Swarming on Random Graphs II

Abstract

We consider an individual-based model where agents interact over a random network via first-order dynamics that involve both attraction and repulsion. In the case of all-to-all coupling of agents in $\mathbb {R}^{d}$ this system has a lowest energy state in which an equal number of agents occupy the vertices of the $d$-dimensional simplex. The purpose of this paper is to sharpen and extend a line of work initiated in [56], which studies the behavior of this model when the interaction between the $N$ agents occurs according to an Erdős–Rényi random graph $\mathcal {G}(N,p)$ instead of all-to-all coupling. In particular, we study the effect of randomness on the stability of these simplicial solutions, and provide rigorous results to demonstrate that stability of these solutions persists for probabilities greater than $Np = O ( \log N )$. In other words, only a relatively small number of interactions are required to maintain stability of the state. The results rely on basic probability arguments together with spectral properties of random graphs.

Appendix

1.1 Estimates for Generalized Adjacency Matrices

Given a random graph drawn from $\mathcal {G}(n,p),$ let $A \in M_{nd \times nd}(\mathbb {R})$ denote an $nd \times nd$ generalized adjacency matrix using any symmetric matrix $M \in \mathbb {M}_{d \times d}(\mathbb {R})$ as a sub-block. In other words, if $E$ denotes the adjacency matrix of the graph then we have

$$\begin{aligned} A = E \otimes M \end{aligned}$$

for $\otimes$ denoting the Kronecker product. Let $M_{jk} = M_{kj}, \; 1\le j \le n, \; j \le k \le n$ denote the i.i.d. matrix-valued random variables corresponding to the edges in the graph, so that $\mathbb {E}(M_{jk}) = pM$. For a given a vector $\mathbf {x}\in \mathbb {R}^{nd}$ consider the partition $\mathbf {x}= (\mathbf {x}_{1},\ldots ,\mathbf {x}_{n})^{t}$ for $\mathbf {x}_{i} \in \mathbb {R}^{d},$ and recall the “mean-zero” hypothesis

$$\begin{aligned} \sum ^{n}_{i=1} \mathbf {x}_i = \mathbf {0}. \end{aligned}$$

(48)

If we denote the corresponding subset of the unit ball $\mathcal {S}^{nd} \subset \mathbb {R}^{nd}$ as

$$\begin{aligned} \mathcal {S}^{nd}_{0} := \left\{ \mathbf {x}: \; \sum _{i} \mathbf {x}_i = \mathbf {0}, \quad \sum _{i} ||\mathbf {x}_{i}||^{2}_{2} \le 1 \right\} , \end{aligned}$$

our aim lies in proving the following generalization of the theorem due to [19]:

Theorem 6.1

Let $\alpha$ and $c_{0}$ denote arbitrary positive constants. If $np > c_{0} \log n$ then there exists a constant $c = c(\alpha ,c_0,d,||M||_{2}) > 0$ so that the estimate

$$\begin{aligned} \max _{ (\mathbf {x},\mathbf {y}) \in \mathcal {S}^{nd}_{0} \times \mathcal {S}^{nd} } |\langle \mathbf {x}, A \mathbf {y}\rangle | \le c \sqrt{np} \end{aligned}$$

(49)

holds with probability at least $1 - n^{-\alpha }$.

Note carefully that we only require one of $\mathbf {x}$ or $\mathbf {y}$ to satisfy the mean zero property (48). The proof of the theorem essentially reproduces the arguments of [19] by changing a few scalars to vectors and multiplications to inner products. The first ingredient is the following lemma:

Lemma 6.2

Fix $(\mathbf {x},\mathbf {y}) \in \mathcal {S}^{nd}_{0} \times \mathcal {S}^{nd}$ and let $\Lambda = \{ (j,k) : |\langle \mathbf {x}_k , M \mathbf {y}_j \rangle | \le \sqrt{p/n} \}$. Then

$$\begin{aligned} \left| \mathbb {E}\left( \sum _{(j,k) \in \Lambda } \langle \mathbf {x}_{k}, M_{kj} \mathbf {y}_j \rangle \right) \right| \le ||M||^{2}_{2} \sqrt{np}. \end{aligned}$$

(50)

Proof

As $\mathbf {x}\in \mathcal {S}^{nd}_{0}$ it follows that

$$\begin{aligned} 0 = p \sum _{j,k} \langle \mathbf {x}_{k} , M \mathbf {y}_{j} \rangle = \mathbb {E}\left( \sum _{ (j,k) \in \Lambda } \langle \mathbf {x}_{k}, M_{kj} \mathbf {y}_j \rangle \right) + p \sum _{ (j,k) \in \Lambda ^{c} } \langle \mathbf {x}_{k}, M \mathbf {y}_j \rangle . \end{aligned}$$

By definition, whenever $(j,k) \in \Lambda ^{c}$ it follows that $|\langle \mathbf {x}_{k}, M \mathbf {y}_j \rangle | > \sqrt{p/n}$. Thus $p|\langle \mathbf {x}_{k}, M \mathbf {y}_j \rangle | < \sqrt{np}|\langle \mathbf {x}_{k}, M \mathbf {y}_j \rangle |^2 \le \sqrt{np} ||M||^{2}_{2}||\mathbf {x}_k||^{2}_{2} ||\mathbf {y}_j||^{2}_{2}$ for any such $(j,k)$, where the last inequality follows from Cauchy-Schwarz. Combining these facts yields

$$\begin{aligned} \left| \mathbb {E}\left( \sum _{(j,k) \in \Lambda } \langle \mathbf {x}_{k}, M_{kj} \mathbf {y}_j \rangle \right) \right|&= p \left| \sum _{ (j,k) \in \Lambda ^{c} } \langle \mathbf {x}_{k}, M \mathbf {y}_j \rangle \right| \le ||M||^{2}_{2} \sqrt{np} \sum _{j,k} ||\mathbf {x}_{k}||^2_{2} ||\mathbf {y}_{j}||^{2}_{2}\\&= ||M||^{2}_{2} \sqrt{np} ||\mathbf {x}||^{2}_{2} ||\mathbf {y}||^{2}_{2} \end{aligned}$$

as desired. $\square$

For a fixed $(\mathbf {x},\mathbf {y}) \in \mathcal {S}^{nd}_{0} \times \mathcal {S}^{nd}$ let $S(\mathbf {x},\mathbf {y}),L(\mathbf {x},\mathbf {y}),U(\mathbf {x},\mathbf {y})$ denote the random variables defined as

$$\begin{aligned} S(\mathbf {x},\mathbf {y}) \!:=\! \sum _{(j,k) \in \Lambda } \!\langle \mathbf {x}_j, M_{jk} \mathbf {y}_k \rangle \!=\! \sum _{\underset{j\le k}{ (j,k) \in \Lambda } } \langle \mathbf {x}_j , M_{jk} \mathbf {y}_k \rangle \!+\!\!\sum _{\underset{j > k}{ (j,k) \in \Lambda } }\! \langle \mathbf {x}_j , M_{jk} \mathbf {y}_k \rangle \!:=\! U(\mathbf {x},\mathbf {y}) \!+\! L(\mathbf {x},\mathbf {y}). \end{aligned}$$

Although dependencies exist between $L$ and $U$ due to the undirected graph, when considered in isolation each random variable is simply a sum of independent indicator random variables. Indeed, fix any ordering of the indices $\Lambda \cap \{ j \le k \}$ and write $U(\mathbf {x},\mathbf {y}) = \sum _{i} u_{i},$ where the sum ranges from one to the (deterministic) size of $\Lambda \cap \{ j \le k \},$ and note that each $u_{i}$ is either zero or $\langle \mathbf {x}_j , M \mathbf {y}_k \rangle$ with probability $(1-p)$ or $p$, respectively. Obviously for $L(\mathbf {x},\mathbf {y})$ an analogous statement holds.

Note that $|u_{i}| \le \sqrt{p/n}$ by definition of $\Lambda$, and that

$$\begin{aligned} \sum _{i} \mathrm {Var}(u_i) = p(1-p) \sum _{\Lambda \cap \{j \le k\} } |\langle \mathbf {x}_{j} , M\mathbf {y}_{k} \rangle |^2 \le p||M||^{2}_{2} ||\mathbf {x}||^{2}_{2} ||\mathbf {y}||^{2}_{2} \le p||M||^{2}_{2}. \end{aligned}$$

By applying the Chernoff bound 2.1 to the random variables $(u_{i} - \mathbb {E}(u_{i}))/2\sqrt{p/n}$ with the choices $\sigma ^2 = nK||M||^{2}_{2}/4$ and $\lambda ^2 = nK||M||^{2}_{2}$ these facts imply that for any $K \ge 1$ the estimate

$$\begin{aligned} \mathbb {P}\left( \left| U(\mathbf {x},\mathbf {y}) - \mathbb {E}(U(\mathbf {x},\mathbf {y})) \right| \ge K \sqrt{np} ||M||^2_{2} \right) \le 2\mathrm {e}^{-K n ||M||_{2} /4} \end{aligned}$$

holds. A similar argument shows that the same inequality holds for $L(\mathbf {x},\mathbf {y})$ as well. As $S = L+U,$ the triangle inequality and the union bound yield the estimate

$$\begin{aligned} \mathbb {P}\left( \left| S(\mathbf {x},\mathbf {y}) - \mathbb {E}(S(\mathbf {x},\mathbf {y})) \right| \ge 2K \sqrt{np} ||M||^2_{2} \right) \le 4\mathrm {e}^{-K n ||M||_{2} /4}. \end{aligned}$$

(51)

Next, given $0< \delta < 1$ define the finite grid

$$\begin{aligned} T^{\delta } := \left\{ \mathbf {x}\in \left( \frac{\delta }{\sqrt{n}} \mathbb {Z} \right) ^{nd} : ||\mathbf {x}||_{2} \le 1 \right\} \end{aligned}$$

and its mean-zero variant

$$\begin{aligned} T^{\delta }_{0} := \left\{ \mathbf {x}\in \left( \frac{\delta }{\sqrt{n}} \mathbb {Z} \right) ^{nd} : \sum _{i} \mathbf {x}_i = 0 \quad ||\mathbf {x}||_{2} \le 1 \right\} . \end{aligned}$$

The following lemma allows us to control the norm of $A$ on $\mathcal {S}^{nd}_{0}$ by controlling $\langle \mathbf {x}, A \mathbf {y}\rangle$ for pairs of vectors in the finite grid instead:

Lemma 6.3

If $|\langle \mathbf {x}, A \mathbf {y}\rangle | \le c$ for all $(\mathbf {x},\mathbf {y}) \in T^{\delta }_{0} \times T^{\delta }$ then $|\langle \mathbf {x}, A \mathbf {y}\rangle | \le \frac{cd}{(1-\delta )^2}$ for all $(\mathbf {x},\mathbf {y}) \in \mathcal {S}^{nd}_{0} \times \mathcal {S}^{nd}$.

Proof

Let $\mathbf {z}= (1-\delta ) \mathbf {x}, \mathbf {u}= (1-\delta ) \mathbf {y}$ and note that

$$\begin{aligned} \sum _{i} \mathbf {z}_i = \mathbf {0}. \end{aligned}$$

(52)

Decompose $\mathbf {z}$ as $\mathbf {z}= \mathbf {z}_{1} + \cdots + \mathbf {z}_{d}$ where the non-zero components of $\mathbf {z}_{k}$ correspond to the $k^\mathrm{{th}}$ equation in (52); that is, $\mathbf {z}_k$ contains the $k^\mathrm{th}$, $(d+k)^\mathrm{th},(2d+k)^\mathrm{th},$ etc. components of $\mathbf {z}$ and has zeros elsewhere. As

$$\begin{aligned} ||\mathbf {z}||^2_{2} = ||\mathbf {z}_1||^2 + \cdots + ||\mathbf {z}_d||^2 \le (1-\delta )^2, \end{aligned}$$

the vector in $\mathbb {R}^n$ comprised of the non-zero locations of $\mathbf {z}_{k}$ has norm less than $(1-\delta )$ and entries that sum to zero. By lemma 2.3 of [19], each $\mathbf {z}_k$ is therefore a convex combination of points of $T^{\delta }_0,$

$$\begin{aligned} \mathbf {z}_k = \sum ^{J_{k} }_{j=1} \theta ^{k}_{j} \mathbf {v}^{k}_{j}, \quad \theta ^{k}_{j} > 0, \quad \sum ^{J_{k} }_{j=1} \theta ^{k}_{j} = 1. \end{aligned}$$

Summing the $\mathbf {z}_{k}$ then shows that there exists an $N = J_{1} + \cdots + J_d \in \mathbb {N}$, $\theta _l > 0$ and $\mathbf {v}_l \in T^{\delta }_{0}$ so that

$$\begin{aligned} \mathbf {z}= \sum ^{N}_{l=1} \theta _{l} \mathbf {v}_{l}, \quad \sum ^{N}_{l=1} \theta _{l} = d. \end{aligned}$$

Lemma 2.3 of [19] also implies that $\mathbf {u}$ is a convex combination of points in $T^{\delta },$ so that there exist $M \in \mathbb {N}$, $\eta _{j} > 0$ and $\mathbf {w}_{j} \in T^{\delta }$ so that

$$\begin{aligned} \mathbf {u}= \sum ^{M}_{j=1} \eta _{j} \mathbf {w}_{j}, \quad \sum ^{M}_{j=1} \eta _{j} = 1. \end{aligned}$$

As a consequence,

$$\begin{aligned} (1-\delta )^2| \langle \mathbf {x}, A \mathbf {y}\rangle | = | \langle \mathbf {z}, A \mathbf {u}\rangle | = \left| \sum ^{N}_{l=1} \theta _{l} \sum ^{M}_{j=1} \eta _j \langle \mathbf {v}_l , A \mathbf {w}_j \rangle \right| \le c\sum ^{N}_{l=1} \theta _{l} \sum ^{M}_{j=1} \eta _j = cd. \end{aligned}$$

$\square$

An estimate of the total number of points $|T^{\delta }|,|T^{\delta }_{0}|$ in the $\delta$-nets follows from a direct appeal to claim 2.9 of [19]. As $T^{\delta }_{0} \subset T^{\delta }$ and $|T^{\delta }| \le \mathrm {e}^{c(nd)}$ for some constant $c$ that depends on $\delta$, which follows from claim 2.9 of [19], it follows that $|T^{\delta }_{0} \times T^{\delta }| \le \mathrm {e}^{2c(nd)}$ as well. Applying the union bound over $T^{\delta }_{0} \times T^{\delta }$, we may therefore summarize the preceeding in the following lemma:

Lemma 6.4

Given any $c > 0$ there exists $c' > 0$ so that with probability at least $1-\mathrm {e}^{-cn}$ the estimate

$$\begin{aligned} \max _{(\mathbf {x},\mathbf {y}) \in T^{\delta }_0 \times T^{\delta }} |S(\mathbf {x},\mathbf {y})| \le c' \sqrt{np} \end{aligned}$$

(53)

holds.

It remains to estimate, for $(\mathbf {x},\mathbf {y}) \in T^{\delta }_0 \times T^{\delta }$, the remaining contribution

$$\begin{aligned} H(\mathbf {x},\mathbf {y}) := \langle \mathbf {x}, A \mathbf {y}\rangle - S(\mathbf {x},\mathbf {y}) = \sum _{(j,k) \in \Lambda ^c} \langle \mathbf {x}_j, M_{jk} \mathbf {y}_k \rangle \end{aligned}$$

where we recall $\Lambda ^{c} := \{(j,k) : |\langle \mathbf {x}_j, M \mathbf {y}_k \rangle | > \sqrt{p/n} \}$. From Cauchy-Schwarz it follows that

$$\begin{aligned} |H(\mathbf {x},\mathbf {y})| \le ||M||_{2}\sum _{(k,l) \in \Lambda ^c} ||\mathbf {x}_k||_{2}|| \mathbf {y}_l||_2 \mathbf {1}_{ \{ e_{kl} = 1 \}}. \end{aligned}$$

(54)

Here and in what follows, for an index subset $W$ the notation $\mathbf {1}_{W}$ denotes the indicator function. If we define the sets

$$\begin{aligned} X_{i}&:= \left\{ k \in \{1\ldots n\}: \frac{\delta }{\sqrt{n}}2^{i-1} \le ||\mathbf {x}_k||_2 < \frac{\delta }{\sqrt{n}}2^{i} \right\} \\ Y_{i}&:= \left\{ k \in \{1\ldots n\}: \frac{\delta }{\sqrt{n}}2^{i-1} \le ||\mathbf {y}_k||_2 < \frac{\delta }{\sqrt{n}}2^{i} \right\} \end{aligned}$$

and fix an edge $(k,l)$ corresponding to a non-zero term in the sum (54), then $k \in X_{i}$ and $l \in Y_j$ for some $(i,j)$ and an edge $(k,l)$ exists between these two vertex sets. Moreover, as $(k,l) \in \Lambda ^{c}$ it follows that

$$\begin{aligned} \sqrt{p/n} < |\langle \mathbf {x}_k , M \mathbf {y}_l \rangle | \le \delta ^{2}||M||_{2}2^{i+j}/n \quad \Rightarrow \quad 2^{i}2^{j} > \sqrt{pn} \end{aligned}$$

provided we take $\delta ||M||_{2} \le 1$. As a consequence,

$$\begin{aligned} |H(\mathbf {x},\mathbf {y})| \le \delta ^2 ||M||_{2}\sum _{\underset{2^i2^j>\sqrt{np}}{i,j} } \mathrm {edge}(X_i,Y_j) \frac{2^{i}2^{j}}{n}, \end{aligned}$$

(55)

where $\mathrm {edge}(X_i,Y_j)$ denotes the number of edges between the sets.

To bound (55), we let $c_{1}$ denote an as-yet-undetermined constant and put

$$\begin{aligned} W := \{ (i,j) : 2^i2^j>\sqrt{np} , \; \max \{|X_i|,|Y_j|\} > (n/\mathrm {e}) \}, \end{aligned}$$

then decompose the sum further as

$$\begin{aligned} \sum _{\underset{2^i2^j>\sqrt{np}}{i,j} } \mathrm {edge}(X_i,Y_j) \frac{2^{i}2^{j}}{n} = \sum _{W} \mathrm {edge}(X_i,Y_j) \frac{2^{i}2^{j}}{n} + \sum _{W^c} \mathrm {edge}(X_i,Y_j) \frac{2^{i}2^{j}}{n}. \end{aligned}$$

(56)

Assuming that each vertex has degree bounded by $c_{1}np$ (c.f. lemma 6.5) then $\mathrm {edge}(X_i,Y_j) \le c_{1}np\min \{|X_i|,|Y_j|\}.$ Thus the first term is bounded by

$$\begin{aligned} \sum _{W} \mathrm {edge}(X_i,Y_j) \frac{2^{i}2^{j}}{n}&\le c_{1}\mathrm {e}p \sum _{W} \min \{ |X_i|,|Y_j| \} \max \{|X_i|,|Y_j| \}\frac{ 2^{i}2^{j} }{n}\\&\le c_{1} \mathrm {e}\sqrt{ np } \sum _{i,j} 2^{2i}|X_i|2^{2j}|Y_j|n^{-2}. \end{aligned}$$

Noting that

$$\begin{aligned} \sum _{i} \frac{ |X_{i}|2^{2i} }{n} \le 4\delta ^{-2} ||\mathbf {x}||^2 \quad \sum _{j} \frac{ |Y_{j}|2^{2j} }{n} \le 4\delta ^{-2} ||\mathbf {y}||^2 \end{aligned}$$

(57)

then shows that the first term is $O(\sqrt{np})$ provided the bounded degree property holds. Fortunately, if $np > c_{0} \log n$ for any $c_{0} > 0$, this property holds with probability at least $1-n^{-\alpha }$ for any $\alpha > 0$ provided $c_{1}$ is sufficiently large:

Lemma 6.5

(Bounded Degree) Let $p \ge c_0 \log n/n$ for any $c_0 > 0$ and $\mathrm{deg}_1,\ldots ,\mathrm{deg}_n$ denote the vertex degrees of an Erdős–Rényi random graph $\mathcal {G}(n,p)$. For any $\alpha > 0$ there exists a $c_1 = c_1(c_0,\alpha ) > 2$ so that

$$\begin{aligned} \mathbb {P}\left( \max _{i} \; \mathrm{deg}_{i} > c_{1}np\right) \le 2 n^{-\alpha }. \end{aligned}$$

(58)

Proof

Write $\mathrm{deg}_{i} = \sum _{j} e_{ij}$ where $e_{ij}$ denote the edges of the graph. As $\mathbb {E}(\mathrm{deg}_{i}) = np$ and $\mathrm {Var}(\mathrm{deg}_{i}) = np(1-p) \le (c_{1}-1)np$ it follows from the Chernoff bound and the union bound that

$$\begin{aligned}&\mathbb {P}\left( |\mathrm{deg}_{i} - np| > \lambda \sqrt{(c_1-1)np} \right) \le 2 \mathrm {e}^{-\lambda ^2/4}\\&\quad \Rightarrow \mathbb {P}\left( \max _i \; |\mathrm{deg}_{i} - np| > \lambda \sqrt{(c_1-1)np} \right) \le 2\mathrm {e}^{-\lambda ^2/4 + \log n}. \end{aligned}$$

The choice $\lambda = \sqrt{(c_{1} - 1)np}$ yields

$$\begin{aligned} \mathbb {P}\left( \max _i \; |\mathrm{deg}_{i} - np| > (c_1-1)np \right) \le 2\mathrm {e}^{-np(c_1-1)/4 + \log n} \le 2 e^{ \log n(1 - c_{0}(c_1-1)/4)}. \end{aligned}$$

Taking $c_1$ sufficiently large gives the desired result. $\square$

Therefore with probability at least $1 - n^{-\alpha }$, the first term in (56) is $O(\sqrt{np})$ uniformly for $(\mathbf {x},\mathbf {y}) \in T^{\delta }_0 \times T^{\delta }$ as long as the bounded degree property holds. It remains to handle the second term. From the definition of $W^c$, if $V = \{1,\ldots ,n\}$ and $2^{|V|}$ its power set we need only consider those sets $X \in 2^{|V|}$ with $|X| \le (n/\mathrm {e})$,

$$\begin{aligned} V_{s} := \{(X,Y) \in 2^{|V|} \times 2^{|V|} : \max \{ |X|,|Y| \} \le n/\mathrm {e}\}. \end{aligned}$$

Before proceeding, we first need to establish an estimate on the random variable $\mathrm {edge}(X,Y)$ when $(X,Y) \in V_{s}$:

Lemma 6.6

For $(X,Y) \in V_s$ let $\mathrm {edge}(X,Y)$ denote the random variable corresponding to the number of edges from an Erdős–Rényi random graph $\mathcal {G}(n,p)$ between $X$ and $Y$. Let $\mu (X,Y) = \mathbb {E}(\mathrm {edge}(X,Y))$ and $\mathcal {M}(X,Y) := \max \{|X|,|Y| \}$. Given any $\alpha > 0$ let $k_{0}(X,Y)$ denote the unique solution to

$$\begin{aligned} k_0 \log (k_0/\mathrm {e}) = \frac{\alpha + 4}{\mu }\mathcal {M} \log \frac{n\mathrm {e}}{\mathcal {M}}. \end{aligned}$$

(59)

Then

$$\begin{aligned} \mathbb {P}\left( \max _{V_s} \left( \mathrm {edge}(X,Y) - k_0(X,Y) \mu (X,Y) \right) > 0 \right) \le n^{-\alpha }. \end{aligned}$$

(60)

Proof

From the union bound, it follows (through slight abuse of notation) that

$$\begin{aligned}&\mathbb {P}\left( \max _{V_s} \left( \mathrm {edge}(X,Y) - k_0(X,Y) \mu (X,Y) \right) > 0 \right) \le \\&\sum ^{n/\mathrm {e}}_{|X|=1} \sum ^{n/\mathrm {e}}_{|Y| = 1} \sum ^{ \left( {\begin{array}{c}n\\ |X|\end{array}}\right) }_{X} \sum ^{ \left( {\begin{array}{c}n\\ |Y|\end{array}}\right) }_{Y} \mathbb {P}\left( \mathrm {edge}(X,Y) > k_0(X,Y) \mu (X,Y) \right) \end{aligned}$$

Using a one-sided concentration inequality (c.f. [32]) for sums of indicator variables,

$$\begin{aligned} \mathbb {P}( \mathrm {edge}(X,Y) > k \mu ( X,Y ) ) \le \mathrm {e}^{-k\log (k/\mathrm {e}) \mu (X,Y)}, \end{aligned}$$

which is valid for any $k > 1$, shows that

$$\begin{aligned} \mathbb {P}( \mathrm {edge}(X,Y)&> k_{0}(X,Y) \mu ( X,Y ) )\\&\le \mathrm {e}^{-(\alpha +4) \mathcal {M} \log (n\mathrm {e}/\mathcal {M}) } \end{aligned}$$

since $k_{0} > 1$. It therefore follows that

$$\begin{aligned}&\sum ^{n/\mathrm {e}}_{|X|=1} \sum ^{n/\mathrm {e}}_{|Y| = 1} \sum ^{ \left( {\begin{array}{c}n\\ |X|\end{array}}\right) }_{X} \sum ^{ \left( {\begin{array}{c}n\\ |Y|\end{array}}\right) }_{Y} \mathbb {P}\left( \mathrm {edge}(X,Y) > k_0(X,Y) \mu (X,Y) \right) \\&\quad \le \sum ^{n/\mathrm {e}}_{|X|=1} \sum ^{n/\mathrm {e}}_{|Y| = 1} \left( {\begin{array}{c}n\\ |X|\end{array}}\right) \left( {\begin{array}{c}n\\ |Y|\end{array}}\right) \mathrm {e}^{-(\alpha +4) \mathcal {M} \log (n\mathrm {e}/\mathcal {M}) } \\&\sum ^{n/\mathrm {e}}_{|X|=1} \sum ^{n/\mathrm {e}}_{|Y| = 1} \mathrm {exp}\left\{ |X|\log ( n\mathrm {e}/|X|) + |Y|\log ( n\mathrm {e}/|Y|)-(\alpha +4) \mathcal {M} \log (n\mathrm {e}/\mathcal {M})\right\} . \end{aligned}$$

The fact that $f(x) := x\log (n\mathrm {e}/x)$ is increasing for $1 \le x \le n$ then implies

$$\begin{aligned} |X|\log ( n\mathrm {e}/|X|) + |Y|\log ( n\mathrm {e}/|Y|) \le 2 \mathcal {M} \log (n\mathrm {e}/\mathcal {M}), \quad \mathcal {M} \log (n\mathrm {e}/\mathcal {M}) \ge 1 + \log n. \end{aligned}$$

Combining this with the previous estimate yields

$$\begin{aligned} \mathbb {P}\left( \max _{V_s} \left( \mathrm {edge}(X,Y) - k_0(X,Y) \mu (X,Y) \right) > 0 \right) \le \sum ^{n/\mathrm {e}}_{|X|=1} \sum ^{n/\mathrm {e}}_{|Y| = 1} n^{-(\alpha +2)} \le n^{-\alpha } \end{aligned}$$

as desired. $\square$

The definition of $k_0(X,Y)$ and the fact $\mu (X,Y) \le p|X||Y|$ then combine to yield the following corollary, which provides the basic estimate for the remaining contribution of $W^c$ to the sum.

Corollary 6.7

With probability at least $1-n^{-\alpha }$, the estimate

$$\begin{aligned} \mathrm {edge}(X,Y) \log \frac{\mathrm {edge}(X,Y)}{\mathrm {e}p |X||Y| } \le (\alpha +4)\mathcal {M}(X,Y) \log \frac{n \mathrm {e}}{\mathcal {M}(X,Y)} \end{aligned}$$

(61)

holds for all $X,Y \in V_s$.

Estimating the second term in (56)

$$\begin{aligned} \sum _{W^c} \mathrm {edge}(X_i,Y_j) \frac{2^{i}2^{j}}{n} \end{aligned}$$

now proceeds by decomposing into the following cases:

$$\begin{aligned} \mathrm {I}&:= W^c \cap \left\{ \mathrm {edge}(X_i,Y_j) \le \mathrm {e}^2 \sqrt{p/n} |X_i| |Y_j| 2^i2^j \right\} \\ \mathrm {II}_{\mathrm {A}}&:= W^c \cap \{ 2^{i} > \sqrt{np} \; 2^j \} \quad \mathrm {II}_{\mathrm {B}} := W^c \cap \{ 2^{j} > \sqrt{np} \; 2^i \}\\ \mathrm {III}&:= W^c \setminus \left( \mathrm {I} \cup \mathrm {II}_{\mathrm {A}} \cup \mathrm {II}_{\mathrm {B}} \right) . \end{aligned}$$

On $\mathrm {I}$ we have that

$$\begin{aligned} \sum _{\mathrm {I}} \le \mathrm {e}^2 \sqrt{ np } \sum _{i,j} 2^{2i}|X_i|2^{2j}|Y_j|n^{-2} = O(\sqrt{np}) \end{aligned}$$

due to (57) as before. On $\mathrm {II}_{\mathrm {A}}$ and $\mathrm {II}_{\mathrm {B}}$ the bounded degree property implies that $\mathrm {edge}(X_i,Y_j) \le c_{1}np|X_i|$ and $\mathrm {edge}(X_i,Y_j) \le c_{1}np|Y_j|,$ respectively. Therefore

$$\begin{aligned} \sum _{\mathrm {II}_{\mathrm {A}}} \le c_{1}\sqrt{np} \sum _{\mathrm {II}_{\mathrm {A}}} \frac{|X_i|2^{2i}}{n} \frac{2^j\sqrt{np}}{2^i} = c_{1}\sqrt{np} \sum _{i} \frac{|X_i|2^{2i}}{n} \sum _{j} \frac{2^j\sqrt{np}}{2^i} \mathbf {1}_{ \mathrm {II}_{\mathrm {A}} }(i,j). \end{aligned}$$

For fixed $i$ let $j_{\mathrm{max}}(i)$ denote the largest $j$ so that $(i,j) \in \mathrm {II}_{\mathrm {A}}$. Then

$$\begin{aligned} \sum _{j} \frac{2^j\sqrt{np}}{2^i} \mathbf {1}_{ \mathrm {II}_{\mathrm {A}} }(i,j)\!\le \! \frac{ 2^{j_{\mathrm{max}}(i)} \sqrt{np} }{2^i} \sum ^{\infty }_{j=0} 2^{-j} \!\le \! 2\,\Rightarrow \, \sum _{\mathrm {II}_{\mathrm {A}}} \!\le \! 2c_{1}\sqrt{np} \sum _{i} \frac{|X_i|2^{2i}}{n} \!=\! O(\sqrt{np})\nonumber \\ \end{aligned}$$

(62)

due to (57) once again. By reversing the roles of $i$ and $j$, a similar argument shows

$$\begin{aligned} \sum _{\mathrm {II}_{\mathrm {B}}} = O(\sqrt{np}) \end{aligned}$$

as well. For the remaining set $\mathrm {III},$ let $\mathrm {III} = \mathrm {III}^{i>j} \cup \mathrm {III}^{j>i}$ where the notation signifies $|Y_j| \ge |X_i|$ on $\mathrm {III}^{j>i}$ and vice-versa. We treat the first set and leave the second as an exercise since it follows analogously. Using the probabilistic estimate (61) as a guide, we decompose further into

$$\begin{aligned} \mathrm {III}^{j>i}_{\mathrm {A}}&:= \mathrm {III}^{j>i} \cap \left\{ 4 \log \frac{\mathrm {edge}(X_i,Y_j)}{\mathrm {e}p |X_i||Y_j| } \ge \log \frac{n \mathrm {e}}{|Y_j|} \right\} \\ \mathrm {III}^{j>i}_{\mathrm {B}}&:= \mathrm {III}^{j>i} \cap \left\{ 4 \log \frac{\mathrm {edge}(X_i,Y_j)}{\mathrm {e}p |X_i||Y_j| } < \log \frac{n \mathrm {e}}{|Y_j|} \right\} \cap \left\{ \frac{ne}{|Y_j|} \le 2^{4j} \right\} \\ \mathrm {III}^{j>i}_{\mathrm {C}}&:= \mathrm {III}^{j>i} \cap \left\{ 4 \log \frac{\mathrm {edge}(X_i,Y_j)}{\mathrm {e}p |X_i||Y_j| } < \log \frac{n \mathrm {e}}{|Y_j|} \right\} \cap \left\{ \frac{ne}{|Y_j|} > 2^{4j} \right\} \end{aligned}$$

For the first set the estimate (61) implies

$$\begin{aligned} \sum _{\mathrm {III}^{j>i}_{\mathrm {A}} } \le 4(\alpha +4)\sqrt{np} \sum _{ j}\frac{|Y_j|2^{2j}}{n}\sum _{i} \frac{ 2^{i} }{ 2^{j} \sqrt{np} }\mathbf {1}_{\mathrm {III}^{j>i}_{\mathrm {A}} }(i,j) = O(\sqrt{np}) \end{aligned}$$

due to the fact that $2^{j}\sqrt{np} > 2^{i}$, the geometric series argument from (62) and the estimate (57). For the second set note that if $(i,j) \in \mathrm {III}$ then $(i,j) \notin \mathrm {I},$ so that the inequality

$$\begin{aligned} \mathrm {edge}(X_i,Y_j) > \mathrm {e}^2 \sqrt{p/n} |X_i| |Y_j| 2^i2^j \end{aligned}$$

holds on $\mathrm {III}$ by definition. This combines with the definition of $\mathrm {III}^{j>i}_{\mathrm {B}}$ to imply that

$$\begin{aligned} \mathrm {e}\le \mathrm {e}\frac{2^i2^j}{\sqrt{np}} \le \frac{\mathrm {edge}(X_i,Y_j)}{\mathrm {e}p |X_i||Y_j| } \le 2^j, \end{aligned}$$

which shows that $1 \le \log \frac{\mathrm {edge}(X_i,Y_j)}{\mathrm {e}p |X_i||Y_j| }$ and that $2^i < \sqrt{np}$ as well. The estimate (61) therefore implies

$$\begin{aligned} \sum _{\mathrm {III}^{j>i}_{\mathrm {B}} } \!\le \! 4(\alpha \!+\!4)\sum _{\mathrm {III}^{j>i}_{\mathrm {B}} } \frac{ |Y_j| \log (2^j) 2^i 2^j }{n} \!\le \! 4(\alpha \!+\!4)\sqrt{np} \sum _{j} \frac{|Y_j|2^{2j}}{n}\sum _{i} \frac{2^i}{\sqrt{np}} \mathbf {1}_{ \mathrm {III}^{j>i}_{\mathrm {B}} } (i,j) \!=\! O(\sqrt{np}) \end{aligned}$$

due to the fact that $2^{i} > \sqrt{np},$ the geometric series argument (62) and the estimate (57), exactly as before. On the final set the facts that

$$\begin{aligned} \frac{ne}{|Y_j|} < \left( \frac{ne}{|Y_j|2^{2j}}\right) ^2, \quad 4 \le 4 \log \frac{\mathrm {edge}(X_i,Y_j)}{\mathrm {e}p |X_i||Y_j| } < 2 \log \frac{ne}{|Y_j|2^{2j}} < 4 \log \frac{ne}{|Y_j|2^{2j}} \end{aligned}$$

imply that $\mathrm {edge}(X_i,Y_j) \le \mathrm {e}^2 np |X_i|2^{-2j},$ whence

$$\begin{aligned} \sum _{\mathrm {III}^{j>i}_{\mathrm {C}} } \le e^2 \sqrt{np} \sum _{i}\frac{ |X_i| 2^{2i}}{n} \sum _{j} \frac{ \sqrt{np} }{2^i2^j } \mathbf {1}_{ \mathrm {III}^{j>i}_{\mathrm {C}} }(i,j). \end{aligned}$$

Arguing as before, the fact that $2^i2^j > \sqrt{np}$ gives that each sum over $j$ is bounded by two, so that applying (57) one final time demonstrates that the sum is $O(\sqrt{np})$.

The following lemma summarizes the preceeding arguments:

Lemma 6.8

Let $np > c_{0} \log n$ for any $c_{0} > 0$ and $\alpha$ denote an arbitrary positive constant. Then there exists a $c>0$ independent of $n$ so that with probability at least $1 - n^{-\alpha }$ the estimate

$$\begin{aligned} |H(\mathbf {x},\mathbf {y})| \le c \sqrt{np} \end{aligned}$$

(63)

holds for all $(\mathbf {x},\mathbf {y}) \in T^{\delta }_0 \times T^{\delta }$.

Combining this with the estimate for $S(\mathbf {x},\mathbf {y})$ and the reduction to the discrete set $T^{\delta }_0 \times T^{\delta }$ yields the theorem.