Kesten-McKay law for random subensembles of Paley equiangular tight frames

TL;DR

This paper proves a conjecture about the singular value distribution of random subensembles of Paley equiangular tight frames using the method of moments, with potential for broader applications.

Contribution

It introduces a novel proof of a conjecture on singular value distribution for Paley equiangular tight frames and extends the analysis to more general real equiangular tight frames.

Findings

Confirmed the conjecture for Paley equiangular tight frames

Extended analysis to real equiangular tight frames of redundancy 2

Suggests potential for broader generalizations in frame theory

Abstract

We apply the method of moments to prove a recent conjecture of Haikin, Zamir and Gavish (2017) concerning the distribution of the singular values of random subensembles of Paley equiangular tight frames. Our analysis applies more generally to real equiangular tight frames of redundancy 2, and we suspect similar ideas will eventually produce more general results for arbitrary choices of redundancy.

Equations

$$\alpha \Vert x{\Vert }^2\le \sum _{i\in I}|⟨x,{\phi }_i⟩{|}^2\le \beta \Vert x{\Vert }^2$$

$$\{(\tfrac{x}{5})\}_{x=0}^{4}=(0,+,-,-,+)\qquad\Longrightarrow\qquad S=\left[\begin{array}[]{c|ccccc}0&+&+&+&+&+\\ \hline\cr+&\pagecolor{black!10}0&\pagecolor{black!10}+&\pagecolor{black!10}-&\pagecolor{black!10}-&\pagecolor{black!10}+\\ +&+&0&+&-&-\\ +&-&+&0&+&-\\ +&-&-&+&0&+\\ +&+&-&-&+&0\end{array}\right]$$

$${\mu }_Z:=\frac{1}{n}\sum _{i=1}^n{\delta }_{{\lambda }_i}.$$

$$d\mu_{\operatorname{KM}(v)}=\left\{\begin{array}[]{cl}\frac{v\sqrt{4(v-1)-x^{2}}}{2\pi(v^{2}-x^{2})}&\text{if }x^{2}\leq 4(v-1)\\ 0&\text{otherwise}\end{array}\right\}dx.$$

$$\mathbb{P}\big{\{}|f(X)-\mathbb{E}f(X)|\geq bt\big{\}}\lesssim e^{-t^{2}/c\sigma^{2}}.$$

$$|\mathbb{E}V-\mathbb{E}W|{\lesssim }_p\frac{1}{\sqrt{n}},|\mathrm{Var}(V)-\mathrm{Var}(W)|{\lesssim }_p\frac{1}{\sqrt{n}}.$$

$$|V|{\lesssim }_p\frac{1}{n}\sum _i|{\lambda }_i(Z){|}^k\le \Vert Z{\Vert }_{2\to 2}^k{\asymp }_p\frac{1}{n^{k/2}}\Vert X{\Vert }_{2\to 2}^k\le \frac{1}{n^{k/2}}\Vert S{\Vert }_{2\to 2}^k\le 1$$

$$\mathbb{E}|V-W|=\mathbb{E}\big{(}|\tfrac{N}{pn}-1|\cdot|W|\big{)}\lesssim_{p}\mathbb{E}|\tfrac{N}{pn}-1|\leq\frac{1}{pn}\big{(}\mathbb{E}(N-pn)^{2}\big{)}^{1/2}\lesssim_{p}\frac{1}{\sqrt{n}},$$

$$|\mathrm{Var}(V)-\mathrm{Var}(W)|$$

$$\displaystyle\leq\mathbb{E}\big{(}|V+W||V-W|\big{)}+|\mathbb{E}V+\mathbb{E}W||\mathbb{E}V-\mathbb{E}W|$$

$${\lesssim }_p\mathbb{E}|V-W|{\lesssim }_p\frac{1}{\sqrt{n}},$$

$$\mathbb{E}\frac{1}{pn}\mathrm{tr}(Z^k)\to {\int }_{\mathbb{R}}{x}^{k}d{\mu }_{\mathrm{KM}(1/p)}(x).$$

$$C(n,k):=\frac{(n+k)!(n-k+1)}{k!(n+1)!}.$$

$$\int_{\mathbb{R}}x^{k}~{}d\mu_{\operatorname{KM}(v)}(x)=\left\{\begin{array}[]{cl}\displaystyle\sum_{j=1}^{k/2}C(k/2-1,k/2-j)v^{j}(v-1)^{k/2-j}&\text{if $k$ is even}\\ 0&\text{if $k$ is odd.}\end{array}\right.$$

$$\frac{1}{n^{k/2+1}}\mathbb{E}\operatorname{tr}(X^{k})\to\left\{\begin{array}[]{cl}\displaystyle\sum_{t=k/2+1}^{k}(-1)^{t-k/2-1}\cdot B(k/2-1,t-k/2-1)\cdot p^{t}&\text{if $k$ is even}\\ 0&\text{if $k$ is odd,}\end{array}\right.$$

$$B(n,k):=\sum _{j=k}^n\left(\genfrac{}{}{0ex}{}{j}{k}\right)C(n,j).$$

$$\mathrm{tr}(X^k)=\mathrm{tr}((PSP{)}^k)=\mathrm{tr}((PS{)}^k)$$

$$=\sum _{a_1,\dots ,a_k\in [n]}{S}_{a_1{a}_2}{S}_{a_2{a}_3}\cdots S_{a_k{a}_1}\cdot \prod _{i=1}^k{1}_{\{a_i\in \mathcal{I}\}}.$$

$$\frac{1}{n^{k/2+1}}\mathbb{E}\operatorname{tr}(X^{k})=\sum_{t=1}^{k}\bigg{(}\frac{1}{n^{k/2+1}}\sum_{\begin{subarray}{c}a_{1},\ldots,a_{k}\in[n]\\ |\{a_{1},\ldots,a_{k}\}|=t\end{subarray}}S_{a_{1}a_{2}}S_{a_{2}a_{3}}\cdots S_{a_{k}a_{1}}\bigg{)}\cdot p^{t}.$$

$$\Delta (a_1,a_2,a_3,\dots ,a_k):=S_{a_1{a}_2}{S}_{a_2{a}_3}\cdots S_{a_k{a}_1}.$$

$$L_{n}(\pi):=\big{\{}a\colon[k]\to[n]:\{a^{-1}(a(i)):i\in[k]\}=\pi\big{\}}.$$

$$V_n(\pi ):=\frac{1}{n^{k/2+1}}\sum _{a\in L_n(\pi )}\Delta (a(1),\dots ,a(k)).$$

$$\frac{1}{n^{k/2+1}}\mathbb{E}\operatorname{tr}(X^{k})=\sum_{t=1}^{k}\bigg{(}\sum_{\pi\in\Pi(k,t)}V_{n}(\pi)\bigg{)}\cdot p^{t}.$$

$$V_n(\pi )\to (-1{)}^{k/2-m}\cdot C_{s_1-1}\cdots C_{s_m-1}.$$

$$\displaystyle\sum_{\pi\in\Pi(k,t)}V_{n}(\pi)\to\left\{\begin{array}[]{cl}(-1)^{t-k/2-1}\cdot B(k/2-1,t-k/2-1)&\text{if $k$ is even and $t\geq k/2+1$}\\ 0&\text{otherwise.}\end{array}\right.$$

$$(-1)^{t-k/2-1}\cdot V_{n}(\pi)\to\left\{\begin{array}[]{cl}|\operatorname{MD}(\pi)|&\text{if }\pi\in\operatorname{EC}(k,t)\\ 0&\text{otherwise.}\end{array}\right.$$

$$\mathrm{MD}(k,t)=\underset{\pi \in \mathrm{EC}(k,t)}{⨆}\mathrm{MD}(\pi ).$$

$$(-1{)}^{t-k/2-1}\sum _{\pi \in \Pi (k,t)}{V}_n(\pi )\to \sum _{\pi \in \mathrm{EC}(k,t)}|\mathrm{MD}(\pi )|=|\mathrm{MD}(k,t)|=B(k/2-1,t-k/2-1),$$

$$\operatorname{tr}(Y^{k})=\operatorname{tr}(\tfrac{1}{p\sqrt{n}}X^{k}),\qquad\operatorname{tr}\big{(}(Y+\tfrac{1}{p}P)^{k}\big{)}=\operatorname{tr}\big{(}(\tfrac{1}{p}PF^{\top}FP)^{k}\big{)}=\frac{1}{p^{k}}\|FP\|_{S^{2k}}^{2k}.$$

$$\operatorname{tr}\big{(}(Y+\tfrac{1}{p}P)^{k}\big{)}=\operatorname{tr}\sum_{j=0}^{k}\binom{k}{j}Y^{j}(\tfrac{1}{p}P)^{k-j}=\sum_{j=1}^{k}\binom{k}{j}\frac{1}{p^{k-j}}\operatorname{tr}(Y^{j})+\frac{1}{p^{k}}\operatorname{tr}(P),$$

$$\mathrm{tr}(\frac{1}{p\sqrt{n}}{X}^k)=\mathrm{tr}(Y^k)$$

$$=\frac{1}{p^k}\Vert FP{\Vert }_{S^k}^{2k}-\sum _{j=1}^{k-1}\left(\genfrac{}{}{0ex}{}{k}{j}\right)\frac{1}{p^{k-j}}\mathrm{tr}(\frac{1}{p\sqrt{n}}{X}^j)-\frac{1}{p^k}\mathrm{tr}(P).$$

Full text

1 Introduction

Frame theory concerns redundant representation in a Hilbert space. A frame [13] is a sequence $\{\varphi_{i}\}_{i\in I}$ in a Hilbert space $H$ for which there exist $\alpha,\beta\in(0,\infty)$ such that

[TABLE]

for every $x\in H$. If every $\varphi_{i}$ has unit norm, then we say the frame is unit norm, and if $\alpha=\beta$, we say the frame is tight [9]. In the special case where $H=\mathbb{R}^{d}$, a frame is simply a spanning set, but unit norm tight frames are still interesting and useful [4, 21]. For example, equiangular tight frames are unit norm tight frames with the additional property that $|\langle\varphi_{i},\varphi_{j}\rangle|$ is constant over the choice of pair $\{i,j\}$. Equiangular tight frames are important because they necessarily span optimally packed lines, which in turn find applications in multiple description coding [25], digital fingerprinting [22], compressed sensing [1], and quantum state tomography [24]; see [14] for a survey.

Various applications demand control over the singular values of subensembles of frames. In quantum physics, Weaver’s conjecture [28] (equivalent to the Kadison–Singer problem [18, 8], and recently resolved in [19]) concerns the existence of subensembles of unit norm tight frames with appropriately small spectral norm. Compressed sensing [7, 11] has spurred the pursuit of explicit frames with the property that every subensemble is well conditioned [10, 5, 1]. Motivated by applications in erasure-robust analog coding, Haikin, Zamir and Gavish [16, 15] recently launched a new line of inquiry: identify frames for which the singular values of random subensembles exhibit a predictable distribution. (One might consider this to be a more detailed analogue to Tropp’s estimates on the conditioning of random subensembles [27].) Of particular interest are random subensembles of equiangular tight frames, and in this paper, we consider equiangular tight frames comprised of $2d$ vectors in $\mathbb{R}^{d}$, which correspond to symmetric conference matrices. (Note that such frames have already received some attention in the context of compressed sensing [1, 2].)

An $n\times n$ matrix $S$ is said to be a conference matrix if

• (i)

$S_{ii}=0$ for every $i\in[n]$,

• (ii)

$S_{ij}\in\{\pm 1\}$ for every $i,j\in[n]$ with $i\neq j$, and

• (iii)

$S^{\top}S=(n-1)I$.

A symmetric conference matrix of order $n$ exists whenever $n-1\equiv 1\bmod 4$ is a prime power (by a Paley–based construction), and only if $n\equiv 2\bmod 4$ and $n-1$ is a sum of two squares [17]. Explicitly, the Paley conference matrices are obtained by building a circulant matrix from the Legendre symbol and then padding with ones, for example:

[TABLE]

where “$\pm$” denotes $\pm 1$. One may verify that the above example satisfies $S^{2}=5I$. For every $n\times n$ symmetric conference matrix $S$, it holds that $I+\frac{1}{\sqrt{n-1}}S$ is the Gram matrix of an equiangular tight frame consisting of $n$ vectors in $\mathbb{R}^{n/2}$ [25]. In particular, the equiangular tight frames that arise from the Paley conference matrices are known as Paley equiangular tight frames. In what follows, we consider random principal submatrices of symmetric conference matrices with the understanding that they may be identified with the Gram matrix of a random subensemble of the corresponding equiangular tight frame.

Given an $n\times n$ symmetric matrix $Z$ with eigenvalues $\lambda_{1}\leq\cdots\leq\lambda_{n}$, we let $\mu_{Z}$ denote the uniform probability measure over the spectrum of $Z$ (counted with multiplicity):

[TABLE]

This is known as the empirical spectral distribution of $Z$. If $Z$ is a random matrix, then its empirical spectral distribution $\mu_{Z}$ is a random measure. We say a sequence $\{\zeta_{i}\}_{i=1}^{\infty}$ of random measures converges almost surely to a non-random absolutely continuous measure $\mu$ if for every $a,b\in\mathbb{R}$ with $a<b$, it holds that the random variable $\zeta_{i}(a,b)$ converges to $\mu(a,b)$ almost surely.

We are interested in random matrices of a particular form. Let $\mathcal{I}$ denote a random subset of $[n]$ such that the events $\{1\in\mathcal{I}\},\ldots,\{n\in\mathcal{I}\}$ are independent with probability $p$. Then for any fixed $n\times n$ matrix $A$, we write $X\sim\operatorname{Sub}(A,p)$ to denote the (random) principal submatrix of $A$ with rows and columns indexed by $\mathcal{I}$. Following [12], we define the Kesten–McKay distribution with parameter $v\geq 2$ by

[TABLE]

Recall that a lacunary sequence is a set $\{n_{i}:i\in\mathbb{N}\}$ of natural numbers for which there exists $\lambda>1$ such that $n_{i+1}\geq\lambda n_{i}$ for every $i$. We are now ready to state our main result, which corresponds to one of many conjectures posed in [16]; see Figure 1 for an illustration.

Theorem 1**.**

Fix $p\in(0,\frac{1}{2})$, take any lacunary sequence $L$ for which there exists a sequence $\{S_{n}\}_{n\in L}$ of symmetric conference matrices of increasing size $n$, and consider the corresponding random matrices $X_{n}\sim\operatorname{Sub}(S_{n},p)$. Then the empirical spectral distribution of $\frac{1}{p\sqrt{n}}X_{n}$ converges almost surely to the Kesten–McKay distribution with parameter $v=1/p$.

In the next section, we prove this theorem using the method of moments, saving the more technical portions for Section 3.

1.1 Notation

Given $x\in\mathbb{R}^{n}$, let $\operatorname{diag}(x)$ denote the $n\times n$ diagonal matrix whose diagonal entries are the entries of $x$. Given $Z\in\mathbb{R}^{m\times n}$, let $\|Z\|_{2\to 2}$ denote the induced $2$-norm of $Z$ (i.e., the largest singular value of $Z$), and let $\|Z\|_{S^{p}}$ denote the Schatten $p$-norm of $Z$ (i.e., the $p$-norm of the singular values of $Z$). Throughout this paper, we will investigate how quantities relate as $n\to\infty$. For example, suppose we are interested in a quantity $f(n,\theta)\geq 0$ that depends on both $n\in\mathbb{N}$ and some additional parameters $\theta\in\mathbb{R}^{m}$. Then we write $f(n,\theta)=o(g(n,\theta))$ if for every $\theta\in\mathbb{R}^{m}$, it holds that $f(n,\theta)/g(n,\theta)\to 0$ as $n\to\infty$. We write $f(n,\theta)\lesssim g(n,\theta)$ if there exists $c>0$ such that $f(n,\theta)\leq c\cdot g(n,\theta)$ for all $n\in\mathbb{N}$ and $\theta\in\mathbb{R}^{m}$, and we write $f(n,\theta)\lesssim_{\theta}g(n,\theta)$ if for every $\theta\in\mathbb{R}^{m}$, there exists $c(\theta)>0$ such that $f(n,\theta)\leq c(\theta)\cdot g(n,\theta)$ for all $n\in\mathbb{N}$. Finally, we write $f(n,\theta)\asymp g(n,\theta)$ if both $f(n,\theta)\lesssim g(n,\theta)$ and $g(n,\theta)\lesssim f(n,\theta)$.

2 Proof of the main result

Our proof makes use of a standard sufficient condition for the almost sure convergence of random measures, which is a consequence of the moment continuity theorem, the Borel–Cantelli lemma, and Chebyshev’s inequality, cf. Exercise 2.4.6 in [26]:

Proposition 2**.**

Let $\{\zeta_{i}\}_{i=1}^{\infty}$ be a sequence of uniformly subgaussian random probability measures, and let $\mu$ be a non-random subgaussian probability measure. Suppose that for every $k\in\mathbb{N}$, it holds that

• (i)

$\displaystyle\mathbb{E}\int_{\mathbb{R}}x^{k}d\zeta_{i}(x)\to\int_{\mathbb{R}}x^{k}d\mu(x)$, and

• (ii)

\displaystyle\sum_{i=1}^{\infty}\operatorname{Var}\bigg{(}\int_{\mathbb{R}}x^{k}d\zeta_{i}(x)\bigg{)}<\infty.

Then $\zeta_{i}$ converges almost surely to $\mu$.

As we will see, verifying hypothesis (i) in our case reduces to a combinatorics problem, whereas hypothesis (ii) can be treated separately with the help of Talagrand concentration:

Proposition 3** (Talagrand concentration, Theorem 2.1.13 in [26]).**

There exists a universal constant $c>0$ for which the following holds: Suppose $f\colon\mathbb{R}^{n}\to\mathbb{R}$ is both convex and $\sigma$-Lipschitz in $\|\cdot\|_{2}$, and let $X$ be a random vector in $\mathbb{R}^{n}$ with independent coordinates satisfying $|X_{i}|\leq b$ almost surely. Then for every $t\geq 0$, it holds that

[TABLE]

Throughout, $S_{n}$ denotes an $n\times n$ symmetric conference matrix, we draw $X_{n}\sim\operatorname{Sub}(S_{n},p)$ and put $Z_{n}:=\frac{1}{p\sqrt{n}}X_{n}$. We typically suppress the subscript $n$. While the size of $Z$ is random, its average size is $pn$, and so we use $\frac{1}{pn}\operatorname{tr}(Z^{k})$ as a proxy for $\int_{\mathbb{R}}x^{k}d\mu_{Z}(x)$. As one might expect, this is a good approximation:

Lemma 4**.**

Put $V:=\frac{1}{pn}\operatorname{tr}(Z^{k})$ and $W:=\int_{\mathbb{R}}x^{k}d\mu_{Z}(x)$. Then

[TABLE]

Proof.

Since $X$ is a submatrix of $S$, it holds that

[TABLE]

almost surely. Similarly, $|W|\leq\|Z\|_{2\to 2}^{k}\lesssim_{p}1$ almost surely. Next, let $N$ denote the (random) size of $Z$. Then $V=\frac{N}{pn}\cdot W$, and so our bound on $|W|$ gives

[TABLE]

where the last step applies the fact that $N$ has binomial distribution. This immediately implies the desired bound on $|\mathbb{E}V-\mathbb{E}W|$. Finally, since $|V|,|W|\lesssim_{p}1$ almost surely, we have

[TABLE]

which completes the result. ∎

As such, to demonstrate hypothesis (i) from Proposition 2 in our case, it suffices to prove

[TABLE]

The Kesten–McKay moments are implicitly computed in [20], and are naturally expressed in terms of entries of Catalan’s triangle:

[TABLE]

Proposition 5** (Lemma 2.1 in [20]).**

For every $v\geq 2$ and $k\in\mathbb{N}$, it holds that

[TABLE]

Recalling that $Z=\frac{1}{p\sqrt{n}}X$, then Proposition 5 gives that (1) is equivalent to

[TABLE]

where $B(n,k)$ denotes an entry of Borel’s triangle:

[TABLE]

To compute these limits, we first find a convenient expression for $\frac{1}{n^{k/2+1}}\mathbb{E}\operatorname{tr}(X^{k})$. To this end, recall that $X$ is the submatrix of $S$ with index set $\mathcal{I}$, and let $P$ denote the random $n\times n$ diagonal matrix such that $P_{ii}=1_{\{i\in\mathcal{I}\}}$. Then

[TABLE]

Considering $\mathbb{E}\prod_{i=1}^{k}1_{\{a_{i}\in\mathcal{I}\}}=p^{|\{a_{1},\ldots,a_{k}\}|}$, it follows that

[TABLE]

It remains to show that these coefficients converge to the corresponding coefficients in (2).

First, we introduce some additional notation. Taking inspiration from Bargmann invariants [3], it is convenient to write

[TABLE]

Next, we say $\pi$ is a partition of $[k]$ into $t$ blocks if $\pi=\{B_{1},\ldots,B_{t}\}$ such that $B_{1}\sqcup\cdots\sqcup B_{t}=[k]$, and we let $\Pi(k,t)$ denote the set of all such partitions. For each partition $\pi$ of $[k]$, we consider the set of functions $a\colon[k]\to[n]$ whose level sets are the blocks of $\pi$, namely

[TABLE]

With this, we define

[TABLE]

Considering (3), it therefore holds that

[TABLE]

As such, to demonstrate (2), it suffices to determine the limit of $V_{n}(\pi)$ for every partition $\pi$ of $[k]$. We start with a quick calculation:

Lemma 6**.**

For every $\pi\in\Pi(k,t)$ with $t<k/2+1$, it holds that $V_{n}(\pi)\to 0$.

Proof.

Estimate $|V_{n}(\pi)|$ using the triangle inequality to obtain a sum of $|L_{n}(\pi)|\leq n^{t}=o(n^{k/2+1})$ terms, each of size at most $1$. ∎

For each $t<k/2+1$, this establishes that the coefficient of $p^{t}$ in (4) approaches zero, i.e., the corresponding coefficient in (2). Now we wish to tackle the limiting value of $V_{n}(\pi)$ in general. In light of the related literature [23], it comes as no surprise that $V_{n}(\pi)$ depends on whether $\pi$ is a so-called crossing partition. We say a partition $\pi$ of $[k]$ is crossing if there exist $A,B\in\pi$ with $A\neq B$ for which there exist $a_{1},a_{2}\in A$ and $b_{1},b_{2}\in B$ such that $a_{1}<b_{1}<a_{2}<b_{2}$. Otherwise, $\pi$ is said to be non-crossing. Next, for each $x\in[k]$, we let $\pi(x)$ denote the unique member of $\pi$ such that $x\in\pi(x)$. Consider the graph $G_{\pi}$ with vertex set $\pi$ and edges given by $\pi(x)\leftrightarrow\pi(x+1)$ for every $x\in[k]$; here, we interpret $x+1$ modulo $k$ so that $k+1=1$. Let $\operatorname{EC}(k,t)$ denote the set of non-crossing $\pi\in\Pi(k,t)$ for which the edges of $G_{\pi}$ partition into simple even cycles. Finally, let $C_{n}:=\frac{1}{n+1}\binom{2n}{n}$ denote the $n$th Catalan number. With these notions, we can describe the limit of each $V_{n}(\pi)$:

Lemma 7** (Key combinatorial lemma).**

• (i)

Suppose $\pi\in\Pi(k,t)\setminus\operatorname{EC}(k,t)$. Then $V_{n}(\pi)\to 0$.

• (ii)

Suppose $\pi\in\operatorname{EC}(k,t)$ and the edges of $G_{\pi}$ partition into $m$ simple cycles of sizes $2s_{1},\ldots,2s_{m}$. Then $m=k-t+1$ and

[TABLE]

The proof of Lemma 7 is rather technical (involving multiple rounds of induction), and so we save it for Section 3. In the meantime, we demonstrate how Lemma 7 can be applied to prove that the coefficients in (4) converge to the coefficients in (2). Recall that a Dyck path of semi-length $n$ is a path in the plane from $(0,0)$ to $(2n,0)$ consisting of $n$ steps along the vector $(1,1)$, called up-steps, and $n$ steps along the vector $(1,-1)$, called down-steps, that never goes below the $x$-axis. We say a Dyck path is strict if none of the path’s interior vertices reside on the $x$-axis. Each (strict) Dyck path determines a sequence of $2n$ letters from $\{U,D\}$ that represent up- and down-steps in the path; this sequence is known as a (strict) Dyck word. With these notions, we may prove the following result by leveraging the fact that Borel’s triangle counts so-called marked Dyck paths [6]; see Figure 2 for an illustration.

Lemma 8**.**

It holds that

[TABLE]

Proof.

When $t<k/2+1$, the result follows from Lemma 6, and when $k$ is odd, the $k$ edges in each $G_{\pi}$ fail to partition into even simple cycles, and so the result follows from Lemma 7(i). Now suppose $k$ is even and $t\geq k/2+1$. For $\pi\in\operatorname{EC}(k,t)$, recall that the edges of $G_{\pi}$ are indexed by $[k]$ and partitioned into simple even cycles. Define $\operatorname{MD}(\pi)$ to be the words $w\colon[k]\to\{U,U^{\prime},D\}$ such that for every simple cycle in $G_{\pi}$ with edges indexed by $T\subseteq[k]$, the restriction $w|_{T}$ is a strict Dyck word with all but its first up-steps marked (here, $U^{\prime}$ denotes a marked up-step). Note that strict Dyck words of semi-length $s$ are in one-to-one correspondence with Dyck words of semi-length $s-1$, and so there are $C_{s-1}$ of them. As such, Lemma 7 implies that for every $\pi\in\Pi(k,t)$, it holds that

[TABLE]

Let $\operatorname{MD}(k,t)$ denote the set of marked Dyck words $w\colon[k]\to\{U,U^{\prime},D\}$ with $t-k/2-1$ marked up-steps, none of which are at ground level. We observe that

[TABLE]

Then equations (5) and (6) together give

[TABLE]

where the last step applies Theorem 2 in [6]. ∎

At this point, we are in a position to verify hypothesis (i) from Proposition 2 in our case. For hypothesis (ii), we follow the approach suggested by Remark 2.4.5 in [26] of leveraging Talagrand concentration to bound the variance. First, we pass to a setting that is more amenable to analysis with Talagrand concentration. Here and throughout, for each $n\in L$, we fix an $n\times n$ matrix $F$ such that $F^{\top}F=I+\frac{1}{\sqrt{n}}S_{n}$.

Lemma 9**.**

It holds that \displaystyle\operatorname{Var}\big{(}\operatorname{tr}\big{(}(\tfrac{1}{p\sqrt{n}}X)^{k}\big{)}\big{)}\lesssim_{p,k}\max_{j\in[k]}\operatorname{Var}\big{(}\|FP\|_{S^{2j}}^{2j}\big{)}+n^{1/2}.

Proof.

Define $Y:=\frac{1}{p\sqrt{n}}PSP$, and observe that

[TABLE]

Since $Y$ commutes with $P$ and $YP=Y$, the binomial theorem gives

[TABLE]

and so rearranging gives

[TABLE]

The following estimate holds for any choice of random variables $\{X_{i}\}_{i\in[m]}$:

[TABLE]

The lemma follows from applying this estimate to (7) by induction on $k$. ∎

Next, we establish the convexity and Lipschitz continuity required by Talagrand:

Lemma 10**.**

For each $k\in\mathbb{N}$, consider the mapping $f\colon\{x\in\mathbb{R}^{n}:\|x\|_{\infty}<2\}\to\mathbb{R}$ defined by $f(x)=\|F\operatorname{diag}(x)\|_{S^{2k}}^{2k}$. Then $f$ is convex and $(8^{k}kn^{1-1/2k})$-Lipschitz.

Proof.

We adopt the shorthand notation $D_{x}:=\operatorname{diag}(x)$. First, $f$ is convex since $\|\cdot\|_{S^{2k}}$ satisfies the triangle inequality and $t\mapsto t^{2k}$ is convex:

[TABLE]

To compute a Lipschitz bound, we apply the factorization

[TABLE]

with $u:=\|FD_{x}\|_{S^{2k}}$ and $v:=\|FD_{y}\|_{S^{2k}}$ to get

[TABLE]

where the last step follows from the fact that $\|FD_{x}\|_{2\to 2}\leq 2\|F\|_{2\to 2}\leq 2\sqrt{2}$, meaning $u\leq 2\sqrt{2}n^{1/2k}$ (and similarly for $v$). Next, we apply the reverse triangle inequality to get

[TABLE]

which implies the result. ∎

Finally, we apply Talagrand concentration to obtain a variance bound:

Lemma 11**.**

It holds that \displaystyle\operatorname{Var}\big{(}\tfrac{1}{pn}\operatorname{tr}\big{(}(\tfrac{1}{p\sqrt{n}}X)^{k}\big{)}\big{)}\lesssim_{p,k}n^{-1/k}.

Proof.

Given the mapping $f$ from Lemma 10, define $\tilde{f}\colon\mathbb{R}^{n}\to\mathbb{R}$ in terms of subgradients by

[TABLE]

This is known as the smallest convex extension of $f$ to $\mathbb{R}^{n}$, and it is straightforward to verify that $\tilde{f}$ is convex and $(8^{k}kn^{1-1/2k})$-Lipschitz with $\tilde{f}|_{R}=f$. Let $B\in\mathbb{R}^{n}$ have independent entries, each equal to $1$ with probability $p$ and [math] otherwise. Since $B\in R$ almost surely, it holds that $\tilde{f}(B)$ has the same distribution as $\|FP\|_{S^{2k}}^{2k}$, and we let $E$ denote its expectation. By Talagrand concentration (Proposition 3), there exists $c>0$ such that

[TABLE]

Combining with Lemma 9 then gives

[TABLE]

as desired. ∎

We may now verify hypotheses (i) and (ii) from Proposition 2 in our case.

Proof of Theorem 1.

Put $Z_{n}:=\frac{1}{p\sqrt{n}}X_{n}$ and $\mu:=\mu_{\operatorname{KM}(1/p)}$. First, we modify the random measure $\mu_{Z_{n}}$ so that we may apply Proposition 2 to prove the result. Indeed, $\mu_{Z_{n}}$ fails to be a probability measure with probability $(1-p)^{n}$, since $\mu_{Z_{n}}=0$ when $\mathcal{I}=\mathcal{I}_{n}$ is the empty set. To rectify this, we define

[TABLE]

Then it suffices to prove $\zeta_{n}\to\mu$ almost surely, since the Borel–Cantelli lemma implies $1_{\{\mathcal{I}_{n}=\emptyset\}}\to 0$ almost surely, and so

[TABLE]

for every $a,b\in\mathbb{R}$ with $a<b$. Conveniently, for every $n\in L$ and $k\in\mathbb{N}$, it holds that

[TABLE]

almost surely, and so the left-hand side inherits moments from the right-hand side.

To apply Proposition 2, we first observe that

[TABLE]

almost surely, and so $\{\zeta_{n}\}_{n\in L}$ are uniformly bounded, and therefore uniformly subgaussian. Similarly, $\mu$ is bounded and therefore subgaussian. Fix $k\in\mathbb{N}$. As a consequence of Lemma 8, it holds that

[TABLE]

and so by Lemma 4, we have

[TABLE]

As such, $\{\zeta_{n}\}_{n\in L}$ satisfies hypothesis (i) from Proposition 2. Next, Lemma 11 establishes that \operatorname{Var}\big{(}\tfrac{1}{pn}\operatorname{tr}(Z_{n}^{k})\big{)}\lesssim_{p,k}n^{-1/k}, and so Lemma 4 implies

[TABLE]

Writing $L=\{n_{i}:i\in\mathbb{N}\}$, select $\lambda>1$ such that $n_{i+1}\geq\lambda n_{i}$ for every $i\in\mathbb{N}$. Then

[TABLE]

As such, $\{\zeta_{n}\}_{n\in L}$ also satisfies hypothesis (ii) from Proposition 2, and so $\zeta_{n}\to\mu$ almost surely, as desired. ∎

3 Proof of Lemma 7

It remains to compute, for each $\pi\in\Pi(k,t)$, the limit of

[TABLE]

where $L_{n}(\pi)$ is the set of $a\colon[k]\rightarrow[n]$ whose level sets are the blocks of $\pi$ and

[TABLE]

We begin with some basic properties of $\Delta$.

Lemma 12**.**

For every $a_{1},\ldots,a_{k}\in[n]$, each of the following holds:

1. (i)

If $a_{1}\neq a_{2}$, then $\Delta(a_{1},a_{2})=1$. 2. (ii)

If $a_{j}=a_{j+1}$ for any $j\in[k-1]$ or $a_{k}=a_{1}$, then $\Delta(a_{1},\ldots,a_{k})=0$. 3. (iii)

If $\sigma$ is any cyclic permutation of $[k]$, then $\Delta(a_{\sigma(1)},\ldots,a_{\sigma(k)})=\Delta(a_{1},\ldots,a_{k})$. 4. (iv)

If $a_{1}\neq a_{k-1}$, then $\sum_{b\in[n]}\Delta(a_{1},\ldots,a_{k-1},b)=0$. 5. (v)

If $a_{1}=a_{k-1}$ and $a_{1}\neq a_{k}$, then $\Delta(a_{1},\ldots,a_{k})=\Delta(a_{1},\ldots,a_{k-2})$.

Proof.

First, (i) follows from the fact that $S$ is symmetric with off-diagonal entries in $\{\pm 1\}$. Next, (ii) follows from the fact that the diagonal entries of $S$ are 0. Recalling the definition of $\Delta$, then (iii) follows from commutativity. Next suppose $a_{k-1}\neq a_{1}$. Then

[TABLE]

and (iv) follows since $\sum_{b\in[n]}S_{a_{k-1}b}S_{ba_{1}}$ is the $(a_{k-1},a_{1})$ entry of $S^{2}=(n-1)I$. Finally, in the case where $a_{1}=a_{k-1}$, we have

[TABLE]

and (v) follows since $S_{a_{1}a_{k}}S_{a_{k}a_{1}}=1$ provided $a_{1}\neq a_{k}$. ∎

Let $\pi$ be a partition of $[k]$. Recall that for $j\in[k]$, we let $\pi(j)$ denote the block of $\pi$ containing $j$. We extend this notation to any integer $j$ by considering $\pi(j)$ to be the block of $\pi$ containing a representative of $j$ modulo $k$. For convenience, we record the following immediate consequence of Lemma 12(iii).

Lemma 13**.**

Let $\pi$ be a partition of $[k]$ and fix $j\in\mathbb{Z}$. Define $\pi^{\prime}$ to be the partition of $[k]$ with $\pi^{\prime}(i)=\pi(i-j)$ for all $i\in[k]$. Then $V_{n}(\pi^{\prime})=V_{n}(\pi).$

To establish Lemma 7(i), we will show separately that $V_{n}(\pi)\rightarrow 0$ for every crossing partition $\pi\in\Pi(k,t)$ and that $V_{n}(\pi)\rightarrow 0$ for every non-crossing partition $\pi\in\Pi(k,t)$ such that $G_{\pi}$ contains an odd cycle.

Lemma 14**.**

Let $\pi\in\Pi(k,t)$ be a crossing partition. Then $V_{n}(\pi)\rightarrow 0$.

Proof.

For $\pi\in\Pi(k,t)$ to be a crossing partition, it must hold that $t\geq 2$ and $k\geq 4$. Observe that the case $t=2$ follows immediately from Lemma 6 since $k\geq 4$. Now consider $t>2$ and suppose the lemma has been established for every crossing partition on $t-1$ blocks. By Lemma 6, we may further suppose that $k$ satisfies $t\geq k/2+1$. Then for $\pi\in\Pi(k,t)$, the pigeonhole principle guarantees that $\pi$ contains a singleton block $\{j\}\in\pi$. By Lemma 13, we may assume $\{k\}\in\pi$. We proceed in cases:

Case I: $\pi(1)=\pi(k-1)$. We may apply Lemma 12(v) to obtain

[TABLE]

The restriction of $\pi\setminus\{k\}$ to $[k-2]$ results in a crossing partition $\pi^{\prime}$ of $[k-2]$ into $t-1$ blocks. Moreover, the above expression for $V_{n}(\pi)$ implies

[TABLE]

and so our induction hypothesis provides $V_{n}(\pi)\rightarrow 0$.

Case II: $\pi(1)\neq\pi(k-1)$. Writing out $\pi=\{B_{1},\ldots,B_{t-1},\{k\}\}$, we choose representatives $j_{1},\ldots,j_{t-1}\in[k-1]$ with $\pi(j_{i})=B_{i}$. Then by Lemma 12(iv), we have

[TABLE]

For $i,j\in[t-1]$, we define new blocks

[TABLE]

and the corresponding crossing partition $\pi^{i}=\{B_{1}^{i},\ldots,B_{t-1}^{i}\}\in\Pi(k,t-1)$. Then

[TABLE]

Since each $V_{n}(\pi^{i})\rightarrow 0$ by our induction hypothesis, we see $V_{n}(\pi)\rightarrow 0$ as well. ∎

For non-crossing partitions, we will study the structure of the graph $G_{\pi}$ for $\pi\in\Pi(k,t)$, which we recall has vertex set $\pi$ and edges $\pi(j)\leftrightarrow\pi(j+1)$ for all $j\in[k]$. Observe that if $G_{\pi}$ has a loop, then $\pi(j)=\pi(j+1)$ for some $j\in[k]$, and so $V_{n}(\pi)=0$ by Lemma 12(ii). For this reason, we direct our attention to loop-free partitions $\pi$, that is, partitions $\pi$ for which $G_{\pi}$ is loop-free.

Given a loop-free graph $G$ on vertices $V$ with edges $E$, we say $v\in V$ is a cut vertex if the induced subgraph of $G$ on $V\setminus\{v\}$ is disconnected. A graph with no cut vertices is called biconnected, and the biconnected components of a graph are its maximal biconnected subgraphs. When the biconnected components of $G$ are all simple cycles, we call $G$ a cactus.

Lemma 15**.**

If $\pi\in\Pi(k,t)$ is a loop-free non-crossing partition, then $t\geq k/2+1$ and $G_{\pi}$ is a cactus whose edges partition into $k-t+1$ simple cycles.

Proof.

First, suppose $G_{\pi}$ is a cactus whose edges partition into $k-t+1$ simple cycles. Since $G_{\pi}$ has no loops, the number of cycles is at least at least half the number of edges, that is, $k-t+1\geq k/2$. Rearranging then gives $t\geq k/2+1$. It remains to verify that $G_{\pi}$ is, indeed, a cactus whose edges partition into $k-t+1$ simple cycles.

Fixing $k$, we proceed by induction on $k-t$. If $\pi\in\Pi(k,k)$, then $G_{\pi}$ is itself a simple cycle and hence a cactus. For $k-t>0$, we now consider a loop-free non-crossing partition $\pi\in\Pi(k,t)$. By the pigeonhole principle, we may select $B\in\pi$ such that $B$ contains at least two elements of $[k]$. Let $j$ denote the least element of $B$. Writing $B=\{j\}\sqcup B^{\prime}$, we consider $\pi^{\prime}\in\Pi(k,t+1)$ defined by $\pi^{\prime}=(\pi\setminus B)\cup\{B^{\prime},\{j\}\}$. Since $\pi^{\prime}$ is also loop-free and non-crossing, our induction hypothesis guarantees that $G_{\pi^{\prime}}$ is a cactus with $k-t$ simple cycles. Our task is to use this information to show that $G_{\pi}$ is a cactus with $k-t+1$ simple cycles.

Suppose first that $\{j\}$ and $B^{\prime}$ reside in the same simple cycle of $G_{\pi^{\prime}}$. Then the simple cycles of $G_{\pi^{\prime}}$ not containing $\{j\}$ and $B^{\prime}$ remain simple cycles and biconnected components of $G_{\pi}$. Moreover, by identifying $\{j\}$ and $B^{\prime}$, we see that the simple cycle of $G_{\pi^{\prime}}$ containing $\{j\}$ and $B^{\prime}$ corresponds to two simple cycles of $G_{\pi}$ sharing the cut vertex $B=\{j\}\sqcup B^{\prime}$. As such, $G_{\pi}$ contains $k-t+1$ biconnected components, each of which is a simple cycle.

We now claim that $\{j\}$ and $B^{\prime}$ must reside in the same simple cycle of $G_{\pi^{\prime}}$, in which case we are done by the previous paragraph. Suppose instead that there exists a cut vertex $X\in\pi^{\prime}\setminus\{\{j\},B^{\prime}\}$ that separates $\{j\}$ and $B^{\prime}$ within $G_{\pi^{\prime}}$, and select $j^{\prime}\in[k]$ with $\pi(j^{\prime})=B^{\prime}$. Since $j$ is the least element of $B$, we necessarily have $j<j^{\prime}$. Furthermore, since $X$ separates $\{j\}$ and $B^{\prime}$, we can traverse along a trail in $G_{\pi^{\prime}}$ from $\{j\}$ to $X$, to $B^{\prime}$, and back to $X$ to obtain indices $i\in(j,j^{\prime})$ and $i^{\prime}\in(j^{\prime},k+j)$ with $\pi^{\prime}(i)=\pi^{\prime}(i^{\prime})=X$. These indices $j<i<j^{\prime}<i^{\prime}$ satisfy $\pi(j)=\pi(j^{\prime})$ and $\pi(i)=\pi(i^{\prime})$, contradicting our assumption that $\pi$ is non-crossing. Hence, $\{j\}$ and $B^{\prime}$ must reside in the same simple cycle of $G_{\pi^{\prime}}$ as claimed. ∎

Lemma 16**.**

For every loop-free non-crossing $\pi\in\Pi(k,t)$, each of the following holds:

• (i)

If $G_{\pi}$ contains any odd cycles, then $V_{n}(\pi)\rightarrow 0$.

• (ii)

If the edges of $G_{\pi}$ partition into $m$ simple cycles of sizes $2s_{1},\ldots,2s_{m}$, then

[TABLE]

Proof.

Any loop-free non-crossing partition must have at least two blocks. When $t=2$, we may assume $k=2$ by Lemma 6 so that the only partition under consideration is $\{\{1\},\{2\}\}$, in which case $m=1$ and $s_{1}=1$. Lemma 12(i) allows us to verify the result in this case:

[TABLE]

Now consider $t>2$, and suppose the lemma has been established for every loop-free non-crossing partition on $t-1$ blocks. By Lemma 15, we may assume that $k$ satisfies $t\geq k/2+1$. Then for $\pi\in\Pi(k,t)$, the pigeonhole principle guarantees that $\pi$ contains a singleton block $\{j\}\in\pi$. By Lemma 13, we may assume $\{k\}\in\pi$. We proceed in cases:

Case I: $\pi(1)=\pi(k-1)$. We may apply Lemma 12(v) to obtain

[TABLE]

The restriction of $\pi\setminus\{k\}$ to $[k-2]$ results in a loop-free non-crossing partition $\pi^{\prime}$ of $[k-2]$ into $t-1$ blocks. Moreover, the above expression for $V_{n}(\pi)$ implies

[TABLE]

For (i), observe that if $G_{\pi}$ contains any odd cycles, then $G_{\pi^{\prime}}$ must also contain odd cycles. In this case, we may apply our induction hypothesis to $V_{n}(\pi^{\prime})$ to conclude $V_{n}(\pi)\rightarrow 0$. For (ii), the edges of $G_{\pi}$ partition into $m$ simple cycles of sizes $2s_{1},\ldots,2s_{m}$ with $s_{m}=1$, and so the edges of $G_{\pi^{\prime}}$ partition into $m-1$ simple cycles of sizes $2s_{1},\ldots,2s_{m-1}$. Then (8) and our induction hypothesis together imply

[TABLE]

Since $C_{0}=1$, this establishes (ii).

Case II: $\pi(1)\neq\pi(k-1)$. In this case, $\pi(k)$ necessarily resides in a cycle of length $\ell\geq 3$. Select representatives $k,j_{2},\ldots,j_{t}\in[k]$ with $\pi(k)=B_{1}$ and $\pi(j_{i})=B_{i}$ so that the vertices in the cycle are given by $B_{1},\ldots,B_{\ell}$. Then we may apply Lemma 12(iv) to obtain

[TABLE]

For $i,j\in[2,t]$, define new blocks

[TABLE]

and the corresponding partitions $\pi^{i}=\{B_{2}^{i},\ldots,B_{t}^{i}\}$, we have

[TABLE]

By Lemma 14, $V_{n}(\pi^{i})\rightarrow 0$ whenever $\pi^{i}$ is a crossing partition. Since $\pi^{i}$ is obtained from $\pi$ by merging blocks $B_{i}$ and $\{k\}$, we can argue as in the proof of Lemma 15 to conclude that $\pi^{i}$ is crossing if and only if $B_{i}$ and $\{k\}$ do not reside in the same simple cycle of $G_{\pi}$. Hence,

[TABLE]

where each $\pi^{i}$ is non-crossing for $2\leq i\leq\ell$. Both $\pi^{2}$ and $\pi^{\ell}$ contain loops, so $V_{n}(\pi^{2})=V_{n}(\pi^{\ell})=0$ by Lemma 12(ii). When $\ell=3$, this gives $V_{n}(\pi)\rightarrow 0$, as desired by (i). Supposing for the remainder that $\ell\geq 4$, we must still compute the limit of

[TABLE]

Observe that our cycle $\{B_{1},\ldots,B_{\ell}\}$ in $G_{\pi}$ of length $\ell$ corresponds to the two simple cycles in $G_{\pi^{i}}$ of $\{B_{2}^{i},\ldots,B_{i}^{i}\}$ and $\{B_{i}^{i},B_{i+1}^{i},\ldots,B_{\ell}^{i}\}$ with lengths $i-1$ and $\ell-i+1$ and share the cut vertex $B_{i}^{i}$. Moreover, all other simple cycles are identical between the two graphs.

If $\ell$ is odd, then for each $i\in[3,\ell-1]$, either $i-1$ or $\ell-i+1$ is odd, and so $G_{\pi^{i}}$ must have an odd cycle. Since each $\pi^{i}$ has $t-1$ blocks and an odd cycle, we can apply our induction hypothesis to conclude that each $V_{n}(\pi^{i})\rightarrow 0$ so that $V_{n}(\pi)\rightarrow 0$, as desired by (i). Suppose instead that $\ell$ is even, but $G_{\pi}$ has an odd cycle. This odd cycle is also contained in each $G_{\pi^{i}}$ for $i\in[3,\ell-1]$, and again we can apply our induction hypothesis to conclude that $V_{n}(\pi)\rightarrow 0$, thereby establishing (i).

Finally, for (ii), suppose that $\ell$ is even and that the edges of $G_{\pi}$ partition into $m$ cycles of lengths $2s_{1},\ldots,2s_{m}$ with $2s_{m}=\ell$. Notice that if $i-1$ is odd, then $G_{\pi^{i}}$ contains an odd cycle, and $V_{n}(\pi^{i})\rightarrow 0$. Since the contribution of these terms is negligible, we must compute the limit of

[TABLE]

The cycles of lengths $2s_{1},\ldots,2s_{m-1}$ are common to both $G_{\pi}$ and $G_{\pi^{2i+1}}$, while the cycle of length $\ell=2s_{m}$ in $G_{\pi}$ corresponds to two cycles of length $2i$ and $\ell-2i$ in $G_{\pi^{2i+1}}$. Applying our induction hypothesis, we have

[TABLE]

Reindexing and applying the convolution identity for Catalan numbers, we have

[TABLE]

Hence, $V_{n}(\pi)\rightarrow(-1)^{k/2-m}\cdot C_{s_{1}-1}\cdots C_{s_{m}-1}$, thereby establishing (ii). ∎

Proof of Lemma 7.

To prove (i), consider $\pi\in\Pi(k,t)\setminus\operatorname{EC}(k,t)$. Then either $\pi$ is crossing, $G_{\pi}$ contains a loop, or $G_{\pi}$ contains an odd cycle. If $\pi$ is crossing, then $V_{n}(\pi)\rightarrow 0$ by Lemma 14. If $G_{\pi}$ contains a loop, then $V_{n}(\pi)=0$ by Lemma 12(ii). If $\pi$ is loop-free and non-crossing but $G_{\pi}$ contains an odd cycle, then $V_{n}(\pi)\rightarrow 0$ by Lemma 16(i). This establishes (i). Finally, (ii) follows from applying both Lemmas 15 and 16(ii). ∎

Acknowledgments

MM and DGM were partially supported by AFOSR FA9550-18-1-0107. DGM was also supported by NSF DMS 1829955 and the Simons Institute of the Theory of Computing.