Eigenvalues Outside the Bulk of Inhomogeneous Erdős–Rényi Random Graphs

Abstract

In this article, an inhomogeneous Erdős–Rényi random graph on $\{1,\ldots , N\}$ is considered, where an edge is placed between vertices i and j with probability $\varepsilon _N f(i/N,j/N)$, for $i\le j$, the choice being made independently for each pair. The integral operator $I_f$ associated with the bounded function f is assumed to be symmetric, non-negative definite, and of finite rank k. We study the edge of the spectrum of the adjacency matrix of such an inhomogeneous Erdős–Rényi random graph under the assumption that $N\varepsilon _N\rightarrow \infty$ sufficiently fast. Although the bulk of the spectrum of the adjacency matrix, scaled by $\sqrt{N\varepsilon _N}$, is compactly supported, the kth largest eigenvalue goes to infinity. It turns out that the largest eigenvalue after appropriate scaling and centering converges to a Gaussian law, if the largest eigenvalue of $I_f$ has multiplicity 1. If $I_f$ has k distinct non-zero eigenvalues, then the joint distribution of the k largest eigenvalues converge jointly to a multivariate Gaussian law. The first order behaviour of the eigenvectors is derived as a byproduct of the above results. The results complement the homogeneous case derived by [18].

Appendix

Lemma 6.1

The eigenfunctions $\{r_i:1\le i\le k\}$ of the operator $I_f$ are Riemann integrable.

Proof

Let $D_f\subset [0,1]\times [0,1]$ be the set of discontinuity points f. Since f is Riemann integrable, the Lebesgue measure of $D_f$ is 0. Let

$$\begin{aligned} D_f^x=\{y\in [0,\,1]:\, (x,y)\in D_f\}, \, x\in [0,\,1]. \end{aligned}$$

If $\lambda$ is the one dimensional Lebesgue measure, then Fubini’s theorem implies that

$$\begin{aligned} E=\{ x\in [0,1]: \lambda (D_f^x)=0\} \end{aligned}$$

has full measure. Fix an $x\in E$ and consider $x_n\rightarrow x$ and observe that

$$\begin{aligned} f(x_n,y)\rightarrow f(x,y) \quad \text {for all}{ y\notin D_f^x}. \end{aligned}$$

Fix $1\le i\le k$ and let $\theta _i$ be the eigenvalue with corresponding eigenfunction $r_i$, that is,

$$\begin{aligned} r_i(x)= \frac{1}{\theta _i}\int _0^1 f(x,y) r_i(y) \, dy. \end{aligned}$$

(6.1)

Using f is bounded and $r\in L^2[0,1]$, dominated convergence theorem implies

$$\begin{aligned} r_i(x_n)= \frac{1}{\theta _i}\int _{(D_f^x)^c} f(x_n,y) r_i(y)\, dy\rightarrow \frac{1}{\theta _i}\int _0^1 f(x,y) r_i(y)\, dy=r_i(x) \end{aligned}$$

and hence r is continuous at x. So the discontinuity points of $r_i$ form a subset of $E^c$ which has Lebesgue measure 0. Further, (6.1) shows that $r_i$ is bounded and hence Riemann integrability follows. $\square$

The following result is a version of the Perron–Frobenius theorem in the infinite dimensional setting (also known as the Krein–Rutman theorem). Since our integral operator is positive, self-adjoint and finite dimensional so the proof in this setting is much simpler and can be derived following the work of [28]. In what follows, we use for $f, g\in L^2[0,1]$, the inner product

$$\begin{aligned} \langle f, \, g\rangle =\int _0^1 f(x)g(x)dx. \end{aligned}$$

Lemma 6.2

Suppose $f>0$ a.e. on $[0,1]\times [0,1]$. Then largest eigenvalue $\theta _1$ of $T_f$ is positive and the corresponding eigenfunction $r_1$ can be chosen such that $r_1(x)>0$ for almost every $x\in [0,1]$. Further, $\theta _1>\theta _2$.

Proof

First observe that

$$\begin{aligned} 0<\theta _1= \langle r_1,\, \theta _1 r_1\rangle&= \langle r_1, \, I_f(r_1)\rangle = | \langle r_1, \, I_f(r_1)\rangle |\\&\le \langle u_1, I_f(u_1)\rangle \le \theta _1 \end{aligned}$$

where $u_1(x)=|r_1|(x)$ and the last inequality follows from the Rayleigh-Ritz formulation of the largest eigenvalue. Hence note that the string of inequalities is actually an equality, that is,

$$\begin{aligned} \langle r_1, \, I_f(r_1) \rangle = \langle u_1, I_f(u_1)\rangle . \end{aligned}$$

Breaking $r_1= r_1^{+}-r_1^{-}$ implies either $r_1^+=0$ or $r_1^{-}=0$ almost everywhere. Without loss of generality assume that $r_1\ge 0$ almost everywhere. Using

$$\begin{aligned} \theta _1 r_1(x)= \int _0^1 f(x,y) r_1(y)\, dy \end{aligned}$$

Note that if $r_1(x)$ is zero for some x then due to the positivity assumption on f, $r_1(y)=0$ for almost every $y\in [0,1]$ which is a contradiction. Hence we have that $r_1(x)>0$ almost every $x\in [0, 1]$.

For the final claim, without loss of generality assume that $\int _0^1 r_1(x)\, dx\ge 0$. If $\theta _1=\theta _2$, then the previous argument would give us $r_2(x)>0$ and this will contradict the orthogonality of $r_1$ and $r_2$. $\square$

Lemmas 4.1–4.4 are proved in the rest of this section. Therefore, the notations used here should refer to those in Sect. 4 and should not be confused with those in Sect. 5. For example, $e_1$ and $e_2$ are as in Lemma 4.2.

Proof of Lemma 4.1

Note that for any even integer k

$$\begin{aligned} \mathrm{E}(\Vert W\Vert ^k) \le \mathrm{E}({{\,\mathrm{Tr}\,}}(W^k)). \end{aligned}$$

(6.2)

Using $\mathrm{E}(W(i,j)^2)\le \varepsilon M$ and condition (1.2) it is immediate that conditions of Theorem 1.4 of [34] are satisfied. We shall use the following estimate from the proof of that result. It follows from [34, Sect. 4]

$$\begin{aligned} \mathrm{E}({{\,\mathrm{Tr}\,}}(W^k)) \le K_1N (2\sqrt{\varepsilon MN})^k \end{aligned}$$

(6.3)

where $K_1$ is some positive constant and there exists a constant $a>0$ such that k can be chosen as

$$\begin{aligned} k = \sqrt{2} a (\varepsilon M)^{1/4} N^{1/4}. \end{aligned}$$

Using (6.2), (6.3) and $(1-x)^k \le e^{-kx}$ for k, $x > 0$,

$$\begin{aligned}&P \left( \Vert W \Vert \ge 2 \sqrt{MN\varepsilon } + C_1 (N \varepsilon )^{1/4} ( \log N)^{\xi /4}\right) \nonumber \\&= K_1 N \left( 1 - \frac{C_1 (N \varepsilon )^{1/4} ( \log N)^{\xi /4}}{ 2 \sqrt{MN\varepsilon } + C_1 (N \varepsilon )^{1/4} ( \log N)^{\xi /4}} \right) ^k\nonumber \\&\le K_1 N \exp \left( -\frac{k C_1 (N \varepsilon )^{1/4} ( \log N)^{\xi /4}}{ 2 \sqrt{MN\varepsilon } + C_1 (N \varepsilon )^{1/4} ( \log N)^{\xi /4}} \right) . \end{aligned}$$

(6.4)

Now plugging in the value of k in the bound (6.4) and using

$$\begin{aligned} 2 \sqrt{M} + C_1 (N \varepsilon )^{-1/4} ( \log N)^{\xi /4} \le 2 \sqrt{M} +C_1 \end{aligned}$$

we have

$$\begin{aligned} (6.4) \le K_1 N \exp \left( -\frac{C_1 a M^{1/4}\sqrt{2} ( \log N)^{\xi /4}}{ 2 \sqrt{M} + C_1} \right) \le e^{-C_2 (\log N )^{\xi /4}} \end{aligned}$$

for some constant $C_2>0$ and N large enough. This proves (4.1) and hence the lemma.

$\square$

Proof of Lemma 4.2

Let A be the event where Lemma 4.1 holds, that is, $\Vert W\Vert \le C \sqrt{N\varepsilon }$ for some constant C. Since the entries of $e_1$ and $e_2$ are in $[-1/\sqrt{N}, 1/\sqrt{N}]$ so $\Vert e_i\Vert \le 1$ for $i=1,2$. Hence on the high probability event it holds that

$$\begin{aligned} \left| \mathrm{E}\left( e_1^\prime W^ne_2 \mathbf{1} _{A}\right) \right| \le (CN\varepsilon )^{n/2}. \end{aligned}$$

We show that the above expectation on the low probability event $A^c$ is negligible. For that first observe

$$\begin{aligned} |\mathrm{E}[(e_1^\prime W^n e_2)^2]|\le N^{nC^\prime } \end{aligned}$$

for some constant $0<C^\prime <\infty$. Thus using Lemma 4.1 one has

$$\begin{aligned} \left| \mathrm{E}\left( e_1^\prime W^ne_2 \mathbf{1} _{A^c}\right) \right|&\le \left| \mathrm{E}\left[ (e_1^\prime W^ne_2)^2\right] ^{1/2}\right| P(A_N^c)^{1/2} \\&\le \exp \left( nC^{\prime }\log N- 2^{-1}C_2 (\log N)^{\xi /4}\right) \end{aligned}$$

Since $n\le \log N$ and $\xi >8$ the result follows. $\square$

Proof of Lemma 4.3

The proof is similar to the proof of Lemma 6.5 of [18]. The exponent in the exponential decay is crucial, so the proof is briefly sketched. Observe that

$$\begin{aligned}&e_1^\prime W^ne_2-\mathrm{E}\left( e_1^\prime W^ne_2\right) \nonumber \\&\quad =\sum _{i\in \{1,\ldots , N\}^{n+1}}e_1(i_1)e_2(i_{n+1}) \left( \prod _{l=1}^{n}W(i_l, i_{l+1})-\mathrm{E}\left[ \prod _{l=1}^{n}W(i_l, i_{l+1})\right] \right) \end{aligned}$$

(6.5)

To use the independence, one can split the matrix W as $W^{\prime } + W^{{\prime \prime }}$ where the upper triangular matrix $W^{\prime }$ has entries $W^{\prime }(i,j) = W(i,j) \mathbf{1} (i\leqslant j)$ and the lower triangular matrix $W^{{\prime \prime }}$ with entries $W^{{\prime \prime }}(i,j) = W(i,j) \mathbf{1} (i > j)$. Therefore the above quantity under the sum breaks into $2^n$ terms each having similar properties. Denote one such term as

$$\begin{aligned} L_n= \sum _{i\in \{1,\ldots , N\}^{n+1}}e_1(i_1)e_2(i_{n+1}) \left( \prod _{l=1}^{n}W^\prime (i_l, i_{l+1})-\mathrm{E}\left[ \prod _{l=1}^{n}W^\prime (i_l, i_{l+1})\right] \right) . \end{aligned}$$

Using the fact that each entry of $e_1$ and $e_2$ are bounded by $1/\sqrt{N}$, it follows by imitating the proof of Lemma 6.5 of [18] that

$$\begin{aligned} \mathrm{E}[|L_n|^p] \le \frac{\left( Cnp\right) ^{np}(N\varepsilon )^{np/2}}{N^{p/2}}, \end{aligned}$$

where p is an even integer and C is a positive constant, independent of n and p. Rest of the $2^n-1$ terms arising in (6.5) have the same bound and hence

$$\begin{aligned}&P\left( \left| e_1^\prime W^ne_2-\mathrm{E}\left( e_1^\prime W^ne_2\right) \right| >N^{(n-1)/2}\varepsilon ^{n/2}(\log N)^{n\xi /4}\right) \\&\le \frac{\left( 2Cnp\right) ^{np}(N\varepsilon )^{np/2}}{N^{p/2}N^{p(n-1)/2}\varepsilon ^{pn/2}(\log N)^{pn\xi /4}}= \frac{\left( 2Cnp\right) ^{np}}{(\log N)^{pn\xi /4}}. \end{aligned}$$

Choose $\eta \in (1,\, \xi /4)$ and consider

$$\begin{aligned} p=\frac{(\log N)^{\eta }}{2Cn}, \end{aligned}$$

(with N large enough to make p an even integer) to get

$$\begin{aligned}&P\left( \left| e_1^\prime W^ne_2-\mathrm{E}\left( e_1^\prime W^ne_2\right) \right| >N^{(n-1)/2}\varepsilon ^{n/2}(\log N)^{n\xi /4}\right) \\&\le \exp \left( -\frac{1}{2C}(\log N)^{\eta }(\frac{\xi }{4}-\eta )\log \log N\right) . \end{aligned}$$

Note that $n\le L$, ensures that $p>1$. Since the bound is uniform over all $2\le n\le L$, the first bound (4.2) follows.

For (4.3) one can use Hoeffding’s inequality [21, Theorem 2] as follows.

Define

$$\begin{aligned} {\widetilde{A}}(k,l)= A(k,l) e_1(k)e_2(l), \,\,\, 1\le k\le l\le N. \end{aligned}$$

Since A(k, l) are Bernoulli random variables, so one has $\{{\widetilde{A}}(k,l): 1\le k\le l\le N\}$ are independent random variables taking values in $[-1/N, 1/N]$ and hence by Hoeffding’s inequality we have, for any $\delta >0$,

$$\begin{aligned}&P\left( \left| \sum _{1\le k\le l\le N} {\widetilde{A}}(k,l)- E\left( \sum _{1\le k\le l\le N} {\widetilde{A}}(k,l)\right) \right| >\delta N\varepsilon \right) \\&\le 2\exp \left( -\delta ^2(N\varepsilon )^2\right) \le 2 \exp \left( -\delta ^2 (\log N)^{2\xi }\right) . \end{aligned}$$

Dealing with the case $k>l$ similarly, the desired bound on$e_1^\prime We_2$ follows. $\square$

Proof of Lemma 4.4

Follows by a simple moment calculation. $\square$