^{c}\}=\mathbb{P}\{I_{1}|B^{c}\}\mathbb{P}\{I_{2}|B^{c}\}.
Proof.

Note that I1I_{1} holds whenever for each 1ik1\leq i\leq k there are at least kk elements j[n]{1,R1i}j\in[n]\setminus\{1,R_{1}^{i}\} such that V(R1i,1)>V(R1i,j)V(R_{1}^{i},1)>V(R_{1}^{i},j). Assuming BcB^{c}, one can see this is equivalent to existing at least kk elements j1[n]{1,2}{R21,,R2k}j_{1}\in[n]\setminus\{1,2\}\setminus\{R_{2}^{1},\cdots,R_{2}^{k}\} with this property. Repeating this argument for I2I_{2}, there must be at least kk elements j2[n]{1,2}{R11,R12,,R1k}j_{2}\in[n]\setminus\{1,2\}\setminus\{R_{1}^{1},R_{1}^{2},\cdots,R_{1}^{k}\} with V(R2i,2)>V(R2i,j)V(R_{2}^{i},2)>V(R_{2}^{i},j) for all 1jk1\leq j\leq k. Ruling out the possibility of choosing j2,j1j_{2},j_{1} from {1,2}\{1,2\}, {R11,R12,,R1k}\{R_{1}^{1},R_{1}^{2},\cdots,R_{1}^{k}\}, and {R21,,R2k}\{R_{2}^{1},\cdots,R_{2}^{k}\} implies that there is no common edge among (R1i,1),(R1i,j1),(R2i,,2),(R2i,j2)(R_{1}^{i},1),(R_{1}^{i},j_{1}),(R_{2}^{i},,2),(R_{2}^{i},j_{2}). Since the scores are chosen independently, and the inequality conditions for the isolation of vertex 1 and those of vertex 2 are disjoint, we obtain the conditional independence. ∎

Lemma 4.12.

Let t<1/2t^{\prime}<-1/2 be a real number. Assume kk grows to infinity as nn\rightarrow\infty with klog(n)+tloglog(n)log(n)k\leq\log(n)+t^{\prime}\log\log(n)\sqrt{\log(n)}. Then we have

lim supn𝔼{I1I2}𝔼{I1}21.\limsup_{n\rightarrow\infty}\frac{\mathbb{E}\{I_{1}I_{2}\}}{\mathbb{E}\{I_{1}\}^{2}}\leq 1.
Proof.

We begin with this equality

{I1}{Bc}\displaystyle\frac{\mathbb{P}\{I_{1}\}}{\mathbb{P}\{B^{c}\}} ={I1Bc}{Bc}+{I1B}{Bc}\displaystyle=\frac{\mathbb{P}\{I_{1}\cap B^{c}\}}{\mathbb{P}\{B^{c}\}}+\frac{\mathbb{P}\{I_{1}\cap B\}}{\mathbb{P}\{B^{c}\}} (6)

Hence,

1{Bc}\displaystyle\frac{1}{\mathbb{P}\{B^{c}\}} ={I1Bc}{I1}{Bc}+{I1B}{I1}{Bc}\displaystyle=\frac{\mathbb{P}\{I_{1}\cap B^{c}\}}{\mathbb{P}\{I_{1}\}\mathbb{P}\{B^{c}\}}+\frac{\mathbb{P}\{I_{1}\cap B\}}{\mathbb{P}\{I_{1}\}\mathbb{P}\{B^{c}\}}
={I1|Bc}{I1}+{I1B}{I1}{Bc}.\displaystyle=\frac{\mathbb{P}\{I_{1}|B^{c}\}}{\mathbb{P}\{I_{1}\}}+\frac{\mathbb{P}\{I_{1}\cap B\}}{\mathbb{P}\{I_{1}\}\mathbb{P}\{B^{c}\}}. (7)

Lemmas 4.5 and 4.7 imply

{I1B}{I1}{Bc}\displaystyle\frac{\mathbb{P}\{I_{1}\cap B\}}{\mathbb{P}\{I_{1}\}\mathbb{P}\{B^{c}\}} O(k3n)O(ek(1+6log(k)k))\displaystyle\leq O(\frac{k^{3}}{n})O(e^{k(1+\sqrt{\frac{6\log(k)}{k}})})
=O(exp(k+6klog(k)+3log(k)log(n)))\displaystyle=O(\exp(k+\sqrt{6k\log(k)}+3\log(k)-\log(n)))
O(exp(tloglog(n)log(n)+6klog(k)+3log(k)))\displaystyle\leq O(\exp(t^{\prime}\log\log(n)\sqrt{\log(n)}+\sqrt{6k\log(k)}+3\log(k)))
O(exp(tlog(k)k+6klog(k)+3log(k)))\displaystyle\leq O(\exp(t^{\prime}\log(k)\sqrt{k}+\sqrt{6k\log(k)}+3\log(k))) (8)
k0,\displaystyle\rightarrow_{k\rightarrow\infty}0,

where we use the fact that klog(n)k\leq\log(n). Considering the fact 1{Bc}1\frac{1}{\mathbb{P}\{B^{c}\}}\rightarrow 1, equation (7) implies that

{I1|Bc}{I1}1.\frac{\mathbb{P}\{I_{1}|B^{c}\}}{\mathbb{P}\{I_{1}\}}\rightarrow 1. (9)

It follows from Lemmas 4.10, 4.11 that

{I1I2}=\displaystyle\mathbb{P}\{I_{1}I_{2}\}= {I1I2Bc}+{I1I2B}\displaystyle\mathbb{P}\{I_{1}I_{2}\cap B^{c}\}+\mathbb{P}\{I_{1}I_{2}\cap B\}
\displaystyle\leq {I1I2Bc}+exp(log(n)k+12klog(k)+O(k))\displaystyle\mathbb{P}\{I_{1}I_{2}\cap B^{c}\}+\exp\left(-\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right)
\displaystyle\leq {I1I2|Bc}+exp(log(n)k+12klog(k)+O(k))\displaystyle\mathbb{P}\{I_{1}I_{2}|B^{c}\}+\exp\left(-\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right)
=\displaystyle= {I1|Bc}{I2|Bc}+exp(log(n)k+12klog(k)+O(k)).\displaystyle\mathbb{P}\{I_{1}|B^{c}\}\mathbb{P}\{I_{2}|B^{c}\}+\exp\left(-\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right). (10)

Dividing both sides of (10) by {I1}{I2}={I1}2\mathbb{P}\{I_{1}\}\mathbb{P}\{I_{2}\}=\mathbb{P}\{I_{1}\}^{2} and using Lemma 4.5, we have

{I1I2}{I1}{I2}\displaystyle\frac{\mathbb{P}\{I_{1}I_{2}\}}{\mathbb{P}\{I_{1}\}\mathbb{P}\{I_{2}\}}
{I1|Bc}{I2|Bc}{I1}{I2}+exp(log(n)k+12klog(k)+O(k)){I1}{I2}\displaystyle\leq\frac{\mathbb{P}\{I_{1}|B^{c}\}\mathbb{P}\{I_{2}|B^{c}\}}{\mathbb{P}\{I_{1}\}\mathbb{P}\{I_{2}\}}+\frac{\exp\left(-\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right)}{{\mathbb{P}\{I_{1}\}\mathbb{P}\{I_{2}\}}}
{I1|Bc}{I2|Bc}{I1}{I2}+exp(log(n)k+12klog(k)+O(k))exp(2k(1+6log(k)k))\displaystyle\leq\frac{\mathbb{P}\{I_{1}|B^{c}\}\mathbb{P}\{I_{2}|B^{c}\}}{\mathbb{P}\{I_{1}\}\mathbb{P}\{I_{2}\}}+\frac{\exp\left(-\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right)}{\exp\left(-2k(1+\sqrt{\frac{6\log(k)}{k}})\right)}
{I1|Bc}{I2|Bc}{I1}{I2}+exp(log(n)+k+12klog(k)+24klog(k)+O(k))\displaystyle\leq\frac{\mathbb{P}\{I_{1}|B^{c}\}\mathbb{P}\{I_{2}|B^{c}\}}{\mathbb{P}\{I_{1}\}\mathbb{P}\{I_{2}\}}+\exp\left(-\log(n)+k+\frac{1}{2}\sqrt{k}\log(k)+\sqrt{24k\log(k)}+O(\sqrt{k})\right)
{I1|Bc}{I2|Bc}{I1}{I2}\displaystyle\leq\frac{\mathbb{P}\{I_{1}|B^{c}\}\mathbb{P}\{I_{2}|B^{c}\}}{\mathbb{P}\{I_{1}\}\mathbb{P}\{I_{2}\}}
+exp(tloglog(n)log(n)+12klog(k)+24klog(k)+O(k))\displaystyle\quad+\exp\left(t^{\prime}\log\log(n)\sqrt{\log(n)}+\frac{1}{2}\sqrt{k}\log(k)+\sqrt{24k\log(k)}+O(\sqrt{k})\right)
{I1|Bc}{I2|Bc}{I1}{I2}+exp((12+t)log(k)k+24klog(k)+O(k)).\displaystyle\leq\frac{\mathbb{P}\{I_{1}|B^{c}\}\mathbb{P}\{I_{2}|B^{c}\}}{\mathbb{P}\{I_{1}\}\mathbb{P}\{I_{2}\}}+\exp\left((\frac{1}{2}+t^{\prime})\log(k)\sqrt{k}+\sqrt{24k\log(k)}+O(\sqrt{k})\right). (11)

Equation (9) guarantees the first term in right hand side converges to 1. The second term vanishes as kk\rightarrow\infty as a result of t<0.5t^{\prime}<-0.5. This completes the proof. ∎

Theorem 4.1.

The random graph model 𝔾(n,k)\mathbb{G}(n,k) with klog(n)+tloglog(n)log(n)k\leq\log(n)+t^{\prime}\log\log(n)\sqrt{\log(n)} for t<12t^{\prime}<-\frac{1}{2} is not connected with high probability. In particular, if k=tlog(n)k=\lfloor t\log(n)\rfloor for 0<t<10<t<1, the graph is disconnected with high probability.

Proof.

Having both Lemmas 4.6 and 4.12 one can apply the second moment method. We begin with

𝔼{(i=1nIi)2}𝔼{i=1nIi}2\displaystyle\mathbb{E}\left\{\left(\sum_{i=1}^{n}I_{i}\right)^{2}\right\}-\mathbb{E}\left\{\sum_{i=1}^{n}I_{i}\right\}^{2} =Var{i=1nIi}\displaystyle=\mathrm{Var}\left\{\sum_{i=1}^{n}I_{i}\right\}
(0𝔼{i=1nIi})2{i=1nIi=0}.\displaystyle\geq\left(0-\mathbb{E}\left\{\sum_{i=1}^{n}I_{i}\right\}\right)^{2}\mathbb{P}\left\{\sum_{i=1}^{n}I_{i}=0\right\}. (12)

Therefore,

{i=1nIi=0}\displaystyle\mathbb{P}\left\{\sum_{i=1}^{n}I_{i}=0\right\} 𝔼{(i=1nIi)2}𝔼{i=1nIi}2𝔼{i=1nIi}2\displaystyle\leq\frac{\mathbb{E}\{(\sum_{i=1}^{n}I_{i})^{2}\}-\mathbb{E}\{\sum_{i=1}^{n}I_{i}\}^{2}}{\mathbb{E}\{\sum_{i=1}^{n}I_{i}\}^{2}}
𝔼{(i=1nIi)2}𝔼{i=1nIi}21\displaystyle\leq\frac{\mathbb{E}\{(\sum_{i=1}^{n}I_{i})^{2}\}}{\mathbb{E}\{\sum_{i=1}^{n}I_{i}\}^{2}}-1
=i=1n𝔼{Ii2}+i,j,ij𝔼{IiIj}n2𝔼{I1}21\displaystyle=\frac{\sum_{i=1}^{n}\mathbb{E}\{I_{i}^{2}\}+\sum_{i,j,i\neq j}\mathbb{E}\{I_{i}I_{j}\}}{n^{2}\mathbb{E}\{I_{1}\}^{2}}-1
=n𝔼{I1}+n(n1)𝔼{I1I2}n2𝔼{I1}21\displaystyle=\frac{n\mathbb{E}\{I_{1}\}+n(n-1)\mathbb{E}\{I_{1}I_{2}\}}{n^{2}\mathbb{E}\{I_{1}\}^{2}}-1
=1n𝔼{I1}+(n1)𝔼{I1I2}n𝔼{I1}21.\displaystyle=\frac{1}{n\mathbb{E}\{I_{1}\}}+\frac{(n-1)\mathbb{E}\{I_{1}I_{2}\}}{n\mathbb{E}\{I_{1}\}^{2}}-1. (13)

Now it follows from Lemma 4.6 that the first fraction vanishes and from Lemma 4.12 that the second fraction will be sufficiently close to 1 (or less than 1) for large nn. Hence, the probability of having no isolated vertex converges to 0. ∎

5 Connectivity for t>1t>1

A common method to prove connectivity of random graphs is by locating (with high-probability) a Erdős-Rényi sub-graph consisting of all the vertices. Then, the celebrated result of Erdős and Rényi [6] guarantees connectivity if the connectivity probability is p=tlog(n)np=\frac{t\log(n)}{n} where t>1t>1. This method used in [9] proves the connectivity of 𝔾(n,k)\mathbb{G}(n,k) for k=tlog(n)k=t\log(n) where t>C=2.4625t>C=2.4625. The connection between Erdős-Rényi and 𝔾(n,k)\mathbb{G}(n,k) is made through a concentration property of order statistics. The authors of [9] assume that the edges are independently scored by exp(1)\exp(1) random variables which results in having a connection for all edges of scores greater than equal log((n1)/(tlog(n)))+2/t\log((n-1)/(t\log(n)))+\sqrt{2/t} with high probability. Therefore, if Li,jL_{i,j} is a random variable of law exp(1)\exp(1), then

{Li,j>log(n1tlog(n))+2t}=tlog(n)n1e2t.\mathbb{P}\left\{L_{i,j}>\log\left(\frac{n-1}{t\log(n)}\right)+\sqrt{\frac{2}{t}}\right\}=\frac{t\log(n)}{n-1}e^{-\sqrt{\frac{2}{t}}}.

It can be easily shown that te2t>1te^{-\sqrt{\frac{2}{t}}}>1 for t>2.4625t>2.4625. This provides an independent possibility of connection for all edges with the probability of tlog(n)/(n1)t^{\prime}\log(n)/(n-1) for a t=te2t>1t^{\prime}=te^{-\sqrt{\frac{2}{t}}}>1. Thereafter, the connectivity result of Erdős-Rényi random graphs [6, 4, 12] implies the connectivity of 𝔾(n,k)\mathbb{G}(n,k).

We prove the connectivity of 𝔾(n,k)\mathbb{G}(n,k) for all t>1t>1 in two steps. First, we rule out the possibility of having components of size O(1)O(1). Second, we apply the idea from [9] described above to find an Erdős-Rényi graph with all vertices contained in 𝔾(n,k)\mathbb{G}(n,k). As opposed to [9], we do not restrict tt to find a t>1t^{\prime}>1. Hence, the resulting Erdős-Rényi graph is not necessarily connected. However, by modifying the original proof of the connectivity of Erdős-Rényi graphs for t>1t^{\prime}>1, we can prove a weaker result for an arbitrary t<1t^{\prime}<1 which then helps us deduce the connectivity result.

Theorem 5.1.

Let κ0\kappa\geq 0 be a non-negative integer. Consider random graphs of class 𝔾(n,k)\mathbb{G}(n,k) where klog(n)+tloglog(n)log(n)k\geq\log(n)+t^{\prime}\log\log(n)\sqrt{\log(n)} for a real number t>12t^{\prime}>\frac{1}{2} . Then

{ a vertex of degree less than or equal κ}0\mathbb{P}\{\exists\textrm{ a vertex of degree less than or equal }\kappa\}\rightarrow 0

as nn\rightarrow\infty.

Proof.

We can assume vertex 11 has degree less than κ\kappa. We will use the same technique as in Lemma 4.9. Let s=kks=\lfloor\sqrt{k}\rfloor\leq k. For i1si\in\mathcal{R}^{\leq s}_{1} define i=i{{i,j}:j1s}\mathcal{E^{\prime}}_{i}=\mathcal{E}_{i}\setminus\{\{i,j\}:j\in\mathcal{R}^{\leq s}_{1}\}. Therefore, if we consider the following union

=1i1si,\mathcal{E}^{\prime}=\mathcal{E}_{1}\bigcup_{i\in\mathcal{R}^{\leq s}_{1}}\mathcal{E^{\prime}}_{i},

it will be a partition in which |1|=n1|\mathcal{E}_{1}|=n-1, and |i|=ns1|\mathcal{E^{\prime}}_{i}|=n-s-1 for all i1si\in\mathcal{R}^{\leq s}_{1}. Define 1\mathcal{E}_{1} objects of type 0 and i\mathcal{E^{\prime}}_{i} objects of type ii for i1si\in\mathcal{R}_{1}^{\leq s}. One can easily see that each realization of 𝔾(n,k)\mathbb{G}(n,k) induces a uniformly random permutation over \mathcal{E}^{\prime} which has only a fixed order on the first ss or s+1s+1 elements of 1\mathcal{E}_{1}, i.e., type 0.

A necessary condition for vertex 1 to be of degree at most κ\kappa is that for at least sκs-\kappa elements i1si\in\mathcal{R}^{\leq s}_{1}, there should be at least kk edges of i\mathcal{E}_{i} appearing before the sths^{th} element of 1\mathcal{E}_{1}. This implies there must be at least ks1k-s-1 edges of i\mathcal{E^{\prime}}_{i} before the sths^{th} element of 1\mathcal{E^{\prime}}_{1}. As a result, in the first (sκ)(ks1)(s-\kappa)(k-s-1) elements of this permutation there are at most s1s-1 elements of type 0. Lemma 3.3 bounds the probability of the event above by

exp(k+12klog(k)+O(k))\exp\left(-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right)

Then, using the union bound for sufficiently large nn we have:

{v:deg(v)κ}\displaystyle\mathbb{P}\{\exists v:\deg(v)\leq\kappa\} n×{vertex 1 is of degree at most κ}\displaystyle\leq n\times\mathbb{P}\{\textrm{vertex $1$ is of degree at most }\kappa\}
nexp(k+12klog(k)+O(k))\displaystyle\leq n\exp\left(-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right)
=exp(log(n)k+12klog(k)+O(k))\displaystyle=\exp\left(\log(n)-k+\frac{1}{2}\sqrt{k}\log(k)+O(\sqrt{k})\right)

If k>O(log(n))k>O(\log(n)), then the right hand side obviously vanishes. In case of k=O(log(n))k=O(\log(n)), we use the fact that k12klog(k)k-\frac{1}{2}\sqrt{k}\log(k) is increasing for sufficiently large kk, and replace kk by log(n)+tloglog(n)log(n)\log(n)+t^{\prime}\log\log(n)\sqrt{\log(n)} to obtain

exp((12t)loglog(n)log(n)+O(log(n)))0.\displaystyle\exp\left((\frac{1}{2}-t^{\prime})\log\log(n)\sqrt{\log(n)}+O(\sqrt{\log(n)})\right)\rightarrow 0.

Hence, the result holds. ∎

Theorem 5.1 excludes the possibility of having components of size O(1)O(1). In other words, the following corollary holds.

Corollary 5.1.

Consider the random graphs model of class 𝔾(n,k)\mathbb{G}(n,k) with klog(n)+tloglog(n)log(n)k\geq\log(n)+t^{\prime}\log\log(n)\sqrt{\log(n)} for a real number t>12t^{\prime}>\frac{1}{2}. Then

{ a component of size at most κ}0\mathbb{P}\{\exists\textrm{ a component of size at most }\kappa\}\rightarrow 0

for any fixed κ\kappa as nn\rightarrow\infty. In particular when k=tlog(n)k=\lfloor t\log(n)\rfloor for t>1t>1 the limit above is still true.

Proof.

In case of having a component of size κ\kappa there most be a vertex of degree at most κ\kappa which is impossible with high probability due to Theorem 5.1. ∎

Next, we establish a proof based on finding an Erdős-Rényi sub-graph and super-graph of 𝔾(n,k)\mathbb{G}(n,k) with both containing all the vertices. Recall that the order statistics of a set of i.i.d. random variables {X1,X2,,Xn}\{X_{1},X_{2},\cdots,X_{n}\} is defined as their non-increasing rearrangement X(1)X(2)X(n)X^{(1)}\geq X^{(2)}\geq\cdots\geq X^{(n)}. Recall that the kthk^{th} order statistic of 𝒱i\mathcal{V}_{i} is denoted by V(i,Rik)V(i,R_{i}^{k}).

Theorem 5.2.

For the random graph model 𝔾(n,k)\mathbb{G}(n,k) with k=tlog(n)k=\lfloor t\log(n)\rfloor there is no component of size 8.24trn/2\lceil\frac{8.24}{t}\rceil\leq r\leq\lfloor n/2\rfloor with high probability.

Proof.

First, we note Lemma A.1 in [9] which states that if we choose scores V(i,j)V(i,j) independently distributed as exponential random variables with parameter 11, and define:

An={ 1in:V(i,Rik)(log(n1tlog(n))2,log(n1tlog(n))+2)},A_{n}=\left\{\forall\;1\leq i\leq n:\;V(i,R_{i}^{k})\in\left(\log\left(\frac{n-1}{t\log(n)}\right)-\sqrt{2},\log\left(\frac{n-1}{t\log(n)}\right)+\sqrt{2}\right)\right\},

then {An}1\mathbb{P}\{A_{n}\}\rightarrow 1 as nn\rightarrow\infty. We let l¯=log(n1tlog(n))2\underline{l}=\log(\frac{n-1}{t\log(n)})-\sqrt{2} and l¯=log(n1tlog(n))+2\bar{l}=\log(\frac{n-1}{t\log(n)})+\sqrt{2}. Next, note that

p¯\displaystyle\bar{p} :={V(i,j)>l¯}=tlog(n)n1e20.2431tlog(n)n1,\displaystyle:=\mathbb{P}\{V(i,j)>\bar{l}\}=\frac{t\log(n)}{n-1}e^{-\sqrt{2}}\approx\frac{0.2431t\log(n)}{n-1},
p¯\displaystyle\underline{p} :={V(i,j)>l¯}=tlog(n)n1e24.1133tlog(n)n1.\displaystyle:=\mathbb{P}\{V(i,j)>\underline{l}\}=\frac{t\log(n)}{n-1}e^{\sqrt{2}}\approx\frac{4.1133t\log(n)}{n-1}.

Moreover, if the graph satisfies AnA_{n} condition (which holds whp), then V(i,j)>l¯V(i,j)>\bar{l} implies that the edge {i,j}\{i,j\} exists and V(i,j)<l¯V(i,j)<\underline{l} implies that the edge {i,j}\{i,j\} does not exist.

Let the graph satisfy AnA_{n}. In order to have a component of size rr, one must choose rr vertices. The connectivity within this component implies that it should contains at least a tree. Moreover, according to Cayley’s formula there are rr2r^{r-2} trees with rr vertices. Hence, the r1r-1 edges of the tree must have scores not less than l¯\underline{l} which happens with probability p¯r1\underline{p}^{r-1} due to independence. On the other hand, we require those rr vertices not to be connected to the rest of graph. This implies the scores for the intermediate edges between the component and the rest of graph must be less than l¯\bar{l}. Note that both of these are necessary conditions. Counting the number of possible components yields the following:

Π\displaystyle\Pi ={ a component of size between 8.24t and n/2}\displaystyle=\mathbb{P}\{\exists\textrm{ a component of size between }\lceil\frac{8.24}{t}\rceil\textrm{ and }\lfloor n/2\rfloor\}
r=8.24tn/2(nr)rr2p¯r1(1p¯)r(nr).\displaystyle\leq\sum_{r=\lceil\frac{8.24}{t}\rceil}^{\lfloor n/2\rfloor}\binom{n}{r}r^{r-2}\underline{p}^{r-1}(1-\bar{p})^{r(n-r)}.

Substituting (nr)<nrr!\binom{n}{r}<\frac{n^{r}}{r!}, r!2πr(r/e)rr!\approx\sqrt{2\pi r}(r/e)^{r} implies

Π\displaystyle\Pi <12πr=8.24tn/2nrrr1/2errr2p¯r1(1p¯)r(nr)\displaystyle<\frac{1}{\sqrt{2\pi}}\sum_{r=\lceil\frac{8.24}{t}\rceil}^{\lfloor n/2\rfloor}n^{r}r^{-r-1/2}e^{r}r^{r-2}\underline{p}^{r-1}(1-\bar{p})^{r(n-r)}
<12πr=8.24tn/2ernrp¯r5/2p¯rer(nr)p¯(using 1x<ex)\displaystyle<\frac{1}{\sqrt{2\pi}}\sum_{r=\lceil\frac{8.24}{t}\rceil}^{\lfloor n/2\rfloor}e^{r}\frac{n^{r}}{\underline{p}}r^{-5/2}\underline{p}^{r}e^{-r(n-r)\bar{p}}\quad(\textrm{using}\;1-x<e^{-x})
<n2πr=8.24tn/2nrerp¯rerp¯n/2(using 1/p¯<n,nr>n/2,r5/2<1)\displaystyle<\frac{n}{\sqrt{2\pi}}\sum_{r=\lceil\frac{8.24}{t}\rceil}^{\lfloor n/2\rfloor}n^{r}e^{r}\underline{p}^{r}e^{-r\bar{p}n/2}\quad(\textrm{using}\;1/\underline{p}<n,\;n-r>n/2,\;r^{-5/2}<1)
<n2πr=8.24tn/2er(1+log(np¯)np¯/2).\displaystyle<\frac{n}{\sqrt{2\pi}}\sum_{r=\lceil\frac{8.24}{t}\rceil}^{\lfloor n/2\rfloor}e^{r(1+\log(n\underline{p})-n\bar{p}/2)}.

As tt is a fixed number, the dominant term in 1+log(np¯)np¯/21+\log(n\underline{p})-n\bar{p}/2 is np¯/20.1216tlog(n)-n\bar{p}/2\approx-0.1216t\log(n). As a result, for sufficiently large nn, 1+log(np¯)np¯/2<0.1215tlog(n)1+\log(n\underline{p})-n\bar{p}/2<-0.1215t\log(n). Thus, we have

Π\displaystyle\Pi <n2πr=8.24te0.1215trlog(n)\displaystyle<\frac{n}{\sqrt{2\pi}}\sum_{r=\lceil\frac{8.24}{t}\rceil}^{\infty}e^{-0.1215tr\log(n)}
=n2πr=8.24tn0.1215tr\displaystyle=\frac{n}{\sqrt{2\pi}}\sum_{r=\lceil\frac{8.24}{t}\rceil}^{\infty}n^{-0.1215tr}
=n10.1215t8.24t2π(1n0.1215t)\displaystyle=\frac{n^{1-0.1215t\lceil\frac{8.24}{t}\rceil}}{\sqrt{2\pi}(1-n^{-0.1215t})}
n0.001162π(1n0.1215t)n0.\displaystyle\leq\frac{n^{-0.00116}}{\sqrt{2\pi}(1-n^{-0.1215t})}\rightarrow_{n\rightarrow\infty}0.

Therefore, with high probability there is no component of size between 10 and n/2n/2. ∎

The interesting fact about Theorem 5.2 is that it holds for any positive tt. This means that if ktlog(n)=Θ(log(n))k\approx t\log(n)=\Theta(\log(n)), one should focus attention on components of size less than 8.24t=O(1)\lceil\frac{8.24}{t}\rceil=O(1) to study the disconnectivity. Since Corollary 5.1 addresses components of size O(1)O(1) and we conclude the main theorem.

Theorem 5.3.

The random graph model 𝔾(n,k)\mathbb{G}(n,k) with klog(n)+tloglog(n)log(n)k\geq\log(n)+t^{\prime}\log\log(n)\sqrt{\log(n)} for a real number t>12t^{\prime}>\frac{1}{2} is connected with high probability. In particular, if k=tlog(n)k=\lfloor t\log(n)\rfloor for t>1t>1, connectivity holds with high probability.

Proof.

If a graph from class 𝔾(n,k)\mathbb{G}(n,k) is disconnected, then it should have at least a component of size ss where 1sn/21\leq s\leq\lfloor n/2\rfloor. Since 8.241=9\lceil\frac{8.24}{1}\rceil=9, Theorem 5.2 rules out the possibility of having components of size between 9 and n/2\lfloor n/2\rfloor with high probability. Theorem 5.1 guarantees that the probability of having components of size at most 88 vanishes, when nn grows to infinity. Therefore, the graph will be connected with high probability. ∎

6 Average Degree

Here, we discuss another set of results for 𝔾(n,k)\mathbb{G}(n,k) random graphs. The degree sequence is bounded above by kk. Therefore, average degree is also bounded by kk. However, we will show that this number is very close to kk. We also specify the error term. Finally we compare our result to the sparse case in [10]. Before presenting the main theorem we recall the negative binomial distribution. Consider an urn having infinite number of red and blue balls and each time we draw red with probability pp. The probability of having jj blue balls appeared before the kthk^{th} red ball is a negative binomial distribution given by the following formula {Xk=j}=(k+j1j)pk(1p)j\mathbb{P}\{X_{k}=j\}=\binom{k+j-1}{j}p^{k}(1-p)^{j}. The theorems below provide an approximation for average degree.

Lemma 6.1.
j=0k1(k+j1j)12k+j=12\sum_{j=0}^{k-1}\binom{k+j-1}{j}\frac{1}{2^{k+j}}=\frac{1}{2}
Proof.

Let XkX_{k} be a negative binomial random variable with p=12p=\frac{1}{2}. Note that {Xk<k}=j=0k1(k+j1j)12k+j\mathbb{P}\{X_{k}<k\}=\sum_{j=0}^{k-1}\binom{k+j-1}{j}\frac{1}{2^{k+j}} is the probability of observing the kthk^{th} red ball earlier than the kthk^{th} blue ball in the infinite urn model. By symmetry this value must be 1/21/2. ∎

Lemma 6.2.
j=0k1(k+j1j)(j2k+j)=k2(2k1)22k1(2k2k1)\sum_{j=0}^{k-1}\binom{k+j-1}{j}\left(\frac{j}{2^{k+j}}\right)=\frac{k}{2}-\frac{(2k-1)}{2^{2k-1}}\binom{2k-2}{k-1}
Proof.

Using Lemma 6.1, one can write the sum above as follows

A\displaystyle A =j=0k1(k+j1j)(j2k+j)\displaystyle=\sum_{j=0}^{k-1}\binom{k+j-1}{j}\left(\frac{j}{2^{k+j}}\right)
=j=1k1(k+j2j1)(k+j12k+j)\displaystyle=\sum_{j=1}^{k-1}\binom{k+j-2}{j-1}\left(\frac{k+j-1}{2^{k+j}}\right)
=(k2)j=1k1(k+j2j1)12j+k1+(12)j=1k1(j1)(k+j2j1)12j+k1\displaystyle=\left(\frac{k}{2}\right)\sum_{j=1}^{k-1}\binom{k+j-2}{j-1}\frac{1}{2^{j+k-1}}+\left(\frac{1}{2}\right)\sum_{j=1}^{k-1}(j-1)\binom{k+j-2}{j-1}\frac{1}{2^{j+k-1}}
=(k2)[12(2k2k1)122k1]+(12)[A(k1)(2k2k1)122k1]\displaystyle=\left(\frac{k}{2}\right)\left[\frac{1}{2}-\binom{2k-2}{k-1}\frac{1}{2^{2k-1}}\right]+\left(\frac{1}{2}\right)\left[A-(k-1)\binom{2k-2}{k-1}\frac{1}{2^{2k-1}}\right]
=k4+A2(2k1)22k2(2k2k1).\displaystyle=\frac{k}{4}+\frac{A}{2}-\frac{(2k-1)}{2^{2k-2}}\binom{2k-2}{k-1}.

We now solve the resulting equation for AA to get the desired expression. ∎

Theorem 6.1.

In the model 𝔾(n,k)\mathbb{G}(n,k) suppose DD represents the degree of a randomly chosen vertex. If k=o(n)k=o(\sqrt{n}) then we have the following asymptotic

𝔼{D}=k[(2k1)22k1(2k2k1)](O(k2/n)+1)\mathbb{E}\{D\}=k-\left[\frac{(2k-1)}{2^{2k-1}}\binom{2k-2}{k-1}\right](O(k^{2}/n)+1) (14)
Proof.

Without loss of generality assume the randomly chosen vertex is vertex 1 and R1i=i+1R_{1}^{i}=i+1 for all 1in11\leq i\leq n-1. Denote the indicator function of vertex 1 being connected to vertex i+1i+1 by EiE_{i}. In order for vertex 1 not to be connected to vertex i+1i+1 there must be at least kk elements of i+1\mathcal{E}_{i+1} before the ithi^{th} element of 1\mathcal{E}_{1} which is {1,i+1}\{1,i+1\}. Hence, there could be 0,1,,i10,1,\cdots,i-1 elements of 1\mathcal{E}_{1} before the kthk^{th} element of i+1\mathcal{E}_{i+1}. With the same idea used in Lemma 3.1 one can prove that the probability of jj elements of 1\mathcal{E}_{1} appear before the kthk^{th} element of i+1\mathcal{E}_{i+1} for jkj\leq k is:

(k+j1j)(n+O(k)2n+O(k))k+j=(k+j1j)(12)k+j(1+O(k2/n))\binom{k+j-1}{j}\left(\frac{n+O(k)}{2n+O(k)}\right)^{k+j}=\binom{k+j-1}{j}\left(\frac{1}{2}\right)^{k+j}(1+O(k^{2}/n))

Varying 0ji10\leq j\leq i-1 we conclude that for each 1ik1\leq i\leq k one has

{Ei=0}=[j=0i1(k+j1j)12k+j](1+O(k2/n))\mathbb{P}\{E_{i}=0\}=\left[\sum_{j=0}^{i-1}\binom{k+j-1}{j}\frac{1}{2^{k+j}}\right](1+O(k^{2}/n))

We now compute 𝔼{D}\mathbb{E}\{D\} using Lemmas 6.1 and 6.2 as follows

𝔼{D}\displaystyle\mathbb{E}\{D\} =i=1n1𝔼{Ei}\displaystyle=\sum_{i=1}^{n-1}\mathbb{E}\{E_{i}\}
=i=1n1{Ei=1}\displaystyle=\sum_{i=1}^{n-1}\mathbb{P}\{E_{i}=1\}
=i=1k{Ei=1}\displaystyle=\sum_{i=1}^{k}\mathbb{P}\{E_{i}=1\}
=ki=1k{Ei=0}\displaystyle=k-\sum_{i=1}^{k}\mathbb{P}\{E_{i}=0\}
=ki=1k[j=0i1(k+j1j)12k+j](O(k2/n)+1)\displaystyle=k-\sum_{i=1}^{k}\left[\sum_{j=0}^{i-1}\binom{k+j-1}{j}\frac{1}{2^{k+j}}\right](O(k^{2}/n)+1)
=k[j=0k1(k+j1j)kj2k+j](O(k2/n)+1)\displaystyle=k-\left[\sum_{j=0}^{k-1}\binom{k+j-1}{j}\frac{k-j}{2^{k+j}}\right](O(k^{2}/n)+1)
=k[k2j=0k1(k+j1j)j2k+j](O(k2/n)+1)\displaystyle=k-\left[\frac{k}{2}-\sum_{j=0}^{k-1}\binom{k+j-1}{j}\frac{j}{2^{k+j}}\right](O(k^{2}/n)+1)
=k[(2k1)22k1(2k2k1)](O(k2/n)+1).\displaystyle=k-\left[\frac{(2k-1)}{2^{2k-1}}\binom{2k-2}{k-1}\right](O(k^{2}/n)+1).

This completes the proof. ∎

Remark:

As mentioned earlier, the work in [10] considers the preference threshold to be a random variable independently chosen per vertex using a distribution PP over 0\mathbb{N}^{0} with finite mean instead of just a fixed kk for the potential number of neighbors of a randomly chosen vertex. Further, [10, Theorem 5.1] specifies the following formula for the average degree in the limit of nn going to infinity

𝔼{D}=i=1j=1P(i)P(j)0F¯i(x)F¯j(x)𝑑x\mathbb{E}\{D\}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}P(i)P(j)\int_{0}^{\infty}\bar{F}_{i}(x)\bar{F}_{j}(x)dx (15)

where F¯i\bar{F}_{i} denotes the complementary cumulative distribution function of Erlang(;i,1)\textrm{Erlang}(\cdot\,;i,1) (the Erlang distribution of shape ii and rate 11).

In the case of bilateral agreement with preference threshold parameter to be a fixed kk the probability distribution PP becomes a delta mass function on kk, i.e., P(i)=1P(i)=1 if i=ki=k, and P(i)=0P(i)=0 otherwise. In addition, the mean being finite as nn goes to infinity, implies that kk must be finite and cannot grow to infinity. The sum 15 now simplifies to

𝔼{D}=0F¯k(x)2𝑑x,\mathbb{E}\{D\}=\int_{0}^{\infty}\bar{F}_{k}(x)^{2}dx, (16)

Then using F¯k(x)=i=0k11i!exxi\bar{F}_{k}(x)=\sum_{i=0}^{k-1}\frac{1}{i!}e^{-x}x^{i}, distributing the square, and exchanging the finite sum with integral, we obtain

𝔼{D}\displaystyle\mathbb{E}\{D\} =0i,jk10xi+ji!j!e2x𝑑x\displaystyle=\sum_{0\leq i,j\leq k-1}\int_{0}^{\infty}\frac{x^{i+j}}{i!j!}e^{-2x}dx
=0i,jk10(2x)i+j2i+ji!j!e2x𝑑x\displaystyle=\sum_{0\leq i,j\leq k-1}\int_{0}^{\infty}\frac{(2x)^{i+j}}{2^{i+j}i!j!}e^{-2x}dx
=0i,jk1Γ(i+j+1)2i+j+1i!j!\displaystyle=\sum_{0\leq i,j\leq k-1}\frac{\Gamma(i+j+1)}{2^{i+j+1}i!j!}
=120i,jk112i+j(i+ji).\displaystyle=\frac{1}{2}\sum_{0\leq i,j\leq k-1}\frac{1}{2^{i+j}}\binom{i+j}{i}. (17)

We substitute s=i+js=i+j and use the notation of XkX_{k} for a negative binomial random variable with parameter p=1/2p=1/2 representing the probability of appearing kthk^{th} red ball after observing XkX_{k} blue balls. Then we arrive at

𝔼{D}\displaystyle\mathbb{E}\{D\} =12[s=0k1i=0s12s(si)+s=k2k2i=sk+1k112s(si)]\displaystyle=\frac{1}{2}\left[\sum_{s=0}^{k-1}\sum_{i=0}^{s}\frac{1}{2^{s}}\binom{s}{i}+\sum_{s=k}^{2k-2}\sum_{i=s-k+1}^{k-1}\frac{1}{2^{s}}\binom{s}{i}\right]
=12[k+s=0k2(12i=0s12s+k(s+ki))]\displaystyle=\frac{1}{2}\left[k+\sum_{s=0}^{k-2}\left(1-2\sum_{i=0}^{s}\frac{1}{2^{s+k}}\binom{s+k}{i}\right)\right]
=12[k+s=0k2(12{Xks})]\displaystyle=\frac{1}{2}\left[k+\sum_{s=0}^{k-2}\left(1-2\mathbb{P}\{X_{k}\leq s\}\right)\right]
=12[k+s=0k2(2{Xk>s}1)]\displaystyle=\frac{1}{2}\left[k+\sum_{s=0}^{k-2}\left(2\mathbb{P}\{X_{k}>s\}-1\right)\right]
=12[1+2𝔼{Xk1Xkk1}+2(k1){Xkk}]\displaystyle=\frac{1}{2}\left[1+2\mathbb{E}\{X_{k}1_{X_{k}\leq k-1}\}+2(k-1)\mathbb{P}\{X_{k}\geq k\}\right]
=12[1+j=0k1(k+j1j)2j2k+j+2(k1)(1{Xk<k})]\displaystyle=\frac{1}{2}\left[1+\sum_{j=0}^{k-1}\binom{k+j-1}{j}\frac{2j}{2^{k+j}}+2(k-1)(1-\mathbb{P}\{X_{k}<k\})\right]
=k2+j=0k1(k+j1j)j2k+j(Using Lemma 6.1)\displaystyle=\frac{k}{2}+\sum_{j=0}^{k-1}\binom{k+j-1}{j}\frac{j}{2^{k+j}}\quad(\text{Using Lemma }\ref{negative-binom-mean1})
=k[(2k1)22k1(2k2k1)].(Using Lemma 6.2)\displaystyle=k-\left[\frac{(2k-1)}{2^{2k-1}}\binom{2k-2}{k-1}\right].\quad(\text{Using Lemma }\ref{negative-binom-mean2}) (18)

Therefore, we reproduce the same formula for the average degree as in Theorem 14 using the result in [10]. We discussed this remark with the assumption of sparseness. However, Theorem 14 only needs k=o(n)k=o(\sqrt{n}) which covers both the sparse and non-sparse regimes.

Next we determine the asymptotic behavior of the mean degree as both nn and kk go to infinity.

Theorem 6.2.

In the model 𝔾(n,k)\mathbb{G}(n,k), assume that k=o(n1/3)k=o(n^{1/3}) grows to infinity as nn goes to infinity. Suppose DD represents the degree of a randomly chosen vertex. We have the following asymptotic

𝔼{D}=kkπ+18πk+o(1k).\mathbb{E}\{D\}=k-\sqrt{\frac{k}{\pi}}+\frac{1}{8\sqrt{\pi k}}+o\left(\frac{1}{\sqrt{k}}\right).
Proof.

We simply use Stirling’s approximation n!=2πn(ne)n(1+112n+o(1n))n!=\sqrt{2\pi n}(\frac{n}{e})^{n}\left(1+\frac{1}{12n}+o(\frac{1}{n})\right) to find an asymptotic expression for (2k2k1)\binom{2k-2}{k-1}

(2k2k1)\displaystyle\binom{2k-2}{k-1} =(2k2)!(k1)!2\displaystyle=\frac{(2k-2)!}{(k-1)!^{2}}
=2π(2k2)(2k2e)2k2(1+112(2k2)+o(1k))2π(k1)(k1e)2k2(1+112(k1)+o(1k))2\displaystyle=\frac{\sqrt{2\pi(2k-2)}(\frac{2k-2}{e})^{2k-2}\left(1+\frac{1}{12(2k-2)}+o(\frac{1}{k})\right)}{2\pi(k-1)(\frac{k-1}{e})^{2k-2}\left(1+\frac{1}{12(k-1)}+o(\frac{1}{k})\right)^{2}}
=2π(2k2)22k2(1+112(2k2)+o(1k))2π(k1)(1+112(k1)+o(1k))2\displaystyle=\frac{\sqrt{2\pi(2k-2)}2^{2k-2}\left(1+\frac{1}{12(2k-2)}+o(\frac{1}{k})\right)}{2\pi(k-1)\left(1+\frac{1}{12(k-1)}+o(\frac{1}{k})\right)^{2}}
=22k2(1+124(k1)+o(1k))π(k1)(1+16(k1)+o(1k))\displaystyle=\frac{2^{2k-2}\left(1+\frac{1}{24(k-1)}+o(\frac{1}{k})\right)}{\sqrt{\pi(k-1)}\left(1+\frac{1}{6(k-1)}+o(\frac{1}{k})\right)}
=(22k2π(k1))(118(k1)+o(1k)).\displaystyle=\left(\frac{2^{2k-2}}{\sqrt{\pi(k-1)}}\right)\left(1-\frac{1}{8(k-1)}+o\left(\frac{1}{k}\right)\right).

Now Theorem 6.1 implies

𝔼{D}\displaystyle\mathbb{E}\{D\} =k[(2k1)22k1(2k2k1)](O(k2/n)+1)\displaystyle=k-\left[\frac{(2k-1)}{2^{2k-1}}\binom{2k-2}{k-1}\right](O(k^{2}/n)+1)
=k[(2k1)22k1(22k2π(k1))(118(k1)+o(1k))](O(k2/n)+1)\displaystyle=k-\left[\frac{(2k-1)}{2^{2k-1}}\left(\frac{2^{2k-2}}{\sqrt{\pi(k-1)}}\right)\left(1-\frac{1}{8(k-1)}+o\left(\frac{1}{k}\right)\right)\right]\left(O(k^{2}/n)+1\right)
=k[(k1π+12π(k1))(118(k1)+o(1k))](O(k2/n)+1)\displaystyle=k-\left[\left(\sqrt{\frac{k-1}{\pi}}+\frac{1}{2\sqrt{\pi(k-1)}}\right)\left(1-\frac{1}{8(k-1)}+o\left(\frac{1}{k}\right)\right)\right]\left(O(k^{2}/n)+1\right)
=kk1π38π(k1)+o(1k)\displaystyle=k-\sqrt{\frac{k-1}{\pi}}-\frac{3}{8\sqrt{\pi(k-1)}}+o\left(\frac{1}{\sqrt{k}}\right)
=kkπ+18πk+o(1k),\displaystyle=k-\sqrt{\frac{k}{\pi}}+\frac{1}{8\sqrt{\pi k}}+o\left(\frac{1}{\sqrt{k}}\right),

which proves the result. ∎

7 Conclusions

We studied bilateral agreement random graph family and presented a complete proof of the connectivity threshold conjecture for this family by La and Kabkab [9], namely, if k=tlog(n)k=t\log(n), then (with high probability) connectivity holds if t>1t>1, and the graph is disconnected if t<1t<1. The proof for disconnectivity result is established using the second moment method in which we proved that the expected number of isolated vertices grows to infinity and showed that asymptotically any pair of vertices is isolated in an independent fashion, i.e. 𝔼{I1I2}𝔼{I1}𝔼{I2}\mathbb{E}\{I_{1}I_{2}\}\approx\mathbb{E}\{I_{1}\}\mathbb{E}\{I_{2}\}. Therefore, with high probability there exists an isolated vertex which yields the disconnectivity result. For the connectivity part, we took the same approach as in La and Kabkab [9] to find Erdős-Réyni sub and super random graphs with all vertices associated with the original graph. This analysis then excludes the possibility of having components with rr vertices for 10rn/210\leq r\leq n/2. Independently, we also ruled out the existence of components of size O(1)O(1). At the end, we discussed the average degree for this random graph family and presented an asymptotic for it.

As discussed earlier, our results are refinements of the connectivity conjecture of La and Kabkab [9] by establishing higher orders for the connectivity threshold. In particular, when k=log(n)+tloglog(n)log(n)k=\log(n)+t^{\prime}\log\log(n)\sqrt{\log(n)}, we show that connectivity holds for t>12t^{\prime}>\frac{1}{2}, and the graph is disconnected when t<12t^{\prime}<-\frac{1}{2} (both with high probability). However, these results still do not completely determine the connectivity threshold, and determining the precise connectivity threshold, even in terms of the precise value of t(1/2,1/2)t^{\prime}\in(-1/2,1/2), is an interesting avenue for future research.

Acknowledgements

Vijay Subramanian and Hossein Dabirian acknowledge support from NSF via grant CCF-200813. We are also grateful to Alan Frieze, Bruce Hajek, Richard La, Remco van der Hofstad and Mehrdad Moharrami for helpful comments.

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