Anticoncentration and Berry–Esseen bounds for random tensors

Abstract

We obtain estimates for the Kolmogorov distance to appropriately chosen gaussians, of linear functions

$$\begin{aligned} \sum _{i\in [n]^d} \theta _i X_i \end{aligned}$$

of random tensors $\varvec{X}=\langle X_i:i\in [n]^d\rangle$ which are symmetric and exchangeable, and whose entries have bounded third moment and vanish on diagonal indices. These estimates are expressed in terms of intrinsic (and easily computable) parameters associated with the random tensor $\varvec{X}$ and the given coefficients $\langle \theta _i:i\in [n]^d\rangle$, and they are optimal in various regimes. The key ingredient—which is of independent interest—is a combinatorial CLT for high-dimensional tensors which provides quantitative non-asymptotic normality under suitable conditions, of statistics of the form

$$\begin{aligned} \sum _{(i_1,\dots ,i_d)\in [n]^d} \varvec{\zeta }\big (i_1,\dots ,i_d,\pi (i_1),\dots ,\pi (i_d)\big ) \end{aligned}$$

where $\varvec{\zeta }:[n]^d\times [n]^d\rightarrow \mathbb {R}$ is a deterministic real tensor, and $\pi$ is a random permutation uniformly distributed on the symmetric group $\mathbb {S}_n$. Our results extend, in any dimension d, classical work of Bolthausen who covered the one-dimensional case, and more recent work of Barbour/Chen who treated the two-dimensional case.

Appendix A. Proof of Lemma 7.3

The argument is an interesting and nontrivial instance of the Stein/Chen method of normal approximation via concentration (see, e.g., [4, Section 6]).

A.1. Recall that n, d are positive integers with $d\geqslant 2$ and $n\geqslant 4d^2$. We start by setting

$$\begin{aligned} \delta :=16\, \frac{\Lambda }{n} \ \ \ \text { and } \ \ \ \eta _\delta :=\frac{n}{4}\, \mathbb {E}\big [|\Xi _1-\Xi _1'|\cdot \min \{|\Xi _1-\Xi _1'|,\delta \}\, |\, \pi _1\big ]. \end{aligned}$$

(A.1)

Notice that, by (7.6), we have

$$\begin{aligned} \delta \geqslant \frac{n}{4} \, \mathbb {E}\big [|\Xi _1-\Xi _1'|^3\big ]. \end{aligned}$$

(A.2)

Using (7.5), (7.6), (A.2) and the fact that $\min (\alpha ,\beta )\geqslant \alpha - \frac{\alpha ^2}{4\beta }$ for every $\alpha \geqslant 0$ and $\beta >0$, we obtain that

$$\begin{aligned} \mathbb {E}[\eta _\delta ]&= \frac{n}{4}\, \mathbb {E}\big [|\Xi _1-\Xi _1'|\cdot \min \{|\Xi _1-\Xi _1'|,\delta \}\big ] \\&\geqslant \frac{n}{4}\, \mathbb {E}\big [(\Xi _1-\Xi _1')^2\big ] - \frac{n}{16\delta }\, \mathbb {E}\big [|\Xi _1-\Xi _1'|^3\big ] = 1- \frac{n^2}{16^2\Lambda }\, \mathbb {E}\big [|\Xi _1-\Xi _1'|^3\big ] \geqslant \frac{3}{4}. \nonumber \end{aligned}$$

(A.3)

Sublemma A.1 1

We have

$$\begin{aligned} \textrm{Var}(\eta _\delta ) \leqslant 2^9\delta ^2. \end{aligned}$$

(A.4)

We postpone the proof of Sublemma A.1 to the end of this appendix.

A.2. Now let $z\in \mathbb {R}$ be arbitrary. We will show that the probabilities ${\mathbb {P}}(z-|\Theta |\leqslant \Xi _1\leqslant z)$ and ${\mathbb {P}}\big (z\leqslant \Xi _1 \leqslant z+|\Theta |\big )$ are both upper-bounded by the quantity appearing in the right-hand-side of (7.20); clearly, this is enough to complete the proof. The argument is symmetric, and so we shall focus on bounding the probability ${\mathbb {P}}\big (z\leqslant \Xi _1 \leqslant z+|\Theta |\big )$.

Let $f_\Theta :\mathbb {R}\rightarrow \mathbb {R}$ be defined by

$$\begin{aligned} f_\Theta (x) :={\left\{ \begin{array}{ll} -\big (\frac{1}{2}|\Theta |+\delta \big ) &{} \text {if } x\leqslant z-\delta , \\ -\frac{1}{2}|\Theta |+x-z &{} \text {if } z-\delta<x<z+|\Theta |+\delta , \\ \frac{1}{2}|\Theta |+\delta &{} \text {if } x\geqslant z+|\Theta |+\delta . \end{array}\right. } \end{aligned}$$

(A.5)

By ($\mathcal {E}$4), the pair $(\pi _1,\pi _2)$ is exchangeable. Hence, by (7.1) and (7.19), we have

$$\begin{aligned} \mathbb {E}\big [(\Xi _1'-\Xi _1)\big (f_\Theta (\Xi _1') + f_{\Theta '}(\Xi _1)\big )\big ]=0 \end{aligned}$$

(A.6)

which in turn implies, after adding $2\,\mathbb {E}\big [(\Xi _1-\Xi _1')f_\Theta (\Xi _1)\big ]$ and multiplying by $\frac{n}{4}$, that

$$\begin{aligned} \frac{n}{2}\,&\mathbb {E}\big [(\Xi _1-\Xi '_1) f_\Theta (\Xi _1)\big ] \\&\quad = \frac{n}{4}\,\mathbb {E}\big [(\Xi _1'-\Xi _1)\big (f_{\Theta '}(\Xi _1) - f_{\Theta }(\Xi _1)\big )\big ] + \frac{n}{4}\, \mathbb {E}\big [(\Xi _1'-\Xi _1) \big (f_\Theta (\Xi _1') - f_{\Theta }(\Xi _1)\big )\big ]. \nonumber \end{aligned}$$

(A.7)

Moreover, by (7.4), we have

$$\begin{aligned} \frac{n}{2}\, \mathbb {E}\big [(\Xi _1'-\Xi _1)f_\Theta (\Xi _1)\big ] = \frac{n}{2}\, \mathbb {E}\big [ f_\Theta (\Xi _1)\, \mathbb {E}[\Xi _1'-\Xi _1\,|\,\pi _1]\big ] =\mathbb {E}\big [f_\Theta (\Xi _1)\Xi _1\big ]\nonumber \\ \end{aligned}$$

(A.8)

and so, by (7.3), the Cauchy–Schwarz inequality and the fact that $\Vert f_\Theta \Vert _{L_\infty }\leqslant \delta +\frac{1}{2}|\Theta |$,

$$\begin{aligned} \Big |\frac{n}{2}\,\mathbb {E}\big [(\Xi _1'-\Xi _1)f_\Theta (\Xi _1)\big ]\Big | \leqslant \delta \, \mathbb {E}\big [|\Xi _1|\big ] +\frac{1}{2}\,\mathbb {E}\big [|\Xi _1\Theta |\big ] \leqslant \delta + \frac{1}{2}\, \Vert \Theta \Vert _{L_2}. \end{aligned}$$

(A.9)

By the definition of $\Theta$ in (7.19), the triangle inequality and (7.7), we thus have

$$\begin{aligned} \Big |\frac{n}{2}\,\mathbb {E}\big [(\Xi _1'-\Xi _1)f_\Theta (\Xi _1)\big ]\Big | \leqslant \delta + e^{d} (2d)!\, \sum _{s=2}^{d}\sqrt{\frac{\beta _2}{n^s}}. \end{aligned}$$

(A.10)

Next observe that $\Vert f_\Theta - f_{\Theta '}\Vert _{L_{\infty }}\leqslant \frac{1}{2}\,\big ||\Theta |-|\Theta '|\big | \leqslant \frac{1}{2}\,|\Theta -\Theta '|$. Hence,

$$\begin{aligned}&\Big |\frac{n}{4}\,\mathbb {E}\big [(\Xi _1'-\Xi _1)(f_{\Theta '}(\Xi _1) - f_{\Theta }(\Xi _1))\big ]\Big | \leqslant \Big | \frac{n}{8}\, \mathbb {E}\big [|\Xi _1'-\Xi _1|\cdot |\Theta - \Theta '|\big ] \Big |\nonumber \\&\quad \leqslant \frac{n}{8}\, \Vert \Xi _1'-\Xi _1\Vert _{L_2} \Vert \Theta - \Theta '\Vert _{L_2}\nonumber \\ {}&{\mathop {\leqslant }\limits ^{(7.5),(7.19)}}\, \frac{\sqrt{n}}{4} \sum _{s=2}^{d}\Vert \Xi _s-\Xi _s'\Vert _{L_2} \leqslant 2\,d^2e^{d}(2d)!\, \sum _{s=2}^{d}\sqrt{\frac{\beta _2}{n^s}} \end{aligned}$$

(A.11)

where we have used the Cauchy–Schwarz inequality, the triangle inequality and (7.8).

From this point on, the proof proceeds exactly as in [3, Lemma 2.2]. More precisely, in order to estimate the second term of the right-hand-side of (A.7), by the fundamental theorem of calculus, we have

$$\begin{aligned} A:= & {} \frac{n}{4}\, \mathbb {E}\big [(\Xi _1'-\Xi _1) \big (f_\Theta (\Xi _1') - f_{\Theta }(\Xi _1)\big )\big ] = \frac{n}{4}\,\mathbb {E}\Big [(\Xi _1'-\Xi _1) \int _{0}^{\Xi _1'-\Xi _1} \!\! f'_\Theta (\Xi _1+t)\, dt\Big ] \nonumber \\= & {} \frac{n}{4}\, \mathbb {E}\left[ (\Xi _1'-\Xi _1) \int _{-\infty }^{+\infty } f'_\Theta (\Xi _1+t)\, \big (\textbf{1}_{[0,\Xi _1'-\Xi _1]}(t)-\textbf{1}_{[\Xi _1'-\Xi _1,0]}(t)\big )\, dt \right] \end{aligned}$$

(A.12)

with the convention that $\textbf{1}_{[\beta ,\alpha ]}$ is constantly 0 if $\alpha <\beta$. Also observe that, by (A.5), we have $f'_\Theta =\textbf{1}_{[z-\delta ,z+\delta +|\Theta |]}$ which implies that for every $t\in \mathbb {R}$,

• $(\Xi _1'-\Xi _1)\,f'_\Theta (\Xi _1+t)\, \big (\textbf{1}_{[0,\Xi _1'-\Xi _1]}(t)-\textbf{1}_{[\Xi _1'-\Xi _1,0]}(t)\big ) \geqslant 0$.

Therefore,

$$\begin{aligned} A \geqslant \frac{n}{4}\, \mathbb {E}\Big [\int _{|t|\leqslant \delta } (\Xi _1'-\Xi _1)\, \textbf{1}_{[z-\delta ,z+\delta +|\Theta |]}(\Xi _1+t)\, \big (\textbf{1}_{[0,\Xi _1'-\Xi _1]}(t)-\textbf{1}_{[\Xi _1'-\Xi _1,0]}(t)\big )\, dt\Big ].\nonumber \\ \end{aligned}$$

(A.13)

Moreover, for every $|t|\leqslant \delta$ we have

• $(\Xi _1'-\Xi _1)\, \big (\textbf{1}_{[0,\Xi _1'-\Xi _1]}(t)-\textbf{1}_{[\Xi _1'-\Xi _1,0]}(t)\big )\geqslant 0$ and

• $\textbf{1}_{[z-\delta ,z+\delta +|\Theta |]}(\Xi _1+t) \geqslant \textbf{1}_{[z,z+|\Theta |]}(\Xi _1)$,

and so, by (A.13),

$$\begin{aligned}&A \geqslant \frac{n}{4}\, \mathbb {E}\Big [ \textbf{1}_{[z,z+|\Theta |]}(\Xi _1)\, (\Xi _1'-\Xi _1) \int _{|t|\leqslant \delta } \big (\textbf{1}_{[0,\Xi _1'-\Xi _1]}(t)-\textbf{1}_{[\Xi _1'-\Xi _1,0]}(t)\big )\, dt\Big ] \\&\quad = \frac{n}{4}\, \mathbb {E}\big [ \textbf{1}_{[z,z+|\Theta |]}(\Xi _1)\, |\Xi _1'-\Xi _1|\cdot \min \big \{\delta ,|\Xi _1'-\Xi _1|\big \}\big ] \nonumber \\&\quad {\mathop {=}\limits ^{A.1}} \mathbb {E}\big [\textbf{1}_{[z,z+|\Theta |]}(\Xi _1)\,\eta _\delta \big ] \geqslant \mathbb {E}\big [\textbf{1}_{[z,z+|\Theta |]}(\Xi _1)\big ]\, \mathbb {E}[\eta _\delta ] - \big |\mathbb {E}\big [\textbf{1}_{[z,z+|\Theta |]}(\Xi _1)\big (\eta _\delta -\mathbb {E}[\eta _\delta ]\big )\big ]\big | \nonumber \\&\quad {\mathop {\geqslant }\limits ^{(A.3)}} \frac{3}{4}\,{\mathbb {P}}\big (z\leqslant \Xi _1\leqslant z+|\Theta |\big ) - \big |\mathbb {E}\big [\textbf{1}_{[z,z+|\Theta |]}(\Xi _1)\big (\eta _\delta -\mathbb {E}[\eta _\delta ]\big )\big ]\big |. \nonumber \end{aligned}$$

(A.14)

On the other hand, by Sublemma 1 and inequality $\alpha \beta \leqslant \frac{1}{2}\,(\alpha ^2+\beta ^2)$ applied for “$\alpha =\frac{1}{\sqrt{2}}\,\textbf{1}_{[z,z+|\Theta |]}(\Xi _1)$" and “$\beta = \sqrt{2}\,(\eta _\delta -\mathbb {E}[\eta _\delta ])$", we see that

$$\begin{aligned} \mathbb {E}\big [\textbf{1}_{[z,z+|\Theta |]}(\Xi _1)\big (\eta _\delta -\mathbb {E}[\eta _\delta ]\big )\big ]&\leqslant \frac{1}{2}\, \Big ( \frac{1}{2}\, {\mathbb {P}}\big (z\leqslant \Xi _1\leqslant z+|\Theta |\big ) + 2 \textrm{Var}(\eta _\delta ) \Big ) \\&\leqslant \frac{1}{4}\, {\mathbb {P}}\big (z\leqslant \Xi _1\leqslant z+|\Theta |\big ) + 2^9\delta ^2. \nonumber \end{aligned}$$

(A.15)

Thus, by (A.14) and (A.15), we have

$$\begin{aligned} A\geqslant \frac{1}{2}\, {\mathbb {P}}\big (z\leqslant \Xi _1\leqslant z+|\Theta |\big ) - 2^9\delta ^2. \end{aligned}$$

(A.16)

Invoking the choice of $\delta$ in (A.1) and the choice of A in (A.12) and combining (A.7), (A.10), (A.11) and (A.16), we conclude that

$$\begin{aligned} {\mathbb {P}}\big (z\leqslant \Xi _1\leqslant z + |\Theta |\big ) \leqslant 2^5\frac{\Lambda }{n}+2^{17} \frac{\Lambda ^2}{n^2} + 5d^2 e^d (2d)! \sum _{s=2}^{d}\sqrt{\frac{\beta _2}{n^s}}. \end{aligned}$$

(A.17)

1.1 A.3 Proof of Sublemma A.1

Let $\varvec{\zeta }:[n]^2\times [n]^2\rightarrow \mathbb {R}$ be defined by setting

$$\begin{aligned}&\varvec{\zeta }(i,j,p,q) :=|\varvec{\xi }_1(i,p) + \varvec{\xi }_1(j,q) - \varvec{\xi }_1(i,q) - \varvec{\xi }_1(j,p)| \ \times \\&\quad \times \min \big \{|\varvec{\xi }_1(i,p) + \varvec{\xi }_1(j,q) -\varvec{\xi }_1(i,q) - \varvec{\xi }_1(j,p)|, \delta \big \} \nonumber \end{aligned}$$

(A.18)

for every $i,j,p,q\in [n]$ with $i\ne j$ and $p\ne q$; otherwise, set $\varvec{\zeta }(i,j,p,q)=0$. Then observe that, by the choice of $\eta _{\delta }$ in (A.1), we have

$$\begin{aligned} \eta _\delta = \frac{n}{4}\, \mathbb {E}\big [ |\Xi _1-\Xi _1'| \cdot \min \big \{|\Xi _1-\Xi _1'|,\delta \big \}\, |\, \pi _1\big ] = \frac{1}{4n}\, \sum _{\begin{array}{c} i,j\in [n]\\ i\ne j \end{array}} \varvec{\zeta }\big (i,j,\pi _1(i),\pi _1(j)\big ).\nonumber \\ \end{aligned}$$

(A.19)

Also recall that, by ($\mathcal {E}$3), $\pi _1$ is uniformly distributed on $\mathbb {S}_n$. Thus, (A.19) asserts the random variable $\eta _\delta$ can be expressed as the Z-statistic of order two associated with the tensor $\varvec{\zeta }$ which has a rather special form.

The proof is based on a specific decomposition of this Z-statistic. This decomposition, also introduced by Barbour and Chen [3], is less symmetric than the decomposition described in Sect. 9 (yet it has enough symmetries in order to be computationally useful), but it has the advantage that its non-constant components are mean zero random variables. Of course, this property is very useful for computing the variance. For the convenience of the reader we shall briefly recall this decomposition; for more information we refer to [3].

Specifically, for every $i,j,p,q\in [n]$ set

$$\begin{aligned}{} & {} \varvec{\zeta }(i,j,\cdot ,q) :=\frac{1}{n-1}\sum _{\begin{array}{c} r\in [n]\\ r\ne q \end{array}} \varvec{\zeta }(i,j,r,q), \\ {}{} & {} \varvec{\zeta }(i,j,p,\cdot ) :=\frac{1}{n-1}\sum _{\begin{array}{c} r\in [n]\\ r\ne p \end{array}} \varvec{\zeta }(i,j,p,r), \\{} & {} \varvec{\zeta }(i,j,\cdot ,\cdot ) :=\frac{1}{n(n-1)} \sum _{\begin{array}{c} r,s\in [n]\\ r\ne s \end{array}} \varvec{\zeta }(i,j,r,s), \\{} & {} \varvec{\zeta }(i,\cdot ,p,\cdot ) :=\frac{1}{(n-1)^2} \sum _{\begin{array}{c} r,s\in [n]\\ r\ne i,s\ne p \end{array}} \varvec{\zeta }(i,r,p,s),\\{} & {} \! \varvec{\zeta }(i,\cdot ,\cdot ,\cdot ) :=\frac{1}{n(n-1)^2} \sum _{\begin{array}{c} r,s,\ell \in [n]\\ \ell \ne i,r\ne s \end{array}} \! \varvec{\zeta }(i,\ell ,r,s), \\{} & {} \varvec{\zeta }(\cdot ,j,\cdot ,q) :=\frac{1}{(n-1)^2} \sum _{\begin{array}{c} r,s\in [n]\\ r\ne j,s\ne q \end{array}} \! \varvec{\zeta }(r,j,s,q), \\{} & {} \varvec{\zeta }(\cdot ,j,\cdot ,\cdot ) :=\frac{1}{n(n-1)^2} \sum _{\begin{array}{c} r,s,\ell \in [n]\\ \ell \ne j,r\ne s \end{array}} \!\varvec{\zeta }(\ell ,j,r,s), \\{} & {} \varvec{\zeta }(\cdot ,\cdot ,\cdot ,\cdot ) :=\frac{1}{n^2(n-1)^2} \sum _{\begin{array}{c} r,s,k,\ell \in [n]\\ r\ne s,k\ne \ell \end{array}} \! \varvec{\zeta }(r,s,k,\ell ). \end{aligned}$$

Then, for every $i,j,p,q\in [n]$ we define

$$\begin{aligned} \varvec{\zeta }_D(i,j,p,q)&:=\varvec{\zeta }(i,j,p,q) - \varvec{\zeta }(i,j,\cdot ,q)-\varvec{\zeta }(i,j,p,\cdot ) +\varvec{\zeta }(i,j,\cdot ,\cdot ) \end{aligned}$$

(A.20)

$$\begin{aligned} \varvec{\zeta }_D(i,\cdot ,p,\cdot )&:=\varvec{\zeta }(i,\cdot ,p,\cdot ) - \varvec{\zeta }(i,\cdot ,\cdot ,\cdot ) \end{aligned}$$

(A.21)

$$\begin{aligned} \varvec{\zeta }_D(\cdot ,j,\cdot ,q)&:=\varvec{\zeta }(\cdot ,j,\cdot ,q) - \varvec{\zeta }(\cdot ,j,\cdot ,\cdot ). \end{aligned}$$

(A.22)

Notice that for every $i,p,q\in [n]$ we have $\varvec{\zeta }_D(i,i,p,q)=0$; moreover, for every $i,j\in [n]$,

$$\begin{aligned} \sum _{\begin{array}{c} p,q\in [n]\\ p\ne q \end{array}}\varvec{\zeta }_D(i,j,p,q) = 0, \ \ \ \ \sum _{p=1}^{n}\varvec{\zeta }_D(i,\cdot ,p,\cdot )=0 \ \ \text { and } \ \ \sum _{p=1}^{n}\varvec{\zeta }_D(\cdot ,i,\cdot ,p)=0.\nonumber \\ \end{aligned}$$

(A.23)

By (A.19) and the previous definitions, we may decompose $\eta _\delta$ as

$$\begin{aligned} \eta _\delta&= \frac{1}{4n} \sum _{\begin{array}{c} i,j\in [n]\\ i\ne j \end{array}} \varvec{\zeta }_D\big (i,j,\pi _1(i),\pi _1(j)\big ) + \frac{n-1}{4n}\sum _{i=1}^{n}\varvec{\zeta }_D \big (\cdot ,i,\cdot ,\pi _1(i)\big ) \ \\&\quad + \frac{n-1}{4n}\sum _{i=1}^{n}\varvec{\zeta }_D\big (i,\cdot ,\pi _1(i),\cdot \big ) +\frac{n-1}{4}\varvec{\zeta }(\cdot ,\cdot ,\cdot ,\cdot ). \nonumber \end{aligned}$$

(A.24)

Invoking (A.23) and the fact that $\pi _1$ is uniformly distributed on $\mathbb {S}_n$ we see, in particular, that $\mathbb {E}[\eta _\delta ] = \frac{n-1}{4}\varvec{\zeta }(\cdot ,\cdot ,\cdot ,\cdot )$. Therefore,

$$\begin{aligned} \textrm{Var}(\eta _\delta )&\leqslant \frac{3}{2^4n^2}\, \mathbb {E}\Big [\Big (\sum _{\begin{array}{c} i,j\in [n]\\ i\ne j \end{array}} \varvec{\zeta }_D\big (i,j,\pi _1(i),\pi _1(j)\big )\Big )^2\Big ] \, \\&\quad + \frac{3}{2^4}\, \mathbb {E}\Big [\Big (\sum _{i=1}^{n}\varvec{\zeta }_D \big (\cdot ,i,\cdot ,\pi _1(i)\big )\Big )^2\Big ] + \frac{3}{2^4}\, \mathbb {E}\Big [\Big (\sum _{i=1}^{n} \varvec{\zeta }_D\big (i,\cdot ,\pi _1(i),\cdot \big )\Big )^2\Big ] \nonumber \\&:=E_1 +E_2 +E_3. \nonumber \end{aligned}$$

(A.25)

Finally, using the Cauchy–Schwarz inequality, (A.23) and arguing²⁰ as in Sect. 5.6, it is not hard to verify that

$$\begin{aligned} E_1\leqslant \Big (\frac{2^63}{n}+2^63\Big )\delta ^2, \ \ \ \ E_2\leqslant 2^3 3\delta ^2 \ \ \text { and } \ \ E_3\leqslant 2^3 3\delta ^2. \end{aligned}$$

(A.26)

By (A.25), (A.26) and the fact that $n\geqslant 3$, we conclude that (A.4) is satisfied.