Large deviations for the largest eigenvalue of Gaussian networks with constant average degree

Abstract

Large deviation behavior of the largest eigenvalue $\lambda _1$ of Wigner matrices including those arising from an Erdős-Rényi random graph ${\mathcal {G}}_{n,p}$ with i.i.d. random conductances on the edges has been the topic of considerable interest. However, despite several recent advances, not much is known when the underlying graph is sparse i.e., $p\rightarrow 0$, except the recent works (Bhattacharya et al., Ann Probab 49(4):1847–1885, 2021and Bhattacharya and Ganguly, SIAM J Discret Math, 2020) which consider the simpler case of the graph without additional edge weights. Under sufficiently general conditions on the conductance distribution, one expects the ‘dense’ behavior as long as the average degree np is at least logarithmic in n. In this article we focus on the case of constant average degree i.e., $p=\frac{d}{n}$ for some fixed $d>0$ with standard Gaussian weights. Results in Bandeira and Van Handel (Ann Probab 44(4):2479–2506, 2016) about general non-homogeneous Gaussian matrices imply that in this regime $\lambda _1$ scales like $\sqrt{\log n}.$ We prove the following results towards a precise understanding of the large deviation behavior in this setting.

• 1. (Upper tail probabilities and structure theorem): For $\delta >0,$ we pin down the exact exponent $\psi (\delta )$ such that

  $$\begin{aligned} {\mathbb {P}}(\lambda _1\ge \sqrt{2(1+\delta )\log n})=n^{-\psi (\delta )+o(1)}. \end{aligned}$$

  Further, we show that conditioned on the upper tail event, with high probability, a unique maximal clique emerges with a very precise $\delta$ dependent size (takes either one or two possible values) and the Gaussian weights are uniformly high in absolute value on the edges in the clique. Finally, we also prove an optimal localization result for the leading eigenvector, showing that it allocates most of its mass on the aforementioned clique which is spread uniformly across its vertices.

• 2. (Lower tail probabilities): The exact stretched exponential behavior

  $$\begin{aligned} {\mathbb {P}}(\lambda _1\le \sqrt{2(1-\delta )\log n})=\exp \left( -n^{\ell (\delta )+o(1)}\right) \end{aligned}$$

  is also established.

As an immediate corollary, one obtains that $\lambda _1$ is typically $(1+o(1))\sqrt{2\log n}$, a result which surprisingly appears to be new. A key ingredient in our proofs is an extremal spectral theory for weighted graphs obtained by an $\ell _1-$reduction of the standard $\ell _2-$variational formulation of the largest eigenvalue via the classical Motzkin-Straus theorem [37], which could be of independent interest.

Appendix A. Key estimates

In this appendix, we include the outstanding proofs of basic properties about Gaussian random variables. as well as the proof of Lemma 4.2 involving a straightforward application of Chernoff’s bound.

Proof of Lemma 4.1

Recalling the basic tail bounds from (24), for some constant $c_1>0$,

$$\begin{aligned} {\mathbb {P}}( \max _{i=1,\cdots ,m} X_i\ge \sqrt{2(1+\delta ) \log n} )&= 1- (1- {\mathbb {P}}(X_1 \ge \sqrt{2(1+\delta ) \log n} )) ^m \\&\ge c_1 \frac{1}{n^{\delta } \sqrt{\log n}}. \end{aligned}$$

Similarly, for some constant $c_2>0$,

$$\begin{aligned} {\mathbb {P}}( \max _{i=1,\cdots ,m} X_i\le \sqrt{2(1-\delta ) \log n} )&= ( 1 - {\mathbb {P}}(X_1 \ge \sqrt{2(1-\delta ) \log n} ) )^m \le e^{-c_2 \frac{n^\delta }{\sqrt{\log n}}}. \end{aligned}$$

$\square$

Proof of Lemma 4.2

We use the Chernoff’s bound for Bernoulli variables for $q>p$:

$$\begin{aligned} {\mathbb {P}}( \text {Bin}(m,p) \ge mq) \le e^{-m I_p(q)}, \end{aligned}$$

(216)

where $I_p(x): = x\log \frac{x}{p}+(1-x) \log \frac{1-x}{1-p}$ is the relative entropy function. Thus,

$$\begin{aligned} {\mathbb {P}}\Big ( \text {Bin}\Big ( \frac{n(n-1)}{2}, 1- \frac{d}{n} \Big ) \ge \frac{n(n-1)}{2} \Big (1-\frac{d}{4n}\Big )\Big ) \le e^{-\frac{n(n-1)}{2} I_{1- \frac{d}{n} }(1-\frac{d}{4n} )}, \end{aligned}$$

(217)

Using $\log (1+x) \ge \frac{x}{2}$ for small positive x,

$$\begin{aligned} I_{1-\frac{d}{n}}\Big (1-\frac{d}{4n} \Big ) \ge \Big ( 1-\frac{d}{4n} \Big ) \frac{3d}{8(n-d)} + \frac{d}{4n} \log \frac{1}{4} \ge \frac{C_1}{n-d} - \frac{C_2}{n^2}. \end{aligned}$$

(218)

Hence, by (217) and (218), there exists a constant $c>0$ such that for sufficiently large n,

$$\begin{aligned} {\mathbb {P}}\Big ( \text {Bin}\Big ( \frac{n(n-1)}{2}, 1- \frac{d}{n} \Big ) \ge \frac{n(n-1)}{2} \Big (1-\frac{d}{4n}\Big )\Big ) \le e^{-cn}. \end{aligned}$$

This implies that

$$\begin{aligned} {\mathbb {P}}\Big ( \text {Bin}\Big ( \frac{n(n-1)}{2}, \frac{d}{n} \Big ) \le \frac{n(n-1)}{2} \frac{d}{4n}\Big ) \le e^{-cn}, \end{aligned}$$

which concludes the proof. $\square$

Proof of Lemma 5.1

Recall that we are aiming to show

$$\begin{aligned} {\mathbb {P}}({\tilde{Y}}_1^2+\cdots +{\tilde{Y}}_m^2 \ge L ) \le C^m e^{-\frac{1}{2}L} e^{\frac{1}{2}m} \Big (\frac{L}{m}\Big )^{m} e^{\frac{1}{2} \varepsilon m \log \log n}, \end{aligned}$$

and in particular, for any $a,b,c>0$, if $m \le b\frac{\log n}{\log \log n}+c$ and $L = a\log n$, then, for any $\gamma >0$, for sufficiently large n,

$$\begin{aligned} {\mathbb {P}}({\tilde{Y}}_1^2+\cdots +{\tilde{Y}}_m^2 \ge a\log n ) \le n^{-\frac{a}{2} + \frac{\varepsilon b}{2} + \gamma } . \end{aligned}$$

(219)

By exponential Chebyshev’s bound, for any $t>0$,

$$\begin{aligned} {\mathbb {P}}({\tilde{Y}}_1^2+\cdots +{\tilde{Y}}_m^2 \ge L ) \le e^{- tL } ({\mathbb {E}}e^{t{\tilde{Y}}_1^2})^m. \end{aligned}$$

(220)

Using the lower bound for the tail (24), the probability density function of ${\tilde{Y}}$, denoted by ${\tilde{f}}(x)$ for $|x| \ge \sqrt{\varepsilon \log \log n}$, satisfies

$$\begin{aligned} {\tilde{f}}(x) \le \frac{C}{ ( \sqrt{\varepsilon \log \log n} ) ^{-1} e^{-\frac{1}{2}\varepsilon \log \log n} } e^{-\frac{1}{2}x^2} = C\sqrt{\varepsilon \log \log n} e^{\frac{1}{2}\varepsilon \log \log n} e^{-\frac{1}{2}x^2} . \end{aligned}$$

Hence, using the upper bound for the tail (24), by making a change of variable $x = \frac{1}{\sqrt{1-2t}}y$,

$$\begin{aligned} {\mathbb {E}}e^{t{\tilde{Y}}_1^2}&\le C \sqrt{\varepsilon \log \log n} e^{\frac{1}{2}\varepsilon \log \log n} \int _{\sqrt{\varepsilon \log \log n}}^\infty e^{tx^2} e^{-\frac{1}{2}x^2} dx \\&= C \sqrt{\varepsilon \log \log n} e^{\frac{1}{2}\varepsilon \log \log n} \frac{1}{\sqrt{1-2t}} \int _{ \sqrt{1-2t} \sqrt{\varepsilon \log \log n}}^\infty e^{-\frac{1}{2}y^2} dy \\&\le C \sqrt{\varepsilon \log \log n} e^{\frac{1}{2}\varepsilon \log \log n} \frac{1}{\sqrt{1-2t}} \frac{1}{\sqrt{1-2t} \sqrt{\varepsilon \log \log n} } e^{-\frac{1}{2}(1-2t) \varepsilon \log \log n}\\&=C \frac{1}{1-2t}e^{t \varepsilon \log \log n}. \end{aligned}$$

Applying this to (220),

$$\begin{aligned} {\mathbb {P}}({\tilde{Y}}_1^2+\cdots +{\tilde{Y}}_m^2 \ge L ) \le C^m e^{-tL} \frac{1}{(1-2t)^m} e^{t\varepsilon m \log \log n}. \end{aligned}$$

We take $t = \frac{1}{2} (1- \frac{m}{ L } ) < \frac{1}{2}$ (recall that $L>m$) in order to balance two terms $e^{-tL}$ and $\frac{1}{(1-2t)^m}$. We conclude the proof of (38).

We now show (219). We first check that for any $L>0$, a function $x\mapsto (\frac{L}{x})^x$ is increasing on $(0, \frac{L}{e})$. This is because the derivative of $x \log (\frac{L}{x})$, which is given by $\log (\frac{L}{x}) - 1$, is positive for $x\in (0, \frac{L}{e})$. Hence, for any $\gamma >0$, for sufficiently large n, the LHS of (219) is bounded by

$$\begin{aligned} C^{b\frac{\log n}{\log \log n} +c} n^{-\frac{a}{2} + \frac{\varepsilon b}{2}} n^{\frac{b}{2\log \log n}} \Big ( \frac{a }{b} \log \log n \Big )^{b\frac{\log n}{\log \log n} +c} \le n^{-\frac{a}{2} + \frac{\varepsilon b}{2}+\gamma }. \end{aligned}$$

Here, we used the fact that for large n, $(c_1 \log \log n)^{c_2 \frac{\log n}{\log \log n}} \le n^{\frac{\gamma }{2}}.$$\square$