2.1. The number of removed edges

The number of removed edges $Z_n$ in the ECM is given by

\begin{equation*} Z_n = \sum_{i = 1}^n X_{ii} + \sum_{1 \le i < j \le n}\left( X_{ij} - {\mathds{1}}_{\left\{{X_{ij} > 0}\right\}}\right)\!,\end{equation*}

where $X_{ij}$ again denotes the number of edges between vertices i and j. For the analysis of the ECM it is important to understand the behavior of this number. In particular we are interested in the scaling of $Z_n$ with respect to n. Here we give an asymptotic upper bound, which implies that, up to sub-linear terms, $Z_n$ scales no faster than $n^{2 - \gamma}$. The proof can be found in Section A.1.

Theorem 2.1. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$ and $\widehat{G}_n = {\texttt{ECM}}(\textbf{D}_n)$. Then for any $\delta > 0$,

\begin{equation*} \frac{Z_n}{n^{2 - \gamma + \delta}} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

The scaling $n^{2-\gamma}$ is believed to be sharp, up to some slowly-varying function. We therefore conjecture that for any $\delta > 0$,

\begin{equation*} \frac{Z_n}{n^{2 - \gamma - \delta}} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} \infty.\end{equation*}

From Theorem 2.1 we obtain several useful results, summarized in the corollary below. The proof can be found in Section A.1. Let $Z_{ij}$ be the number of edges between i and j that have been removed, and $Y_i$ the number of removed stubs of node i. Then we have

\begin{equation*} Y_i = \sum_{j = 1}^n Z_{ij} = X_{ii} + \sum_{j \ne i}\left( X_{ij} - {\mathds{1}}_{\left\{{X_{ij} > 0}\right\}}\right)\!.\end{equation*}

Corollary 2.1. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$ and $G_n = {\texttt{ECM}}(\textbf{D}_n)$. Then, for any integer $p \ge 0$ and $\delta > 0$,

\begin{equation*} \frac{\sum_{i = 1}^n D_i^p Y_i}{n^{\frac{p}{\gamma} + 2 - \gamma + \delta}} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0 \quad \text{and} \quad \frac{\sum_{1 \le i < j \le n} Z_{ij} D_i D_j}{n^{\frac{2}{\gamma} + 2 - \gamma + \delta}} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

The first result of Corollary 2.1 gives the scaling of the difference of the sum of powers of degrees, between CM and ECM. To see why, note that since $\widehat{D}_i^q = (D_i - Y_i)^q$ and $Y_i \le D_i$, by the mean value theorem we have, for any integer $q \ge 1$,

\begin{equation*} \left|\sum_{i=1}^n D_i^q - \sum_{i=1}^n \widehat{D}_i^q\right| \le q \sum_{i = 1}^n D_i^{q-1} Y_i.\end{equation*}

Hence, for any $q \ge 1$ and $\delta > 0$,

\begin{equation*} \left|\sum_{i=1}^n D_i^q - \sum_{i=1}^n \widehat{D}_i^q\right| = o_\mathbb{P}\left(n^{\frac{q-1}{\gamma} + 2 -\gamma +\delta}\right)\!.\end{equation*}

2.2. Results for regularly-varying random variables

In addition to the number of edges, we shall make use of several results regarding the scaling of expressions with regularly-varying random variables. We summarize them here, starting with a concentration result on the sum of i.i.d. samples, which is a direct consequence of the Kolmogorov–Marcinkiewicz–Zygmund strong law of large numbers.

Lemma 2.1. Let $(X_i)_{i \ge 1}$ be independent copies of a nonnegative regularly-varying random variable X with exponent $\gamma > 1$ and mean $\mu$. Then, with $\kappa = (\gamma - 1)/(1 + \gamma)$,

\begin{equation*} \frac{\left|\mu n - \sum_{i = 1}^n X_i\right|}{n^{1 - \kappa}} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

In particular, since $L_n = \sum_{i = 1}^n D_i$, with all $D_i$ being i.i.d. with regularly-varying distribution (1) and mean $\mu$, it holds that $n^{\kappa - 1}\left|L_n - \mu n\right| \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0$. Therefore, the above lemma allows us to replace $L_n$ with $\mu n$ in our expressions.

The next proposition gives the scaling of sums of different powers of independent copies of a regularly-varying random variable. Recall that $(x \vee y)$ denotes the maximum of x and y.

Finally we have following lemma, where we write $f(t) \sim g(t)$ as $t \to \infty$ to denote that $\lim_{t \to \infty} f(t)/g(t) = 1$. Recall that $(x \wedge y)$ denotes the minimum of x and y.

Lemma 2.2. ([Reference Van der Hoorn20, Lemma 2.6].) Let X be a nonnegative regularly-varying random variable with exponent $1 < \gamma < 2$ and slowly-varying function $\mathcal{L}$. Then

\begin{equation*} {\mathbb{E}}\left[{X \left(1 \wedge \frac{X}{t}\right)}\right] \sim \frac{\gamma}{3\gamma - 2 - \gamma^2} \mathcal{L}(t) t^{1 - \gamma} \quad \text{as } t \to \infty.\end{equation*}

2.3. Heuristics of the proof for Pearson in the ECM

Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$, consider $G_n = \texttt{CM}(\textbf{D}_n)$, and let us write $r_n = r^+_n - r^-_n$, where $r^\pm_n$ are positive functions given by

(11) \begin{align} r^+_n = \frac{\sum_{i,j = 1}^n X_{ij} D_i D_j}{\sum_{i = 1}^n D_i^3 - \frac{1}{L_n} \left(\sum_{i = 1}^n D_i^2\right)^2}, \quad r^-_n = \frac{\frac{1}{L_n} \left(\sum_{i = 1}^n D_i^2\right)^2}{\sum_{i = 1}^n D_i^3 - \frac{1}{L_n} \left(\sum_{i = 1}^n D_i^2\right)^2}.\end{align}

First note that by the stable-law central limit theorem (see for instance [Reference Whitt42]), there exist two slowly-varying functions $\mathcal{L}_0$ and $\mathcal{L}_0^\prime$ such that

(12) \begin{align} \frac{\sum_{i = 1}^n D_i^2}{\mathcal{L}_0(n) n^{\frac{2}{\gamma}}} \ensuremath{\stackrel{d}{\longrightarrow}} \mathcal{S}_{\gamma/2} \quad \text{and} \quad \frac{\sum_{i = 1}^n D_i^3}{\mathcal{L}_0^\prime(n) n^{\frac{3}{\gamma}}} \ensuremath{\stackrel{d}{\longrightarrow}} \mathcal{S}_{\gamma/3} \quad \text{as } n \to \infty.\end{align}

Applying this and using that $L_n \approx \mu n$,

(13) \begin{equation} \mathcal{L}_1(n) \mu n^{1 - \frac{1}{\gamma}} r^-_n \approx \frac{\mathcal{L}_0^\prime(n) n^{\frac{3}{\gamma}}}{\left((\mathcal{L}_0(n) n^{\frac{2}{\gamma}}\right)^2} \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{\sum_{i = 1}^n D_i^3} \ensuremath{\stackrel{d}{\longrightarrow}} \frac{\mathcal{S}_{\gamma/2}^2}{\mathcal{S}_{\gamma/3}},\end{equation}

with $\mathcal{L}_1(n) = \mathcal{L}_0^\prime(n)/(\mathcal{L}_0(n))^2$. Note that $r^-_n$ scales roughly as $n^{\frac{1}{\gamma}-1}$ and thus tends to zero. This extends the results in the literature that $r_n$ has a nonnegative limit [Reference Van der Hofstad and Litvak17].

Next, we need to show that this result also holds when we move to the erased model, i.e. when all degrees $D_i$ are replaced by $\widehat{D}_i$. To understand how this works, consider the sum of the squares of the degrees in the erased model $\sum_{i = 1}^n \widehat{D}_i^2$. Recall that $Y_i$ is the number of removed stubs of node i. Then we have

\begin{equation*} \sum_{i = 1}^n \widehat{D}_i^2 = \sum_{i = 1}^n (D_i - Y_i)^2 = \sum_{i = 1}^n D_i^2 + \sum_{i = 1}^n\left( Y_i^2 - 2D_iY_i\right)\!,\end{equation*}

and hence

(14) \begin{equation} \left|\sum_{i = 1}^n \widehat{D}_i^2 - \sum_{i = 1}^n D_i^2\right| \le 2 \sum_{i = 1}^n Y_i D_i.\end{equation}

Therefore we only need to show that the error vanishes when we divide by $n^{\frac{2}{\gamma}}$. For this we can use Corollary 2.1 to get, heuristically,

(15) \begin{equation} n^{-\frac{2}{\gamma}}\left|\sum_{i = 1}^n \widehat{D}_i^2 - \sum_{i = 1}^n D_i^2\right| \le 2n^{-\frac{2}{\gamma}} \sum_{i = 1}^n Y_i D_i \approx n^{-\frac{1}{\gamma} + 2 - \gamma} \to 0.\end{equation}

These results will be used to prove that when $\widehat{G}_n = \texttt{ECM}(\textbf{D}_n)$,

\begin{equation*} n^{1 - \frac{1}{\gamma}} \left|r^{-}_n - \widehat{r}^{\,-}_n\right| \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{equation*}

so that by (13),

\begin{equation*} \mathcal{L}_1(n) \mu n^{1 - \frac{1}{\gamma}} \widehat{r}^{\,-}_n \ensuremath{\stackrel{d}{\longrightarrow}} \frac{\mathcal{S}_{\gamma/2}^2}{\mathcal{S}_{\gamma/3}} \quad \text{as } n \to \infty.\end{equation*}

The final ingredient is Proposition 3.3, where we show that for some $\delta > 0$, as $n \to \infty$,

(16) \begin{equation} n^{1 - \frac{1}{\gamma} + \delta} \widehat{r}^{\,+}_n \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation}

The result then follows, since for n large enough $\mathcal{L}_1(n) \le C n^{\delta}$ and hence

\begin{equation*} \mu \mathcal{L}_1(n) n^{1 - \frac{1}{\gamma}} \widehat{r}_n = \mu \mathcal{L}_1(n) n^{1 - \frac{1}{\gamma}} \widehat{r}^{\,+}_n - \mu \mathcal{L}_1(n) n^{1 - \frac{1}{\gamma}}\widehat{r}^{\,-}_n \ensuremath{\stackrel{d}{\longrightarrow}} -\frac{\mathcal{S}^2_{\gamma/2}}{\mathcal{S}_{\gamma/3}}.\end{equation*}

To establish (16), let $\widehat{X}_{ij}$ denote the number of edges between i and j in the erased graph, and note that since we remove self-loops, $\widehat{X}_{ii} = 0$, while in the other cases $\widehat{X}_{ij} = {\mathds{1}}_{\left\{{X_{ij} > 0}\right\}}$. We consider the numerator of $\widehat{r}^{\,+}_n$,

\begin{equation*} \sum_{1 \le i < j \le n} \widehat{X}_{ij}D_i D_j,\end{equation*}

and will show that as $n \to \infty$,

\begin{equation*} n^{1 - \frac{4}{\gamma} + \delta}\sum_{1 \le i < j \le n} \widehat{X}_{ij}D_i D_j \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{equation*}

by approximating ${\mathbb{E}}_n\left[{\widehat{X}_{ij}}\right]$ by $1 - e^{-D_iD_j/L_n}$; see Lemma 3.2. Since the denominator of $\widehat{r}^{\,+}_n$ scales as $n^{3/\gamma}$ we get that $n^{1 - 1/\gamma + \delta} \, \widehat{r}^{\,+}_n \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0$.

2.4. Intuition behind Conjecture 1.1

We note that, by (2) and since $\Big(\sum_{i\in[n]} D_i^2\Big)^2=o_{\scriptscriptstyle{\mathbb{P}}}\Big(L_n \sum_{i\in[n]} D_i^3\Big),$

(17) \begin{equation}{ r_n=\frac{\sum_{i,j\in[n]} D_iD_j \Big[X_{ij}-\frac{D_iD_j}{L_n}\Big]}{\sum_{i\in[n]} D_i^3}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)).}\end{equation}

We rewrite this as

(18) \begin{equation}{ r_n=\frac{\sum_{i,j\in[n]} D_iD_j \Big[X_{ij}-\frac{D_iD_j}{L_n-1}\Big]}{\sum_{i\in[n]} D_i^3}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))+\frac{\sum_{i,j,\in[n]} \frac{D_i^2D_j^2}{L_n(L_n-1)}}{\sum_{i\in[n]} D_i^3}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)).}\end{equation}

The second term is

(19) \begin{equation}{ O_{\scriptscriptstyle\mathbb{P}}\left({n^{4\gamma-2-3\gamma}}\right)=O_{\scriptscriptstyle\mathbb{P}}\left({n^{\gamma-2}}\right)=o_{\scriptscriptstyle{\mathbb{P}}}(n^{-1/2}),}\end{equation}

since $\gamma\in \big(\tfrac{1}{2},1\big)$, and can thus be ignored. We are thus left to study the first term.

Since $\mathbb{E}_n[X_{ij}]=D_iD_j/(L_n-1)$, this term is centered. Furthermore, the probability that any half-edge incident to vertex i is connected to vertex j equals $D_j/(L_n-1)$. These indicators are weakly dependent, so we assume that we can replace the conditional law of $X_{ij}$ given the degrees by a binomial random variable with $D_i$ experiments and success probability $D_j/(L_n-1)$. We will also assume that these random variables are asymptotically independent. These are the two main assumptions made in this heuristic explanation.

Since a binomial random variable is close to a normal when the number of experiments tends to infinity, these assumptions then suggest that

(20) \begin{equation}{ r_n\approx \mathcal{N}\big(0,\sigma_n^2\big),}\end{equation}

where, with $\text{Var}_n$ denoting the conditional variance given the degrees,

(21) \begin{equation}{ \sigma_n^2=\frac{\sum_{i,j\in[n]} D_i^2D_j^2 \text{Var}_n(X_{ij})}{\Big(\sum_{i\in[n]} D_i^3\Big)^2}.}\end{equation}

Furthermore, again using that $X_{ij}$ is close to a binomial, $\text{Var}_{n}(X_{ij})\approx D_i (D_j/(L_n-1))\break (1-D_j/(L_n-1))\approx D_iD_j/L_n$. This suggests that

(22) \begin{equation}{ \sigma_n^2\approx \frac{\sum_{i<j\in[n]} D_i^3D_j^3/L_n}{\Big(\sum_{i\in[n]} D_i^3\Big)^2}=\frac{1}{L_n},}\end{equation}

which supports the conjecture in (7) but now with $\sigma^2=1/\mu$.

It turns out that the above analysis is not quite correct, as $X_{ij}=X_{ji}$ when $i\leq j$, which means that these terms are highly correlated. Since terms with $i<j$ also appear several times, whereas $i=j$ does not, this turns out to change the variance formula slightly, as we discuss in the proof of Lemma 1.1 in Section 3.5.

2.5. Proofs for clustering in CM and ECM

The general idea behind the proof for both Theorems 1.3 and 1.4 is that, conditioned on the degrees, the clustering coefficients are concentrated around their conditional mean. We then proceed by analyzing this term using stable laws for regularly-varying random variables to obtain the results.

2.5.1 Configuration model

To construct a triangle, six different half-edges at three distinct vertices need to be paired into a triangle. For a vertex with degree $D_i$, there are $D_i(D_i-1)/2$ ways to choose two half-edges incident to it. The probability that any two half-edges are paired in the CM can be approximated by $1/L_n$. Thus, the probability that a given set of six half-edges forms a triangle can be approximated by $1/L_n^3$. We then investigate $\mathcal{I}$, the set of all sets of six half-edges that could possibly form a triangle together. The expected number of triangles can then be approximated by $|\mathcal{I}|/L_n^3$. By computing the size of the set $\mathcal{I}$, we obtain that the conditional expectation for the clustering coefficient can be written as

\begin{equation*} {\mathbb{E}}_n\left[{C_n}\right] \approx \frac{|\mathcal{I}|}{L_n^3\sum_{i \in[n]}D_i(D_i-1)}\approx \frac{1}{L_n^3}\left(\left(\sum_{i = 1}^n D_i^2\right)^3 - 3\sum_{i = 1}^n D_i^4+2\frac{\sum_{i=1}^nD_i^6}{\sum_{i=1}^nD_i^2}\right)\!.\end{equation*}

The full details can be found in Section 4.1. Here the first term describes the expected number of times six half-edges are paired into a triangle. The last two terms exclude triangles containing either multi-edges or self-loops. Then by the stable-law central limit theorem [Reference Whitt42] we have that there exists a slowly-varying function $\mathcal{L}_2$ such that

\begin{equation*} \frac{\sum_{i = 1}^n D_i^2}{\mathcal{L}_2(n)n^{\frac{2}{\gamma}}} \ensuremath{\stackrel{d}{\longrightarrow}} \mathcal{S}_{\gamma/2}, \quad \frac{\sum_{i = 1}^n D_i^4}{\mathcal{L}_2(n)^2n^{\frac{4}{\gamma}}} \ensuremath{\stackrel{d}{\longrightarrow}} \mathcal{S}_{\gamma/4}, \quad \text{and}\quad \frac{\sum_{i=1}^nD_i^6}{\mathcal{L}_2(n)^6n^{\frac{2}{\gamma}}\sum_{i=1}^{n}D_i^2}\ensuremath{\stackrel{d}{\longrightarrow}} \frac{\mathcal{S}_{\gamma/6}}{\mathcal{S}_{\gamma/2}}.\end{equation*}

Hence, using that $L_n \approx \mu n$ we obtain that

\begin{equation*} \frac{{\mathbb{E}}_n\left[{C_n}\right]}{\mathcal{L}_2(n) n^{\frac{4}{\gamma} - 3}} \ensuremath{\stackrel{d}{\longrightarrow}} \frac{1}{\mu^3} \left(\mathcal{S}_{\gamma/2}^2 - 3\mathcal{S}_{\gamma/4}+\frac{2\mathcal{S}_{\gamma/6}}{\mathcal{S}_{\gamma/2}}\right)\!,\end{equation*}

where $\mathcal{S}_{\gamma/2}$, $\mathcal{S}_{\gamma/4}$, and $\mathcal{S}_{\gamma/6}$ are given by (6). To complete the proof we establish a concentration result for $C_n$, conditioned on the degrees. To be more precise, we employ a careful counting argument, following the approach in the proof of [Reference Van der Hofstad13, Proposition 7.13], to show (see Lemma 4.1) that there exists a $\delta > 0$ such that

\begin{equation*} \frac{n^\delta \textrm{Var}_n(C_n)}{n^{\frac{8}{\gamma} - 6}} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{equation*}

where $\textrm{Var}_n$ denotes the conditional variance given the degrees. Then it follows from Chebyshev’s inequality, conditioned on the degrees, that

\begin{equation*} \frac{\left|C_n - {\mathbb{E}}_n\left[{C_n}\right]\right|}{\mathcal{L}_2(n) n^{\frac{4}{\gamma} - 3}} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{equation*}

and we conclude that

\begin{equation*} \frac{C_n}{\mathcal{L}_2(n) n^{\frac{4}{\gamma} - 3}} \ensuremath{\stackrel{d}{\longrightarrow}} \frac{1}{\mu^3} \left(\mathcal{S}_{\gamma/2}^2 - 3\mathcal{S}_{\gamma/4}+2\mathcal{S}_{\gamma/6}/\mathcal{S}_{\gamma/2}\right)\!.\end{equation*}

2.5.2 Erased configuration model

The difficulty for clustering in ECM, compared to CM, is in showing that $\widehat{C}_n$ behaves as its conditional expectation, as well as in establishing its scaling. To compute this we first fix an $\varepsilon > 0$ and show in Lemma 4.2 that the main contribution is given by triples of nodes with degrees $\varepsilon \sqrt{n} \le D \le \frac{\sqrt{n}}{\varepsilon}$; i.e.,

\begin{align*} \sum_{1 \le i < j < k \le n} \widehat{X}_{ij}\widehat{X}_{jk}\widehat{X}_{ik} = & \sum_{1 \le i < j < k \le n} \widehat{X}_{ij}\widehat{X}_{jk}\widehat{X}_{ik} {\mathds{1}}_{\left\{{\varepsilon \sqrt{n} \le D_i, D_j, D_k \le \frac{\sqrt{n}}{\varepsilon}}\right\}}\\ &+ O\Big(\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)}\Big) \mathcal{E}_1(\varepsilon),\end{align*}

where $\mathcal{E}_1(\varepsilon)$ is an error function, independent of n, with $\lim_{\varepsilon \to 0} \mathscr{E}_1(\varepsilon) = 0$. Then we use that approximately ${\mathbb{E}}_n\left[{\widehat{X}_{ij}}\right] \approx 1 - \textup{e}^{-{D_i D_j}/{L_n}}$ to show that

\begin{equation*}{\mathbb{E}}_n\left[{\widehat{C}_n}\right]\approx \frac{6\sum_{1\leq i< j < k\leq n} g_{n, \varepsilon}(D_i, D_j, D_k) +O\Big(\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)}\Big) \mathcal{E}_1(\varepsilon)}{\sum_{i = 1}^n \widehat{D}_i(\widehat{D}_i - 1)},\end{equation*}

where

\begin{equation*}g_{n, \varepsilon}(x,y,z) = \left(1 - \textup{e}^{-\frac{xy}{L_n}}\right)\left(1 - \textup{e}^{-\frac{yz}{L_n}}\right)\left(1 - \textup{e}^{-\frac{zx}{L_n}}\right){\mathds{1}}_{\left\{{\varepsilon \sqrt{n} \le x, y, z \le \frac{\sqrt{n}}{\varepsilon}}\right\}}.\end{equation*}

Here $\mathbb{E}_n$ again denotes expectation conditioned on $\textbf{\textit{D}}_n$, hence conditional on the sampled degrees of the underlying CM. The precise statement can be found in Lemma 4.3. After that, we show in Lemma 4.6 that $\widehat{C}_n$ concentrates around its expectation conditioned on the sampled degrees, so that conditioned on the sampled degree sequence, we can approximate $\widehat{C}_n\approx {\mathbb{E}}_n\left[{\widehat{C}_n}\right]$. We then replace $L_n$ by $\mu n$ in Lemma 4.4, so that, conditioned on the degree sequence,

\begin{equation*}\widehat{C}_n \approx \frac{6\sum_{1\leq i< j < k\leq n} f_{n, \varepsilon}(D_i, D_j, D_k)+O\Big(\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)}\Big) \mathcal{E}_1(\varepsilon)}{\sum_{i = 1}^n \widehat{D}_i(\widehat{D}_i - 1)},\end{equation*}

with

\begin{equation*} f_{n, \varepsilon}(x,y,z) = \left(1 - \textup{e}^{-\frac{xy}{\mu n}}\right)\left(1 - \textup{e}^{-\frac{yz}{\mu n}}\right)\left(1 - \textup{e}^{-\frac{zx}{\mu n}}\right) {\mathds{1}}_{\left\{{\varepsilon \sqrt{n} \le x, y, z \le \frac{\sqrt{n}}{\varepsilon}}\right\}}.\end{equation*}

We then take the random degrees into account, by showing that

\begin{equation*} \frac{1}{\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)}}\sum_{1\leq i< j < k\leq n}f_{n, \varepsilon}(D_1, D_2, D_3) \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} \frac{1}{6}\mu^{-\frac{3\gamma}{2}}A_\gamma(\varepsilon) + \mathscr{E}_2(\varepsilon),\end{equation*}

where

\begin{equation*} A_\gamma(\varepsilon) = \int_\varepsilon^{1/\varepsilon} \int_\varepsilon^{1/\varepsilon} \int_\varepsilon^{1/\varepsilon} \frac{1}{(xyz)^{\gamma + 1}} (1 -\textup{e}^{-xy}) (1 - \textup{e}^{-xy}) (1 -\textup{e}^{-xy}) \, \,\text{d} x \,\text{d} y \,\text{d} z,\end{equation*}

and $\mathcal{E}_2(\varepsilon)$ is a deterministic error function, with $\lim_{\varepsilon \to 0} \mathscr{E}_2(\varepsilon) = 0$. Finally, we again replace the $\widehat{D}_i$ with $D_i$ and use the stable-law central limit theorem to obtain a slowly-varying function $\mathcal{L}_3$ such that

\begin{equation*} \frac{\mathcal{L}_3(n) n^{\frac{2}{\gamma}}}{\sum_{i = 1}^n \widehat{D}_i(\widehat{D}_i - 1)} \ensuremath{\stackrel{d}{\longrightarrow}} \frac{1}{\mathcal{S}_{\gamma/2}}.\end{equation*}

Combining all these results implies that, for any $\varepsilon > 0$,

\begin{equation*} \frac{\mathcal{L}_3(n) \widehat{C}_n}{\mathcal{L}(\sqrt{\mu n})^3 n^{-\frac{3 \gamma}{2} + 3 - \frac{2}{\gamma}}} \ensuremath{\stackrel{d}{\longrightarrow}} \mu^{-\frac{3\gamma}{2}}\frac{A_\gamma(\varepsilon)}{\mathcal{S}_{\gamma/2}^2} + \frac{\mathscr{E}_1(\varepsilon)+\mathcal{E}_2(\varepsilon)}{\mathcal{S}_{\gamma/2}^2},\end{equation*}

from which the result follows by taking $\varepsilon \downarrow0$.

3.1. Limit theorem for $r^-_n$

We first prove a limit theorem for $r^-_n$, when $G_n = \texttt{CM}(\textbf{D}_n)$. Recall that

\begin{equation*} r^-_n = \frac{\frac{1}{L_n} \left(\sum_{i = 1}^n D_i^2\right)^2}{\sum_{i = 1}^n D_i^3 - \frac{1}{L_n} \left(\sum_{i = 1}^n D_i^2\right)^2}.\end{equation*}

Proposition 3.1. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$ and ${\mathbb{E}}\left[{\mathscr{D}}\right] = \mu$. Then, if $G_n = {\texttt{CM}}(\textbf{D}_n)$, there exists a slowly-varying function $\mathcal{L}_0$ such that

\begin{equation*} \mu \mathcal{L}_0(n)n^{1-\frac{1}{\gamma}}r^-_n \ensuremath{\stackrel{d}{\longrightarrow}} \frac{\mathcal{S}_{\gamma/2}^2}{\mathcal{S}_{\gamma/3}}\end{equation*}

as $n \to \infty$. Here $\mathcal{S}_{\gamma/2}$ and $\mathcal{S}_{\gamma/3}$ are given by (6).

Proof. We will first show that there exists a slowly-varying function $\mathcal{L}_0$ such that

(23) \begin{equation} \mu\mathcal{L}_0(n)n^{1 - \frac{1}{\gamma}} \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{\mu n \sum_{i = 1}^n D_i^3} \ensuremath{\stackrel{d}{\longrightarrow}} \frac{\mathcal{S}_{\gamma/2}^2}{\mathcal{S}_{\gamma/3}}\end{equation}

as $n \to \infty$, with $\mathcal{S}_{\gamma/2}$ and $\mathcal{S}_{\gamma/3}$ defined by (6).

Let $D_{(i)}$ denote the ith largest degree in $\textbf{D}_n$, i.e. $D_{(1)} \ge D_{(2)} \ge \dots D_{(n)}$, and let $\Gamma_i$ be defined as in (5). Then, since $\sum_{i = 1}^n D_{(i)}^p = \sum_{i = 1}^n D_i^p$ for any $p \ge 0$, it follows from [Reference Van der Hofstad13, Theorem 2.33] that for some slowly-varying functions $\mathcal{L}_2$, $\mathcal{L}_3$, $\mathcal{L}_4$, and $\mathcal{L}_6$, as $n \to \infty$,

(24) \begin{equation}\begin{aligned} &\left(\frac{n^{-\frac{4}{\gamma}}}{\mathcal{L}_2(n)^2} \left(\sum_{i = 1}^n D_i^2\right)^2, \, \frac{n^{-\frac{3}{\gamma}}}{\mathcal{L}_3(n)} \sum_{i = 1}^n D_i^3, \, \frac{n^{-\frac{4}{\gamma}}}{\mathcal{L}_4(n)}\sum_{i = 1}^n D_i^4, \, \frac{n^{-\frac{6}{\gamma}}}{\mathcal{L}_6(n)}\sum_{i = 1}^n D_i^6\right)\\ &\qquad\ensuremath{\stackrel{d}{\longrightarrow}} \left(\left(\sum_{i = 1}^\infty \Gamma_i^{-\frac{2}{\gamma}}\right)^2, \, \sum_{i = 1}^\infty \Gamma_i^{-\frac{3}{\gamma}}, \, \sum_{i = 1}^\infty \Gamma_i^{-\frac{4}{\gamma}}, \, \sum_{i = 1}^\infty \Gamma_i^{-\frac{6}{\gamma}}\right)\!.\end{aligned}\end{equation}

Here we include the fourth and sixth moments, since these will be needed later for proving Theorem 1.3.

Note that $\mathcal{L}_0(n) \,:\!=\mathcal{L}_2(n)/\mathcal{L}_1(n)^2$ is slowly varying and $\mathbb{P}\left({\sum_{i = 1}^\infty \Gamma_i^{-t/\gamma} \le 0}\right) = 0$ for any $t \ge 2$. Hence (23) follows from (24) and the continuous mapping theorem. Hence, to prove the main result, it is enough to show that

\begin{equation*} \mathcal{L}_0(n)n^{1 - \frac{1}{\gamma}}\left|r^-_n - \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{\mu n \sum_{i = 1}^n D_i^3}\right| \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

We will prove the stronger statement

(25) \begin{equation} n^{1 - \frac{1}{\gamma} + \frac{\kappa}{2}}\left|r^-_n - \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{\mu n \sum_{i = 1}^n D_i^3}\right| \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{equation}

where $\kappa = (\gamma - 1)/(\gamma + 1) > 0$ is the same as in Lemma 2.1.

Note that by Lemma 2.1 we have that $\mu n/ L_n \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 1$. Hence, by (23), we have that for any $\delta > 0$,

\begin{equation*} n^{1 - \frac{1}{\gamma} - \delta}\frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{L_n \sum_{i = 1}^n D_i^3} = n^{1 - \frac{1}{\gamma} - \delta} \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{\mu n \sum_{i = 1}^n D_i^3}\left(\frac{\mu n}{L_n}\right) \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{equation*}

from which we conclude that

(26) \begin{equation} n^{1 - \frac{1}{\gamma} - \delta} \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation}

To show (25), we write

\begin{align*} \left|\frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{\mu n\sum_{i = 1}^n D_i^3} - r^-_n\right| &= \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{L_n \sum_{i = 1}^n D_i^3} \left|\frac{(L_n - \mu n)\sum_{i = 1}^n D_i^3 + \mu n \left(\sum_{i = 1}^n D_i^2\right)^2} {\mu n \left(L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2\right)}\right|\\ &\le \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2}\frac{\left|\mu n - L_n\right|}{\mu n} \\ &\quad + \left(\frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2}\right)^2.\end{align*}

For the first term we have, using (26) and Lemma 2.1,

\begin{align*} &n^{1 - \frac{1}{\gamma} + \frac{\kappa}{2}}\frac{\left(\sum_{i = 1}^n D_i^2\right)^2} {L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2} \frac{\left|\mu n - L_n\right|}{\mu n}\\ &= n^{1 - \frac{1}{\gamma} - \frac{\kappa}{2}}\left(\frac{\left(\sum_{i = 1}^n D_i^2\right)^2} {L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2}\right) \left(\frac{\left|\mu n - L_n\right|}{\mu n^{1 - \kappa}}\right) \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{align*}

Now, let $\delta = 1 - 1/\gamma - \kappa/2 > 0$. Then, since $1 - 1/\gamma + \kappa/2 = 2 - 2/\gamma - \delta$, it follows from (26) that

\begin{equation*} n^{1 - \frac{1}{\gamma} + \frac{\kappa}{2}}\left(\frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2}\right)^2 = \left(\frac{n^{1 - \frac{1}{\gamma} - \frac{\delta}{2}} \left(\sum_{i = 1}^n D_i^2\right)^2}{L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2}\right)^2 \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{equation*}

which finishes the proof of (25).

3.2. Limit theorem for $\widehat{r}^{\,-}_n$

We now turn to the ECM. Observe that for $\widehat{G}_n = \texttt{ECM}(\textbf{D}_n)$, (14) and Corollary 2.1 imply that, for any $\delta > 0$,

\begin{equation*} \frac{\left|\sum_{i = 1}^n D_i^2 - \sum_{i = 1}^n \widehat{D}_i^2\right|}{n^{\frac{1}{\gamma} + 2 - \gamma + \delta}} \le \frac{2\sum_{i = 1}^n D_i Y_i}{n^{\frac{1}{\gamma} + 2 - \gamma + \delta}} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

Since $\frac{2}{\gamma} > \frac{1}{\gamma} + 2 - \gamma$ for all $\gamma > 1$ this result implies that for any $\delta > 0$,

(27) \begin{equation} \frac{\sum_{i = 1}^n \widehat{D}_i^2}{n^{\frac{2}{\gamma} + \delta}} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation}

This line of reasoning can be extended to sums of $\widehat{D}_i^p$ for any $p > 0$, proving that the degrees $\widehat{D}_i$ in the ECM satisfy the same scaling results as those for $D_i$. In particular we have the following extension of (26) to the ECM. Recall that $\mathscr{D}$ denotes an integer-valued random variable with a regularly-varying distribution defined by

\begin{equation*} \mathbb{P}\left({\mathscr{D} > t}\right) = \mathcal{L}(t) t^{-\gamma}\end{equation*}

(see (1)).

Lemma 3.1. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1 < \gamma < 3$ and $\widehat{G}_n = {\texttt{ECM}}(\textbf{D}_n)$. Then, for any $\delta > 0$,

\begin{equation*} n^{1 - \frac{1}{\gamma} - \delta} \frac{\left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2}{L_n \sum_{i = 1}^n \widehat{D}_i^3 - \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

Now recall that $\widehat{r}^{\,-}_n$ denotes the negative part of Pearson’s correlation coefficient for the CM, i.e.

\begin{equation*} \widehat{r}^{\, -}_n = \frac{\frac{1}{L_n} \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2}{\sum_{i = 1}^n \widehat{D}_i^3 - \frac{1}{\widehat{L}_n} \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2}.\end{equation*}

The next proposition shows that $|r^{-}_n-\widehat{r}^{\,-}_n| = o_\mathbb{P}\left(n^{\frac{1}{\gamma} - 1}\right)$.

Proposition 3.2. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$. Let $\widehat{G}_n = {\texttt{ECM}}(\textbf{D}_n)$, denote by $G_n$ the graph before the removal of edges, and recall that $r_n^-$ and $\widehat{r}^{\,-}_n$ denote the negative parts of Pearson’s correlation coefficient in $G_n$ and $\widehat{G}_n$, respectively. Then

\begin{equation*} n^{1 - \frac{1}{\gamma} + \frac{(\gamma-1)^2}{4 \gamma}}\left|r^{-}_n - \widehat{r}^{\,-}_n\right| \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

Proof. The proof consists of splitting the main term into separate terms, which can be expressed in terms of erased stubs or edges, and showing that each of these terms converges to zero. Throughout the proof we let

\begin{equation*} \delta = \frac{(\gamma - 1)^2}{4\gamma}.\end{equation*}

We start by splitting the main term as

(28) \begin{align} \left|r^{-}_n - \widehat{r}^{\,-}_n\right| &\le \frac{\left|\left(\sum_{i = 1}^n D_i^2\right)^2 - \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2\right|} {L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2} \end{align}

(29) \begin{align} &\quad + \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2\left|\frac{1}{L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2} - \frac{1}{\widehat{L}_n \sum_{i = 1}^n \widehat{D}_i^3 - \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2}\right|.\end{align}

For (28) we use that

\begin{align*} \left|\left(\sum_{i = 1}^n D_i^2\right)^2 - \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2\right| &= \left(\sum_{i = 1}^n D_i^2 - \widehat{D}_i^2\right) \left(\sum_{i = 1}^n D_i^2 + \widehat{D}_i^2\right)\\ &\le \left(\sum_{i = 1}^n 2 D_i Y_i\right)\left(\sum_{i = 1}^n 2D_i^2 + Y_i^2\right) \le 6 \sum_{i = 1}^n D_i^2 \sum_{j = 1}^n Y_j D_j\end{align*}

to obtain

(30) \begin{align*} \frac{\left|\left(\sum_{i = 1}^n D_i^2\right)^2 - \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2\right|} {L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2} &\le \frac{6 \sum_{i = 1}^n D_i^2 \sum_{i = 1}^n Y_i D_i} {L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2} \\ &= \frac{6 \sum_{i = 1}^n Y_i D_i}{\sum_{i = 1}^n D_i^2} \frac{\left(\sum_{i = 1}^n D_i^2\right)^2} {L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2}. \addtocounter{equation}{1}\tag{\theequation} \end{align*}

Now observe that

(31) \begin{equation} 2 - \frac{1}{\gamma} - \gamma = -(\gamma - 1)^2/\gamma = -4\delta,\end{equation}

and hence

\begin{align*} 1 - \frac{1}{\gamma} + \delta = \gamma - 1 -3\delta = -\left(\frac{1}{\gamma} + 2 -\gamma + \delta\right) + \left(\frac{2}{\gamma} - \frac{\delta}{2}\right) +\left(1 - \frac{1}{\gamma} - \frac{\delta}{2}\right) - \delta,\end{align*}

with all three terms inside the brackets positive. Therefore, it follows from (30), together with Corollary 2.1, Proposition 2.1, and (26), that

(32) \begin{align*} &n^{1 -\frac{1}{\gamma} + \delta}\frac{\left|\left(\sum_{i = 1}^n D_i^2\right)^2 - \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2\right|}{L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2}\\[6pt] &\quad \le n^{-\delta}\left(\frac{6\sum_{i = 1}^n D_i Y_i}{n^{\frac{1}{\gamma} + 2 - \gamma + \delta}}\right) \left(\frac{n^{\frac{2}{\gamma} - \frac{\delta}{2}}}{\sum_{i = 1}^n D_i^2}\right) \left( \frac{n^{1 - \frac{1}{\gamma} - \frac{\delta}{2}}\left(\sum_{i = 1}^n D_i^2\right)^2} {L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2}\right) \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0. \addtocounter{equation}{1}\tag{\theequation} \end{align*}

The second term, (29), requires more work. Let us first write

\begin{align*} &\left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2 \left|\frac{1}{L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2} - \frac{1}{\widehat{L}_n \sum_{i = 1}^n \widehat{D}_i^3 - \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2}\right|\\[5pt] &\qquad :\!= \frac{\left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2}{\widehat{L}_n \sum_{i = 1}^n \widehat{D}_i^3 - \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2} \left(I_n^{(1)} + I_n^{(2)} + I_n^{(3)}\right)\!,\end{align*}

with

\begin{align*} I_n^{(1)} &\,:\!= \frac{\left(\sum_{i = 1}^n D_i^2\right)^2 - \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2} {L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2},\\[4pt] I_n^{(2)} &\,:\!= \frac{Z_n \sum_{i = 1}^n \widehat{D}_i^3}{L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2},\\[4pt] I_n^{(3)} &\,:\!= \frac{L_n \left|\sum_{i = 1}^n D_i^3 - \sum_{i = 1}^n \widehat{D}_i^3\right|} {L_n \sum_{i = 1}^n D_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2},\end{align*}

and we recall that $Z_n = L_n - \widehat{L}_n$ denotes the total number of removed edges. Note that

\begin{equation*} n^{1 - \frac{1}{\gamma} - \frac{\delta}{2}}\frac{\left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2}{\widehat{L}_n \sum_{i = 1}^n \widehat{D}_i^3 - \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

Therefore, in order to complete the proof, it suffices to show that

\begin{equation*} n^{\frac{3\delta}{2}} I_n^{(t)} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0 \quad \text{for } t = 1,2, 3.\end{equation*}

For $t=1$ this follows from (32), since

\begin{equation*} \frac{1}{\gamma} - 1 + \frac{\delta}{2} = \frac{\gamma^2 - 10\gamma + 9}{8 \gamma} = \frac{(\gamma - 1)(\gamma - 9)}{8\gamma} < 0,\end{equation*}

and hence

\begin{equation*} \frac{3\delta}{2} < 1 - \frac{1}{\gamma} + \delta.\end{equation*}

For $t = 2$ we use that $\widehat{D}_i \le D_i$ to obtain

\begin{align*} I_n^{(2)}&\le \frac{E_n \sum_{i = 1}^n {D}_i^3}{{L}_n \sum_{i = 1}^n {D}_i^3 - \left(\sum_{i = 1}^n D_i^2\right)^2} = \frac{E_n}{L_n}\left(1 - \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{L_n \sum_{i = 1}^n {D}_i^3}\right)^{-1}.\end{align*}

By Proposition 3.1 it follows that

(33) \begin{equation} \left(1 - \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{L_n \sum_{i = 1}^n {D}_i^3}\right)^{-1} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 1.\end{equation}

In addition we have that $\varepsilon \,:\!= \gamma - 1 - 3\delta/2 > 0$ and hence, by Theorem 2.1 and the strong law of large numbers,

\begin{equation*} n^{\frac{3\delta}{2}}I_n^{(2)} \le \left(n^{\gamma - 2 - \varepsilon} \frac{E_n}{\mu}\right) \frac{\mu n}{L_n} \left(1 - \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{L_n \sum_{i = 1}^n {D}_i^3}\right)^{-1} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

Finally, for $I_n^{(3)}$ we first compute

\begin{equation*} \left|\sum_{i = 1}^n D_i^3 - \sum_{i = 1}^n \widehat{D}_i^3\right| = \sum_{i = 1}^n Y_i^3 + 3D_i^2 Y_i - 3D_iY_i^2 \le 4 \sum_{i = 1}^n Y_i D_i^2,\end{equation*}

and hence,

\begin{align*} I_n^{(3)} \le \frac{4 \sum_{i = 1}^n Y_i D_i^2}{\sum_{i = 1} D_i^3} \left(1 - \frac{\left(\sum_{i = 1}^n D_i^2\right)^2}{L_n \sum_{i = 1}^n {D}_i^3}\right)^{-1}.\end{align*}

By (33) the last term converges in probability to one. Finally, by (31),

\begin{equation*} \frac{3\delta}{2} \le 2\delta = 4\delta - 2\delta = \left(\gamma - 2 - \frac{2}{\gamma} - \delta\right) + \left(\frac{3}{\gamma} - \delta\right)\!,\end{equation*}

and hence, by Corollary 2.1 and Proposition 2.1,

\begin{equation*} n^{\frac{3\delta}{2}}I_n^{(3)} \le \left(4n^{\gamma - 2 - \frac{2}{\gamma} - \delta} \sum_{i = 1}^n Y_i D_i^2\right) \left(\frac{n^{\frac{3}{\gamma} - \delta}}{\sum_{i = 1} D_i^3}\right) \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{equation*}

which finishes the proof.

3.3. Convergence of $\ \widehat{r}^{\,+}_n$

The next step towards the proof of Theorem 1.1 is to show that, for some $\delta > 0$,

\begin{equation*} n^{1 - \frac{1}{\gamma} + \delta} \widehat{r}^{\, +}_n \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{equation*}

where

\begin{equation*} \widehat{r}^{\, +}_n = \frac{\sum_{i,j = 1}^n \widehat{X}_{ij} \widehat{D}_i \widehat{D}_j}{\sum_{i = 1}^n \widehat{D}_i^3 - \frac{1}{\widehat{L}_n} \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2}\end{equation*}

denotes the positive part of Pearson’s correlation coefficient in the ECM. Here $\widehat{X}_{ij} = {\mathds{1}}_{\left\{{X_{ij} > 0}\right\}}$ denotes the event that i and j are connected by at least one edge in the CM graph $G_n$. The main ingredient of this result is the following lemma, which gives an approximation for $\sum_{1 \le i < j \le n} \mathbb{P}_n\left({X_{ij} = 0}\right)D_i D_j$.

Lemma 3.2. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$ and ${\mathbb{E}}\left[{\mathscr{D}}\right] = \mu$. Consider graphs $G_n = {\texttt{CM}}(\textbf{D}_n)$ and define

\begin{equation*} M_n = \sum_{1 \le i < j \le n}\left|\mathbb{P}_n\left({X_{ij} = 0}\right) - \exp\left\{-\frac{D_i D_j}{L_n}\right\}\right|D_i D_j.\end{equation*}

Then, for any $K > 0$ and $0 < \delta < \left(\frac{2 - \gamma}{\gamma} \wedge \frac{\gamma - 1}{\gamma}\right)$,

\begin{equation*} n^{1 - \frac{4}{\gamma} + \delta} Z_n \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

In our proofs $M_n$ will be divided by

\begin{equation*} \sum_{i = 1}^n D_i^3 - \frac{1}{L_n} \left(\sum_{i = 1}^n D_i^2\right)^2,\end{equation*}

which is of the order $n^{3/\gamma}$. Hence the final expression will be of the order $n^{\frac{1}{\gamma} - 1 - \delta} =o\big(n^{\frac{1}{\gamma} - 1}\big)$, which is enough to prove the final result.

To prove Lemma 3.2, we will use the following technical result.

Lemma 3.3. ([Reference Van der Hoorn and Olvera-Cravioto24, Lemma 6.7].) For any nonnegative $x, x_0 > 0$, $y_i, z_i \geq 0$ with $z_i < x$ for all i, and any $m \geq 1$, we have

\begin{equation*} - \frac{x_0}{x^2} (x_0-x)^+ - \frac{x_0}{2} \max_{1 \leq i \leq m} \frac{z_i}{(x-z_i)^2} \leq \prod_{i=1}^{m} \left( 1- \frac{z_i}{x} \right)^{y_i} - \exp\left\{ - \frac{1}{x_0} \sum_{i=1}^{m} y_i z_i \right\} \leq \frac{|x-x_0|}{(x \wedge x_0)}.\end{equation*}

Proof of Lemma 3.2. We will first consider the term $\left|\mathbb{P}_n\left({X_{ij} = 0}\right) - \exp\left\{-\frac{D_i D_j}{L_n}\right\}\right|$. It follows from computations done in [Reference Van der Hofstad, Hooghiemstra and van Mieghem14] that

(34) \begin{equation} 0 \le \mathbb{P}_n\left({X_{ij} = 0}\right) - \prod_{t = 0}^{D_i - 1} \left(1 - \frac{D_j}{L_n -2t -1}\right) \le \frac{D_i^2 D_j}{(L_n - 2D_i)^2}.\end{equation}

For the product term in (34) we have the bounds

\begin{equation*} \left(1 - \frac{D_j}{L_n -2D_i + 1}\right)^{D_i} \le \prod_{t = 0}^{D_i - 1} \left(1 - \frac{D_j}{L_n -2t -1}\right) \le \left(1 - \frac{D_j}{L_n}\right)^{D_i},\end{equation*}

and therefore, using Lemma 3.3 with $m = 1$, we can bound the difference between $\mathbb{P}_n\left({X_{ij} = 0}\right)$ and $\exp\left\{-D_i D_j/L_n\right\}$. For the lower bound we take $x = L_n$, $x_0 = L_n + 1 - 2D_i$, $y = D_i$, and $z = D_j$ to get

(35) \begin{equation} -\frac{L_n(2D_i - 1)}{(L_n - 2D_i + 1)^2} - \frac{D_j}{L_n + 1 - 2D_1 - D_j} \le \mathbb{P}_n\left({X_{ij} = 0}\right) - \exp\left\{-\frac{D_i D_j}{L_n}\right\},\end{equation}

while changing $x_0$ to $L_n$ yields

(36) \begin{equation} \mathbb{P}_n\left({X_{ij} = 0}\right) - \exp\left\{-\frac{D_i D_j}{L_n}\right\} \le \frac{D_i^2 D_j}{(L_n - 2D_i)^2}.\end{equation}

Combining (35) and (36) gives

\begin{align*} &\sum_{1 \le i < j \le n}\left|\mathbb{P}_n\left({X_{ij} = 0}\right) - \exp\left\{-\frac{D_i D_j}{L_n}\right\}\right|D_i D_j \\ &\quad\le \sum_{1 \le i < j \le n} \frac{2L_nD_i^2 D_j}{(L_n - 2D_i + 1)^2} + \sum_{1 \le i < j \le n} \frac{D_i D_j^2}{L_n + 1 - 2D_1 - D_j} + \sum_{1 \le i < j \le n} \frac{D_i^3 D_j^2}{(L_n - 2D_i)^2}\\[4pt] &\quad :\!= I_n^{(1)} + I_n^{(2)} + I_n^{(3)}.\end{align*}

We will now show that

(37) \begin{equation} n^{1 - \frac{4}{\gamma} + \delta} I_n^{(t)} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0 \quad \text{for } t = 1, 2, 3,\end{equation}

which proves the result.

For the remainder of the proof we define

\begin{equation*} D_n^{\text{max}} \,:\!= \max_{1 \le i \le n} D_i\end{equation*}

and observe that by our choice of $\delta$,

\begin{equation*} \varepsilon_1 \,:\!= \frac{2}{\gamma} - 1 - \delta = \frac{2 - \gamma}{\gamma} - \delta > 0 \quad \text{and} \quad \varepsilon_2 \,:\!= 1 - \frac{1}{\gamma} - \delta = \frac{\gamma - 1}{\gamma} - \delta > 0.\end{equation*}

For $t = 1$, we have

\begin{align*} I_n^{(1)} = \sum_{1 \le i < j \le n} \frac{2L_nD_i^2 D_j}{(L_n - 2D_i + 1)^2} \le \frac{2L_n^2 \sum_{i = 1}^n D_i^2}{\left(L_n^2 - 4 L_n D_n^{\text{max}}\right)} = \left(2\sum_{i = 1}^n D_i^2 \right)\left(1 - \frac{4D_n^{\text{max}}}{L_n}\right)^{-1}.\end{align*}

By the strong law of large numbers and Proposition 2.1, it follows that

\begin{equation*} \frac{M_n}{L_n} = \frac{\mu n}{L_n} \frac{D_n^{\text{max}}}{\mu n} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{equation*}

and hence

(38) \begin{equation} \left(1 - \frac{4D_n^{\text{max}}}{L_n}\right)^{-1} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 1.\end{equation}

Proposition 2.1 then implies

\begin{equation*} n^{1 + \delta - \frac{4}{\gamma}}I_n^{(1)} \le \left(2n^{-\frac{2}{\gamma} - \varepsilon_1}\sum_{i = 1}^n D_i^2 \right)\left(1 - \frac{4D_n^{\text{max}}}{L_n}\right)^{-1} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

The analysis for $I_n^{(2)}$ is similar, so that we are left with $I_n^{(3)}$. Here, we have

\begin{align*} I_n^{(3)} &= \sum_{1 \le i < j \le n} \frac{D_i^3 D_j^2}{(L_n - 2D_i)^2} \le \frac{\sum_{i = 1}^n D_i^2 \sum_{j = 1}^n D_j^3}{\left(L_n^2 - 4 L_n D_n^{\text{max}}\right)}\\ &=\frac{1}{L_n^2}\left(\sum_{i = 1}^n D_i^2\right)\left(\sum_{j = 1}^n D_j^3\right)\left(1 - \frac{4D_n^{\text{max}}}{L_n}\right)^{-1}.\end{align*}

The last term again converges in probability to one, by (38). For the remaining terms we use the definition of $\varepsilon_2$ and Proposition 2.1 to obtain

\begin{align*} n^{1 + \delta - \frac{4}{\gamma}}I_n^{(3)} &\le n^{2 - \frac{5}{\gamma} + \varepsilon_2}\frac{1}{L_n^2} \left(\sum_{i = 1}^n D_i^2\right)\left(\sum_{j = 1}^n D_j^3\right)\left(1 - \frac{4D_n^{\text{max}}}{L_n}\right)^{-1} \\ &= \frac{n^2}{L_n^2}\left(n^{-\frac{2}{\gamma} - \frac{\varepsilon_2}{2}}\sum_{i = 1}^n D_i^2\right) \left(n^{-\frac{3}{\gamma} - \frac{\varepsilon_2}{2}}\sum_{j = 1}^n D_j^3\right) \left(1 - \frac{4D_n^{\text{max}}}{L_n}\right)^{-1} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{align*}

which finishes the proof of (37).

We proceed to prove the convergence of $\widehat{r}_n^+$.

Proposition 3.3. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$, ${\mathbb{E}}\left[{\mathscr{D}}\right] = \mu$, and let $\widehat{G}_n = {\texttt{ECM}}(\textbf{D}_n)$. Then, for any slowly-varying function $\mathcal{L}$,

(39) \begin{equation} \mathcal{L}(n) n^{1 - \frac{1}{\gamma}}\widehat{r}^{\, +}_n \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation}

Proof. Let

\begin{equation*} \delta = \left(\frac{2 - \gamma}{4\gamma} \wedge \frac{\gamma - 1}{4\gamma}\right)\!.\end{equation*}

We will show that

(40) \begin{equation} n^{1 - \frac{1}{\gamma} + \delta} \, \widehat{r}^{\, +}_n \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0,\end{equation}

which then implies (39), since by Potter’s theorem $\mathcal{L}(n) n^{-\delta} \to 0$ for any $\delta > 0$.

The main part of the proof of (40) will be to show that

(41) \begin{equation} n^{1 - \frac{4}{\gamma} + 2\delta} \sum_{1 \le i < j \le n} \widehat{X}_{ij} D_i D_j \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation}

To see that (41) implies (40), we write

\begin{align*} \widehat{r}^{\,+}_n &\le \frac{\sum_{1 \le i < j \le n} \widehat{X}_{ij} D_i D_j}{\sum_{i = 1}^n \widehat{D}_i^3 - \frac{1}{\widehat{L}_n} \left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2}\\ &= \sum_{1 \le i < j \le n} \widehat{X}_{ij} D_i D_j \left(\frac{1}{\sum_{i = 1}^n \widehat{D}_i^3}\right) \left(1 - \frac{\left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2}{\widehat{L}_n \sum_{i = 1}^n \widehat{D}_i^3}\right)^{-1}.\end{align*}

By Lemma 3.1,

\begin{equation*} \left(1 - \frac{\left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2} {\widehat{L}_n\sum_{i = 1}^n \widehat{D}_i^3}\right)^{-1} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 1,\end{equation*}

and hence, using (41) and Proposition 2.1,

\begin{equation*} n^{1 - \frac{1}{\gamma} + \delta} \, \widehat{r}^{\,+}_n \le \left(n^{1 - \frac{4}{\gamma} + 2\delta} \sum_{1 \le i < j \le n} \widehat{X}_{ij} D_i D_j\right) \left(\frac{n^{-\frac{3}{\gamma} - \delta}}{\sum_{i = 1}^n \widehat{D}_i^3}\right) \left(1 - \frac{\left(\sum_{i = 1}^n \widehat{D}_i^2\right)^2}{\widehat{L}_n \sum_{i = 1}^n \widehat{D}_i^3}\right)^{-1} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

To prove (41) let $\kappa = (\gamma - 1)/(\gamma + 1)$ and define the events

(42) \begin{align} \mathcal{A}_n &= \left\{\left|L_n - \mu n\right| \le n^{1 - \kappa}\right\}\!,\nonumber\\[5pt] \mathcal{B}_n &= \left\{\sum_{1 \le i < j \le n}\left|\mathbb{P}_n\left({X_{ij = 0}}\right) - \exp\left\{-\frac{D_i D_j}{L_n}\right\}\right|D_i D_j \le n^{\frac{4}{\gamma} - 1- 3\delta} \right\}\!.\end{align}

Then, if we set $\Lambda_n = \mathcal{A}_n \cap \mathcal{B}_n$, it follows from Lemmas 2.1 and 3.2 that $\mathbb{P}\left({\Lambda_n}\right)\to 1$, and hence it is enough to prove that for any $K > 0$,

(43) \begin{equation} \lim_{n \to \infty} \mathbb{P}\left({n^{1 - \frac{4}{\gamma} + 2\delta}\sum_{1 \le i < j \le n} \widehat{X}_{ij} D_i D_j > K, \Lambda_n}\right) = 0.\end{equation}

First, since ${\mathbb{E}}_n\left[{\widehat{X}_{ij}}\right] = \mathbb{P}_n\left({X_{ij} > 0}\right)$ and $\Lambda_n$ is completely determined by the degree sequence,

\begin{align*} {\mathbb{E}}_n\left[{\sum_{1 \le i < j \le n} \widehat{X}_{ij} D_i D_j{\mathds{1}}_{\Lambda_n}}\right] &= \sum_{1 \le i < j \le n} \mathbb{P}_n\left({X_{ij} > 0}\right) D_i D_j {\mathds{1}}_{\Lambda_n}\\ &= \sum_{1 \le i < j \le n} \left(1 - \mathbb{P}_n\left({X_{ij} = 0}\right)\right)D_i D_j {\mathds{1}}_{\Lambda_n}\\ &\le \sum_{1 \le i < j \le n} \left(1 - \exp\left\{-\frac{D_iD_j}{L_n}\right\}\right)D_i D_j{\mathds{1}}_{\Lambda_n} + n^{\frac{4}{\gamma} - 1 - 3\delta}\\ &\le \sum_{1 \le i < j \le n} \left(1 - \exp\left\{-\frac{D_iD_j}{\mu n - n^{1 - \kappa}}\right\}\right) D_i D_j + n^{\frac{4}{\gamma} - 1 - 3\delta}.\end{align*}

From this we obtain, using Markov’s inequality, that

(44) \begin{align} &\mathbb{P}\left({n^{1 - \frac{4}{\gamma} + 2\delta}\sum_{1 \le i < j \le n} \widehat{X}_{ij} D_i D_j > K, \Lambda_n}\right) \notag \\ &\qquad\le \frac{n^{3 - \frac{4}{\gamma} + 2\delta}} {K} {\mathbb{E}}\left[{\left(1 - \exp\left\{-\frac{\mathscr{D}_1 \mathscr{D}_2}{\mu n - n^{1 - \kappa}}\right\}\right)\mathscr{D}_1 \mathscr{D}_2}\right] + O\left({n^{-\delta}}\right)\!,\end{align}

where $\mathscr{D}_1$ and $\mathscr{D}_2$ are two independent copies of $\mathscr{D}$. It follows that $\mathscr{D}_1\mathscr{D}_2$ is again regularly varying with exponent $1 < \gamma < 2$. Therefore, since $1 - e^{-x} \le (1 \wedge x)$, and using Lemma 2.2,

(45) \begin{align*} &\frac{n^{3 - \frac{4}{\gamma} + 2\delta}} {K} {\mathbb{E}}\left[{\left(1 - \exp\left\{-\frac{\mathscr{D}_1 \mathscr{D}_2}{\mu n - n^{1 - \kappa}}\right\} \right)\mathscr{D}_1 \mathscr{D}_2}\right] \\[5pt] &\quad\le \frac{n^{3 - \frac{4}{\gamma} + 2\delta}} {K} {\mathbb{E}}\left[{\left(1 \wedge \frac{\mathscr{D}_1 \mathscr{D}_2}{\mu n - n^{1 - \kappa}}\right) \mathscr{D}_1 \mathscr{D}_2}\right] \\[5pt] &\quad = O\left({n^{4 - \frac{4}{\gamma} - \gamma + 2\delta}}\right)\!.\addtocounter{equation}{1}\tag{\theequation} \end{align*}

Now observe that by our choice of $\delta > 0$ and since $2 - \gamma < 1$, we get

\begin{equation*} 4 - \frac{4}{\gamma} - \gamma + 2\delta = \frac{-(\gamma - 2)^2}{\gamma} + 2\delta \le \frac{-(\gamma - 2)^2}{\gamma} + \frac{2 - \gamma}{2\gamma} \le \frac{-(\gamma - 2)^2}{2\gamma} < 0.\end{equation*}

Plugging this into (45), it follows from (44) that

\begin{equation*} \mathbb{P}\left({n^{1 - \frac{4}{\gamma} + 2\delta}\sum_{1 \le i < j \le n} \widehat{X}_{ij} D_i D_j > K , \Lambda_n}\right) = O\left({n^{-\frac{(\gamma - 2)^2}{2 \gamma}} + n^{-\delta}}\right)\!,\end{equation*}

and hence (43) follows.

3.4. Proving Theorem 1.1

We are now ready to prove the central limit theorem for the ECM.

Proof of Theorem 1.1. Let $\widehat{G}_n = \texttt{ECM}(\textbf{D}_n)$, let $\mathcal{S}_{\gamma/2}$ and $\mathcal{S}_{\gamma/3}$ be defined as in (6), and let $\mathcal{L}_0$ be given by Proposition 3.1. Now we write

\begin{align*} \mu\mathcal{L}_0(n)n^{1 - \frac{1}{\gamma}} r(\widehat{G}_n) &= \mu\mathcal{L}_0(n)n^{1 - \frac{1}{\gamma}} \widehat{r}^{\,+}_n - \mu\mathcal{L}_0(n)n^{1 - \frac{1}{\gamma}} \widehat{r}^{\,-}_n.\end{align*}

By Proposition 3.3 it follows that the first part converges to zero in probability, as $n \to \infty$. For the second part, let $\delta = (\gamma - 1)^2/(4 \gamma)$ and note that by Potter’s theorem [Reference Bingham, Goldie and Teugels4, Theorem 1.5.6] we have that $\mathcal{L}_0(n) \le n^{\delta}$ for all large enough n. Then, if we denote by $G_n$ the graph before the removal of edges, it follows by Proposition 3.2 that

\begin{equation*} \mu\mathcal{L}_0(n)n^{1 - \frac{1}{\gamma}} \left|r^{-}_n - \widehat{r}^{\,-}_n\right| \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

Finally, we remark that the graph $G_n$ is generated by the CM, so that the above and Proposition 3.1 now imply

\begin{equation*} \mu\mathcal{L}_0(n)n^{1 - \frac{1}{\gamma}} \widehat{r}^{\,-}_n \ensuremath{\stackrel{d}{\longrightarrow}} \frac{\mathcal{S}_{\gamma/2}^2}{\mathcal{S}_{\gamma/3}}\end{equation*}

as $n \to \infty$, from which the result follows.

3.5. Pearson in the CM: proof of Lemma 1.1

In this section, we prove Lemma 1.1 on the tightness of the conditional variance of Pearson in the CM.

Proof of Lemma 1.1. Since, conditionally on the degrees, the only randomness in $r_n$ is in $(X_{ij})_{1\leq i<j\leq n}$, we use the covariance formula for sums of random variables to compute that

(46) \begin{align} \text{Var}_n(r_n)&=\frac{\sum_{i,j,k,l\in[n]} D_iD_jD_kD_l\text{Cov}_n(X_{ij}, X_{kl})}{\Big[\sum_{i\in[n]} D_i^3-\Big(\sum_{i\in[n]} D_i^2\Big)^2/L_n\Big]^2}\\[4pt] &=\frac{\sum_{i,j,k,l\in[n]} D_iD_jD_kD_l\text{Cov}_n(X_{ij}, X_{kl})}{\Big(\sum_{i\in[n]} D_i^3\Big)^2}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)),\nonumber \end{align}

since $\sum_{i\in[n]} D_i^3\gg \Big(\sum_{i\in[n]} D_i^2\Big)^2/L_n$, where we write $\text{Cov}_n$ for the conditional variance in the CM given the i.i.d. degrees. We next compute $\text{Cov}_n(X_{ij}, X_{kl})$, depending on the precise form of $\{i,j,k,l\}$. For this, we note that

(47) \begin{equation}{ X_{ij}=\sum_{s=1}^{D_i} \sum_{t=1}^{D_j} I_{st}, }\end{equation}

with $I_{st}$ the indicator that the sth half-edge incident to i pairs to the tth half-edge incident to j.

Case: $(i,j)=(k,l)$ with $i<j$. We compute that

(48) \begin{align} \text{Var}_n(X_{ij})&=\sum_{(s,t)}\sum_{(s^{\prime},t^{\prime})} \text{Cov}_n(I_{st}, I_{s^{\prime}t^{\prime}})\\ &=\frac{D_i(D_i-1)D_j(D_j-1)}{(L_n-1)(L_n-3)}+\frac{D_iD_j}{(L_n-1)}-\frac{D_i^2D_j^2}{(L_n-1)^2}\nonumber\\ &=\frac{D_iD_j}{(L_n-1)}+2\frac{D_i^2D_j^2}{(L_n-1)^2(L_n-3)}-\frac{D_iD_j(D_j-1)}{(L_n-1)(L_n-3)} -\frac{D_i^2D_j}{(L_n-1)(L_n-3)}\nonumber\\ &=\frac{D_iD_j}{L_n}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)).\nonumber\end{align}

Thus,

(49) \begin{align} n \frac{\sum_{i<j\in[n]} D_i^2D_j^2\text{Var}_n(X_{ij})}{\Big(\sum_{i\in[n]} D_i^3\Big)^2} &=\frac{n}{L_n}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)) \frac{\sum_{i<j\in[n]} D_i^3D_j^3}{\Big(\sum_{i\in[n]} D_i^3\Big)^2}\\[5pt] &=\frac{1}{\mu}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)) \frac{\sum_{i<j\in[n]} D_i^3D_j^3}{\Big(\sum_{i\in[n]} D_i^3\Big)^2}.\nonumber\end{align}

Since we sum over all $i,j\in[n]$ and not just $i<j$, this term appears four times.

Case: $|\{i,j,k,l\}|=1$. In this case, we obtain

(50) \begin{align} \text{Var}_n(X_{ii})&=\frac{D_i(D_i-1)}{(L_n-1)}+\frac{D_i(D_i-1)(D_i-2)(D_i-3)}{(L_n-1)(L_n-3)}-\frac{D_i^4}{(L_n-1)^2}\\[3pt] &=\frac{D_i^2}{L_n}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))+\frac{D_i^4}{L_n^3}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))-6\frac{D_i^3}{L_n^2}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))\nonumber\\[3pt] &=\frac{D_i^2}{L_n}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)),\nonumber\end{align}

since $D_i=O_{\scriptscriptstyle\mathbb{P}}\left({n^{\gamma}}\right)$ with $\gamma\in\big(\tfrac{1}{2},1\big)$. Therefore,

\begin{align} n \frac{\sum_{i\in[n]} D_i^4\text{Var}_n(X_{ii})}{\Big(\sum_{i\in[n]} D_i^3\Big)^2} &=\frac{n(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))}{L_n}\frac{\sum_{i\in[n]} D_i^6}{\Big(\sum_{i\in[n]} D_i^3\Big)^2}.\nonumber\end{align}

The above two computations show that these contributions sum up to

(51) \begin{equation}{ \frac{1}{\mu}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)) \frac{4\sum_{i<j\in[n]} D_i^3D_j^3}{\Big(\sum_{i\in[n]} D_i^3\Big)^2} +\frac{1}{\mu}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)) \frac{\sum_{i\in[n]} D_i^6}{\Big(\sum_{i\in[n]} D_i^3\Big)^2} \ensuremath{\stackrel{d}{\longrightarrow}} \frac{2-\mathcal{S}_{\gamma/6}/\mathcal{S}_{\gamma/3}^2}{\mu}. }\end{equation}

We are left to show that all other terms constitute error terms.

Case: $|\{i,j,k,l\}|=4$. When $|\{i,j,k,l\}|=4$, we compute that

(52) \begin{align} \text{Cov}_n(X_{ij}, X_{kl})&=D_iD_jD_lD_k \Big(\frac{1}{(L_n-1)(L_n-3)}-\frac{1}{(L_n-1)^2}\Big)\\[5pt] &=\frac{2D_iD_jD_lD_k}{(L_n-1)^2(L_n-3)}.\nonumber\end{align}

Thus, the contribution to the variance of $r_n$ of this case equals

(53) \begin{equation}{ \frac{\Big(\sum_{i\in[n]} D_i^2\Big)^4}{L_n^3\Big(\sum_{i\in[n]} D_i^3\Big)^2}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)) =O_{\scriptscriptstyle\mathbb{P}}\left({n^{2\gamma-3}}\right)=o_{\scriptscriptstyle{\mathbb{P}}}(n^{-1}), }\end{equation}

since $\gamma\in\big(\tfrac{1}{2},1\big)$.

Case: $|\{i,j,k,l\}|=3$. When $|\{i,j,k,l\}|=3$, we compute that

(54) \begin{align} \text{Cov}_n(X_{ij}, X_{il})&= \frac{D_i(D_i-1)D_jD_l}{(L_n-1)(L_n-3)}-\frac{D_i^2D_jD_l}{(L_n-1)^2}\nonumber\\ &=2\frac{D_i^2D_jD_l}{(L_n-1)^2(L_n-3)}-\frac{D_iD_jD_l}{(L_n-1)(L_n-3)}.\end{align}

Thus, the contribution of this case to the variance of $r_n$ equals

(55) \begin{align} &2\frac{\Big(\sum_{i\in[n]} D_i^2\Big)^2}{L_n^3\sum_{i\in[n]} D_i^3}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))- \frac{\Big(\sum_{i\in[n]} D_i^2\Big)^3}{L_n^2\Big(\sum_{i\in[n]} D_i^3\Big)^2}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))\\[4pt] &\qquad=O_{\scriptscriptstyle\mathbb{P}}\left({n^{\gamma-3}}\right)+O_{\scriptscriptstyle\mathbb{P}}\left({n^{-2}}\right)=o_{\scriptscriptstyle{\mathbb{P}}}(n^{-1}),\nonumber\end{align}

since $\gamma\in\big(\tfrac{1}{2},1\big)$.

This completes the proof.

4.1. Clustering in the configuration model

In this section, we compute the clustering coefficient for a CM with a power-law degree distribution with $\gamma\in(1,2)$. To prove Theorem 1.3, we first use a second moment method to show that the number of triangles $\Delta_n$ concentrates on its expected value conditioned on the degrees. Then we take the random degrees into account and show that the rescaled clustering coefficient converges to the stable distributions from Theorem 1.3.

4.1.1 Concentration for the number of triangles

The concentration result is formally stated and proved in the next lemma.

Lemma 4.1. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$, and ${G}_n = {\texttt{CM}}(\textbf{D}_n)$. Let $\Delta_n$ denote the number of triangles in $G_n$. Then, for any $\varepsilon>0$,

\begin{equation*} \lim_{n\to\infty}\mathbb{P}_n\left({\left|{\Delta_n-{\mathbb{E}}_n\left[{\Delta_n}\right]}\right|>\varepsilon{\mathbb{E}}_n\left[{\Delta_n}\right]}\right)=0.\end{equation*}

Proof. Fix $0<\delta<\big(\tfrac{2}{\gamma}-1\big)/6$. Define the event

(56) \begin{equation} \mathcal{B}_n=\Bigg\{\sum_{i=1}^{n}D_i^2\geq Kn^{ 2/\gamma-\delta}, \sum_{i=1}^{n}D_i^3\leq Kn^{ 3/\gamma+\delta}\Bigg\} .\end{equation}

Let $\mathcal{A}_n$ be the event defined in (42), and let $\Lambda_n=\mathcal{B}_n\cap \mathcal{A}_n$. We have $\lim_{n\to\infty}\mathbb{P}_n\left({\Lambda_n}\right)=1$; thus, we only need to prove the result on the event $\Lambda_n$. The proof is similar to the proof of [Reference Van der Hofstad13, Proposition 7.13].

Define

\begin{align*}\mathcal{I}&=\{(s_1t_1,s_2u_1,u_2t_2,i,j,k) \,:\,1\leq i<j<k\leq n, 1\leq s_1\neq s_2\leq { D}_i,1\leq t_1\neq t_2\leq { D}_j,\\ & \quad\ \, 1\leq u_1\neq u_2\leq { D}_k\}.\end{align*}

Thus, $\mathcal{I}$ contains all combinations of six labeled half-edges that could possibly form a triangle on three distinct vertices. Then

(57) \begin{align}|\mathcal{I}|=\sum_{1\leq i<j<k\leq n}D_i(D_i-1)D_j(D_j-1)D_k(D_k-1)\end{align}

denotes the number of ways six half-edges could form a triangle. For $m\in\mathcal{I}$, let ${\mathds{1}}_{m}$ denote the indicator variable that the six half-edges of m form a triangle in the way specified by m. Then

\begin{equation*}\Delta_n = \sum_{m\in\mathcal{I}}{\mathds{1}}_{m}.\end{equation*}

The probability that the half-edges in m form a triangle can be written as

\begin{equation*} \mathbb{P}_n\left({{\mathds{1}}_m=1}\right)=\prod_{j=1}^{3}(L_n+1-2j)^{-1}.\end{equation*}

This results in

(58) \begin{equation} {\mathbb{E}}_n\left[{\Delta_n}\right]=\sum_{m\in\mathcal{I}}\mathbb{P}_n\left({{\mathds{1}}_m=1}\right) = \frac{|\mathcal{I}|}{\prod_{j=1}^{3}(L_n+1-2j)}.\end{equation}

Furthermore, by [Reference Van der Hofstad13, Theorem 2.5],

\begin{equation*}{\mathbb{E}}_n\left[{\Delta_n(\Delta_n-1)}\right]=\sum_{m_1\neq m_2\in\mathcal{I}}\mathbb{P}_n\left({{\mathds{1}}_{m_1}={\mathds{1}}_{m_2}=1}\right)\!.\end{equation*}

When all six pairs of half-edges involved in $m_1$ and $m_2$ are distinct, the probability that these pairs of half-edges that form $m_1$ and $m_2$ are paired in the correct way is

\begin{equation*} \mathbb{P}_n\left({{\mathds{1}}_{m_1}={\mathds{1}}_{m_2}=1}\right)=\prod_{j=1}^{6}(L_n+1-2j)^{-1}.\end{equation*}

If $m_1$ and $m_2$ contain one pair of half-edges that is the same (so that $m_1$ and $m_2$ form two triangles merged on one edge), then

\begin{equation*}\mathbb{P}_n\left({{\mathds{1}}_{m_1}={\mathds{1}}_{m_2}=1}\right)=\prod_{j=1}^{5}(L_n+1-2j)^{-1}.\end{equation*}

Let $\mathcal{I}_2$ denote the set of combinations of 10 half-edges that could possibly form two triangles merged by one edge. Then, similarly to (57),

(59) \begin{align}|\mathcal{I}_2| &= \sum_{1\leq i < j\leq n} D_i(D_i-1)(D_i-2)D_j(D_j-1)(D_j-2) \sum_{1\leq k < l\leq n}D_k(D_k-1)D_l(D_l-1)\nonumber\\& \leq \Bigg(\sum_{i \in[n]}D_i^3\Bigg)^2\Bigg(\sum_{i \in[n]}D_i^2\Bigg)^2.\end{align}

Similarly, if $m_1$ and $m_2$ overlap at two pairs of half-edges (so that $m_1$ and $m_2$ form two triangles merged by two edges), then

\begin{equation*}\mathbb{P}_n\left({{\mathds{1}}_{m_1}={\mathds{1}}_{m_2}=1}\right)=\prod_{j=1}^{4}(L_n+1-2j)^{-1}.\end{equation*}

Let $\mathcal{I}_3$ denote the set of combinations of eight half-edges that could possibly form two triangles merged by two edges. Then, similarly to (57),

(60) \begin{align}|\mathcal{I}_3| &= \sum_{1\leq i < j\leq n} D_i(D_i-1)(D_i-2)D_j(D_j-1)(D_j-2) \sum_{1\leq k \leq n}D_k(D_k-1)\nonumber\\& \leq \Bigg(\sum_{i \in[n]}D_i^3\Bigg)^2\sum_{i \in[n]}D_i^2.\end{align}

In all other cases the probability of the event ${\mathds{1}}_{m_1}={\mathds{1}}_{m_2}=1$ then equals zero. These are cases where $m_1$ prescribes some half-edge to be merged to half-edge $j_1$, whereas $m_2$ prescribes it to be merged to some other half-edge $j_2$. Therefore,

(61) \begin{align}{\mathbb{E}}_n\left[{\Delta_n(\Delta_n-1)}\right]\leq\frac{|\mathcal{I}|^2}{\prod_{j=1}^{6}(L_n+1-2j)}+\frac{|\mathcal{I}_2|}{\prod_{j=1}^{5}(L_n+1-2j)}+\frac{|\mathcal{I}_3|}{\prod_{j=1}^{4 }(L_n+1-2j)}.\nonumber\\[5pt]\end{align}

On the event $\mathcal{B}_n$ defined in (56),

\begin{align*}\frac{|\mathcal{I}_2|}{\prod_{j=1}^{5}(L_n+1-2j)}\Big/\frac{|\mathcal{I}|^2}{\prod_{j=1}^{6}(L_n+1-2j)}& =O\Bigg( \frac{\Big(\sum_{i\in[n]}D_i^3\Big)^2/\Big(\sum_{i\in[n]}D_i^2\Big)^4}{(L_n+1-12)^{-1}}\Bigg)\\& = O\left({n \cdot n^{6/\gamma+2\delta-8/\gamma+4\delta}}\right) = O\left({n^{1-2/\gamma+6\delta }}\right)\\& = o(1),\end{align*}

by the choice of $\delta$. In a similar way, we can show that the third term is small compared to the first term of (61). Therefore,

\begin{equation*}{\mathbb{E}}_n\left[{\Delta_n(\Delta_n-1)}\right]\leq\frac{|\mathcal{I}|^2}{\prod_{j=1}^{6}(L_n+1-2j)}(1+o(1)).\end{equation*}

Finally, on the event $\mathcal{B}_n$, we have

\begin{equation*}|\mathcal{I}|= \Theta\Bigg(\Bigg(\sum_i D_i^2\Bigg)^3\Bigg)= \Omega\big(n^{6/\gamma-3\delta}\big).\end{equation*}

Using that $L_n=\mu n(1+o(1))$ on the event $\mathcal{A}_n$ results in

\begin{align*}\frac{ \textup{Var}_n\left({\Delta_n}\right)}{({\mathbb{E}}_n\left[{\Delta_n}\right])^2}&=\frac{ {\mathbb{E}}_n\left[{\Delta_n(\Delta_n-1)}\right]}{({\mathbb{E}}_n\left[{\Delta_n}\right])^2}-1+\frac{{\mathbb{E}}_n\left[{\Delta_n}\right]}{({\mathbb{E}}_n\left[{\Delta_n}\right])^2}\\& \leq \frac{(L_n-1)(L_n-3)(L_n-5)}{(L_n-7)(L_n-9)(L_n-11)}(1+o(1))-1+ \frac{\prod_{j=1}^{3}(L_n+1-2j)}{|\mathcal{I}|}\\&=1+o(1)-1+O(n^{3 + 3\delta - \frac{6}{\gamma}})=o(1),\end{align*}

for $\gamma\in(1,2)$. Then by Chebyshev’s inequality, on the event $\Lambda_n$,

\begin{equation*}\mathbb{P}_n\left({\left|{\Delta_n-{\mathbb{E}}_n\left[{\Delta_n}\right]}\right|>\varepsilon{\mathbb{E}}_n\left[{\Delta_n}\right]}\right)\leq \frac{\textup{Var}_n\left({\Delta_n}\right)}{({\mathbb{E}}_n\left[{\Delta_n}\right])^2\varepsilon^2}=o(1),\end{equation*}

which gives the result.

4.1.2 Proof of Theorem 1.3

We again prove the result under the event $\Lambda_n=\mathcal{B}_n\cap \mathcal{A}_n$, where $\mathcal{B}_n$ and $\mathcal{A}_n$ are given in (56) and (42), respectively.

By (57) and (58),

\begin{align*}{\mathbb{E}}_n\left[{\Delta_n}\right]& = \frac{1}{\mu^3 n^3}\sum_{1\leq i<j<k\leq n}D_i(D_i-1)D_j(D_j-1)D_k(D_k-1)(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))\\&= \frac{1}{\mu^3 n^3}\Bigg(\frac 16\Bigg(\sum_{i=1}^nD_i(D_i-1)\Bigg)^3-\frac{1}{2}\sum_{i=1}^{n}D_i(D_i-1)\sum_{j=1}^{n}D_j^2(D_j-1)^2\\&\quad +\frac 13 \sum_{i=1}^{n}D_i^3(D_i-1)^3\Bigg)(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)).\end{align*}

Then, the definition of the clustering coefficient in (4) yields

(62) \begin{equation}\begin{aligned} {\mathbb{E}}_n\left[{C_n}\right]& = \frac{1}{\mu^3 n^3}\Bigg(\Bigg(\sum_{i=1}^nD_i(D_i-1)\Bigg)^2-3 \sum_{i=1}^{n}D_i^2(D_i-1)^2\\ &\quad +2\frac{\sum_{i=1}^{n}D_i^3(D_i-1)^3}{\sum_{i=1}^{n}D_i(D_i-1)}\Bigg)(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))\\ & = \frac{1}{\mu^3 n^3}\Bigg(\Bigg(\sum_{i=1}^nD_i^2\Bigg)^2-3\sum_{i=1}^{n}D_i^4+2\sum_{i=1}^{n}D_i^6/\sum_{i=1}^{n}D_i^2\Bigg)(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)).\end{aligned}\end{equation}

Lemma 4.1 then gives that conditioned on the degree sequence,

\begin{equation*} C_n= \frac{1}{\mu^3 n^3}\Bigg(\Bigg(\sum_{i=1}^nD_i^2\Bigg)^2-3\sum_{i=1}^{n}D_i^4+2\sum_{i=1}^{n}D_i^6/\sum_{i=1}^{n}D_i^2\Bigg)(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)).\end{equation*}

By (24), there exist slowly-varying functions $\mathcal{L}_1(n)$, $\mathcal{L}_2(n)$, and $\mathcal{L}_3(n)$ such that

(63) \begin{align} \Bigg({n^{-2/\gamma}}\frac{\sum_{i=1}^{n}{D}_i^2}{\mathcal{L}_1(n)}, {n^{-4/\gamma}}\frac{\sum_{i=1}^{n}{D}_i^4}{\mathcal{L}_2(n)},{n^{-6/\gamma}}\frac{\sum_{i=1}^{n}D_i^6}{\mathcal{L}_3(n)}\Bigg)\ensuremath{\stackrel{d}{\longrightarrow}} \left(\mathcal{S}_{\gamma/2},\mathcal{S}_{\gamma/4},{\mathcal{S}_{\gamma/6}}\right)\!,\end{align}

where

\begin{equation*} \mathcal{S}_{\gamma/2} = \sum_{i=1}^{\infty}\Gamma_i^{-2/\gamma},\quad \mathcal{S}_{\gamma/4} = \sum_{i=1}^{\infty}\Gamma_i^{-4/\gamma},\quad \mathcal{S}_{\gamma/6} = \sum_{i=1}^{\infty}\Gamma_i^{-6/\gamma},\quad\end{equation*}

for the same random variables $\Gamma_i$. Furthermore, by [Reference Whitt42, Eq. (5.23)], the slowly-varying functions in (63) satisfy, for some slowly-varying function $\mathcal{L}_0(n)$,

\begin{equation*}\mathcal{L}_1(n)=\sqrt{\mathcal{L}_0(n)}(K_{\gamma/2})^{2/\gamma},\quad \mathcal{L}_2(n)=\mathcal{L}_0(n)(K_{\gamma/4})^{4/\gamma},\quad \mathcal{L}_3(n)=\mathcal{L}_0(n)^{3/2}(K_{\gamma/6})^{6/\gamma},\end{equation*}

where

\begin{equation*}K_\alpha = \frac{1-\alpha}{\Gamma(2-\alpha)\cos(\pi\alpha/2)},\end{equation*}

with $\Gamma$ the gamma function. Therefore,

\begin{align*} &\frac{n^{-4/\gamma}}{\mathcal{L}_0(n)}\Bigg(\Bigg(\sum_{i=1}^{n}{D}_i^2\Bigg)^2,\sum_{i=1}^{n}{D}_i^4,\frac{\sum_{i=1}^{n}D_i^6}{\sum_{i=1}^{n}{D}_i^2}\Bigg)\\&\qquad\quad\ensuremath{\stackrel{d}{\longrightarrow}}\left((K_{\gamma/2})^{\gamma/4}\mathcal{S}_{\gamma/2}^2,(K_{\gamma/4})^{\gamma/4}\mathcal{S}_{\gamma/4},\frac{(K_{\gamma/6})^{\gamma/6}\mathcal{S}_{\gamma/6}}{(K_{\gamma/2})^{\gamma/2}\mathcal{S}_{\gamma/2}}\right)\!.\end{align*}

Combining this with (62) results in (9).

4.2. Clustering coefficient in the erased configuration model

In this section, we study the clustering coefficient in the ECM. Again, we start with the expectation and the variance of the clustering coefficient conditioned on the sampled degree sequence, i.e. the sequence $\textbf{D}_n = \{D_1, D_2, \dots, D_n\}$ sampled from the distribution (1). Note that this is not the eventual degree sequence of the graph constructed by the ECM.

Structure of the proof of Theorem 1.4. We prove Theorem 1.4 in four steps:

Step 1. We show in Lemma 4.2 that the expected contribution to the number of triangles from vertices with sampled degrees larger than $\sqrt{n}/\varepsilon$ and smaller than $\varepsilon\sqrt{n}$ is small for fixed $0<\varepsilon<1$. Therefore, in the rest of the proof we focus only on counting triangles between vertices of degrees in $[\varepsilon\sqrt{n},\sqrt{n}/\varepsilon]$.

Step 2. We calculate the expected number of triangles between vertices of sampled degrees proportional to $\sqrt{n}$, conditioned on the degree sequence. In Lemma 4.3, we show that this expectation can be written as the sum of a function of the degrees.

Step 3. We show that the variance of the number of triangles between vertices of sampled degree proportional to $\sqrt{n}$ is small in Lemma 4.6. Thus, we can replace the number of triangles conditioned on the degrees by its expected value, which we computed in Step 2.

Step 4. We show that the expected number of triangles conditioned on the sampled degrees converges to the value given in Theorem 1.4, when taking the random degrees into account.

We will start by proving the three lemmas described above. Let $B_n(\varepsilon)$ denote the interval $[\varepsilon\sqrt{\mu n}, \sqrt{\mu n}/\varepsilon]$ for some $\varepsilon>0$. Furthermore, let $\widehat{X}_{ij}$ denote the number of edges between vertices i and j in the ECM. Then we can write the number of triangles as

(64) \begin{align*}\Delta_n &= \sum_{1\leq i< j< k\leq n} \widehat{X}_{ij}\widehat{X}_{jk}\widehat{X}_{ik}{\mathds{1}}_{\{D_i,D_j, D_k\in B_n(\varepsilon)\}} + \sum_{1\leq i< j < k\leq n} \widehat{X}_{ij}\widehat{X}_{jk}\widehat{X}_{ik}{\mathds{1}}_{\{D_i, D_j \text{ or }D_k \notin B_n(\varepsilon)\}}\\&=:\Delta_n(B_n(\varepsilon))+\Delta_n(\bar{B}_n(\varepsilon)). \addtocounter{equation}{1}\tag{\theequation} \end{align*}

We want to show that the major contribution to $\Delta_n$ comes from $\Delta_n(B_n(\varepsilon))$. The following lemma shows that the expected contribution of $\Delta_n(\bar{B}_n(\varepsilon))$ to the number of triangles is small.

Lemma 4.2. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$, and let $\widehat{G}_n = {\texttt{ECM}}(\textbf{D}_n)$. Let $\Delta_n(\bar{B}_n(\varepsilon))$ denote the number of triangles in $\widehat{G}_n$ with at least one of the sampled degrees not in $B_n(\varepsilon)$. Then

\begin{equation*}\limsup_{n\to\infty}\frac{{\mathbb{E}}\left[{\Delta_n(\bar{B}_n(\varepsilon))}\right]}{\mathcal{L}(\sqrt{\mu n})^3n^{\frac 32 (2-\gamma)}}=O\left({\mathcal{E}_1(\varepsilon)}\right)\end{equation*}

for some function $\mathcal{E}_1(\varepsilon)$ not depending on n such that $\mathcal{E}_1(\varepsilon)\to 0$ as $\varepsilon\to 0$.

Proof. Let $\Delta_{i,j,k}$ denote the event that a triangle is present on vertices i, j, and k. By [Reference Van der Hofstad, van Leeuwaarden and Stegehuis18, Lemma 4.1],

\begin{align*} \mathbb{P}_n\left({\Delta_{i,j,k}}\right)& =\Theta\Bigg(\prod_{(u,v)\in\{(i,j),(j,k),(i,k)\}}(1-\textup{e}^{-D_uD_v/L_n}){\mathds{1}}_{\left\{{D_uD_v<L_n}\right\}}\Bigg)\\ & = \Theta\left(\left(\frac{D_iD_j}{L_n} \wedge 1\right) \left(\frac{D_iD_k}{L_n} \wedge 1\right)\left(\frac{D_jD_k}{L_n}\wedge 1\right)\right)\!. \end{align*}

Therefore, for some $\tilde{K}>0$,

\begin{align*}&{\mathbb{E}}_n\left[{\Delta_n(\bar{B}_n(\varepsilon))}\right]\\[4pt] &\quad\leq \tilde{K}\sum_{\mathclap{1\leq i< j < k\leq n}}\,\,\, \left(\frac{D_iD_j}{L_n} \wedge 1\right) \left(\frac{D_iD_k}{L_n} \wedge 1\right)\left(\frac{D_jD_k}{L_n}\wedge 1\right){\mathds{1}}_{\left\{{D_i,D_j,\text{ or }D_k \in \bar{B}_n(\varepsilon)}\right\}}.\end{align*}

Thus,

(65) \begin{equation}\begin{aligned}[b]&{\mathbb{E}}\left[{\Delta_n(\bar{B}_n(\varepsilon))}\right]\\[4pt] &\quad \leq \tilde{K}\tfrac12 n^3{\mathbb{E}}\left[{\left(\frac{{D}_1{D}_2}{\mu n}\wedge 1\right)\left(\frac{{D}_1{D}_3}{\mu n} \wedge 1\right)\left(\frac{{D}_2{D}_3}{\mu n}\wedge 1\right){\mathds{1}}_{\left\{{{D}_1 \in \bar{B}_n(\varepsilon)}\right\}}}\right].\end{aligned}\end{equation}

We now show that the contribution of (65) when $D_1<\varepsilon\sqrt{\mu n}$ is small. We write

(66) \begin{align*}&{\mathbb{E}}\left[{\left(\frac{{D}_1{D}_2}{\mu n}\wedge 1\right)\left(\frac{{D}_1{D}_3}{\mu n}\wedge 1\right)\left(\frac{{D}_2{D}_3}{\mu n} \wedge 1\right){\mathds{1}}_{\left\{{{D}_1<\varepsilon\sqrt{\mu n}}\right\}}}\right]\\ &\quad\leq \tilde{K}\sum_{t_1=1}^{\varepsilon\sqrt{\mu n}}\sum_{t_2=1}^{n}\sum_{t_3=1}^{n}\mathbb{P}\left({D_1=t_1,D_2=t_2,D_3=t_3}\right)\left(\frac{t_1t_2}{\mu n} \wedge 1\right)\left(\frac{t_1t_3}{\mu n} \wedge 1\right)\left(\frac{t_2t_3}{\mu n} \wedge 1\right)\\ &\quad\leq \tilde{K}K^3 \sum_{t_1=1}^{\varepsilon\sqrt{\mu n}}\sum_{t_2,t_3=1}^{n} \mathcal{L}(t_1)\mathcal{L}(t_2)\mathcal{L}(t_3)(t_1t_2t_3)^{-\gamma-1}\left(\frac{t_1t_2}{\mu n} \wedge 1\right) \left(\frac{t_1t_3}{\mu n} \wedge 1\right) \left(\frac{t_2t_3}{\mu n} \wedge 1\right)\!, \addtocounter{equation}{1}\tag{\theequation} \end{align*}

where we have used Assumption 1.1 and the fact that $D_1$, $D_2$, and $D_3$ are independent. As

(67) \begin{align} \int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}(xyz)^{-\gamma-1} \left(xy \wedge 1\right) \left(xz \wedge 1\right) \left(yz \wedge 1\right)\,\text{d} x \,\text{d} y \,\text{d} z<\infty, \end{align}

for all $\gamma\in(1,2)$, by [Reference Karamata28, Theorem 2],

\begin{align*}&\int_{0}^{\varepsilon\sqrt{\mu n}}\!\!\int_{0}^{\infty}\int_{0}^{\infty}\!\!\mathcal{L}(t_1)\mathcal{L}(t_2)\mathcal{L}(t_3)(t_1t_2t_3)^{-\gamma-1} \left(\frac{t_1t_2}{\mu n} \wedge 1\right) \!\!\left(\frac{t_1t_3}{\mu n} \wedge 1\right)\!\! \left(\frac{t_2t_3}{\mu n} \wedge1\right)\!\!\,\text{d} t_3 \,\text{d} t_2 \,\text{d} t_1\\&= (\mu n)^{-\frac{3}{2}\gamma}\int_{0}^{\varepsilon}\!\!\int_{0}^{\infty}\int_{0}^{\infty}\!\!\mathcal{L}(x\sqrt{\mu n})\mathcal{L}(y\sqrt{\mu n})\mathcal{L}(z\sqrt{\mu n})(xyz)^{-\gamma-1} \left(xy\wedge 1\right) \!\!\left(xz\wedge 1\right)\!\! \left(yz\wedge 1\right)\!\!\,\text{d} x \,\text{d} y \,\text{d} z\\& = (1+o(1))\mathcal{L}(\sqrt{\mu n})^3\!\!\int_{0}^{\varepsilon}\!\!\int_{0}^{\infty}\int_{0}^{\infty}(xyz)^{-\gamma-1} \left(xy\wedge 1\right) \!\!\left(xz\wedge 1\right)\!\! \left(yz\wedge 1\right)\!\!\,\text{d} x \,\text{d} y \,\text{d} z.\end{align*}

We then bound the sum in (66) as

\begin{align*}&\sum_{t_1=1}^{\varepsilon\sqrt{\mu n}}\sum_{t_2=1}^{n}\sum_{t_3=1}^{n}\mathcal{L}(t_1)\mathcal{L}(t_2)\mathcal{L}(t_3)(t_1t_2t_3)^{-\gamma-1}\left(\frac{t_1t_2}{\mu n} \wedge 1\right) \left(\frac{t_1t_3}{\mu n} \wedge 1\right) \left(\frac{t_2t_3}{\mu n} \wedge 1\right)\\& \leq 2\int_{0}^{\varepsilon\sqrt{\mu n}+1}\int_{0}^{\infty}\int_{0}^{\infty}\mathcal{L}(t_1)\mathcal{L}(t_2)\mathcal{L}(t_3)(t_1t_2t_3)^{-\gamma-1}\\& \quad \times \left(\frac{t_1t_2}{\mu n} \wedge 1\right) \left(\frac{t_1t_3}{\mu n} \wedge 1\right) \left(\frac{t_2t_3}{\mu n} \wedge1\right)\,\text{d} t_3 \,\text{d} t_2 \,\text{d} t_1\\ & =2(1+o(1))(\mu n)^{-\frac{3}{2}\gamma}\mathcal{L}(\sqrt{\mu n})^3\int_{0}^{\varepsilon} \int_{0}^{\infty} \int_{0}^{\infty}(xyz)^{-\gamma-1} \left(xy \wedge 1\right) \left(xz \wedge 1\right) \left(yz \wedge 1\right)\,\text{d} x \,\text{d} y \,\text{d} z.\end{align*}

Therefore,

(68) \begin{equation}\begin{aligned}[b]&{\mathbb{E}}\left[{\left(\frac{{D}_1{D}_2}{\mu n} \wedge 1\right)\left(\frac{{D}_1{D}_3}{\mu n} \wedge 1\right) \left(\frac{{D}_2{D}_3}{\mu n} \wedge 1\right){\mathds{1}}_{\left\{{{D}_1<\varepsilon\sqrt{\mu n}}\right\}}}\right]\\[4pt] &\leq K_2 n^{\frac{3}{2}(1-\gamma)}\mathcal{L}(\sqrt{\mu n})^3\int_{0}^{\varepsilon} \int_{0}^{\infty} \int_{0}^{\infty}(xyz)^{-\gamma-1} \left(xy \wedge 1\right) \left(xz \wedge 1\right) \left(yz \wedge 1\right)\,\text{d} x \,\text{d} y \,\text{d} z,\end{aligned}\end{equation}

for some constant $K_2$. Thus, we only need to prove that the last triple integral in (68) tends to zero as $\varepsilon\to 0$. Using (67), we obtain

\begin{equation*}\int_{0}^{\varepsilon}\int_{0}^{\infty}\int_{0}^{\infty}(xyz)^{-\gamma-1} \left(xy \wedge 1\right) \left(xz \wedge 1\right) \left(yz \wedge 1\right)\,\text{d} x \,\text{d} y \,\text{d} z \,:\!= \mathcal{E}_0(\varepsilon),\end{equation*}

where $\mathcal{E}_0(\varepsilon)$ is such that $\mathcal{E}_0(\varepsilon)\to 0$ as $\varepsilon\to 0$. Thus, by (65) and (66), the contribution to the expectation where one of the degrees is smaller than $\varepsilon\sqrt{\mu n}$ is bounded by $O_{\scriptscriptstyle\mathbb{P}}\left({\mathcal{L}(\sqrt{\mu n})^3n^{3-\frac 3 2 \gamma}\mathcal{E}_0(\varepsilon)}\right)$. Similarly,

\begin{equation*}\int_{1/\varepsilon}^\infty\int_{0}^{\infty}\int_{0}^{\infty}(xyz)^{-\gamma-1} \left(xy \wedge 1\right) \left(xz \wedge 1\right) \left(yz \wedge 1\right)\,\text{d} x \,\text{d} y \,\text{d} z \,:\!= \mathcal{E}_0^\prime(\varepsilon),\end{equation*}

where again $\mathcal{E}_0^\prime(\varepsilon)$ satisfies $\mathcal{E}_0^\prime(\varepsilon)\to 0$ as $\varepsilon\to 0$. Therefore, we can show in a similar way that the contribution to the expected number of triangles where one of the degrees is larger than $\sqrt{\mu n}/\varepsilon$ is $O_{\scriptscriptstyle\mathbb{P}}\left({\mathcal{L}(\sqrt{\mu n})^3 n^{3-\frac 3 2 \gamma}\mathcal{E}_0^\prime(\varepsilon)}\right)$. Then, taking $\mathcal{E}_1(\varepsilon)=\max(\mathcal{E}_0(\varepsilon),\mathcal{E}_0^\prime(\varepsilon))$ proves the lemma.

The next lemma computes the expected contribution of the vertices of sampled degree proportional to $\sqrt{n}$ to the number of triangles. Define

(69) \begin{equation}g_{n, \varepsilon} (D_i,D_j,D_k)\,:\!=\left(1-\textup{e}^{-\frac{D_iD_j}{L_n}}\right)\left(1-\textup{e}^{-\frac{D_iD_k}{L_n}}\right)\left(1-\textup{e}^{-\frac{D_kD_j}{L_n}}\right){\mathds{1}}_{\left\{{(D_i,D_j,D_k)\in B_n(\varepsilon)}\right\}},\end{equation}

and let $\sum^\prime$ denote a sum over distinct indices, such that $i<j<k$.

Lemma 4.3. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$, and let $\widehat{G}_n = {\texttt{ECM}}(\textbf{D}_n)$. Let $\Delta_n(B_n(\varepsilon))$ denote the number of triangles in $\widehat{G}_n$ with sampled degrees in $B_n(\varepsilon)$. Then, on the event $\mathcal{A}_n$ as defined in (42),

\begin{equation*} {\mathbb{E}}_n\left[{\Delta_n(B_n(\varepsilon))}\right]=(1+o(1))\sideset{}{'}\sum\limits_{i,j,k} g_{n, \varepsilon}(D_i,D_j,D_k). \end{equation*}

Proof. We can write the expectation as

(70) \begin{align} {\mathbb{E}}_n\left[{\Delta_n(B_n(\varepsilon))}\right]= \sideset{}{'}\sum\limits_{i,j,k}\mathbb{P}_n\left({\Delta_{i,j,k}}\right){\mathds{1}}_{\left\{{(D_i,D_j,D_k)\in B_n(\varepsilon)}\right\}},\\[-24pt] \nonumber\end{align}

where $\mathbb{P}_n\left({\Delta_{i,\,j,\,k}}\right)$ denotes the probability of a triangle between vertices i, j, k being present. This probability can be written as

(71) \begin{equation}\begin{aligned}[b]\mathbb{P}_n\left({\Delta_{i,j,k}}\right) &=1-\mathbb{P}_n\left({X_{ij}=0}\right)-\mathbb{P}_n\left({X_{ik}=0}\right)-\mathbb{P}_n\left({X_{jk}=0}\right)+ \mathbb{P}_n\left({X_{ij}=X_{ik}=0}\right)\\&\quad +\mathbb{P}_n\left({X_{ij}=X_{jk}=0}\right)+\mathbb{P}_n\left({X_{ik}=X_{jk}=0}\right)\\&\quad -\mathbb{P}_n\left({X_{ij}=X_{ik}=X_{jk}=0}\right)\!.\end{aligned}\end{equation}

Because $D_i,D_j,D_k\leq \sqrt{n}/\varepsilon$ and $L_n=\mu n(1+o(1))$ under the event $\mathcal{A}_n$, we can use [Reference Van der Hofstad, van Leeuwaarden and Stegehuis18, Lemma 3.1], which calculates the probability that an edge is present conditionally on the presence of other edges in CMs with arbitrary degree distributions. This results in

\begin{align*}&\mathbb{P}_n\left({X_{ij}=X_{ik}=X_{jk}=0}\right)\\ & \quad =\mathbb{P}_n\left({X_{ij}=0}\right)\mathbb{P}_n\left({X_{jk}=0\mid X_{ij}=0}\right)\mathbb{P}_n\left({X_{ik}=0\mid X_{ij}=X_{jk}=0}\right)\\ & \quad = \textup{e}^{-\frac{D_i D_j}{L_n}}\textup{e}^{-\frac{D_j D_k}{L_n}}\textup{e}^{-\frac{D_i D_k}{L_n}}(1+o(1)),\end{align*}

and similarly

\begin{equation*}\mathbb{P}_n\left({X_{ij}=X_{jk}=0}\right)=\textup{e}^{-\frac{D_i D_j}{L_n}}\textup{e}^{-\frac{D_j D_k}{L_n}}(1+o(1)).\end{equation*}

Combining this with (71) yields

(72) \begin{align*}\mathbb{P}_n\left({\Delta_{i,j,k}}\right)&=1-\Big(\textup{e}^{-\frac{D_i D_j}{L_n}}+\textup{e}^{-\frac{D_i D_j}{L_n}}+\textup{e}^{-\frac{D_j D_k}{L_n}}\Big)(1+o(1))\\& \quad +\Big(\textup{e}^{-\frac{D_i D_j}{L_n}}\textup{e}^{-\frac{D_j D_k}{L_n}}+\textup{e}^{-\frac{D_i D_j}{L_n}}\textup{e}^{-\frac{D_i D_k}{L_n}}+\textup{e}^{-\frac{D_i D_k}{L_n}}\textup{e}^{-\frac{D_j D_k}{L_n}}\Big)(1+o(1))\\&\quad -\textup{e}^{-\frac{D_i D_j}{L_n}}\textup{e}^{-\frac{D_j D_k}{L_n}}\textup{e}^{-\frac{D_i D_k}{L_n}}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))\\& = (1+o(1))\left(1-\textup{e}^{-\frac{D_i D_j}{L_n}}\right)\left(1-\textup{e}^{-\frac{D_j D_k}{L_n}}\right)\left(1-\textup{e}^{-\frac{D_j D_k}{L_n}}\right)\!, \addtocounter{equation}{1}\tag{\theequation}\vspace*{-6pt} \end{align*}

where the second equality follows because $\varepsilon\sqrt{n}<D_i,D_j,D_k<\sqrt{n}/\varepsilon$ and $L_n=\mu n(1+o(1))$ under $\mathcal{A}_n$. For $D_i,D_j,D_k\in[\varepsilon\sqrt{n},1/\varepsilon\sqrt{n}]$, the main term in (72) can be uniformly bounded from above and from below by some functions $f_1(\varepsilon)$ and $f_2(\varepsilon)$ not depending on n. Combining this with (70) shows that

(73) \begin{equation} {\mathbb{E}}_n\left[{\Delta_n(B_n(\varepsilon))}\right]=(1+o(1))\sideset{}{'}\sum\limits_{i,j,k}g_{n,\varepsilon}(D_i,D_j,D_k),\end{equation}

on the event $\mathcal{A}_n$, which proves the lemma.

We now replace the $L_n$ inside the definition of $g_{n,\varepsilon}$ by $\mu n$. In particular, define

(74) \begin{equation}f_{n, \varepsilon}(x,y,z) = \Big(1 - \textup{e}^{-\frac{xy}{\mu n}}\Big) \Big(1 - \textup{e}^{-\frac{yz}{\mu n}}\Big) \Big(1 - \textup{e}^{-\frac{zx}{\mu n}}\Big){\mathds{1}}_{\left\{{x, y, z \in B_n(\varepsilon)}\right\}};\end{equation}

then we have the following result.

Lemma 4.4. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$ and $\kappa = (\gamma - 1)/(1 + \gamma) > 0$. Then, for all $\varepsilon > 0$ and $\delta < \kappa$,

\begin{equation*} n^{\frac{3 \gamma}{2} - 3 + \delta} \left|\sum_{1\leq i< j < k\leq n} g_{n, \varepsilon}(D_i, D_j, D_k) - f_{n, \varepsilon}(D_i, D_j, D_k)\right| \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0. \end{equation*}

Proof. Let $\mathcal{A}_n$ be as in (42) and note that $\kappa$ is the same as in the statement of the lemma. Then, since $\mathbb{P}\left({\mathcal{A}_n}\right) \to 1$, it is enough to prove the result conditioned on $\mathcal{A}_n$. Next, note that $1 - \textup{e}^{-a} \le 1$ for all $a \ge 0$. Hence, by symmetry it suffices to prove

\begin{equation*}n^{\frac{3 \gamma}{2} - 3 + \delta} \sum_{1\leq i< j < k\leq n} \left|\textup{e}^{-\frac{D_i D_j}{L_n}} -\textup{e}^{-\frac{D_i D_j}{\mu n}}\right| {\mathds{1}}_{\left\{{D_i, D_j, D_k \in B_n(\varepsilon)}\right\}} \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

For this we compute that, on the event $\mathcal{A}_n$,

\begin{equation*}\left|\textup{e}^{-\frac{D_i D_j}{L_n}} - \textup{e}^{-\frac{D_i D_j}{\mu n}}\right| \le D_i D_j \frac{|L_n - \mu n|}{\mu n L_n}\le D_i D_j \frac{n^{- 1 - \kappa}}{(\mu^2 - n^{-\kappa})} = D_i D_j O\left({n^{-1-\kappa}}\right)\!.\end{equation*}

We recall that by Karamata’s theorem [Reference Bingham, Goldie and Teugels4, Theorem 1.5.11] it follows that, as $n \to \infty$,

\begin{align*}{\mathbb{E}}\left[{D_1 {\mathds{1}}_{\left\{{D_1\in B_n(\varepsilon)}\right\}}}\right] &\le {\mathbb{E}}\left[{D_1 {\mathds{1}}_{\left\{{D_1 > \varepsilon \sqrt{\mu n}}\right\}}}\right]\sim \frac{\mathcal{L}\left(\varepsilon\sqrt{\mu n}\right)\varepsilon^{1-\gamma}(\mu n)^{\frac{1-\gamma}{2}}}{\gamma - 1} \\&= O\left(\mathcal{L}\left(\varepsilon\sqrt{\mu n}\right)n^{\frac{1-\gamma}{2}}\right)\!,\end{align*}

where $a_n \sim b_n$ as $n \to \infty$ means $\lim_{n\to\infty} a_n/b_n = 1$.

Therefore, using Lemma A.1,

\begin{align*}&{\mathbb{E}}\left[{\left|\textup{e}^{-\frac{D_1 D_2}{L_n}} - \textup{e}^{-\frac{D_1 D_2}{\mu n}}\right| {\mathds{1}}_{\left\{{D_1, D_2, D_3 \in B_n(\varepsilon)}\right\}}}\right]\\ &\quad\le O\left({n^{-1-\kappa}}\right) {\mathbb{E}}\left[{D_1 {\mathds{1}}_{\left\{{D_1 \in B_n(\varepsilon)}\right\}}}\right]^2 \mathbb{P}\left({D_3 \in B_n(\varepsilon)}\right) \\ &\quad = O\left({\mathcal{L}\left(\varepsilon\sqrt{\mu n}\right)^2\mathcal{L}(\sqrt{\mu n}) n^{-(\kappa + \frac{3\gamma}{2})}}\right)\!,\end{align*}

so that by Markov’s inequality, we obtain that for any $K > 0$ and $\varepsilon > 0$,

\begin{align*}\hspace*{50pt}&\mathbb{P}\left({n^{\frac{3 \gamma}{2} - 3 + \delta} \sum_{1\leq i< j < k\leq n} \left|\textup{e}^{-\frac{D_i D_j}{L_n}} - \textup{e}^{-\frac{D_i D_j}{\mu n}}\right| {\mathds{1}}_{\left\{{D_i, D_j, D_k \in B_n(\varepsilon)}\right\}} > K, \mathcal{A}_n}\right) \\ &\quad\le \frac{\binom{n}{3} n^{\frac{3 \gamma}{2} - 3 + \delta}}{K}{\mathbb{E}}\left[{\left|\textup{e}^{-\frac{D_1 D_2}{L_n}} - \textup{e}^{-\frac{D_1 D_2}{\mu n}}\right| {\mathds{1}}_{\mathcal{A}_n} {\mathds{1}}_{\left\{{D_1, D_2, D_3 \in B_n(\varepsilon)}\right\}}}\right] \\ &\quad= O\left({\mathcal{L}\left(\varepsilon\sqrt{\mu n}\right)^2\mathcal{L}(\sqrt{\mu n})n^{\delta - \kappa}}\right) = o(1).\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qed\end{align*}

We now again study the expected number of triangles of vertices with sampled degrees in $B_n(\varepsilon)$, and investigate the behavior of the expression of Lemma 4.3.

Lemma 4.5. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$, and let $\widehat{G}_n = {\texttt{ECM}}(\textbf{D}_n)$. Let $\Delta_n(B_n(\varepsilon))$ denote the number of triangles in $\widehat{G}_n$ with sampled degrees in $B_n(\varepsilon)$. Then, as $n\to\infty$,

(75) \begin{equation} \begin{aligned} &\frac{{\mathbb{E}}_n\left[{\Delta_n(B_n(\varepsilon))}\right]}{\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)} \mu^{-\frac{3}{2}\gamma}} \\ &\quad= (1+o_{\scriptscriptstyle{\mathbb{P}}}(1))\frac{1}{6}\int_{\varepsilon}^{1/\varepsilon} \int_{\varepsilon}^{1/\varepsilon} \int_{\varepsilon}^{1/\varepsilon} \frac{\gamma^3}{(t_1t_2t_3)^{\gamma+1}}h(t_1,t_2,t_3)\,\text{d} t_1\,\text{d} t_2\,\text{d} t_3+O_{\scriptscriptstyle\mathbb{P}}\left({\varepsilon}\right)\!, \end{aligned} \end{equation}

where $h(x,y,z)\,:\!=(1-\text{e}^{-xy})(1-\text{e}^{-xz})(1-\text{e}^{-yz})$.

Proof. Combining Lemmas 4.3 and 4.4 yields that conditionally on the sampled degrees $(D_i)_{i\in[n]}$,

(76) \begin{equation}{\mathbb{E}}_n\left[{\Delta_n(B_n(\varepsilon))}\right]=(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)) \sum_{1\leq i< j < k\leq n}f_{n,\varepsilon}(D_i,D_j,D_k) + o_\mathbb{P}\left(\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)}\right)\!,\end{equation}

where $f_{n,\varepsilon}(D_i,D_j,D_k)$ is as in (74). To investigate the convergence of (81) when taking the random degrees into account, define for $b>a\geq0$ the random measure

\begin{equation*}N_1^{\scriptscriptstyle{(n)}}([a,b])=\frac{(\mu n)^{\frac{\gamma}{2}}}{\mathcal{L}(\sqrt{\mu n})}\frac{1}{n}\sum_{i=1}^n{\mathds{1}}_{\left\{{D_i\in[a\sqrt{\mu n},b\sqrt{\mu n}]}\right\}},\end{equation*}

so that $N_1^{\scriptscriptstyle{(n)}}$ counts the number of vertices with degrees in the interval $[a\sqrt{\mu n},b\sqrt{\mu n}]$. Since every $D_i$ is drawn i.i.d. from (1), the number of vertices with degrees between $a\sqrt{\mu n}$ and $b \sqrt{\mu n}$ is binomially distributed, and therefore this number concentrates around its mean value, which is large. Combining this with Lemma A.1 yields

\begin{align*}N_1^{\scriptscriptstyle{(n)}}\left([a,b]\right) &=\frac{(\mu n)^{\frac{\gamma}{2}}}{\mathcal{L}(\sqrt{\mu n})}\frac{1}{n}\sum_{i=1}^n{\mathds{1}}_{\left\{{D_i\in[a\sqrt{\mu n},b\sqrt{\mu n}]}\right\}}\\&=\frac{(\mu n)^{\frac{\gamma}{2}}}{\mathcal{L}(\sqrt{\mu n})}\mathbb{P}\left({D_i\in[a\sqrt{\mu n},b\sqrt{\mu n}]}\right) (1+o_{\scriptscriptstyle{\mathbb{P}}}(1))\\& = (1+o_{\scriptscriptstyle{\mathbb{P}}}(1))\gamma \int_{a}^{b}t^{-\gamma-1}\,\text{d} t= (1+o_{\scriptscriptstyle{\mathbb{P}}}(1))(a^{-\gamma}-b^{-\gamma}).\end{align*}

Therefore,

(77) \begin{equation}\begin{aligned}[b]N_1^{\scriptscriptstyle{(n)}}\left([a,b]\right)& \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} \int_{a}^{b}t^{-\gamma-1}\,\text{d} t = (a^{-\gamma}-b^{-\gamma})\, =\!:\, \lambda([a,b]).\end{aligned}\end{equation}

Let $N^{\scriptscriptstyle{(n)}}$ be the product measure $N_1^{\scriptscriptstyle{(n)}}\times N_1^{\scriptscriptstyle{(n)}}\times N_1^{\scriptscriptstyle{(n)}}$ and let $F=[\varepsilon,1/\varepsilon]^3$. Thus, $N^{\scriptscriptstyle{(n)}}$ counts the number of triples with all three degrees proportional to $\sqrt{\mu n}$.

By (74),

\begin{align*} & \sum_{1\leq i< j< k\leq n}f_{n,\varepsilon}(D_i,D_j,D_k) \\ & \quad = \sum_{1\leq i< j < k\leq n}\bigg(1 - \textup{e}^{-\frac{D_iD_j}{\mu n}}\bigg) \bigg(1 - \textup{e}^{-\frac{D_iD_k}{\mu n}}\bigg) \bigg(1 - \textup{e}^{-\frac{D_jD_k}{\mu n}}\bigg) {\mathds{1}}_{\left\{{D_i,D_j,D_k \in B_n(\varepsilon)}\right\}}\\ & \quad =\frac{1}{6}\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)} \mu^{-\frac{3}{2}\gamma} \int_F (1-\text{e}^{-t_1t_2})(1-\text{e}^{-t_1t_3})(1-\text{e}^{-t_2t_3})\,\text{d} N^{\scriptscriptstyle{(n)}}(t_1,t_2,t_3)\\ & \qquad -\frac{1}{3}\mathcal{L}(\sqrt{\mu n})^2n^{(2-\gamma)} \mu^{-\gamma} \int_F (1-\text{e}^{-t_1t_2})^2\Big(1-\text{e}^{-t_1^2}\Big)\,\text{d} N^{\scriptscriptstyle{(n)}}(t_1,t_2)\\ & \quad = \frac{(1+o(1))}{6}\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)} \mu^{-\frac{3}{2}\gamma} \int_F (1-\text{e}^{-t_1t_2})(1-\text{e}^{-t_1t_3})(1-\text{e}^{-t_2t_3})\,\text{d} N^{\scriptscriptstyle{(n)}}(t_1,t_2,t_3), \end{align*}

where the second term in the second equality excludes the summation terms where at least two indices are equal.

The function h(x, y, z) is bounded and continuous on $F=[\varepsilon, 1/\varepsilon]^3$. Therefore, writing the Taylor expansion of $\textup{e}^{-x}$ yields

\begin{equation*}(1-\textup{e}^{-xy})=\sum_{i=1}^k\frac{(xy)^i}{i!}(-1)^i+O\left(\frac{\varepsilon^{-2k}}{(k+1)!}\right)\end{equation*}

on $x,y\in [\varepsilon,1/\varepsilon]$, where the error term goes to zero as $k\to\infty$. Therefore, for any $\delta>0$, we can choose $k_1,k_2,k_3$ such that

\begin{align*}h(x,y,z)& =\sum_{l_1=1}^{k_1}\sum_{l_2=1}^{k_2}\sum_{l_3=1}^{k_3}\left(\frac{(xy)^{l_1}}{l_1!}(-1)^{l_1}\frac{(yz)^{l_2}}{l_2!}(-1)^{l_2}\frac{(xz)^{l_3}}{l_3!}(-1)^{l_3}\right)+O\left(\delta\right)\\&=\sum_{l_1=1}^{k_1}\sum_{l_2=1}^{k_2}\sum_{l_3=1}^{k_3}\left(\frac{x^{l_1+l_3}y^{l_1+l_2}z^{l_2+l_3}}{l_1!l_2!l_3!}(-1)^{l_1+l_2+l_3}\right)+O\left(\delta\right)\!.\end{align*}

Choosing $\delta=\varepsilon^{3\gamma+1}$ gives

\begin{align*}& \frac{\sum_{1\leq i< j < k\leq n}f_{n,\varepsilon}(D_i,D_j,D_k)}{\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)} \mu^{-\frac{3}{2}\gamma}}=\frac{1}{6} \int_F h(t_1,t_2,t_3)\,\text{d} N^{\scriptscriptstyle{(n)}}(t)\\[3pt] &=\frac 16 \int_{F}\sum_{l_1=1}^{k_1}\sum_{l_2=1}^{k_2}\sum_{l_3=1}^{k_3}\left(\frac{x^{l_1+l_3}y^{l_1+l_2}z^{l_2+l_3}}{l_1!l_2!l_3!}(-1)^{l_1+l_2+l_3}\right)+O\left(\varepsilon^{3\gamma+1}\right)\,\text{d} N^{\scriptscriptstyle{(n)}}(t)\\[3pt] &=\frac 16 \int_{F}\sum_{l_1=1}^{k_1}\sum_{l_2=1}^{k_2}\sum_{l_3=1}^{k_3}\frac{x^{l_1+l_3}y^{l_1+l_2}z^{l_2+l_3}}{l_1!l_2!l_3!}(-1)^{l_1+l_2+l_3}\,\text{d} N^{\scriptscriptstyle{(n)}}(t)+O_{\scriptscriptstyle\mathbb{P}}\left({\varepsilon}\right)\\[3pt] &=\frac 16\sum_{l_1=1}^{k_1}\sum_{l_2=1}^{k_2}\sum_{l_3=1}^{k_3}\frac{(-1)^{l_1+l_2+l_3}}{l_1!l_2!l_3!}\int_{\varepsilon}^{1/\varepsilon}x^{l_1+l_3}\,\text{d} N_1^{\scriptscriptstyle{(n)}}(x)\int_{\varepsilon}^{1/\varepsilon}y^{l_1+l_2}\,\text{d} N_1^{\scriptscriptstyle{(n)}}(y)\\[3pt] &\quad \cdot \int_{\varepsilon}^{1/\varepsilon}{z^{l_2+l_3}}\,\text{d} N_1^{\scriptscriptstyle{(n)}}(z)+O_{\scriptscriptstyle\mathbb{P}}\left({\varepsilon}\right)\!.\end{align*}

Here we used that because $N_1^{\scriptscriptstyle{(n)}}[a,b]\ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} \lambda[a,b]$,

\begin{align*} \int_{F}\sum_{l_1=1}^{k_1}\sum_{l_2=1}^{k_2}\sum_{l_3=1}^{k_3}\varepsilon^4\,\text{d} N^{\scriptscriptstyle{(n)}}(t)& = k_1k_2k_3 \varepsilon^{3\gamma+1} \big(N_1^{\scriptscriptstyle{(n)}}([\varepsilon,1/\varepsilon])\big)^3\\& = \varepsilon^{3\gamma+1}O_{\scriptscriptstyle\mathbb{P}}\left({\lambda([\varepsilon,1/\varepsilon])^3}\right) = \varepsilon^{3\gamma+1}O_{\scriptscriptstyle\mathbb{P}}\left({(\varepsilon^{-\gamma}-\varepsilon^{\gamma})^3}\right) =O_{\scriptscriptstyle\mathbb{P}}\left({\varepsilon}\right)\!. \end{align*}

Since $x^t$ is bounded and continuous on $[\varepsilon,1/\varepsilon]$, we may use [Reference Van der Hofstad, van Leeuwaarden and Stegehuis19, Lemma 5] to conclude that

\begin{equation*}\int_{\varepsilon}^{1/\varepsilon}x^{t}\,\text{d} N_1^{\scriptscriptstyle{(n)}}(x)\ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}}\int_{\varepsilon}^{1/\varepsilon}x^{t}\,\text{d} \lambda(x)\,:\!= \varphi_\varepsilon(t).\end{equation*}

Thus,

\begin{align*}& \frac{\sum_{1\leq i< j < k\leq n}f_{n,\varepsilon}(D_i,D_j,D_k)}{\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)} \mu^{-\frac{3}{2}\gamma}}\\& =(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)) \frac 16\sum_{l_1=1}^{k_1}\sum_{l_2=1}^{k_2}\sum_{l_3=1}^{k_3}\frac{(-1)^{l_1+l_2+l_3}}{l_1!l_2!l_3!} \varphi_\varepsilon(l_1 + l_3) \varphi_\varepsilon(l_1 + l_2) \varphi_\varepsilon(l_2 + l_3)+O_{\scriptscriptstyle\mathbb{P}}\left({\varepsilon}\right)\\&=(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))\frac 16\int_F\sum_{l_1=1}^{k_1}\sum_{l_2=1}^{k_2}\sum_{l_3=1}^{k_3}\frac{x^{l_1+l_3}y^{l_1+l_2}z^{l_2+l_3}(-1)^{l_1+l_2+l_3}}{l_1!l_2!l_3!}\,\text{d} \lambda(x)\,\text{d} \lambda(y)\,\text{d} \lambda(z)+O_{\scriptscriptstyle\mathbb{P}}\left({\varepsilon}\right)\\&=(1+o_{\scriptscriptstyle{\mathbb{P}}}(1))\frac 16 \int_F h(x,y,z)\,\text{d} \lambda(x)\,\text{d} \lambda(y)\,\text{d} \lambda(z)+O_{\scriptscriptstyle\mathbb{P}}\left({\varepsilon}\right)\!,\end{align*}

which concludes the proof together with (76).

The following lemma bounds the variance of the number of triangles between vertices of sampled degrees proportional to $\sqrt{n}$.

Lemma 4.6. Let $\textbf{D}_n$ be sampled from $\mathscr{D}$ with $1< \gamma < 2$, and let $\widehat{G}_n = {\texttt{ECM}}(\textbf{D}_n)$. Let $\Delta_n(B_n(\varepsilon))$ denote the number of triangles in $\widehat{G}_n$ with sampled degrees in $B_n(\varepsilon)$. Then, as $n\to\infty$,

\begin{equation*} \frac{\textup{Var}_n\left({\Delta_n(B_n(\varepsilon))}\right)}{{\mathbb{E}}_n\left[{\Delta_n(B_n(\varepsilon))}\right]^2}\ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 0.\end{equation*}

Proof. Choose $0<\delta<\tfrac{1}{5}(1-\tfrac{1}{2}\gamma)$. Define

\begin{equation*} \mathcal{B}_n=\Bigg\{\sum_{i=1}^{n}{\mathds{1}}_{\left\{{D_i>\varepsilon\sqrt{\mu n}}\right\}} < n^{\frac{1}{2}(2-\gamma)+\delta}\Bigg\},\end{equation*}

let $\mathcal{A}_n$ be as defined in (42), and set $\Lambda_n=\mathcal{A}_n\cap\mathcal{B}_n$. Because the sampled degrees are i.i.d. samples from (1), $\mathbb{P}\left({\Lambda_n}\right)\to 1$. Therefore, we work on the event $\Lambda_n$ in the rest of the proof. By Lemma 4.5, ${\mathbb{E}}_n\left[{\Delta_n(B_n(\varepsilon))}\right]=\Theta_{\scriptscriptstyle{\mathbb{P}}}(n^{3-\frac{3}{2}\gamma}\mathcal{L}(\sqrt{n}))^3$. Thus, we need to prove that

\begin{equation*} \textup{Var}_n\left({\Delta_n(B_n(\varepsilon))}\right)=o_{\mathbb{P}}\left(n^{6-3\gamma}\mathcal{L}(\sqrt{n})^6\right)\!.\end{equation*}

We can write the variance $\textup{Var}_n\left({\Delta_n(B_n(\varepsilon))}\right)$ as

(78) \begin{align} &\sum_{(i,j,k),(s,t,u)} \left(\mathbb{P}_n\left({\Delta_{i,j,k},\Delta_{s,t,u}}\right)-\mathbb{P}_n\left({\Delta_{i,j,k}}\right)\mathbb{P}_n\left({\Delta_{s,t,u}}\right)\right){\mathds{1}}_{\left\{{D_i,D_j,D_j,D_s,D_t,D_u\in B_n(\varepsilon)}\right\}},\end{align}

where $\Delta_{i,j,k}$ denotes the event that there is a triangle between vertices i, j, and k. This splits into several cases, depending on $\left|{\{i,j,k,s,t,u\}}\right|$. Let the part of the variance where $\left|{\{i,j,k,s,t,u\}}\right|=m$ be denoted by $V^{\scriptscriptstyle{(m)}}$, so that

\begin{equation*}\textup{Var}_n\left({\Delta_n(B_n(\varepsilon))}\right)=V^{\scriptscriptstyle{(6)}}+V^{\scriptscriptstyle{(5)}}+V^{\scriptscriptstyle{(4)}}+V^{\scriptscriptstyle{(3)}}.\end{equation*}

Let $M_n(\varepsilon)=\sum_i{\mathds{1}}_{\left\{{D_i\in B_n(\varepsilon)}\right\}}$. Then

\begin{equation*}V^{\scriptscriptstyle{(m)}}\leq \sum_{i,j,k,s,t\,:\, |\{i,j,k,s,t\}|=m}{\mathds{1}}_{\left\{{D_i,D_j,D_j,D_s,D_t\in B_n(\varepsilon)}\right\}}= M_n(\varepsilon)^m, \qquad m=3,4,5.\end{equation*}

Under $\mathcal{B}_n$,

\begin{equation*}M_n(\varepsilon)\leq \sum_{i=1}^{n}{\mathds{1}}_{\left\{{D_i>\varepsilon\sqrt{\mu n}}\right\}}<n^{\frac{1}{2}(2-\gamma)+\delta}.\end{equation*}

Thus, by the choice of $\delta$,

\begin{equation*} V^{\scriptscriptstyle{(5)}} \leq M_n(\varepsilon)^5=O\left({n^{5-5\gamma/2+5\delta}}\right)=o(n^{6-3\gamma}),\end{equation*}

as required. Similar bounds show that $V^{\scriptscriptstyle{(4)}}$ and $V^{\scriptscriptstyle{(3)}}$ are $o(n^{6-3\gamma})$. Thus, the contributions of $V^{\scriptscriptstyle{(m)}}$, $m=3,4,5$, to the variance are sufficiently small.

Now we investigate the case where six different indices are involved and show that $V^{\scriptscriptstyle{(6)}}=o_{\scriptscriptstyle{\mathbb{P}}}(n^{6-3\gamma}\mathcal{L}(\sqrt{n})^6)$. Equation (73) computes the second term inside the brackets in (78). To compute the first term inside the brackets, we make a very similar computation that leads to (73). A similar computation as in (72) yields that on the event $\mathcal{A}_n$,

\begin{equation*}\mathbb{P}_n\left({\Delta_{i,j,k}, \Delta_{s,t,u}}\right) = (1+o_{\scriptscriptstyle{\mathbb{P}}}(1))g_{n,\varepsilon}(D_i,D_j,D_k)g_{n,\varepsilon}(D_s,D_t,D_u).\end{equation*}

Hence,

\begin{equation*}\mathbb{P}_n\left({\Delta_{i,j,k},\Delta_{s,t,u}}\right) -\mathbb{P}_n\left({\Delta_{i,j,k}}\right)\mathbb{P}_n\left({\Delta_{s,t,u}}\right)= o_{\scriptscriptstyle{\mathbb{P}}}(1)g_{n,\varepsilon}(D_i,D_j,D_k)g_{n,\varepsilon}(D_s,D_t,D_u).\end{equation*}

When $D_i,D_j,D_k\in[\varepsilon,1/\varepsilon]\sqrt{n}$, we have $g_{n,\varepsilon}(D_i,D_j,D_k)\in[f_1(\varepsilon),f_2(\varepsilon)]$, uniformly in i, j, k. Therefore, by Lemma 4.3,

\begin{align*} V^{\scriptscriptstyle{(6)}} &= \sum_{(i,j,k),(s,t,u)} o(1)g_{n,\varepsilon}(D_i,D_j,D_k)g_{n,\varepsilon}(D_s,D_t,D_u) \\ &=o_{\scriptscriptstyle{\mathbb{P}}} \left({\mathbb{E}}_n\left[{\Delta_n(B_n(\varepsilon))}\right]^2\right) = o_{\scriptscriptstyle{\mathbb{P}}} \left(n^{6-3\gamma}\mathcal{L}(\sqrt{n})^6\right)\!,\end{align*}

and therefore also the contribution to the variance where $\left|{\{i,j,k,s,t,u\}}\right|=6$ is small enough.

Proof of Theorem 1.4. First, we look at the denominator of the clustering coefficient in (4). By (15) and the stable-law central limit theorem [Reference Whitt42, Theorem 4.5.2], there exists a slowly-varying function $\mathcal{L}_0$ such that

(79) \begin{equation}\frac{\sum_{i=1}^n\widehat{D}_i(\widehat{D}_i-1)}{\mathcal{L}_0(n)n^{2/\gamma}}=\frac{\sum_{i=1}^n{D}_i^2}{\mathcal{L}_0(n)n^{2/\gamma}}(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)) \ensuremath{\stackrel{d}{\longrightarrow}} \mathcal{S}_{\gamma/2},\end{equation}

where $\mathcal{S}_{\gamma/2}$ is a stable distribution.

Now we consider the numerator of the clustering coefficient. We prove the convergence of the number of triangles in several steps. First, we show that the major contribution to the number of triangles comes from the triangles between vertices with degrees proportional to $\sqrt{n}$. Fix $\varepsilon>0$. We use (64), where we want to show that the contribution of $\Delta_n(\bar{B}_n(\varepsilon))$ is negligible. Applying Lemma 4.2 with the Markov inequality yields, for every $\delta > 0$,

\begin{equation*}\limsup_{n\to\infty} \mathbb{P}\left({\Delta_n(\bar{B}_n(\varepsilon))> \delta{ \mathcal{L}(\sqrt{\mu n})^3}n^{\frac32 (2-\gamma)}}\right)= O\left(\frac{\mathcal{E}_1(\varepsilon)}{\delta}\right)\!.\end{equation*}

Therefore,

(80) \begin{equation}\begin{aligned}[b]\Delta_n(\bar{B}_n(\varepsilon))=O_{\scriptscriptstyle\mathbb{P}}\left({\mathcal{E}_1(\varepsilon) \mathcal{L}(\sqrt{\mu n})^3n^{\frac32 (2-\gamma)}}\right)\!.\end{aligned}\end{equation}

Because $\mathcal{E}_1(\varepsilon)$ tends to zero as $\varepsilon\to 0$, we now focus on $\Delta_n(B_n(\varepsilon))$. The number of triangles consists of two sources of randomness: the random pairing of the edges, and the random degrees. First we show that $\Delta_n(B_n(\varepsilon))$ concentrates around its mean when conditioned on the degrees. By Lemma 4.6 and Chebyshev’s inequality,

\begin{equation*} \frac{\Delta_n(B_n(\varepsilon))}{{\mathbb{E}}_n\left[{\Delta_n(B_n(\varepsilon))}\right]}\ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} 1, \end{equation*}

conditionally on the degree sequence $(D_i)_{i\in[n]}$. Combining this with Lemma 4.5 yields that conditionally on $(D_i)_{i\in[n]}$,

(81) \begin{align} &\Delta_n(B_n(\varepsilon)) =\mathcal{L}(\sqrt{\mu n})^{{3n}^{\frac{3}{2}(2-\gamma)}} \mu^{-\frac{3}{2}\gamma}\nonumber \\ &\quad \times\left((1+o_{{\mathbb{P}}}(1))\frac{1}{6}\int_{\varepsilon}^{1/\varepsilon} \int_{\varepsilon}^{1/\varepsilon} \int_{\varepsilon}^{1/\varepsilon} \!\!\! \frac{\gamma^3}{(t_{1}t_{2}t_{3})^{\gamma+1}}h(t_1,t_2,t_3)\,\text{d} t_1\,\text{d} t_2\,\text{d} t_{{3+O}_{\mathbb{P}}}\left({\varepsilon}\right)\right)\!,\end{align}

where $h(x,y,z)\,:\!=\,(1-\text{e}^{-xy})(1-\text{e}^{-xz})(1-\text{e}^{-yz})$ and $\varepsilon$ is the same as in (80).

Combining (80) and (81) gives

\begin{align*}&\frac{\Delta_n}{\mathcal{L}(\sqrt{\mu n})^3n^{\frac 32(2-\gamma)}} =\frac{\Delta_n(B_n(\varepsilon))+\Delta_n(\bar{B}_n(\varepsilon))}{\mathcal{L}(\sqrt{\mu n})^3n^{\frac 32(2-\gamma)}}\nonumber\\ &\qquad =(1+o_{\scriptscriptstyle{\mathbb{P}}}(1)) \mu^{-\frac 32 \gamma}\frac{1}{6}\int_{\varepsilon}^{1/\varepsilon} \int_{\varepsilon}^{1/\varepsilon} \int_{\varepsilon}^{1/\varepsilon} \frac{\gamma^3}{(t_1t_2t_3)^{\gamma+1}}h(t_1,t_2,t_3)\,\text{d} t_1\,\text{d} t_2\,\text{d} t_3 \\ &\qquad\quad +O_{\scriptscriptstyle\mathbb{P}}\left({\mathcal{E}_1(\varepsilon)}\right)+O_{\scriptscriptstyle\mathbb{P}}\left({\varepsilon}\right)\!.\end{align*}

Taking the limit as $n\to\infty$ and then $\varepsilon\to0$ combined with (79) and the definition of the clustering coefficient in (4) then results in

(82) \begin{align}\frac{\hat{C}_n\mathcal{L}_0(n)}{\mathcal{L}(\sqrt{\mu n})^3n^{(-3\gamma^2+6\gamma-4)/(2\gamma)}}&=6\frac{\mathcal{L}_0(n)n^{2/\gamma}}{\sum_{i\in[n]}\hat{D}_i(\hat{D}_i-1)}\frac{\Delta_n}{\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)}}\end{align}

(83) \begin{align}& \ensuremath{\stackrel{d}{\longrightarrow}} \mu^{-\frac 32 \gamma}\frac{ \int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\gamma^3}{(t_1t_2t_3)^{\gamma+1}}h(t_1,t_2,t_3)\,\text{d} t_1\,\text{d} t_2\,\text{d} t_3}{\mathcal{S}_{\gamma/2}}.\end{align}

We then compute the triple integral by making the change of variables $t_1=\sqrt{ac/b}$, $t_2=\sqrt{ab/c}$, $t_3=\sqrt{bc/a}$. The Jacobian J(a, b, c) of this change of variables equals

\begin{equation*} J(a,b,c)=\frac{1}{2\sqrt{abc}}. \end{equation*}

Then

(84) \begin{align}&\int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty}\frac{\gamma^3}{(t_1t_2t_3)^{\gamma+1}}h(t_1,t_2,t_3)\,\text{d} t_1\,\text{d} t_2\,\text{d} t_3\nonumber\\ &\quad =\int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty}\frac{\gamma^3}{(t_1t_2t_3)^{\gamma+1}}(1-\textup{e}^{-t_1t_2})(1-\textup{e}^{-t_1t_3})(1-\textup{e}^{-t_2t_3})\,\text{d} t_1\,\text{d} t_2\,\text{d} t_3\nonumber\\ &\quad = \gamma^3\int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty}(\sqrt{abc})^{-1-\gamma}(1-\textup{e}^{-a})(1-\textup{e}^{-b})(1-\textup{e}^{-c})\frac{1}{2\sqrt{abc}}\,\text{d} a\,\text{d} b \,\text{d} c\nonumber\\ &\quad = \frac{\gamma^3}{2}\left( \int_{0}^{\infty}{{a}^{-1-\gamma/2}}(1-\textup{e}^{-a})\,\text{d} a\right)^3 =\frac{\gamma^3}{2}\left(-\Gamma\left(-\frac{\gamma}{2}\right)\right)^3,\end{align}

where we have used that

\begin{equation*}\begin{aligned}[b]\int_{0}^{\infty}a^{-1-\gamma/2}(1-\textup{e}^{-a})\,\text{d} a & = \int_{0}^{\infty}\int_{0}^{a}a^{-1-\gamma/2}\textup{e}^{-y}\,\text{d} y \,\text{d} a = \int_{0}^{\infty}\int_{y}^{\infty}a^{-1-\gamma/2}\textup{e}^{-y}\,\text{d} a \,\text{d} y\\& = -\frac{2}{\gamma}\int_{0}^{\infty}y^{-\gamma/2}\textup{e}^{-y}\,\text{d} y = -\frac{\Gamma(\gamma/2+1)}{\gamma/2}=-\Gamma(\gamma/2).\end{aligned}\end{equation*}

Combining this with (82) proves Theorem 1.4.

5.1 Pearson in the rank-1 inhomogeneous random graph

The analysis of the term $r_n^-$ in (11) is the same as in the ECM, since it only depends on the degrees, and (27) also holds for the inhomogeneous random graph. We therefore only need to show that Proposition 3.3 also holds for the rank-1 inhomogeneous random graph. This means that we need to show that (41) also holds for the rank-1 inhomogeneous random graph. For all models satisfying Condition 1.1, $p(w_i,w_j)\leq (w_iw_j/(\mu n) \wedge 1)$. Because ${\mathbb{E}}_n\left[{D_i}\right]=w_i(1+o(1))$, $D_i=O_{\scriptscriptstyle\mathbb{P}}\left({w_i}\right)$, by Markov’s inequality. Thus,

\begin{equation*}\sum_{1 \le i < j \le n}D_iD_jX_{ij}=O_{\scriptscriptstyle\mathbb{P}}\left({\sum_{1 \le i < j \le n}w_iw_jX_{ij}}\right)= O_{\scriptscriptstyle\mathbb{P}}\left({\sum_{1 \le i < j \le n}w_iw_j \left(\frac{w_iw_j}{\mu n} \wedge 1\right)}\right)\!.\end{equation*}

Because the weights are sampled from (1), this is exactly the same bound as in (45), so that from there we can follow the same lines as the proof of Proposition 3.3. Thus, Proposition 3.3 also holds for rank-1 inhomogeneous random graphs satisfying Condition 1.1. Then we can follow the same lines as the proof of Theorem 1.1 to prove Theorem 1.2.

5.2. Clustering in the rank-1 inhomogeneous random graph

For the clustering coefficient, note that conditioned on the weights, $\mathbb{P}_n\left({\Delta_{i,j,k}}\right)=p(w_i,w_j)p(w_j,w_k)p(w_i,w_k)$. Furthermore, Lemma 4.2 only requires the bound $p(w_i,w_j)$$\leq (w_iw_j/(\mu n) \wedge 1)$, which also holds for all rank-1 inhomogeneous random graphs satisfying Condition 1.1, so that Lemma 4.2 also holds for these rank-1 inhomogeneous random graphs. Furthermore, conditioned on the weights, the probabilities of distinct edges being present are independent, so that

\begin{equation*}{\mathbb{E}}_n\left[{\Delta_n(B_n(\varepsilon))}\right]=\sum_{1 \le i < j < k \le n}q\left(\frac{w_iw_j}{\mu n}\right)q\left(\frac{w_iw_k}{\mu n}\right)q\left(\frac{w_jw_k}{\mu n}\right){\mathds{1}}_{\left\{{(w_i,w_j,w_k)\in B_n(\varepsilon)}\right\}},\end{equation*}

similarly to Lemma 4.3, with q as defined in Condition 1.1. Furthermore, the bound on the variance of the number of triangles in the ECM in Lemma 4.6 for three, four, or five contributing vertices only depends on the degrees, so that it also holds for the rank-1 inhomogeneous random graph satisfying Condition 1.1, since the weights are also sampled from (1). The contribution of six different vertices to the variance is zero, because the presence of distinct edges is independent. Thus, Lemma 4.6 also holds for the rank-1 inhomogeneous random graph. Thus, we can follow the lines of the proof of Theorem 1.4 up to Equation (77). From there, note that

\begin{align*}& \frac{\sum_{1 \le i < j < k \le n}q\left(\frac{w_iw_j}{\mu n}\right)q\left(\frac{w_iw_k}{\mu n}\right)q\left(\frac{w_jw_k}{\mu n}\right){\mathds{1}}_{\left\{{(w_i,w_j,w_k)\in B_n(\varepsilon)}\right\}}}{\mathcal{L}(\sqrt{\mu n})^3n^{\frac{3}{2}(2-\gamma)}\mu^{-\tfrac{3}{2}\gamma}} \\& \quad = \frac{1}{6}\int_Fq(t_1t_2)q(t_1t_3)q(t_2t_3)\,\text{d} N^{\scriptscriptstyle{(n)}}_1(t_1)\,\text{d} N^{\scriptscriptstyle{(n)}}_1(t_2)\,\text{d} N^{\scriptscriptstyle{(n)}}_1(t_3).\end{align*}

Then we use that the function $q(t_1t_2)q(t_1t_3)q(t_2t_3)$ is a bounded, continuous function by Condition 1.1, so that by [Reference Van der Hofstad, van Leeuwaarden and Stegehuis19, Lemma 5],

\begin{align*}& \int_Fq(t_1t_2)q(t_1t_3)q(t_2t_3)\,\text{d} N^{\scriptscriptstyle{(n)}}_1(t_1)\,\text{d} N^{\scriptscriptstyle{(n)}}_1(t_2)\,\text{d} N^{\scriptscriptstyle{(n)}}_1(t_3)\\& \quad \ensuremath{\stackrel{\mathbb{P}}{\longrightarrow}} \int_Fq(t_1t_2)q(t_1t_3)q(t_2t_3)\,\text{d} \lambda(t_1)\,\text{d}\lambda(t_2)\,\text{d} \lambda(t_3).\end{align*}

Thus, we can follow the same lines as the proof of the clustering coefficient in the ECM, replacing the term $(1-\textup{e}^{-t_1t_2})(1-\textup{e}^{-t_1t_3})(1-\textup{e}^{-t_2t_3})$ by $q(t_1t_2)q(t_1t_3)q(t_2t_3)$, which then proves Theorem 1.5.