2. Jaccard index of a vertex pair

Let us denote by G(n, p) an Erdös–Rényi random graph on the vertex set [n], where each edge is present independently with probability p. In this paper we consider $p = p(n)$ as a function of the graph size n. For any two vertices $i, j \in [n]$ , let $\mathbf{1}_{ij}$ be the indicator that takes the value 1 if an edge between i and j is present in G(n, p), and takes the value 0 otherwise. It follows that $\mathbf{1}_{ii} = 0$ , $\mathbf{1}_{ij} = \mathbf{1}_{ji}$ , and $\{\mathbf{1}_{ij} \,:\, 1 \leq i < j \leq n\}$ is a sequence of independent Bernoulli variables with success rate p. The $n\times n$ matrix $\boldsymbol{A}=(\mathbf{1}_{ij})$ is usually called the adjacent matrix of G(n, p), and is a symmetric matrix with all diagonal entries equal to zero.

For any vertex $i\in[n]$ , let the set ${\mathcal N}_i$ be its neighborhood, i.e., ${\mathcal N}_{i}=\{k\,:\,\mathbf{1}_{ik}=1, k\in [n]\}$ . For any pair of vertices $i,j\in[n]$ , we also define their union neighborhood as

\begin{equation*} {\mathcal N}_{ij} = \{k\,:\,\mathbf{1}_{ik}\vee \mathbf{1}_{jk}=1,\,k\in[n]\,\mbox{and}\,k\neq i,j\}, \quad i\neq j.\end{equation*}

Notice that here the neighborhood set ${\mathcal N}_{ij}$ does not contain vertices i and j themselves, even if $\mathbf{1}_{ij}=1$ . Then the Jaccard index of vertices i and j in G(n, p) is formally defined as

(1) \begin{equation} J_{ij}^{(n)} = \frac{|{\mathcal N}_{i}\cap {\mathcal N}_{j}|}{|{\mathcal N}_{ij}|} \,=\!:\, \frac{S_{ij}^{(n)}}{T_{ij}^{(n)}}, \quad i\neq j.\end{equation}

We can see that the index $J_{ij}^{(n)}$ given in (1) is not well defined when ${\mathcal N}_{ij}$ is an empty set or $T_{ij}^{(n)}=0$ . For convenience, following the idea in [Reference Chung, Miasojedow, Startek and Gambin6], we define $J_{ij}^{(n)}=p/(2-p)$ in this special case. Indeed, it is shown later that the conditional expectation of $J_{ij}^{(n)}$ is exactly $p/(2-p)$ given that $T_{ij}^{(n)}>0$ . In terms of the adjacent matrix $\boldsymbol{A}$ , the numerator and denominator in (1) can be rewritten as

(2) \begin{equation} S_{ij}^{(n)} = \sum\limits_{k\neq i,j}\mathbf{1}_{ik}\mathbf{1}_{jk}, \qquad T_{ij}^{(n)} = \sum\limits_{k\neq i,j}\mathbf{1}_{ik} \vee \mathbf{1}_{jk}.\end{equation}

Due to the independence of the elements in $\boldsymbol{A}$ , it is clear that the random variables $S_{ij}^{(n)}$ and $T_{ij}^{(n)}$ follow the binomial distributions $\textrm{Bin}(n-2,p^2)$ and $\textrm{Bin}(n-2,p(2-p))$ , respectively. Hence, the Jaccard index of a vertex pair in G(n, p) is a quotient of two dependent binomial random variables.

2.1. Mean and variance

We first calculate the mean and variance of the Jaccard index of any pair of vertices in G. By (1) and (2), we can see that $\big\{J_{ij}^{(n)},\, 1\le i<j\le n\big\}$ is a sequence of random variables that are pairwise dependent but identically distributed. Without loss of generality, we only consider $J_{12}^{(n)}$ .

For any vertex $3\le k\le n$ , the conditional probability

\begin{equation*} \mathbb{P}(\mathbf{1}_{1k}\mathbf{1}_{2k}=1 \mid \mathbf{1}_{1k} \vee \mathbf{1}_{2k} = 1) = \frac{\mathbb{P}(\mathbf{1}_{1k}\mathbf{1}_{2k} = 1)}{\mathbb{P}(\mathbf{1}_{1k} \vee \mathbf{1}_{2k} = 1)} = \frac{p}{2-p},\end{equation*}

which is independent of k. Then, for any positive integer $1\le m\le n-2$ , given the event $T_{12}^{(n)}=m$ , the conditional distribution of $S_{12}^{(n)}$ is $\textrm{Bin}(m,p/(2-p))$ , due to independence of the indicators $\{\mathbf{1}_{1k},\mathbf{1}_{2k},\,3\le k\le n\}$ . Consequently, we have

(3) \begin{align} {\mathbb E}\Big[S_{12}^{(n)} \mid T_{12}^{(n)} = m\Big] & = \frac{mp}{2-p}, \end{align}

(4) \begin{align} \textrm{Var}\Big[S_{12}^{(n)} \mid T_{12}^{(n)} = m\Big] & = \frac{2mp(1-p)}{(2-p)^{2}}. \end{align}

Noting that $J_{12}^{(n)}=p/(2-p)$ in the special case $T_{12}^{(n)}=0$ , by (1) and (3) we thus have ${\mathbb E}\Big[J_{12}^{(n)} \mid T_{12}^{(n)}\Big] = {p}/({2-p})$ , which implies that ${\mathbb E}\Big[J_{12}^{(n)}\Big]=p/(2-p)$ and $\textrm{Var}\Big[{\mathbb E}\Big(J_{12}^{(n)} \mid T_{12}^{(n)}\Big)\Big] = 0$ . Using the law of total variance, it follows from this and (4) that

(5) \begin{align} \textrm{Var}\Big[J_{12}^{(n)}\Big] = {\mathbb E}\Big[\textrm{Var}\Big(J_{12}^{(n)} \mid T_{12}^{(n)}\Big)\Big] & = \sum_{m=1}^{n-2}{\mathbb P}\Big(T_{12}^{(n)} = m\Big) \textrm{Var}\bigg[\frac{S_{12}^{(n)}}{m} \mid T_{12}^{(n)} = m\bigg] \notag \\ & = \frac{2p(1-p)}{(2-p)^{2}}\sum_{m=1}^{n-2}\frac{1}{m}{\mathbb P}\Big(T_{12}^{(n)}=m\Big),\end{align}

which involves the first inverse moment of the binomial distribution. Recalling that $T_{12}^{(n)}$ has the distribution $\textrm{Bin}(n-2,p(2-p))$ , it follows by [Reference Wuyungaowa and Wang28, Corollary 1] that

(6) \begin{equation} \sum_{m=1}^{n-2}\frac{1}{m}{\mathbb P}\Big(T_{12}^{(n)} = m\Big) = \frac{1}{np(2-p)}\bigg(1 + O\bigg(\frac{1}{np}\bigg)\bigg)\end{equation}

as $np\to\infty$ . By (5) and (6), we thus have

\begin{equation*} \textrm{Var}\Big[J_{12}^{(n)}\Big] = \frac{2(1-p)}{n(2-p)^{3}}\bigg(1 + O\bigg(\frac{1}{np}\bigg)\bigg).\end{equation*}

Collecting the above findings, by Chebyshev’s inequality we thus prove the following result.

Proposition 1. Let $J_{ij}^{(n)}$ be the Jaccard index of any distinct vertices $i,j\in[n]$ in G(n,p). Then ${\mathbb E}\Big[J_{ij}^{(n)}\Big]=p/(2-p)$ for all $n\ge 2$ . In particular, as $np\to\infty$ , it further follows that

\begin{equation*} \textrm{Var}\Big[J_{ij}^{(n)}\Big] = \frac{2(1-p)}{n(2-p)^{3}}\bigg(1 + O\bigg(\frac{1}{np}\bigg)\bigg), \end{equation*}

and $J_{ij}^{(n)}-p/(2-p)$ converges to 0 in probability.

2.2. Asymptotic distribution

We now establish the asymptotic distribution of the Jaccard index of any vertex pair in G(n, p).

Theorem 1. Let $J_{ij}^{(n)}$ be the Jaccard index of any distinct vertices $i,j\in[n]$ in G(n,p).

1. (i) If $np^{2}(1-p)\to\infty$ , then

   \begin{equation*} \sqrt{\frac{n(2-p)^{3}}{2(1-p)}}\bigg(J_{ij}^{(n)}-\frac{p}{2-p}\bigg) \buildrel \textrm{D} \over \longrightarrow Z, \end{equation*}
   where Z denotes a standard normal random variable.

2. (ii) If $np^2\to \lambda$ for some constant $\lambda>0$ , then $2npJ_{ij}^{(n)}\buildrel \textrm{D} \over \longrightarrow \textrm{Poi}(\lambda)$ , where $\textrm{Poi}(\lambda)$ denotes the Poisson distribution with parameter $\lambda$ .

3. (iii) If $np^2\to0$ , then $npJ_{ij}^{(n)}\buildrel \textrm{P} \over \longrightarrow 0$ .

4. (iv) If $n(1-p)\to c$ for some constant $c>0$ , then $n\Big(1-J_{ij}^{(n)}\Big)\buildrel \textrm{D} \over \longrightarrow \textrm{Poi}(2c)$ .

5. (v) If $n(1-p)\to 0$ , then $n\big(1-J_{ij}^{(n)}\big)\buildrel \textrm{P} \over \longrightarrow 0$ .

Proof. As presented in the previous subsection, it is sufficient to consider a single index $J_{12}^{(n)}$ . To prove (i), we first rewrite

(7) \begin{align} \sqrt{\frac{(n-2)(2-p)^{3}}{2(1-p)}}\bigg(J_{12}^{(n)}-\frac{p}{2-p}\bigg) & = \sqrt{\frac{(n-2)(2-p)}{2(1-p)}}\cdot\frac{(2-p)S_{12}^{(n)}-pT_{12}^{(n)}}{T_{12}^{(n)}} \notag \\ & = \frac{(2-p)S_{12}^{(n)}-pT_{12}^{(n)}}{\sqrt{2(n-2)p^{2}(1-p)(2-p)}\,}\cdot\frac{(n-2)p(2-p)}{T_{12}^{(n)}}. \end{align}

For any distinct vertices $i,j,k\in[n]$ , we write

(8) \begin{equation} V_{ij,k} = (2-p)\mathbf{1}_{ik}\mathbf{1}_{jk}-p(\mathbf{1}_{ik}\vee \mathbf{1}_{jk}). \end{equation}

Then, for any fixed two vertices $i,j\in[n]$ , the random variables $\{V_{ij,k},\,k\in[n],k\neq i,j\}$ are independent and identically distributed with common mean 0 and common variance $2p^{2}(1-p)$ $(2-p)$ . Since it follows by (2) that

(9) \begin{equation} (2-p)S_{12}^{(n)} - pT_{12}^{(n)} = \sum_{k=3}^{n} V_{12,k}, \end{equation}

a direct application of the Lindeberg–Feller central limit theorem yields

(10) \begin{equation} \frac{(2-p)S_{12}^{(n)}-pT_{12}^{(n)}}{\sqrt{2(n-2)p^{2}(1-p)(2-p)}\,} \buildrel \textrm{D} \over \longrightarrow Z \end{equation}

whenever $np^2(1-p)\to\infty$ . By Chebyshev’s inequality, the fact that $T_{12}^{(n)}\sim \textrm{Bin}((n-2),$ $p(2-p))$ gives us that, as $np\to\infty$ ,

(11) \begin{equation} \frac{T_{12}^{(n)}}{(n-2)p(2-p)}\buildrel \textrm{P} \over \longrightarrow 1, \end{equation}

which, together with (7) and (10), proves (i) by Slutsky’s lemma.

If $np^{2}\to \lambda$ for some constant $\lambda>0$ , we must have that $p\to 0$ and $np\to\infty$ . Since $S_{12}^{(n)}\sim \textrm{Bin}(n-2,p^2)$ , the Poisson limit theorem yields that $S_{12}^{(n)}\buildrel \textrm{D} \over \longrightarrow \textrm{Poi}(\lambda)$ . By (11), again using Slutsky’s lemma, we obtain

\begin{equation*} 2npJ_{12}^{(n)} = \frac{2np}{T_{12}^{(n)}}\cdot S_{12}^{(n)}\buildrel \textrm{D} \over \longrightarrow \textrm{Poi}(\lambda), \end{equation*}

which proves (ii).

If $np^2\to0$ , to prove (iii) we only need to prove that the probability ${\mathbb P}\Big(npJ_{ij}^{(n)}>np^2/$ $(2-p)\Big)$ tends to 0. Note that if the event $\Big\{S_{12}^{(n)}=0\Big\}$ occurs, the Jaccard index $J_{ij}^{(n)}$ must be 0 or $p/(2-p)$ . Therefore,

\begin{equation*} {\mathbb P}\bigg(npJ_{ij}^{(n)}>\frac{np^2}{2-p}\bigg) = {\mathbb P}\bigg(J_{ij}^{(n)}>\frac{p}{2-p}\bigg) \le {\mathbb P}\Big(S_{12}^{(n)}\neq0\Big) = 1 - {\mathbb P}\Big(S_{12}^{(n)}=0\Big). \end{equation*}

Again by the fact that $S_{12}^{(n)}\sim \textrm{Bin}(n-2,p^2)$ , we have $1-{\mathbb P}\big(S_{12}^{(n)}=0\big)\to 0$ if $np^2\to0$ . This implies (iii).

Analogously to (7), we have

(12) \begin{equation} \frac{p}{2-p}-J_{12}^{(n)} = \frac{-(2-p)S_{12}^{(n)}+pT_{12}^{(n)}}{(n-2)p(2-p)^2} \cdot \frac{(n-2)p(2-p)}{T_{12}^{(n)}}. \end{equation}

Note that (11) still holds, and $n[1-p/(2-p)]$ has the limit 2c if $n(1-p)\to c$ for some constant $c>0$ . To prove (iv), by (12) it is now sufficient to show that

(13) \begin{equation} -(2-p)S_{12}^{(n)}+pT_{12}^{(n)}\buildrel \textrm{D} \over \longrightarrow \textrm{Poi}(2c)-2c. \end{equation}

For any distinct vertices $i,j,k\in[n]$ , we can directly obtain from (8) that the characteristic function of $-V_{ij,k}$ is

\begin{equation*} f_n(t) = {\mathbb E}\big[\textrm{e}^{-\textrm{i}tV_{ij,k}}\big] = p^2\textrm{e}^{-2\textrm{i}t(1-p)} + 2p(1-p)\textrm{e}^{\textrm{i}tp}+(1-p)^2, \end{equation*}

where $\textrm{i}=\sqrt{-1}$ denotes the imaginary unit. Then, by (9) and independence, we have that the characteristic function of $-(2-p)S_{12}^{(n)}+pT_{12}^{(n)}$ is equal to

\begin{equation*} f_n^{n-2}(t) = \big[p^2\textrm{e}^{-2\textrm{i}t(1-p)}+2p(1-p)\textrm{e}^{\textrm{i}tp}+(1-p)^2\big]^{n-2}, \quad t\in\mathbb{R}. \end{equation*}

Note that

\begin{align*} \lim_{n\to\infty}n\big[p^2\textrm{e}^{-2\textrm{i}t(1-p)}+2p(1-p)\textrm{e}^{\textrm{i}tp}+(1-p)^2-1\big] & = 2c\textrm{e}^{\textrm{i}t}+\lim_{n\to\infty}n\big[p^2\textrm{e}^{-2\textrm{i}t(1-p)}-1\big] \\ & = 2c\textrm{e}^{\textrm{i}t}+\lim_{n\to\infty}n\big[p^2\big(\textrm{e}^{-2\textrm{i}t(1-p)}-1\big)\\ & \quad +\big(p^2-1\big)\big] \\ & = 2c\big(\textrm{e}^{\textrm{i}t}-\textrm{i}t-1\big) \end{align*}

if $n(1-p)\to c$ . Therefore, the limit of the characteristic function of $-(2-p)S_{12}^{(n)}+pT_{12}^{(n)}$ satisfies

\begin{equation*} \lim_{n\to\infty}f_n^{n-2}(t)=\exp\big\{2c\big(\textrm{e}^{\textrm{i}t}-\textrm{i}t-1\big)\big\}, \end{equation*}

which implies (13) and completes the proof of (iv).

We only sketch the proof of (v), since it is very similar to (iv). By (11) and (12), it is sufficient to show that $-(2-p)S_{12}^{(n)}+pT_{12}^{(n)}$ converges in probability to 0 under the condition $n(1-p)\to 0$ . In fact, following the proof of (13), in this case we can deduce that its characteristic function $f_n^{n-2}(t)\to 1$ .

3. Average Jaccard index

In this section we derive asymptotic properties of the average Jaccard index of G(n, p), which is given by

(14) \begin{equation} J_n = \frac{2}{n(n-1)}\sum_{i=1}^{n-1} \sum_{j=i+1}^n J_{ij}^{(n)}.\end{equation}

That is, the average Jaccard index $J_n$ is the average of the Jaccard indices over all vertex pairs in the Erdös–Rényi random graph G(n, p). An immediate consequence of Proposition 1 is that the expectation of $J_n$ is equal to $p/(2-p)$ .

We now state the main results of this paper.

Theorem 2. Let $J_n$ be the average Jaccard index of G(n,p). If $np\to\infty$ and $n^2(1-p)\to\infty$ , then

\begin{equation*} \frac{n(2-p)^2}{\sqrt{8p(1-p)}\,}\bigg(J_n-\frac{p}{2-p}\bigg)\buildrel \textrm{D} \over \longrightarrow Z, \end{equation*}

where Z denotes a standard normal random variable.

It is remarkable that the quantity $n^2(1-p)/2$ is the asymptotic expected number of unoccupied edges in G(n, p). Theorem 2 suggests that if G(n, p) is neither too sparse (see, e.g., [Reference van der Hofstad25, Section 1.8]) nor close to a complete graph, then its average Jaccard index $J_n$ has asymptotic normality. In order to prove Theorem 2, we first introduce two auxiliary lemmas, both of which involve the inverse moments of the binomial distribution.

Lemma 1. If the random variable $X_n$ has a binomial distribution with parameters n and p, then, for any fixed constants $a\ge0$ and $b>0$ , as $np\to\infty$ ,

\begin{equation*} {\mathbb E}\bigg[\bigg(\frac{(n+a)p}{b+X_n}-1\bigg)^2\bigg] = O\bigg(\frac{1}{np}\bigg). \end{equation*}

Proof. For any $\varepsilon\in (0,1)$ , we define ${\mathcal A}_n \,:\!=\,\{(1-\varepsilon)(n+a)p \le b + X_n \le (1+\varepsilon)(n+a)p\}$ . Applying Chernoff’s bound for the binomial distribution (see, e.g., [Reference Janson, Luczak and Rucinski14, Corollary 2.3]) gives, for sufficiently large n,

\begin{equation*} {\mathbb P}\big(\overline{{\mathcal A}}_n\big) \le {\mathbb P}\bigg(\bigg|\frac{X_n}{np}-1\bigg| > \frac{\varepsilon}{2}\bigg) \le 2\exp\bigg\{{-}\frac{\varepsilon^2}{12}np\bigg\} = O\bigg(\frac{1}{n^3p^3}\bigg). \end{equation*}

Then, for sufficiently large n, by noting that

\begin{equation*} \bigg(\frac{(n+a)p}{b+X_n}-1\bigg)^2 \le \bigg(\frac{(n+a)p}{b}\bigg)^2 + 1 \le 2n^2p^2, \end{equation*}

we have

\begin{align*} {\mathbb E}\bigg[\bigg(\frac{(n+a)p}{b+X_n}-1\bigg)^2\bigg] & = {\mathbb E}\bigg[\bigg(\frac{(n+a)p}{b+X_n}-1\bigg)^2\mathbf{1}({\mathcal A}_n)\bigg] + {\mathbb E}\bigg[\bigg(\frac{(n+a)p}{b+X_n}-1\bigg)^2\mathbf{1}\big(\overline{{\mathcal A}}_n\big)\bigg] \\ & \le {\mathbb E}\bigg[\frac{(n+a)^2p^2}{(b+X_n)^2}\bigg(\frac{b+X_n}{(n+a)p}-1\bigg)^2 \mathbf{1}({\mathcal A}_n)\bigg] + 2n^2p^2\mathbb{P}\big(\overline{{\mathcal A}}_n\big) \\ & \le (1-\varepsilon)^{-2}{\mathbb E}\bigg[\bigg(\frac{b+X_n}{(n+a)p}-1\bigg)^2\bigg] + 2n^2p^2\mathbb{P}\big(\overline{{\mathcal A}}_n\big) \\ & = O\bigg(\frac{1}{np}\bigg), \end{align*}

where we used the inequality

\begin{equation*} {\mathbb E}\bigg[\bigg(\frac{b+X_n}{(n+a)p}-1\bigg)^2\bigg] \le \frac{2}{(n+a)^2p^2}\big(\mathbb{E}\big[(X_n-np)^2\big] + (b-ap)^2\big) = O\bigg(\frac{1}{np}\bigg). \end{equation*}

Lemma 2. If the random variable $X_n$ has a binomial distribution with parameters n and p, then, for any fixed positive constants b and $\alpha$ , as $np\to\infty$ ,

\begin{equation*} {\mathbb E}\bigg[\frac{1}{(b+X_n)^{\alpha}}\bigg] = \frac{1}{(np)^{\alpha}}(1+o(1)). \end{equation*}

We now give a formal proof of Theorem 2.

Proof of Theorem 2. By (12), the index $J_{ij}^{(n)}$ can be expressed as

(15) \begin{equation} J_{ij}^{(n)} = \frac{p}{2-p} + \frac{(2-p)S_{ij}^{(n)}-pT_{ij}^{(n)}}{(n-2)p(2-p)^2} + R_{ij}^{(n)}, \quad 1 \le i \neq j \le n, \end{equation}

where the remainder term is

(16) \begin{equation} R_{ij}^{(n)} = \frac{(2-p)S_{ij}^{(n)}-pT_{ij}^{(n)}}{(n-2)p(2-p)^2}\bigg(\frac{(n-2)p(2-p)}{T_{ij}^{(n)}}-1\bigg). \end{equation}

Note the special case in (15) that the remainder term $R_{ij}^{(n)}$ vanishes if $T_{ij}^{(n)}=0$ . Taking expectation on both sides of (15) gives, for any distinct vertices $i,j\in[n]$ ,

(17) \begin{equation} {\mathbb E}\Big[R_{ij}^{(n)}\Big]=0. \end{equation}

Denote by $R_n$ the sum of all the remainder terms, i.e.,

(18) \begin{equation} R_{n} = \sum_{i=1}^{n-1}\sum_{j=i+1}^nR_{ij}^{(n)}. \end{equation}

Then it follows by (17) that ${\mathbb E}[R_n]=0$ . By (2) and the simple fact that $\mathbf{1}_{ik}\vee \mathbf{1}_{jk}=\mathbf{1}_{ik}+\mathbf{1}_{jk}-\mathbf{1}_{ik}\mathbf{1}_{jk}$ , we have, for any $1\le i\neq j\le n$ ,

\begin{equation*} (2-p)S_{ij}^{(n)} - pT_{ij}^{(n)} = \sum_{k\neq i,j}\big[(2-p)\mathbf{1}_{ik}\mathbf{1}_{jk}-p\mathbf{1}_{ik}\vee \mathbf{1}_{jk}\big] = \sum_{k\neq i,j}\big[2\mathbf{1}_{ik}\mathbf{1}_{jk}-p\big(\mathbf{1}_{ik}+\mathbf{1}_{jk}\big)\big], \end{equation*}

which, together with (14) and (15), implies that

(19) \begin{align} J_{n} & = \frac{p}{2-p} + \frac{2}{n(n-1)} \sum_{i=1}^{n-1}\sum_{j=i+1}^n\bigg(\frac{(2-p)S_{ij}^{(n)}-pT_{ij}^{(n)}}{(n-2)p(2-p)^2} + R_{ij}^{(n)}\bigg) \notag \\ & = \frac{p}{2-p} + \frac{2}{n(n-1)(n-2)(2-p)^{2}}\sum_{i=1}^{n-1} \sum_{j=i+1}^n\sum_{k\neq i,j}\bigg(\frac{2\mathbf{1}_{ik}\mathbf{1}_{jk}}{p} - (\mathbf{1}_{ik}+\mathbf{1}_{jk})\bigg) + \frac{2}{n(n-1)}R_{n} \notag \\ & = \frac{p}{2-p} - \frac{4P_{1,n}}{n(n-1)(2-p)^{2}} + \frac{4P_{2,n}}{n(n-1)(n-2)p(2-p)^{2}} + \frac{2}{n(n-1)}R_{n}, \end{align}

where

\begin{equation*} P_{1,n} = \sum_{i=1}^{n-1}\sum_{j=i+1}^n\mathbf{1}_{ij}, \qquad P_{2,n} = \sum_{i=1}^{n-1}\sum_{j=i+1}^n\sum_{k\neq i,j}\mathbf{1}_{ij}\mathbf{1}_{ik} \end{equation*}

denote the number of edges and the number of paths of length two in G(n, p), respectively. Further, we can rewrite (19) as

(20) \begin{equation} \frac{(n-1)(2-p)^2}{\sqrt{8p(1-p)}}\bigg(J_n-\frac{p}{2-p}\bigg) = \sqrt{\frac{2p}{1-p}}\bigg({-}\frac{P_{1,n}}{np} + \frac{P_{2,n}}{n(n-2)p^2}\bigg) + \frac{(2-p)^2}{n\sqrt{2p(1-p)}\,}R_n. \end{equation}

For $P_{1,n}$ and $P_{2,n}$ , it is not hard to obtain that their expectations are given by ${\mathbb E}[P_{1,n}] = \frac12n(n-1)p$ and ${\mathbb E}[P_{2,n}] = \frac12n(n-1)(n-2)p^2$ . Applying [Reference Feng, Hu and Su10, Theorem 3(iii)] yields that if $np\to\infty$ and $n^2(1-p)\to \infty$ ,

\begin{equation*} \sqrt{\frac{2p}{1-p}}\bigg(\frac{P_{1,n}-\frac12n(n-1)p}{np}, \frac{P_{2,n}-\frac12n(n-1)(n-2)p^2}{2n(n-2)p^2}\bigg)\buildrel \textrm{D} \over \longrightarrow (Z,Z), \end{equation*}

which implies that

(21) \begin{equation} \sqrt{\frac{2p}{1-p}}\bigg({-}\frac{P_{1,n}}{np}+\frac{P_{2,n}}{n(n-2)p^2}\bigg) \buildrel \textrm{D} \over \longrightarrow Z. \end{equation}

To prove Theorem 2, by (20) and (21) it is sufficient to show that

\begin{equation*} \frac{R_n}{n\sqrt{p(1-p)}\,}\buildrel \textrm{P} \over \longrightarrow 0. \end{equation*}

That is, by Chebyshev’s inequality and the fact that ${\mathbb E}[R_n]=0$ , we only need to prove that

(22) \begin{equation} \textrm{Var}[R_n] = o\big(n^2p(1-p)\big). \end{equation}

On the other hand, by symmetry, it follows by (18) that

(23) \begin{align} \textrm{Var}[R_{n}] & = \textrm{cov}\Bigg(\sum_{i=1}^{n-1}\sum_{j=i+1}^n R_{ij}^{(n)},\sum_{i=1}^{n-1}\sum_{j=i+1}^n R_{ij}^{(n)}\Bigg) \notag \\ & = \frac{n(n-1)}{2}\textrm{cov}\Bigg(R_{12}^{(n)},\sum_{i=1}^{n-1}\sum_{j=i+1}^n R_{ij}^{(n)}\Bigg) \notag \\ & = \frac12n(n-1)\textrm{Var}\Big[R_{12}^{(n)}\Big] + n(n-1)(n-2)\textrm{cov}\Big(R_{12}^{(n)},R_{13}^{(n)}\Big) \notag \\ & \quad + \frac14n(n-1)(n-2)(n-3)\textrm{cov}\Big(R_{12}^{(n)},R_{34}^{(n)}\Big). \end{align}

To prove (22), we next estimate the variance and covariances in (23) separately. By the law of total expectation and (17), for the variance of $R_{12}^{(n)}$ we have

(24) \begin{equation} \textrm{Var}\Big[R_{12}^{(n)}\Big] = {\mathbb E}\Big[\Big(R_{12}^{(n)}\Big)^{2}\Big] = {\mathbb E}\Big[{\mathbb E}\Big[\Big(R_{12}^{(n)}\Big)^{2} \mid T_{12}^{(n)}\Big]\Big] = \sum_{m=0}^{n-2}{\mathbb E}\Big[\Big(R_{12}^{(n)}\Big)^{2} \mid T_{12}^{(n)}=m\Big]{\mathbb P}\Big(T_{12}^{(n)}=m\Big). \end{equation}

Recalling that $T_{12}^{(n)}\sim \textrm{Bin}(n-2,p(2-p))$ , and $R_{12}^{(n)}=0$ if $T_{12}^{(n)}=0$ , By (3), (4), and (16), it follows that

\begin{align*} {\mathbb E}\Big[\Big(R_{12}^{(n)}\Big)^{2} \mid T_{12}^{(n)}=m\Big] & = \frac1{(n-2)^2p^2(2-p)^2}\bigg(\frac{(n-2)p(2-p)}{m}-1\bigg)^2 \textrm{Var}\Big(S_{12}^{(n)} \mid T_{12}^{(n)}=m\Big) \\ & = \frac{2(1-p)}{(n-2)^2p(2-p)^4}\bigg(\frac{(n-2)^2p^2(2-p)^2}{m}-2(n-2)p(2-p)+m\bigg), \end{align*}

which, together with (6) and (24), implies that

(25) \begin{align} \textrm{Var}\Big[R_{12}^{(n)}\Big] & = \frac{2(1-p)}{(n-2)^2p(2-p)^4}\sum_{m=1}^{n-2}\!\bigg(\frac{(n-2)^2p^2(2-p)^2}{m}-2(n-2)p(2-p)+m\bigg) {\mathbb P}\Big(T_{12}^{(n)}=m\Big) \notag \\ & = \frac{2(1-p)}{(n-2)(2-p)^3}\bigg((n-2)p(2-p)\sum_{m=1}^{n-2}\frac1m{\mathbb P}\Big(T_{12}^{(n)}=m\Big) - 1 + 2{\mathbb P}\Big(T_{12}^{(n)}=0\Big)\bigg) \notag \\ & = O\bigg(\frac{1-p}{n^2p}\bigg), \end{align}

where in the last equality we used the simple fact that

\begin{equation*} 0<{\mathbb P}\Big(T_{12}^{(n)}=0\Big)=(1-p)^{2(n-2)}\leq \textrm{e}^{-2(n-2)p}=o\bigg(\frac{1}{np}\bigg). \end{equation*}

To calculate the covariance of $R_{12}^{(n)}$ and $R_{13}^{(n)}$ , by convention we introduce the shorthand notation

\begin{equation*} \widetilde{T}_{ij}^{(n)}\,:\!=\,\frac{(n-2)p(2-p)}{T_{ij}^{(n)}}-1 \end{equation*}

for distinct vertices $i,j\in [n]$ . Recalling (8), we have

\begin{equation*} (2-p)S_{ij}^{(n)}-pT_{ij}^{(n)}=\sum_{k\neq i,j} V_{ij,k}, \quad i,j\in [n]. \end{equation*}

By symmetry and (16), it thus follows that

(26) \begin{align} \textrm{cov}\Big(R_{12}^{(n)},R_{13}^{(n)}\Big) & = \frac{1}{(n-2)^{2}p^{2}(2-p)^{4}}\sum_{k=3}^n\sum_{l\neq1,3} {\mathbb E}\Big[V_{12,k}V_{13,l}\widetilde{T}_{12}^{(n)}\widetilde{T}_{13}^{(n)} \mathbf{1}\Big(T_{12}^{(n)}T_{13}^{(n)}>0\Big)\Big] \notag \\ & = \frac{1}{(n-2)p^{2}(2-p)^{4}} {\mathbb E}\Big[V_{12,3}V_{13,2}\widetilde{T}_{12}^{(n)}\widetilde{T}_{13}^{(n)} \mathbf{1}\Big(T_{12}^{(n)}T_{13}^{(n)}>0\Big)\Big] \notag \\ & \quad + \frac{(n-3)}{(n-2)p^{2}(2-p)^{4}} {\mathbb E}\Big[V_{12,3}V_{13,4}\widetilde{T}_{12}^{(n)}\widetilde{T}_{13}^{(n)} \mathbf{1}\Big(T_{12}^{(n)}T_{13}^{(n)}>0\Big)\Big], \end{align}

which splits the covariance into two parts. Notice that, by (8), the discrete random variable $V_{ij,k}\neq 0$ if and only if $\mathbf{1}_{ik}\vee \mathbf{1}_{jk}=1$ . This implies that if $V_{12,3}V_{13,2}\neq 0$ , we have $T_{12}^{(n)}=1+\sum_{k=4}^{n} \mathbf{1}_{1k}\vee \mathbf{1}_{2k}$ and $T_{13}^{(n)}=1+\sum_{k=4}^{n} \mathbf{1}_{1k}\vee \mathbf{1}_{3k}$ , and they have the same distribution. Since it follows that, by conditioning on $\mathbf{1}_{23}$ ,

\begin{align*} {\mathbb E}[V_{12,3}V_{13,2}] & = {\mathbb E}\big[\big((2-p)\mathbf{1}_{13}\mathbf{1}_{23}-p(\mathbf{1}_{13}\vee \mathbf{1}_{23})\big) \big((2-p)\mathbf{1}_{12}\mathbf{1}_{23}-p(\mathbf{1}_{12}\vee \mathbf{1}_{23})\big)\big] \\ & = p{\mathbb E}\big[\big((2-p)\mathbf{1}_{13}-p\big)\big((2-p)\mathbf{1}_{12}-p\big)\big] + (1-p){\mathbb E}\big[p^2\mathbf{1}_{13}\mathbf{1}_{12}\big] \\ & = p^3(1-p)^2+p^4(1-p) = p^3(1-p), \end{align*}

and that, by Lemma 1 and the Cauchy–Schwarz inequality,

\begin{multline*} \bigg|\mathbb{E}\bigg[\bigg(\frac{(n-2)p(2-p)}{1+\sum_{k=4}^{n}\mathbf{1}_{1k}\vee\mathbf{1}_{2k}}-1\bigg) \bigg(\frac{(n-2)p(2-p)}{1+\sum_{k=4}^{n}\mathbf{1}_{1k}\vee\mathbf{1}_{3k}}-1\bigg)\bigg]\bigg| \\ \leq \mathbb{E}\bigg[\bigg(\frac{(n-2)p(2-p)}{1+\sum_{k=4}^{n}\mathbf{1}_{1k}\vee\mathbf{1}_{2k}}-1\bigg)^2\bigg] = O \bigg(\frac{1}{np}\bigg), \end{multline*}

we have

(27) \begin{align} & {\mathbb E}\Big[V_{12,3}V_{13,2}\widetilde{T}_{12}^{(n)}\widetilde{T}_{13}^{(n)} \mathbf{1}\Big(T_{12}^{(n)}T_{13}^{(n)}>0\Big)\Big] \notag \\ & \qquad = {\mathbb E}\bigg[V_{12,3}V_{13,2} \bigg(\frac{(n-2)p(2-p)}{1+\sum_{k=4}^{n}\mathbf{1}_{1k}\vee\mathbf{1}_{2k}}-1\bigg) \bigg(\frac{(n-2)p(2-p)}{1+\sum_{k=4}^{n}\mathbf{1}_{1k}\vee\mathbf{1}_{3k}}-1\bigg)\bigg] \notag \\ & \qquad = p^{3}(1-p)\mathbb{E}\bigg[ \bigg(\frac{(n-2)p(2-p)}{1+\sum_{k=4}^{n}\mathbf{1}_{1k}\vee\mathbf{1}_{2k}}-1\bigg) \bigg(\frac{(n-2)p(2-p)}{1+\sum_{k=4}^{n}\mathbf{1}_{1k}\vee\mathbf{1}_{3k}}-1\bigg)\bigg] \notag \\ & \qquad = O\bigg(\frac{p^{2}(1-p)}{n}\bigg). \end{align}

Let us define ${\mathcal B}_{ab}\,:\!=\,\{\mathbf{1}_{12}\vee \mathbf{1}_{23}=a,\, \mathbf{1}_{14}\vee \mathbf{1}_{24}=b\}$ , $a,b \in\{0,1\}$ . Noting that $\mathbb{E}[V_{12,3}]=0$ , we have

\begin{equation*} \mathbb{E}\big[V_{12,3}\mathbf{1}(\mathbf{1}_{12}\vee \mathbf{1}_{23}=1)\big] = -\mathbb{E}\big[V_{12,3}\mathbf{1}(\mathbf{1}_{12}\vee \mathbf{1}_{23}=0)\big] = p\mathbb{E}\big[\mathbf{1}_{13}\mathbf{1}(\mathbf{1}_{12}=\mathbf{1}_{23}=0)\big] = p^2(1-p)^2, \end{equation*}

which implies that, for any $a,b=0$ or 1,

\begin{align*} \mathbb{E}[V_{12,3}V_{13,4}\mathbf{1}({\mathcal B}_{ab})] & = \mathbb{E}[V_{12,3}\mathbf{1}(\mathbf{1}_{12}\vee \mathbf{1}_{23}=a)] \mathbb{E}[V_{13,4}\mathbf{1}(\mathbf{1}_{14}\vee \mathbf{1}_{24}=b)] \\ & = \mathbb{E}[V_{12,3}\mathbf{1}(\mathbf{1}_{12}\vee \mathbf{1}_{23}=a)] \mathbb{E}[V_{12,3}\mathbf{1}(\mathbf{1}_{12}\vee \mathbf{1}_{23}=b)] \\ & = ({-}1)^{a+b}p^4(1-p)^4. \end{align*}

Analogously to (27), we have

(28) \begin{align} & \mathbb{E}\Big[V_{12,3}V_{13,4}\widetilde{T}_{12}^{(n)}\widetilde{T}_{13}^{(n)} \mathbf{1}\Big(T_{12}^{(n)}T_{13}^{(n)}>0\Big)\Big] \notag \\ & \qquad = \sum_{a=0}^1\sum_{b=0}^1\mathbb{E} \Big[V_{12,3}V_{13,4}\widetilde{T}_{12}^{(n)}\widetilde{T}_{13}^{(n)}\mathbf{1}({\mathcal B}_{ab})\Big] \notag \\ & \qquad = \sum_{a=0}^1\sum_{b=0}^1 \mathbb{E}\big[V_{12,3}V_{13,4}W_{12}(b)W_{13}(a)\mathbf{1}({\mathcal B}_{ab})\big] \notag \\ & \qquad = \sum_{a=0}^1\sum_{b=0}^1\mathbb{E}\big[V_{12,3}V_{13,4}\mathbf{1}({\mathcal B}_{ab})\big] \mathbb{E}\big[W_{12}(b)W_{13}(a)\big] \notag \\ & \qquad = p^{4}(1-p)^{4}\mathbb{E}\big[(W_{12}(0)-W_{12}(1))(W_{13}(0)-W_{13}(1))\big], \end{align}

where

\begin{equation*} W_{ij}(a) \,:\!=\, \frac{(n-2)p(2-p)}{1+a+\sum_{k=5}^{n}\mathbf{1}_{ik}\vee \mathbf{1}_{jk}}-1, \quad i,j\in [n], \,a=0,1. \end{equation*}

Noting that $W_{12}(0)-W_{12}(1)$ and $W_{13}(0)-W_{13}(1)$ have the same distribution, by Lemma 2 and the Cauchy–Schwarz inequality we have

\begin{align*} {\mathbb E}\big|\big(W_{12}(0)-W_{12}(1)\big)\big(W_{13}(0)-W_{13}(1)\big)\big| & \le {\mathbb E}\big[(W_{12}(0)-W_{12}(1))^2\big] \\ & \le (n-2)^{2}p^{2}(2-p)^{2} {\mathbb E}\left[\frac{1}{\Big(1+\sum_{k=5}^{n}\mathbf{1}_{1k}\vee \mathbf{1}_{2k}\Big)^{4}}\right] \\ & = O\bigg(\frac{1}{n^{2}p^{2}}\bigg), \end{align*}

which, together with (26), (27), and (28), implies that

(29) \begin{equation} \textrm{cov}\Big(R_{12}^{(n)},R_{13}^{(n)}\Big) = O\bigg(\frac{1-p}{n^2}\bigg). \end{equation}

It remains to calculate the second covariance $\mathbf{Cov}\Big(R_{12}^{(n)},R_{34}^{(n)}\Big)$ on the right-hand side of (23), and the procedure is similar to the previous one. We sketch the calculations below, omitting a few specific interpretations. Analogously to (26), we have

(30) \begin{align} \textrm{cov}\Big(R_{12}^{(n)},R_{34}^{(n)}\Big) & = \frac{1}{(n-2)^{2}p^{2}(2-p)^4}\sum_{k=3}^n\sum_{l\neq 3,4} {\mathbb E}\Big[V_{12,k}V_{34,l}\widetilde{T}_{12}^{(n)}\widetilde{T}_{34}^{(n)} \mathbf{1}\Big(T_{12}^{(n)}T_{34}^{(n)}>0\Big)\Big] \notag \\ & = \frac{4}{(n-2)^{2}p^{2}(2-p)^4}{\mathbb E}\Big[V_{12,3}V_{34,1}\widetilde{T}_{12}^{(n)}\widetilde{T}_{34}^{(n)} \mathbf{1}\Big(T_{12}^{(n)}T_{34}^{(n)}>0\Big)\Big] \notag \\ & \quad + \frac{2}{(n-2)p^{2}(2-p)^4}{\mathbb E}\Big[V_{12,3}V_{34,5}\widetilde{T}_{12}^{(n)}\widetilde{T}_{34}^{(n)} \mathbf{1}\Big(T_{12}^{(n)}T_{34}^{(n)}>0\Big)\Big] \notag \\ & \quad + \frac{(n-4)^2}{(n-2)^{2}p^{2}(2-p)^4} {\mathbb E}\Big[V_{12,5}V_{34,5}\widetilde{T}_{12}^{(n)}\widetilde{T}_{34}^{(n)} \mathbf{1}\Big(T_{12}^{(n)}T_{34}^{(n)}>0\Big)\Big]. \end{align}

Define ${\mathcal C}_{ab} \,:\!=\, \{\mathbf{1}_{14}\vee \mathbf{1}_{24}=a,\,\mathbf{1}_{23}\vee \mathbf{1}_{24}=b\}$ , $a,b \in\{0,1\}$ . After straightforward calculations, we have

\begin{align*} {\mathbb E}\big[V_{12,3}V_{34,1}\mathbf{1}({\mathcal C}_{00})\big] & = p^3(1-p)^3, \\ {\mathbb E}\big[V_{12,3}V_{34,1}\mathbf{1}({\mathcal C}_{11})\big] & = p^{3}(1-p)\big(1+3(1-p)^{2}\big), \\ {\mathbb E}\big[V_{12,3}V_{34,1}\mathbf{1}({\mathcal C}_{01})\big] & = {\mathbb E}\big[V_{12,3}V_{34,1}\mathbf{1}({\mathcal C}_{10})\big] = -2p^3(1-p)^3. \end{align*}

Then we can conclude that ${\mathbb E}[V_{12,3}V_{34,1}\mathbf{1}({\mathcal C}_{ab})] = O(p^3(1-p))$ . Hence, by the Cauchy–Schwarz inequality and Lemma 1,

(31) \begin{align} {\mathbb E}\Big[V_{12,3}V_{34,1}\widetilde{T}_{12}^{(n)}\widetilde{T}_{34}^{(n)} \mathbf{1}\Big(T_{12}^{(n)}T_{34}^{(n)}>0\Big)\Big] & = \sum_{a=0}^1\sum_{b=0}^1{\mathbb E}\big[V_{12,3}V_{34,1}W_{12}(a)W_{34}(b)\mathbf{1}({\mathcal C}_{ab})\big] \notag \\ & = \sum_{a=0}^1\sum_{b=0}^1{\mathbb E}[V_{12,3}V_{34,1}\mathbf{1}({\mathcal C}_{ab})] {\mathbb E}[W_{12}(a)]{\mathbb E}[W_{34}(b)] \notag \\ & = O\bigg(\frac{p^2(1-p)}{n}\bigg). \end{align}

Since $V_{12,3}$ and $V_{34,5}$ are independent and have the common mean 0, it follows that

(32) \begin{align} & {\mathbb E}\Big[V_{12,3}V_{34,5}\widetilde{T}_{12}^{(n)}\widetilde{T}_{34}^{(n)} \mathbf{1}\Big(T_{12}^{(n)}T_{34}^{(n)}>0\Big)\Big] \notag \\ & = {\mathbb E}\Big[V_{12,3}V_{34,5}\widetilde{T}_{12}^{(n)}\widetilde{T}_{34}^{(n)} \mathbf{1}(\mathbf{1}_{13}\vee\mathbf{1}_{23}=1)\mathbf{1}(\mathbf{1}_{35}\vee\mathbf{1}_{45}=1)\Big] \notag \\ & = {\mathbb E}[V_{34,5}\mathbf{1}(\mathbf{1}_{35}\vee\mathbf{1}_{45}=1)] {\mathbb E}\bigg[V_{12,3}\widetilde{T}_{12}^{(n)}\mathbf{1}(\mathbf{1}_{13}\vee \mathbf{1}_{23}=1) \bigg(\frac{(n-2)p(2-p)}{1+\sum_{k\neq 3,4,5}(\mathbf{1}_{3k}\vee\mathbf{1}_{4k})}-1\bigg)\bigg] \notag \\ & = {\mathbb E}[V_{34,5}]{\mathbb E}\bigg[V_{12,3}\widetilde{T}_{12}^{(n)} \mathbf{1}(\mathbf{1}_{13}\vee\mathbf{1}_{23}=1) \bigg(\frac{(n-2)p(2-p)}{1+\sum_{k\neq 3,4,5}(\mathbf{1}_{3k}\vee\mathbf{1}_{4k})}-1\bigg)\bigg] = 0. \end{align}

Similarly, we also have ${\mathbb E}\big[V_{12,5}V_{34,5}\widetilde{T}_{12}^{(n)}\widetilde{T}_{34}^{(n)} \mathbf{1}\big(T_{12}^{(n)}T_{34}^{(n)}>0\big)\big] = 0$ . Plugging this, (31), and (32) into (30) yields

\begin{equation*} \textrm{cov}\Big(R_{12}^{(n)},R_{34}^{(n)}\Big) = O\bigg(\frac{1-p}{n^3}\bigg), \end{equation*}

which, together with (23), (25), and (29), implies that

\begin{equation*} \textrm{Var}[R_n] = O\bigg(\frac{1-p}{p}\bigg) + O(n(1-p)) + O(n(1-p)) = O(n(1-p)) \end{equation*}

as $np\to\infty$ . This proves (22), and thus completes the proof of Theorem 2.