Two-sample test of stochastic block models via the maximum sampling entry-wise deviation

Abstract

The paper discusses a statistical problem related to testing for differences between two networks with community structures. While existing methods have been proposed, they encounter challenges and do not perform effectively when the networks become sparse. We propose a test statistic that combines a method proposed by Wu and Hu (2024) and a resampling process. Specifically, the proposed test statistic proves effective under the condition that the community-wise edge probability matrices have entries of order $\Omega (\log n/n)$, where n denotes the network size. We derive the asymptotic null distribution of the test statistic and provide a guarantee of asymptotic power against the alternative hypothesis. To evaluate the performance of the proposed test statistic, we conduct simulations and provide real data examples. The results indicate that the proposed test statistic performs well for both dense and sparse networks.

Appendices

Detailed proofs of Theorem 1 and 2

Firstly we introduce the following lemma before the main proof.

Lemma 1

(Theorem 1.5.1 in Leadbetter et al. (1983)) Let $Z_1, Z_2, \ldots , Z_{n^{*}}$ be a sequence of independent random variables with distribution function F(x) for $-\infty< x < \infty$. Let $z_1, z_2, \ldots , z_{n^{*}}$ be a sequence of real numbers, and let $0< \tau < 1$. Define $M_n = \max \{Z_1, Z_2, \ldots , Z_{n^{*}}\}$. Then, the equality

$$\begin{aligned} P(M_n \le z_n) = e^{-\tau } \end{aligned}$$

holds true if and only if

$$\begin{aligned} \lim _{n^{*} \rightarrow \infty } n^{*}\left( 1 - F(z_n^{*})\right) = \tau . \end{aligned}$$

Proof of Theorem 1

In this section, we present the proofs of Theorem 1.

Proof of Theorem 1

First, we bound the estimation error of the following equation:

$$\begin{aligned} \begin{aligned}&\max \limits _{1 \le u \le K,1 \le v \le K} \vert \frac{\sqrt{B_{1,uv}(1-B_{1,uv})+B_{2,uv}(1-B_{2,uv})}}{\sqrt{\hat{B}_{1,uv}(1-\hat{B}_{1,uv}) +\hat{B}_{2,uv}(1-\hat{B}_{2,uv})}}-1\vert \\&\quad = \max \limits _{1 \le u \le K,1 \le v \le K} \Bigg \vert \frac{\sqrt{B_{1,uv}(1-B_{1,uv})+B_{2,uv}(1-B_{2,uv})}}{\sqrt{\hat{B}_{1,uv} (1-\hat{B}_{1,uv})+\hat{B}_{2,uv}(1-\hat{B}_{2,uv})}}\\&\quad -\frac{\sqrt{\hat{B}_{1,uv}(1-\hat{B}_{1,uv}) +\hat{B}_{2,uv}(1-\hat{B}_{2,uv})}}{\sqrt{\hat{B}_{1,uv}(1-\hat{B}_{1,uv}) +\hat{B}_{2,uv}(1-\hat{B}_{2,uv})}}\Bigg \vert \\&\quad = O_p( \frac{K}{\log n}). \end{aligned} \end{aligned}$$

(A1)

Next, denote that

$$\begin{aligned} \begin{aligned}&F_{n,0} \triangleq \max \limits _{1 \le i \le n,1 \le k \le K} \vert \hat{\gamma }_{mk,0} \vert \\&\quad =\max \limits _{1 \le i \le n,1 \le k \le K} \vert \frac{\hat{\rho }_{m_1k,0}+\hat{\rho }_{m_2k,0}+\cdots +\hat{\rho }_{m_Sk,0}}{\sqrt{S}}\vert \\&\quad =\max \limits _{1 \le i \le n,1 \le k \le K} \Bigg \vert \frac{1}{\sqrt{S}}\sum _{s=1}^{S} \frac{1}{\sqrt{ \#( \ g^{-1}(k)\backslash \left\{ i_{m_s}\right\} )}} \\&\quad \times \sum _{j\in g^{-1}(k)\backslash \left\{ i_{m_s}\right\} }\frac{A_{1,ij}-A_{2,ij}}{\sqrt{ B_{1, g_{i} g_{j}}(1- B_{1, g_{i} g_{j}})+ B_{2,{g}_{i} g_{j}}(1- B_{2, g_{i} g_{j}})}}\Bigg \vert . \end{aligned} \end{aligned}$$

In accordance with Eq. A1, we can derive that

$$\begin{aligned} \begin{aligned} F_{n}&\triangleq \max \limits _{1 \le i \le n,1 \le k \le K} \vert \hat{\gamma }_{mk} \vert \\&=\max \limits _{1 \le i \le n,1 \le k \le K} \vert \frac{\hat{\rho }_{m_1k}+\hat{\rho }_{m_2k} +\cdots +\hat{\rho }_{m_Sk}}{\sqrt{S}}\vert \\&=\max \limits _{1 \le i \le n,1 \le k \le K} \Bigg \vert \frac{1}{\sqrt{S}}\sum _{s=1}^{S} \frac{1}{\sqrt{ \#( \ \hat{g}^{-1}(k)\backslash \left\{ i_{m_s}\right\} )}} \\&\quad \times \sum _{j\in \hat{g}^{-1}(k)\backslash \left\{ i_{m_s}\right\} }\frac{A_{1,ij} -A_{2,ij}}{\sqrt{ \hat{B}_{1, \hat{g}_{i} \hat{g}_{j}}(1- \hat{B}_{1, \hat{g}_{i} \hat{g}_{j}}) + \hat{B}_{2,{\hat{g}}_{i} \hat{g}_{j}}(1- \hat{B}_{2, \hat{g}_{i} \hat{g}_{j}})}}\Bigg \vert \\&=\max \limits _{1 \le i \le n,1 \le k \le K} \Bigg \vert \frac{1}{\sqrt{S}}\sum _{s=1}^{S} \frac{\sum _{j\in g^{-1}(k)\backslash \left\{ i_{m_s}\right\} }\frac{A_{1,ij}-A_{2,ij}}{\sqrt{ B_{1, g_{i} g_{j}}(1- B_{1, g_{i} g_{j}})+ B_{2,{g}_{i} g_{j}}(1- B_{2, g_{i} g_{j}})}}}{\sqrt{ \#( \ g^{-1}(k)\backslash \left\{ i_{m_s}\right\} )}} \\&\quad \quad \times \frac{\sqrt{B_{1,g_{i}g_{j}}(1-B_{1,g_{i}g_{j}})+B_{2,g_{i}g_{j}} (1-B_{2,g_{i}g_{j}})}}{\sqrt{\hat{B}_{1,g_{i}g_{j}}(1-\hat{B}_{1,g_{i}g_{j}})+\hat{B}_{2,g_{i}g_{j}} (1-\hat{B}_{2,g_{i}g_{j}})}}\Bigg \vert \\&\quad = F_{n,0}(1+O_p( \frac{K}{\log n})). \end{aligned} \end{aligned}$$

Under Assumption 3, we have $K = o(\sqrt{\log n})$, $MK=o(n)$ and if $F_{n,0} = O_P(\sqrt{\log MK})$, then we have

$$\begin{aligned} F_n = F_{n,0}+o_p(1). \end{aligned}$$

Thus to prove Theorem 1, it is sufficient to show that

$$\begin{aligned} P\left( F^{2}_{n,0} - 2\log (MK) + \log \log (MK) \le x \right) \rightarrow \exp \left( -\frac{1}{\sqrt{\pi }}e^{-x/2}\right) . \end{aligned}$$

To derive the asymptotic null distribution of $F_{n}$, we define the $\sigma$-field ${\mathcal {G}} =\sigma \{ \hat{\rho }_{ik,0}: 1 \le i \le n; 1 \le k \le K \}$ that is generated from the auxiliary quantity $\hat{\rho }_{ik,0}$ in $F_{n,0}$. Let $\tilde{\rho }_{ik}$ be the observed value of $\hat{\rho }_{ik,0}$. Then, conditional on ${\mathcal {G}}$, the $\hat{\rho }_{ik,0}$s are independent and identically distributed with the probability $P(\hat{\rho }_{ik,0} = \tilde{\rho }_{ik}) = \frac{1}{nK}$ for any $1 \le i \le n$ and $1 \le k \le K$. Note that $\hat{\gamma }_{mk,0}$ is calculated from $\hat{\rho }_{ik,0}$ in $F_{n,0}$, and we denote $\tilde{\rho }_{ik}$ as the observed value of $\hat{\rho }_{ik,0}$ calculated from $\tilde{\rho }_{iv}$. Let $\bar{\rho } = \frac{1}{nK} \sum _{i=1}^n \sum _{k=1}^K \tilde{\rho }_{ik}$ and $\bar{\gamma } = \frac{1}{MK} \sum _{m=1}^M\sum _{k=1}^K \tilde{\gamma }_{mk}$.

As a result, $E(\hat{\rho }_{iv,0} - \bar{\rho } | {\mathcal {G}}) = 0$ and $E(\hat{\gamma }_{mv,0} - \bar{\gamma } | {\mathcal {G}}) = 0$. By Corollary 2.1 in Chernozhukov et al. (2013) and conditional on ${\mathcal {G}}$, as $\min \{n, M, S\} \rightarrow \infty$, we can derive that

$$\begin{aligned}{} & {} \sup _{x \in {\mathbb {R}}} \left| P\left( \max _{1 \le m \le M, 1 \le k \le K} \frac{1}{\sqrt{S}} \sum _{s=1}^{S}\hat{\rho }_{m_sk,0}- \sqrt{B}\bar{\rho } \le x\right) - P\left( \max _{1 \le l \le MK} U_l \le x\right) \right| \\{} & {} \quad \le CS^{-c} \rightarrow 0, \end{aligned}$$

where $U = (U_1, \ldots , U_{MK})^\top \in {\mathbb {R}}^{MK}$ is a Gaussian random vector with mean 0 and covariance matrix $\text {cov}(\hat{\gamma }_{mk,0})$, and c and C are some finite positive constants. Define $\check{\gamma } = \text {vec}(\hat{\gamma }_{mk,0}) = (\check{\gamma }_1, \ldots , \check{\gamma }_{MK})^\top \in {\mathbb {R}}^{MK}$ for $1 \le m \le M$ and $1 \le k \le K$. As a result, the above inequality can be rewritten as

$$\begin{aligned} \sup _{x \in {\mathbb {R}}} \left| P\left( \max _{1 \le l \le MK} \check{\gamma }_l -\bar{\gamma } \le x\right) - P\left( \max _{1 \le l \le MK} U_l\le x\right) \right| \le CS^{-c} \rightarrow 0 \end{aligned}$$

(A2)

as $\min \{n, M, B\} \rightarrow \infty$.

Next, we calculate $\text {cov}(\check{\gamma })$. The diagonal elements of $\text {cov}(\check{\gamma })$ are 1 by definition, it suffices to compute $\text {corr}(\check{\gamma })$. Denote $\hat{\rho }_{.k,0}$ as the vector containing all elements $\hat{\rho }_{ik,0}$ in block k. In addition, let $\Lambda _{m_s} = (\lambda _{m_s,1}, \lambda _{m_s,2}, \ldots , \lambda _{ms,n})^T$ be some random variables that are independently generated from the binomial distribution $\text {Bernoulli}(n, \frac{S}{n})$ for $1 \le m_s \le M$, and they are independent of $\hat{\rho }_{.k,0}$. Thus, for $i = 1, 2, \ldots , n$, $\lambda _{m_s,n}$ follows the Bernoulli distribution with probability $\frac{S}{n}$, which implies that $E(\lambda _{m_s,i}) = E(\lambda ^2_{m_s,i}) = \frac{S}{n}$. As a result, we obtain that $\hat{\gamma }_{m_sk,0} = \Lambda ^T_{m_s}\hat{\rho }_{.k,0}/ \sqrt{S}$. Then, for any $1 \le m_s, m_l \le M$ with subscripts $s \ne l$, it can be shown that

$$\begin{aligned} \delta _{\gamma } \triangleq \text {corr}(\hat{\gamma }_{m_sk,0}, \hat{\gamma }_{m_lk,0})&=\frac{\text {cov}(\hat{\gamma }_{m_sk,0},\hat{\gamma }_{m_lk,0})}{\sqrt{\text {var}(\hat{\gamma }_{m_sk,0})}\sqrt{\text {var}(\hat{\gamma }_{m_lk,0})}} =\frac{E(\Lambda ^T_{m_s}\hat{\rho }_{.k,0}\Lambda ^T_{m_l}\hat{\rho }_{.k,0})}{E(\Lambda ^T_{m_s} \hat{\rho }_{.k,0}\Lambda ^T_{m_s}\hat{\rho }_{.k,0})}\\&=\frac{\sum _{i=1}^{n}E(\lambda _{m_s,i}\lambda _{m_l,i}\hat{\rho }^{2}_{ik,0})}{\sum _{i=1}^{n}E(\lambda ^2_{m_s,i}\hat{\rho }^{2}_{ik,0})}=\frac{\sum _{i=1}^{n}E(\lambda _{m_s,i} \lambda _{m_l,i})E(\hat{\rho }^{2}_{ik,0})}{\sum _{i=1}^{n}E(\lambda ^2_{m_s,i})E(\hat{\rho }^{2}_{ik,0})}\\&=\frac{S}{n}. \end{aligned}$$

It can be seen that correlations between $\check{\gamma }_{s}$ and $\check{\gamma }_{l}$ are all equal for any $s \ne l$. According to Fan and Jiang (2019), we can rewrite $\max _{1 \le l \le MK} U_l$ as following:

$$\begin{aligned} \max _{1 \le l \le MK} U_l = \sqrt{\delta _{\gamma }}\tilde{U}_0 + \sqrt{1-\delta _{\gamma }} \max _{1 \le l \le MK} \tilde{U}_l, \end{aligned}$$

where $\tilde{U}_0, \tilde{U}_1, \ldots , \tilde{U}_{MK}$ are independent and identically distributed standard normal variables. Now we can write (A2) as

$$\begin{aligned}&\sup _{x \in {\mathbb {R}}} \left| P\left( \max _{1 \le l \le MK} \check{\gamma }_l -\bar{\gamma } \le x\right) \right. \nonumber \\&\quad \left. -P\left( \sqrt{\delta _{\gamma }}\tilde{U}_0 + \sqrt{1-\delta _{\gamma }} \max _{1 \le l \le MK} \tilde{U}_l\le x\right) \right| \le CS^{-c} \rightarrow 0 \end{aligned}$$

(A3)

Note that, under the null hypothesis we have ${\mathbb {E}}(\bar{\gamma }) = 0$. Next we derive $\text {var}(\bar{\gamma })$.

$$\begin{aligned} \text {var}(\bar{\gamma })&= E(\bar{\gamma }^2) = E \left[ \frac{1}{MK} \sum _{m=1}^{M} \sum _{v=k}^{K} (\tilde{\gamma }_{mv})^2 \right] \\&= \frac{1}{M^2K^2} \left[ E \sum _{m_1=1}^{M} \sum _{m_2=1}^{M} \sum _{k_1=1}^{K} \sum _{k_2=1}^{K} \tilde{\gamma }_{m_1k_1} \tilde{\gamma }_{m_2k_2} \right] \\&= E[\tilde{\gamma }_{m_1k}\tilde{\gamma }_{m_1k}I(k_1=k_2=k)] +E[\tilde{\gamma }_{mk}I(m_1=m_2=m,k_1=k_2=k)]\\&= \delta _{\gamma } \frac{M(M-1)}{M^2} + \frac{1}{MK}\\&=\frac{S}{n}\frac{M(M-1)}{M^2} + \frac{1}{MK}. \end{aligned}$$

Under Assumption 3, we have $\text {var}(\bar{\gamma })\rightarrow 0$, as $\min \{n, M, B\} \rightarrow \infty$.

This implies $\bar{\gamma } = o_p(1)$. Additionally under Assumption 3, we have $\delta _{\gamma }=o(1)$.

As a result, we can rewrite (A3) as

$$\begin{aligned} \sup _{x \in {\mathbb {R}}} \left| P(\max _{1 \le l \le MK} \check{\gamma }_l\le x)- P( \max _{1 \le l \le MK} \tilde{U}_l\le x)\right| \rightarrow 0, \end{aligned}$$

(A4)

as $\min \{n, M, B\} \rightarrow \infty$.

For any $x \in {\mathbb {R}}$, denote $u = \frac{1}{2} \log (MK) - \log \log (MK) + x$. We then have

$$\begin{aligned} P \left( |N(0, 1) |\ge u \right)&\sim \frac{2}{\sqrt{2 \pi u}} \\&\sim \frac{1}{\sqrt{\pi }}\frac{1}{\sqrt{\log MK}} \exp \left( -\frac{1}{2}\{2\log MK-\log \log MK+x\} \right) \\&\sim \frac{1}{\sqrt{\pi }}\frac{\exp \left( -\frac{x}{2} \right) }{MK}. \end{aligned}$$

Subsequently, by Lemma 1, as $\min \{n, M, B\} \rightarrow \infty$, we have

$$\begin{aligned} P \left( \max _{1 \le l \le MK} \tilde{U}_l^2 - 2 \log (MK) + \log \log (MK)\le x \right) \rightarrow \exp \left( -\frac{1}{\sqrt{\pi }}\exp \left( -\frac{x}{2} \right) \right) , \end{aligned}$$

for any $x \in {\mathbb {R}}$. Combining with (A4), we obtain

$$\begin{aligned} P \left( \max _{1 \le l \le MK} \check{\gamma }_l^2 - 2 \log (MK) + \log \log (MK)\le x \right) \rightarrow \exp \left( -\frac{1}{\sqrt{\pi }}\exp \left( -\frac{x}{2} \right) \right) . \end{aligned}$$

(A5)

Accordingly,

$$\begin{aligned} P \left( F^2_{n,0} - 2 \log (MK) + \log \log (MK)\le x \right) \rightarrow \exp \left( -\frac{1}{\sqrt{\pi }}\exp \left( -\frac{x}{2} \right) \right) . \end{aligned}$$

Since

$$\begin{aligned} F_n = F_{n,0}+o_p(1), \end{aligned}$$

we have

$$\begin{aligned} P \left( F^2_{n} - 2 \log (MK) + \log \log (MK)\le x \right) \rightarrow \exp \left( -\frac{1}{\sqrt{\pi }}\exp \left( -\frac{x}{2} \right) \right) . \end{aligned}$$

This completes the proof of Theorem 1.

$\square$

Proof of Theorem 2

In this section, we present the proofs of Theorem 2.

Proof of Theorem 2

Note that

$$\begin{aligned} \begin{aligned} \hat{\gamma }_{mk}&= \frac{\hat{\rho }_{m_1k}+\hat{\rho }_{m_2k}+\cdots +\hat{\rho }_{m_Sk}}{\sqrt{S}}\\&= \frac{1}{\sqrt{S}}\sum _{s=1}^{S} \frac{1}{\sqrt{ \#( \ \hat{g}^{-1}(k)\backslash \left\{ i_{m_s}\right\} )}} \\&\quad \sum _{j\in \hat{g}^{-1}(k)\backslash \left\{ i_{m_s}\right\} }\frac{A_{1,ij}-A_{2,ij}}{\sqrt{ \hat{B}_{1, \hat{g}_{i} \hat{g}_{j}}(1- \hat{B}_{1, \hat{g}_{i} \hat{g}_{j}})+ \hat{B}_{2,{g}_{i} \hat{g}_{j}}(1- \hat{B}_{2, \hat{g}_{i} \hat{g}_{j}})}} \\&= \frac{1}{\sqrt{S}}\sum _{s=1}^{S} \frac{1}{\sqrt{ \#( \ g^{-1}(k)\backslash \left\{ i_{m_s}\right\} )}} \\&\quad \sum _{j\in g^{-1}(k)\backslash \left\{ i_{m_s}\right\} }\left( \frac{A_{1,ij}-B_{1,g_{i}g_{j}}-(A_{2,ij}-B_{2,g_{i}g_{j}})}{\sqrt{B_{1,g_{i}g_{j}} (1-B_{1,g_{i}g_{j}})+B_{2,g_{i}g_{j}}(1-B_{2,g_{i}g_{j}})}}\right. \\&\quad +\left. \frac{B_{1,g_{i}g_{j}}-B_{2,g_{i}g_{j}}}{\sqrt{B_{1,g_{i}g_{j}}(1-B_{1,g_{i}g_{j}}) +B_{2,g_{i}g_{j}}(1-B_{2,g_{i}g_{j}})}}\right) \\&\quad \quad \quad \times \frac{\sqrt{B_{1,g_{i}g_{j}}(1-B_{1,g_{i}g_{j}})+B_{2,g_{i}g_{j}} (1-B_{2,g_{i}g_{j}})}}{\sqrt{\hat{B}_{1,g_{i}g_{j}}(1-\hat{B}_{1,g_{i}g_{j}}) +\hat{B}_{2,g_{i}g_{j}}(1-\hat{B}_{2,g_{i}g_{j}})}}. \end{aligned} \end{aligned}$$

From the discussion in the proof of Theorem 1, we can obtain that

$$\begin{aligned} \begin{aligned} F_{n}&=\max \limits _{1 \le m \le M,1 \le k \le K} \vert \hat{\gamma }_{mk} \vert \\&=\max \limits _{1 \le m \le M,1 \le k \le K} \vert \frac{1}{\sqrt{S}} \sum _{s=1}^{S} \frac{1}{\sqrt{ \#( \ g^{-1}(k)\backslash \left\{ i_{m_s}\right\} )}}\\&\times \sum _{j\in \hat{g}^{-1}(k)\backslash \left\{ i_{m_s}\right\} } \left( \frac{A_{1,ij}-B_{1,g_{i}g_{j}}-(A_{2,ij}-B_{2,g_{i}g_{j}})}{\sqrt{B_{1,g_{i}g_{j}}(1-B_{1,g_{i}g_{j}})+B_{2,g_{i}g_{j}}(1-B_{2,g_{i}g_{j}})}}\right. \\&\quad \quad +\left. \frac{B_{1,g_{i}g_{j}}-B_{2,g_{i}g_{j}}}{\sqrt{B_{1,g_{i}g_{j}} (1-B_{1,g_{i}g_{j}})+B_{2,g_{i}g_{j}}(1-B_{2,g_{i}g_{j}})}}\right) \vert (1+ o_p(1))\\&\quad \ge (l_1-l_2)(1+o_p(1)),\\ \end{aligned} \end{aligned}$$

where

$$\begin{aligned} l_1= & {} \max \limits _{1 \le m \le M,1 \le k \le K} \vert \frac{1}{\sqrt{S}}\sum _{s=1}^{S} \frac{1}{\sqrt{ \#( \ g^{-1}(k)\backslash \left\{ i_{m_s}\right\} )}} \\{} & {} \sum _{j\in g^{-1}(k)\backslash \left\{ i_{m_s}\right\} }\frac{B_{1,g_{i}g_{j}}-B_{2,g_{i}g_{j}}}{\sqrt{B_{1,g_{i}g_{j}} (1-B_{1,g_{i}g_{j}})+B_{2,g_{i}g_{j}}(1-B_{2,g_{i}g_{j}})}}) \vert ,\\ l_2= & {} \max \limits _{1 \le m \le M,1 \le k \le K} \vert \frac{1}{\sqrt{S}} \sum _{s=1}^{S} \frac{1}{\sqrt{ \#( \ g^{-1}(k)\backslash \left\{ i_{m_s}\right\} )}}\\{} & {} \sum _{j\in g^{-1}(k)\backslash \left\{ i_{m_s}\right\} }(\frac{A_{1,ij}-B_{1,g_{i}g_{j}} -(A_{2,ij}-B_{2,g_{i}g_{j}})}{\sqrt{B_{1,g_{i}g_{j}}(1-B_{1,g_{i}g_{j}})+B_{2,g_{i}g_{j}}(1-B_{2,g_{i}g_{j}})}} \vert . \end{aligned}$$

By by the result of (A5), we have that

$$\begin{aligned} l_2=O_p(\sqrt{\log (MK)}). \end{aligned}$$

Moreover, from Assumptions 1, 2, 3 and $\max \limits _{1 \le i \le n,1 \le j \le n}\vert B_{1,g_{i}g_{j}}-B_{2,g_{i}g_{j}} \vert = \Omega (\frac{\log n}{n}\sqrt{\frac{K}{S}})$, we have that

$$\begin{aligned} \frac{l_1}{\sqrt{\log MK}}\ge & {} \sqrt{\frac{Sn}{K}}\sqrt{\frac{n}{\log n}}\Omega \left( \frac{\log n}{n}\sqrt{\frac{K}{S}}\right) /\sqrt{\log MK}\\= & {} \frac{\sqrt{\log n}}{\sqrt{\log MK}}\rightarrow \infty , \quad as \hspace{5.0pt}n\rightarrow \infty . \end{aligned}$$

Thus, we finally obtain that

$$\begin{aligned} P(T\ge c\log (MK))\rightarrow 1, \end{aligned}$$

for any positive constant c.

This completes the proof of Theorem 2. $\square$