Feature Purification: How Adversarial Training Performs Robust Deep Learning

Zeyuan Allen-Zhu, Yuanzhi Li

Introduction

Large scale neural networks have shown great power to learn from a training data set, and generalize to unseen data sampled from similar distributions for applications across different domains . However, recent study has discovered that these trained large models are extremely vulnerable to small “adversarial attacks” . It has been discovered that small perturbations to the input– often small enough to be invisible to humans– can create numerous errors in prediction. Such slightly perturbed inputs are often referred to as “adversarial examples”.

Since the original discovery of “adversarial examples”, a large body of works have been done emphasizing how to improve the robustness of the deep learning models against such perturbations . One seminal approach is called adversarial training , where one iteratively computes adversarial examples from the training examples, and then retrain the model with these adversarial examples instead of the original examples (a.k.a. the clean examples). This approach was reported in as the only approach that can defend against carefully designed adversarial attacks, and many follow-up works are built upon it .

However, despite the great empirical success on improving the robustness of neural networks over various data sets, the theory of the adversarial examples is much less developed. In particular, we found that the following fundamental questions remain largely unaddressed:

To answer these questions, one sequence of theoretical works try to explain the existence of adversarial examples using the high dimensional nature of the input space and the over-fitting behavior due to the sample size and sample noise , and treat adversarial training from the broader view of min-max optimization . However, recent observations indicate that these adversarial examples can also, and arguably often, arise from features (those that do generalize) rather bugs (those that do not generalize due to effect of poor statistical concentration). To the best of our knowledge, all existing works studying adversarial examples either (1) apply generally to the case of arbitrarily unstructured functions ff and only consider adversarial examples statistically, or (2) apply to a structured setting but only involving linear learners. These theoretical works, while shedding great lights to the study of adversarial examples, do not yet give concrete mathematical answers to the following questions regarding the specific hidden-layer structure of neural networks:

What are the features (i.e. the hidden weights) learned by the neural network via clean training (i.e., over the original data set)? Why are those features “non-robust”?

What are the differences between the features learned by clean training vs adversarial training (i.e., over a perturbed data set consisting of adversarial examples)?

Why do adversarial examples for a network transfer to other independently-trained networks?

Before going into the above questions regarding robustness, it is inevitable to first study what the features are when learned by a neural network during clean training. Theoretical studies are also limited in this direction. Most of existing works (1) only focus on the case when the training data is spherical Gaussian , and some of them require heavy initialization using tensor decomposition, which might fail to capture the specific structure of the input and the property of a random initialization; or (2) only consider the neural tangent kernel regime, where the neural networks are linearized so the features are not learned (they stay at random initialization) .

In this paper, we present a new routine that enables us to formally study the learned features (i.e. the hidden weights) of a neural network, when the inputs are more naturally structured than being Gaussians. Using this routine, we give, to the best of our knowledge, the first theoretical result towards answering the aforementioned fundamental questions of adversarial examples, for certain neural networks with ReLU activation functions.

Our results. We prove, for certain binary classification data set, when we train a two-layer ReLU neural network using gradient descent,Our theory extends to stochastic gradient descent (SGD) at the expense of complicating notations. starting from random initialization,

Given polynomially manly training examples, in polynomially many iterations, the neural network will learn well-generalizing features for the original data set, and the learned network will have close-to-perfect prediction accuracy for the test data sampled from the same distribution.

However, even with a weight-decay regularizer to avoid over-fitting, even with infinitely many training data, and even when super-polynomially many iterations are used to train the neural network to convergence, the learned network still has near-zero robust accuracy against small-norm adversarial perturbations to the data. In other words, those provably well-generalizing features on the original data set are also provably non-robust to adversarial perturbations to the datas, so they cannot be due to having too few training samples .

Adversarial training, using perturbation algorithms such as Fast Gradient Method (FGM) , can provably and efficiently make the learned neural network achieve near-perfect robust accuracy, against even the worst-case norm-bounded adversarial perturbations, using a principle we refer to as “feature purification”. We illustrate “feature purification” in Figure 1 by an experiment, and explain it in mathematical terms next.

Feature purification: How adversarial training can perform robust deep learning. In this work, we also give precise, mathematical characterizations on the difference between learned features by clean training versus adversarial training in the aforementioned setting, leading to (to our best knowledge) the first theory of how, in certain learning tasks using ReLU neural networks, the provably non-robust features after clean training can be “robustified” via adversarial training.

We emphasize that prior theoretical works mainly study adversarial examples in the context of linear models (such as linear regression, linear regression over prescribed feature mappings, or the neural tangent kernels). In those models, the features are not trained, so adversarial training only changes the weights associated with the linear combination of these features, but not the actual features themselves.

In contrast, this paper develops a theory showing that how, over certain learning tasks, adversarial training can actually change the features of certain neural networks to improve their robustness. We abstract this feature change in our setting into a general principle that we call feature purification, which although we only prove it for two-layer ReLU networks (see Theorem thm:intro:cleansa+Theorem thm:adv1_sc), we empirically observe that it occurs more generally to real-world, deep neural networks on real-world data sets. We sketch its high-level idea as follows.

Mathematically, as a provisional step to measure of change of features in a network, let us use (1) wi(0)w_{i}^{(0)} to denote the weight vector of the ii-th neuron at initialization, (2) wiw_{i} to denote its weight after clean training, and (3) wi′w^{\prime}_{i} to denote its weight after adversarial training (using wiw_{i} as initialization). The “feature purification” principle, in math, says if we use θ(z,z′):=∣⟨z,z′⟩∣∥z∥2∥z′∥2\theta(z,z^{\prime}):=\frac{|\langle z,z^{\prime}\rangle|}{\|z\|_{2}\|z^{\prime}\|_{2}} as a provisional measure of the correlation between “features”, then (see Figure 2 for real-life experiments):

for most neurons: θ(wi(0),wi),θ(wi(0),wi′)≤c\theta(w^{(0)}_{i},w_{i}),\theta(w^{(0)}_{i},w^{\prime}_{i})\leq c for a small constant cc (such as 0.20.2);

for most neurons: θ(wi,wi′)≥C\theta(w_{i},w^{\prime}_{i})\geq C for a large constant CC (such as 0.80.8); and

for most pairs of different neurons: θ(wi,wj)≤c\theta(w_{i},w_{j})\leq c for a small constant cc (such as 0.20.2).

In words, this says both clean training and adversarial training discover hidden weights wi,wi′w_{i},w^{\prime}_{i} that are fundamentally different from initialization wi(0)w^{(0)}_{i}. However, since wiw_{i} and wi′w^{\prime}_{i} are close, clean training must have already discovered a big portion of the robust features, and adversarial training merely needs to “purify” some small part of each original feature. In this paper:

we prove this feature purification principle in the case of two-layer ReLU neural networks over certain data sets, with c=o(1)c=o(1) and C=1−o(1)C=1-o(1) (see Theorem thm:intro:cleansa+Theorem thm:adv1_sc); and

we provide empirical evidence that this feature purification principle holds also for deep neural networks used in real-life datasets (see Figure 2 as well as other experiments in the paper).

Why clean training learns non-robust features? Which part of the features are “purified” during adversarial training? In our setting, we also give mathematical characterizations of where the “non-robust” part of each feature comes from during clean training.

As we shall formally discuss in Section 6.2, training algorithms such as gradient descent will, at every step, add to the current parameters a direction that maximally correlates with the labeling function on average. For two-layer ReLU networks, we prove that such simple correlations will accumulate, in each neuron, a small part of its weight that correlates with the average of the training data, and we refer to it as the dense mixture (see Theorem 5.2). However, under natural assumptions of the data such as the sparse coding model — namely inputs come from sparse combinations of hidden dictionary words/vectors— such dense mixtures cannot have high correlation with any individual, clean example. Thus, even with these “dense mixtures” in the features, the network can still generalize well on the original data set. However, we show that these portions of the features are extremely vulnerable to small, adversarial perturbations along the “dense mixture” directions. As a result, one of the main goals of adversarial training, as we show, is to purify the neurons by removing such dense mixtures. This is the supporting theory behind our feature purification principle, as we also measure and verify it empirically in the experiment section.

We believe our result extends the reach of traditional learning theory, where often statistical properties of the model (such as generalization, etc.) is separate from optimization (i.e., how the models are trained). However, to understand adversarial examples in deep learning, one needs to admit that well-generalizing and adversarially robust neural networks do exist (and can even be found efficiently using adversarial training), thus it is also a global optimal solution of the clean training objective. It is rather a property of traditional clean training process using SGD which biases the network towards learning non-robust networks as another global optimal solution of the training objective.

Moreover, in our setting, these dense mixtures in the hidden weights of the network come from the sparse coding structure of the data and the gradient descent algorithm. It is rather independent of the random initialization of the neural network. Thus, we prove that, at least in our scenario, adversarial examples for one network do transfer to other independently trained ones.

Our contribution to computation complexity. We also prove a lower bound that, for the same sparse coding data model, even when the original data is linearly-separable, any linear classifier, any low-degree polynomial, or even the corresponding neural tangent kernel (NTK) of our studied two-layer neural network, cannot achieve meaningful robust accuracy (although they can easily achieve high clean accuracy). Together with our upper bound, we have shown that using a higher-complexity model (such as a two-layer neural network with ReLU activation, comparing to NTK) can in fact achieve better robustness against adversarial perturbations. Thus, our theory strongly supports the experimental finding in , where experts have noticed that robustness against adversarial examples requires a model with higher complexity. The main intuition is that low-complexity models, including the neural tangent kernel, lacks the power to zero out low magnitude signals to improve model robustness, as illustrated in Figure 3 and Section 3.

Our experimental contributions. We present quite a few experimental results supporting our theory. We find that our sparse coding model can indeed capture real-world data to certain degree, and our principle of feature purification also holds for architectures such as AlexNet and ResNet. We showed during clean training, how the features can emerge from random initialization by wining the “lottery tickets”, as predicted by our theory. Perhaps most importantly, we confirmed through experiments that:

Adversarial training using adversarial examples indeed purify “dense mixtures” in practice.

To gain robust accuracy, it suffices to remove such “dense mixtures” (via a low-rank update).

We present our experiments following each of the theorem statements accordingly. We also include a whole Section 8 for more detailed experiments.

Adversarial examples: Empirical study. Since the seminal paper shows the existence of small adversarial perturbations to change the prediction of the neural networks, many empirical studies have been done to make the trained neural networks robust against perturbations (and we refer to the citations therein). The recent study shows that the seminal approach of adversarial training is the most effective way to make the neural networks robust against adversarial perturbations.

Adversarial examples: Theoretical study. Existing theories mostly explain the existence of adversarial examples as the result of finite-sample data set over-fitting to high-dimensional learning problems . Later, it is discovered by Ilyas et al. that well-generalizing features can also be non-robust. Other theories focus on the Fourier perspective of the robustness , showing that adversarial training might be preventing the network from learning the high frequency signals of the input image. Our theoretical work is fundamentally different from the aspect of poor statistical concentration over finite-sample data set, and our Theorem 5.1 and Theorem 5.3 strongly supports that a well-trained, well-generalizing neural network can still be non-robust to adversarial attacks.

Other theories about adversarial examples focus on how adversarial training might require more training data comparing to clean training , and might decrease clean training accuracy . The works by focus on how adversarial training can be performed efficiently in the neural tangent kernel regime. The purpose of these results are also fundamentally different than ours.

Sparse coding (data) model. We use a data model called sparse coding, which is a popular model to model image, text and speech data . There are many existing theoretical works studying algorithm for sparse coding , however, these algorithms share little similarity to training a neural network.

The seminal work by Arora et al. provides a neurally-plausible algorithm for learning sparse coding along with other works using alternative minimization . However, all of these results require a (carefully picking) warm start, while our theory is for training a neural network starting from random initialization.

Threshold degree and kernel lower bound. We also provide, to the best of our knowledge, the first example when the original classification problem is learnable using a linear classifier but no low-degree polynomial can learn the problem robustly against small adversarial perturbations. Yet, the high-complexity neural networks can provably, efficiently and robustly learn the concept class. The lower bound for the classification accuracy using low-degree polynomials has been widely studied as the (approximate) threshold degree of a function or the sign-rank of a matrix . Our paper give the first example of a function with high (approximate) robust threshold degree, yet efficiently and robustly learnable by training a ReLU neural network using gradient descent.

Other related works prove lower bounds for kernel method in the regression case . Generally speaking, such lower bounds are about the actual (approximate) degree of the function, instead of the (approximate) threshold degree. It is well know that for general functions, the the actual degree can be arbitrary larger than the threshold degree.

Preliminaries

We assume the hidden vector zz is “sparse”, in the following sense: for k≤d0.499k\leq d^{0.499}, we have:

The coordinates of zz are independent, symmetric random variables, such that ∣zi∣∈{0}∪[1k,1]|z_{i}|\in\{0\}\cup[\frac{1}{\sqrt{k}},1]. Moreover,

Under Assumption 2.1, w.h.p., ∥z∥0=Θ(k)\|z\|_{0}=\Theta(k) is a sparse vector.

We study the simplest binary-classification problem, where the labeling function is linear over the hidden vector zz:

For simplicity, we assume ∀i∈[D],∣wi⋆∣=Θ(1)\forall i\in[D],|w^{\star}_{i}|=\Theta(1), so all the coordinates of zz have relatively equal contributions. Our theorems extend to other w⋆w^{\star} at the expense of complicating notations.

Remark on sparse coding. The sparse coding model is very natural and is widely used to model image, text and speech data . There certainly exist (provable) algorithms for dictionary learning based on sum of squares, linear programming , but they do not shed light on the training process of neural networks. Even the neural algorithm for sparse coding is still far away from training a neural network using SGD or its variants. The main point of this paper is not to show neural networks can do sparse coding. Instead, our main point is to distinguish the adversarial training and the clean training processes of neural networks using the sparse-coding model as a bridging tool.

Noise model. We have allowed the inputs x=Mz+ξx=\mathbf{M}z+\xi to incorporate a noise vector ξ\xi. Our lower bounds hold even when there is no noise (ξ=0\xi=0). Our upper bound theorems not only apply to ξ=0\xi=0, but more generally to “gaussian noise plus spike noise”:

the noise ξ\xi can be of Euclidean norm O(σx)O(\sigma_{x}), larger than the signal ∥Mz∥2≈1\|\mathbf{M}z\|_{2}\approx 1; and

the spike noise ξi′′\xi^{\prime\prime}_{i} can be 1k0.501\frac{1}{k^{0.501}} which is the maximum possible (because ziz_{i} can be 1k0.5\frac{1}{k^{0.5}}).

Clean and robust error. The goal of clean training is to learn a model ff so that sign(f(x))\mathsf{sign}(f(x)) is as close to yy as possible. We define the classification error on the original data set as:

Warmup Intuitions

Linear learners are not robust. Given the setting of the data set, one direct approach is to use (the sign of) a linear classifier f(x)=⟨w⋆,M⊤x⟩f(x)=\langle w^{\star},\mathbf{M}^{\top}x\rangle to predict the label of xx. There are two issues of using such a classifier:

When σx\sigma_{x} is as large as Θ(1)\Theta(1), such classifier can not even classify xx in good clean accuracy. Recall f(x)=⟨w⋆,M⊤x⟩=⟨w⋆,z⟩+⟨Mw⋆,ξ⟩f(x)=\langle w^{\star},\mathbf{M}^{\top}x\rangle=\langle w^{\star},z\rangle+\langle\mathbf{M}w^{\star},\xi\rangle. By our assumption, typically ∣⟨w⋆,z⟩∣=O(1)|\langle w^{\star},z\rangle|=O(1) and ⟨Mw⋆,ξ′⟩∼N(0,Θ(σx2))\langle\mathbf{M}w^{\star},\xi^{\prime}\rangle\sim\mathcal{N}(0,\Theta(\sigma_{x}^{2})). Thus, when σx≥Θ(1)\sigma_{x}\geq\Theta(1), noise could be much larger than signal, and this linear classifier cannot be used to classify xx correctly. In this case, actually no linear classifier (or even constant-degree polynomials One may think that using for example degree-3 polynomial ∑iwi⋆⟨Mi,x⟩3\sum_{i}w^{\star}_{i}\langle\mathbf{M}_{i},x\rangle^{3} can reduce the level of noise, but due to the diversity in the value of ziz_{i} when zi≠0z_{i}\not=0, one must use something close to linear when ∣zi∣|z_{i}| is large. Applying Markov brothers’ inequality, one can show the low-degree polynomial must be close to a linear function.) can give meaningful clean accuracy.

High-complexity models are more robust. Another choice to learn the labeling function is to use a higher-complexity model f(x)=∑i∈[d]wi⋆⟨Mi,x⟩\mathds1∣⟨Mi,x⟩∣≥12kf(x)=\sum_{i\in[d]}w^{\star}_{i}\langle\mathbf{M}_{i},x\rangle\mathds{1}_{|\langle\mathbf{M}_{i},x\rangle|\geq\frac{1}{2\sqrt{k}}}. Here, the “complexity” of ff is much higher because an indicator function is used.One concrete measure of “higher complexity” is that ff cannot be well-approximated by low degree polynomial. Since ⟨Mi,x⟩=zi+⟨Mi,ξ⟩\langle\mathbf{M}_{i},x\rangle=z_{i}+\langle\mathbf{M}_{i},\xi\rangle, by our noise model, as long as the signal zi≠0z_{i}\not=0 is non-zero, ∣⟨Mi,x⟩∣≥12k|\langle\mathbf{M}_{i},x\rangle|\geq\frac{1}{2\sqrt{k}} with high probability. Thus, this f(x)f(x) is equal to the true labeling function ⟨w⋆,z⟩\langle w^{\star},z\rangle w.h.p. over the original data set, so is (much) more robust to noise comparing to linear models.

To sum up, higher-complexity models (such as those using ReLU) have the power to zero out low-magnitude signals to improve adversarial robustness, as illustrated in Figure 3.

Learning robust classifier using neural network. Motivated by the above discussions between linear vs. high-complexity models, our goal is to show that a two-layer neural networks can (after adversarial training) learn a robust function f(x)f(x) such as

Here, ReLU(y)=max⁡{y,0}\mathsf{ReLU}(y)=\max\{y,0\} is the ReLU function and b≈12kb\approx\frac{1}{2\sqrt{k}}. In this paper, we present a theorem stating that adversarial training of a (wlog. symmetric) two-layer neural network can indeed recover a neural network of this form. In other words, after adversarial training, the features learned by the hidden layer of a neural network can indeed form a basis (namely, M1,…,Md\mathbf{M}_{1},\dots,\mathbf{M}_{d}) of the input xx where the coefficients are sparse. We also present a theorem showing why, clean training will not learn this robust function. We also verify experimentally that the features learned by the first layer of AlexNet (after adversarial training) indeed form a sparse basis of the images, see Figure 4.

Learner Network and Adversarial Training

In this paper we consider a simple, two layer (symmetric) We assume the neurons are symmetric (i.e., with (wi,−wi)(w_{i},-w_{i}) pairs) to simplify proofs. neural network with ReLU activation.

To simplify analysis, we fix ai=1a_{i}=1 throughout the training. We use wi(t)w_{i}^{(t)} to denote the hidden weights at time tt, and use ft(w;x,ρ)f_{t}(w;x,\rho) to denote the network at iteration tt

Given a training set Z={xj,yj}j∈[N]\mathcal{Z}=\{x_{j},y_{j}\}_{j\in[N]} together with one sample of pre-activation noise ρ(j)\rho^{(j)} for each (xj,yj)(x_{j},y_{j}), we define

In our case, the (clean, population) classification error at iteration tt is

At initialization, we let wi(0)∼N(0,σ02I)w_{i}^{(0)}\sim\mathcal{N}\left(0,\sigma_{0}^{2}\mathbf{I}\right) for σ0=1\poly(d)\sigma_{0}=\frac{1}{\poly(d)} and let b(0)=Θ(σ0log⁡d)b^{(0)}=\Theta(\sigma_{0}\sqrt{\log d}). When near initialization, we manually increase the bias b(t+1)=b(t)+ηBb^{(t+1)}=b^{(t)}+\eta\mathfrak{B} where B=cbd\mathfrak{B}=\frac{c_{b}}{d} for some small constant cb>0c_{b}>0— this corresponds to the “lottery ticket winning” phase to be discussed later in Section 6.1; and whenever b(t)b^{(t)} reaches 1k0.5001\frac{1}{k^{0.5001}} we set B=0\mathfrak{B}=0— in this phase, the neurons that have won the “lottery ticket” will keep winning and grow significantly, to be discussed in Section 6.2.

We also choose pre-activation noise σρ(t)=b(t)log⁡d⋅Θ((log⁡log⁡log⁡d)3)\sigma_{\rho}^{(t)}=\frac{b^{(t)}}{\sqrt{\log d}}\cdot\Theta((\log\log\log d)^{3}) for t≤Ta=1\poly(d)ηt\leq T_{\mathsf{a}}=\frac{1}{\poly(d)\eta}, and σρ(t)=b(t)log⁡d⋅Θ((log⁡log⁡log⁡d)3)\sigma_{\rho}^{(t)}=\frac{b^{(t)}}{\log d}\cdot\Theta((\log\log\log d)^{3}) for t>Tat>T_{\mathsf{a}}. The explicit choices of B\mathfrak{B} and TaT_{\mathsf{a}} are given in the proofs.

2 Adversarial Training

We state the adversarial training algorithm in Algorithm 2. It takes as input a perturbation algorithm AA, and repeatedly applies gradient descent over a perturbed data set (that comes from the original data set plus the perturbation given by AA). Formally,

In contrast, the empirical robust classification error against algorithm AA is

Our upper bound theorems apply to all perturbation algorithms under Definition 4.2, and gives small empirical robust error Er^\widehat{\mathcal{E}^{r}}. To obtain small (true) robust error Er\mathcal{E}^{r}, as we shall see, one can for instance let AA be the fast gradient method (FGM).

Initialized from clean training. In this paper, we assume adversarial training (i.e., Algorithm 2) is initialized from a network that is already clean-trained. In contrast, in practice, adversarial training usually begins directly with random initialization. We remark here that:

First, in practice, adversarial training from a clean-trained initialization performs no worse than from a random initialization, see Table 1 on Page 1. In fact, it is sometimes even beneficial to begin with clean training and gradually switch to adversarial training (see e.g. ).

Second, to prove our main conceptual message— feature purification— it is convenient to start from a clean-trained model, and then try to understand which part of the features are changed after robust training. Since neural nets have lots of equivalent transformations that are not very well-understood (even in two-layer case), if we adversarially train it from random initialization, then it is theoretically very hard to quantify how it is related to another clean-trained model learned from random initialization (since we need to understand all the invariants).Even if one performs clean/adversarial training from the same random initialization, the additional randomness in SGD may quickly make the two models diverge from each other.

Yet, our theory still gives support to what happens in adversarial training from random initialization. As we shall prove, this “o(1)o(1) feature change” comes from dense mixtures directions (see Theorem 5.2). Thus, adversarial training from random initialization should directly avoid learning such dense mixtures, as opposed to first learning them (by clean training) and then forgetting (by adversarial training). We illustrate this in Figure 9.

Statements and Explanations of Our Main Results

There exists an absolute constants C,c>0C,c>0 such that for every constant c0∈(0,c]c_{0}\in(0,c], every dd and mm with m=d1+c0m=d^{1+c_{0}}, given N≥Ω(dC)N\geq\Omega(d^{C}) many training data, for every random initialization weight σ0=1\poly(d)\sigma_{0}=\frac{1}{\poly(d)}, for every learning rate \eta\in\big{(}0,\frac{1}{\Omega(d^{C})}\big{]}, if we define Tc:=Θ(d1.01η)T_{\mathsf{c}}:=\Theta(\frac{d^{1.01}}{\eta}), then for every Tf∈[Tc,dlog⁡d/η]T_{\mathsf{f}}\in[T_{\mathsf{c}},d^{\log d}/\eta], the following holds with high probability. The network ftf_{t} with hidden weights {wi(t)}i∈[m]\{w_{i}^{(t)}\}_{i\in[m]} learned by clean training Algorithm 1 satisfies:

Global feature learning: for every t∈[Tc,Tf]t\in[T_{\mathsf{c}},T_{\mathsf{f}}],

Clean training has good clean accuracy: for every t∈[Tc,Tf]t\in[T_{\mathsf{c}},T_{\mathsf{f}}],

Clean training is not robust to small adversarial perturbations: for every t∈[Tc,Tf]t\in[T_{\mathsf{c}},T_{\mathsf{f}}], every τ≥1k0.5+10c\tau\geq\frac{1}{k^{0.5+10c}}, using perturbation δ=−τyMw⋆∥Mw⋆∥2\delta=-\tau\frac{y\mathbf{M}w^{\star}}{\|\mathbf{M}w^{\star}\|_{2}} (which does not depend on ftf_{t}),

Theorem 5.1 indicates that in our setting, clean training of the neural network has good clean accuracy but terrible robust accuracy. Such terrible robust accuracy is not due to over-fitting, as it holds even when a super-polynomially many iterations and infinitely many training examples are used to train the neural network. In the next theorem, we give a precise characterization of what the hidden weights {wi}\{w_{i}\} are after clean training, and why they are not robust.

For every neuron i∈[m]i\in[m], there is a fixed subset Ni\mathcal{N}_{i} of size ∣Ni∣=O(1)|\mathcal{N}_{i}|=O(1) such that, for every t∈[Tc,dlog⁡d/η)t\in[T_{\mathsf{c}},d^{\log d}/\eta),

where (1) ∣βi,j∣<kd1−c|\beta_{i,j}|<\frac{k}{d^{1-c}} for some small constant c∈[0,0.001]c\in[0,0.001], and (2) for at least Ω(d)\Omega(d) many neurons i∈[m]i\in[m], it satisfies ∣Ni∣=1|\mathcal{N}_{i}|=1 and αi,j>d−c\alpha_{i,j}>d^{-c}. Moreover,

Theorem 5.2 says that each neuron wiw_{i} will learn constantly many (allegedly large) components in the directions \big{\{}\mathbf{M}_{j}\colon j\in\mathcal{N}_{j}\big{\}}, and its components in the remaining directions \big{\{}\mathbf{M}_{j}\colon j\not\in\mathcal{N}_{j}\big{\}} are all small. We emphasize that the sets Ni\mathcal{N}_{i} are independent of tt but are solely determined by random initialization. In other words, for each neuron ii, which setset Ni\mathcal{N}_{i} it “wins” is completely determined by the “lottery ticket” (its random initialization). We discuss this in more details in Section 6.1.

Furthermore, Theorem 5.2 shows that instead of learning the pure, robust features {Mj}j∈[d]\{\mathbf{M}_{j}\}_{j\in[d]}, intuitively, ignoring the small dcd^{c} factors, focusing only on those neurons with ∣Ni∣=1|\mathcal{N}_{i}|=1, and assuming for simplicity all the βi,j\beta_{i,j}’s are of similar (positive) magnitude, then, clean training will learn neurons:

Feature purification: mathematical reasoning. Eq. (5.1) says that, after clean training, the neural network will be able to learn a big portion of the robust feature, Θ(1)Mj\Theta(1)\mathbf{M}_{j}, plus some small dense mixture v=\sum_{j^{\prime}\not=j}\big{[}\Theta\left(\frac{k}{d}\right)w_{j^{\prime}}^{\star}\mathbf{M}_{j^{\prime}}\big{]}. In our sparse coding model, each xx is of form x=Mz+ξx=\mathbf{M}z+\xi, where zz is a sparse vector and ξ\xi is the noise. One critical observation is that such dense mixture vv has low correlation with almost all inputs xx from the original distribution, so it has negligible effect for clean accuracy. However, such dense mixture is extremely vulnerable to small but dense adversarial perturbations of the input along this direction vv, making the model non-robust.

As we point out, such “dense adversarial perturbation” directions do not exist in the original data.One can try to add these dense mixtures directly to the training data set, which we conjecture to be similar to the approach in Thus, one has to rely on adversarial training to remove dense mixtures to make the model robust. This is the main spirit of our feature purification principle, and we illustrate it in Figure 6.

Where does dense mixture come from? We shall explain in more details in Section 6.2, but at a high level, in each iteration, the gradient ∇Obj\nabla\mathbf{Obj} will bias towards the direction that correlates with the labeling function y=sign(⟨w⋆,z⟩)y=\mathsf{sign}(\langle w^{\star},z\rangle); and since x=Mz+ξx=\mathbf{M}z+\xi in our model, such direction should be Mw⋆=∑jwj⋆Mj\mathbf{M}w^{\star}=\sum_{j}w_{j}^{\star}\mathbf{M}_{j}, so is a dense mixture direction and will be accumulated across time. The accumulation of dense mixture is consistent with the finding (for linearly-separable data).

However, as we have argued in Section 3, in our setting when the noise level σx≥Ω(1)\sigma_{x}\geq\Omega(1) is large, such dense mixture direction cannot be used to given even good clean accuracy. Therefore, during clean training, the neural network has the incentive to discover features close to {Mj}j∈[d]\{\mathbf{M}_{j}\}_{j\in[d]} because they can “de-noise” ξ\xi better (see discussions in Section 3). Yet, our critical observation is that, even for well-trained neural network which aims to de-noise ξ\xi, even when the neurons are close to being pure features {Mj}j∈[d]\{\mathbf{M}_{j}\}_{j\in[d]}, the “dense direction” still locally correlates with the labeling function yy, and thus can still be accumulated during the course of a local training algorithm such as gradient descent, leading to the small, non-robust part of each feature.

Next, we state the theorem for adversarial training (recall Algorithm 2). It shows that adversarial training indeed purifies the small dense mixtures, leading to local changes of the weights.

Empirical robust accuracy: for t=Tf+Tgt=T_{\mathsf{f}}+T_{\mathsf{g}}

Provable robust accuracy: when AA is the fast gradient method (FGM), for t=Tf+Tgt=T_{\mathsf{f}}+T_{\mathsf{g}}

Feature (local) purification: for every t∈[Tf,Tf+Tg−1]t\in[T_{\mathsf{f}},T_{\mathsf{f}}+T_{\mathsf{g}}-1],

We emphasize that Theorem thm:adv1_sa holds for any perturbation algorithm AA satisfying Definition 4.2, when we only concern the robustness of the network against AA. Meaning that the local feature purification happens regardless of which adversarial perturbation algorithm is used to find the adversarial examples. More surprisingly, Theorem thm:adv1_sb says when a good perturbation algorithm such as FGM is used, then not only the robustness generalizes to unseen examples, it also generalizes to any worst-case perturbation algorithm.We point out that since our results hold for any perturbation algorithm AA, it is impossible to characterize exactly what are the learned features after training (for example showing that they corresponds to the actual dictionary) since AA might be a bad adversarial perturbation finding algorithm and the network does not even need to remove all the dense mixture in order to fool AA. However, the true robustness given by Theorem thm:adv1_sb does imply that the dense mixture should be removed at least in terms of functionality, if one uses a good adversarial perturbation finding algorithm.

Given that this “dense mixture” direction Mw∗\mathbf{M}w^{*} is common across neurons, one may think that during adversarial training, instead of training all the parameters, it may be sufficient to train a low-rank update on top of the clean-trained model. In practice, this indeed works very well, see Section 8.5.

Density of adversarial perturbation. Our previous theorem suggests that one of the main goals of adversarial training is to remove dense mixtures to make the network more robust. Therefore, before adversarial training, the adversarial perturbations are dense in the basis of {Mj}j∈[d]\{\mathbf{M}_{j}\}_{j\in[d]}; and after adversarial training, the adversarial perturbations ought to be more sparse and aligned with inputs from the original data set. Figure 7 has confirmed this theoretical finding using real-life data sets. Later in Section 8.3, we also present concrete measurements of the sparsity of these adversarial perturbations, and compare them in Figure 12.

This gives a gap because c0,c1c_{0},c_{1} can be made arbitrarily small. (see Theorem E.1 and Theorem F.4)

We also show a lower bound that no low-degree polynomial, or even the corresponding neural tangent kernel (NTK), can robustly learn the concept class. Recall for our two-layer ReLU network,

The feature mapping of the neural tangent kernel for our two-layer network ff is

Therefore, given weights {vi}i∈[m]\{v_{i}\}_{i\in[m]}, the NTK function p(x)p(x) is given as

Without loss of generality, assume each wi∼N(0,I)w_{i}\sim\mathcal{N}(0,\mathbf{I}).

In this paper, we consider a wide range of NTK parameters: ρi∼N(0,σρi2)\rho_{i}\sim\mathcal{N}(0,\sigma_{\rho_{i}}^{2}) for arbitrary σρi∈[0,do(1)]\sigma_{\rho_{i}}\in[0,d^{o(1)}] and ∣bi∣≤do(1)|b_{i}|\leq d^{o(1)}.

Our lower bound holds even for a most simple case M=I\mathbf{M}=\mathbf{I} and σx=0\sigma_{x}=0, so the original concept class is linearly separable. We prove the following:

(In contrast, Theorem 5.5 says adversarial training of neural network gives robust radius τ=1k1.76\tau=\frac{1}{k^{1.76}}.)

Since a poly-sized NTK kernel is known to be powerful enough to incorporate any low complexity functions (such as constant-degree polynomials) , we have the following corollary.

In the same setting as Theorem 5.7, if q(x)q(x) is a constant degree polynomial, then we also have the robust error Er(q)≥1−o(1)2\mathcal{E}^{r}(q)\geq\frac{1-o(1)}{2}.

Overview of the Training Process

In this section, we present an overview of the proof for the training process, using gradient descent starting from random initialization. The complete proof is deferred to the Appendix.

Our proof begins by showing how the features in the neural network are emerged from random initialization. In this phase, the loss function is not sufficiently minimized yet, so the classification accuracy remains around 50%50\%. However, we prove in this phase, gradient descent can already drive the neural network to learn a rich set of interesting features out of the random initialization. We call this process “lottery ticket winning” near random initialization, which is related to the study of .

This “lottery ticket winning” process is fundamentally different from the neural tangent kernel analysis (e.g. ). In this phase, although the loss is not sufficiently minimized, the activation patterns of the ReLU activations have changed dramatically, so that they have little correlations with the random initialization. Yet, we develop a new theoretical technique that allows us to control the change of the weights of the neurons, as we summarize below.

We derive the following property at random initialization. At iteration t=0t=0, the hidden weights are initialized as w_{i}^{(0)}\sim\mathcal{N}\big{(}0,\sigma_{0}^{2}\mathbf{I}_{d\times d}\big{)}. Using standard properties of Gaussians, we show the following critical property: as long as m≥d1.01m\geq d^{1.01}, there exists small constants c3>c4>0c_{3}>c_{4}>0 such that

For most of the neurons i∈[m]i\in[m], max⁡j∈[d]{⟨Mj,wi(0)⟩2}≤2σ02log⁡d\max_{j\in[d]}\{\langle\mathbf{M}_{j},w_{i}^{(0)}\rangle^{2}\}\leq 2\sigma_{0}^{2}\log d.

For at most 1dc4\frac{1}{d^{c_{4}}} fraction of of the neurons i∈[m]i\in[m], there is a dimension j∈[d]j\in[d] with ⟨Mj,wi(0)⟩2≥2.01σ02log⁡d\langle\mathbf{M}_{j},w_{i}^{(0)}\rangle^{2}\geq 2.01\sigma_{0}^{2}\log d.

For at least 1dc3\frac{1}{d^{c_{3}}} fraction of of the neurons i∈[m]i\in[m], there is one and only one j∈[d]j\in[d] such that ⟨Mj,wi(0)⟩2≥2.02σ02log⁡d\langle\mathbf{M}_{j},w_{i}^{(0)}\rangle^{2}\geq 2.02\sigma_{0}^{2}\log d, and all the other j′∈[d]j^{\prime}\in[d] satisfies ⟨Mj′,wi(0)⟩2≤2.01σ02log⁡d\langle\mathbf{M}_{j^{\prime}},w_{i}^{(0)}\rangle^{2}\leq 2.01\sigma_{0}^{2}\log d.

In other words, even with very mild over-parameterization m≥d1.01m\geq d^{1.01}, by the property of random gaussian initialization, there will be some “potentially lucky neurons” in (ii), where the maximum correlation to one of the features Mj\mathbf{M}_{j} is slightly higher than usual. Moreover, there will be some “surely lucky neurons” in (iii), where such “slightly higher correlation” appears in one and only one of the target features Mj\mathbf{M}_{j}.

In our proof, we denote the set of the neurons in (iii) whose correlation with Mj\mathbf{M}_{j} is slightly higher than usual as the set Sj,sure(0)\mathcal{S}_{j,sure}^{(0)}, and denote those in (ii) as Sj,pot(0)\mathcal{S}_{j,pot}^{(0)}. We will identify the following process during the training, as given in Theorem C.1:

In other words, if neuron ii wins the lottery ticket at random initialization, then eventually, it will deviate from random initialization and grow to a feature that is more close to (a scaling of) Mj\mathbf{M}_{j}. Our other main observation is that if we slightly over-parameterize the network with m≥d1.001m\geq d^{1.001}, then for each j∈[d]j\in[d], ∣Sj,sure(0)∣≥1|\mathcal{S}_{j,sure}^{(0)}|\geq 1 and ∣Sj,pot(0)∣≤d0.01|\mathcal{S}_{j,pot}^{(0)}|\leq d^{0.01}. Or in words, for each dimension j∈[d]j\in[d], the number of lottery tickets across all neurons is at most d0.01d^{0.01}, but at least one neuron will win a lottery ticket (see Lemma B.2). We also illustrate the lottery ticket winning process experimentally in Figure 8.

2 The Formation of “Dense Mixtures” During Training

The next phase of our analysis begins when all the neurons already won their lottery tickets near random initialization. After that, the loss starts to decrease significantly, so the (clean) classification error starts to drop. We shall prove that in this phase, gradient descent will also accumulate, in each neuron, a small “dense mixture” that is extremely vulnerable to small but adversarial perturbations. To show this, we maintain the following critical property as given in Theorem C.2:

If a neuron ii wins the lottery ticket for feature Mj\mathbf{M}_{j} near random initialization, then it will keep this “lottery ticket” throughout the training.

Or in math words, for each neuron ii, after ⟨Mj,wi(t)⟩2\langle\mathbf{M}_{j},w_{i}^{(t)}\rangle^{2} becomes sufficiently larger than all the other ⟨Mj′,wi(t)⟩2\langle\mathbf{M}_{j^{\prime}},w_{i}^{(t)}\rangle^{2} at the first stage, it will stay much larger than other ⟨Mj′,wi(t)⟩2\langle\mathbf{M}_{j^{\prime}},w_{i}^{(t)}\rangle^{2} for the remaining of the training process. To prove this, we introduce a careful coupling between the (directional) gradient of the neuron, and the (directional) Lipschitz continuity of the network ftf_{t}, this is given in Section C.4.2.

The vulnerable dense mixtures. The most critical observation in this phase is the formation of “dense mixtures”, where we show that even for the “lucky neuron” that wins the lottery ticket, the hidden weight of this neuron will look like (see Theorem 5.2)

In other words, up to scaling, these neurons will look like wi≈Mj+viw_{i}\approx\mathbf{M}_{j}+v_{i}, where viv_{i} is a “dense mixture” vi=Θ(kd)∑j′≠jwj′⋆Mj′v_{i}=\Theta\left(\frac{k}{d}\right)\sum_{j^{\prime}\not=j}w_{j^{\prime}}^{\star}\mathbf{M}_{j^{\prime}}.

The key observation is that viv_{i} is small and dense, in the sense that it is a mixture of all the other features {Mj′}j′∈[d]\{\mathbf{M}_{j^{\prime}}\}_{j^{\prime}\in[d]}, but each of the feature has a much smaller contribution comparing to the leading term Mj\mathbf{M}_{j}. Recall in our sparse coding model, each input x=Mz+ξx=\mathbf{M}z+\xi; so with high probability:

This value is even smaller than 1k\frac{1}{k} when k≤dk\leq\sqrt{d}. Thus, this dense mixture will not be correlated with any particular natural input, and thus the existence of these mixtures will have negligible contribution to the output of ftf_{t} on clean data.

However, if we perturb input xx along the dense direction δ∝∑j′∈[d]Mj′\delta\propto\sum_{j^{\prime}\in[d]}\mathbf{M}_{j^{\prime}}, we can observe that:

Comparing this with Eq (6.2), such “dense perturbation” can change the output of the neural network ftf_{t} by a lot, using a small δ\delta whose norm is much smaller than that of xx. Thus, at this phase, even when the network has a good clean accuracy, it is still non-robust to these small yet dense adversarial perturbations. Moreover, this perturbation direction is “universal”, in the sense that it does not depend on the randomness of the model at initialization, or the randomness we use during the training. This explains transfer attacks in practice: that is, the adversarial perturbation found in one model can also attack other models that are independently trained.

Feature purification. Since Eq. (6.2) suggests most original inputs have negligible correlations with each dense mixture, during clean training, gradient descent will have no incentive to remove those mixtures. Thus, we have to rely on adversarial training to purify those dense mixtures by introducing adversarial examples. Those examples have correlation with viv_{i}’s that are higher than usual. As we prove in Theorem 5.1 and illustrate in Figure 1, such “purifications”, albeit imposing only a small change to each neuron, will greatly improve the robustness of the neural network.

The formation of the dense mixtures. To further help the readers understand how those “dense mixtures” are formed, we sketch the proof of Theorem 5.2, which shows why clean training is provably non-robust. The main observation is that when the dense mixtures are small, the negative gradient of the (say, population) loss with respect to each neuron wiw_{i} is approximately given by (recall x=Mz+ξx=\mathbf{M}z+\xi):

We emphasize that this is indeed a special property of gradient descent. Consider again the case σx=Ω(1)\sigma_{x}=\Omega(1) discussed in Section 3, where x=Mz+ξx=\mathbf{M}z+\xi with ∥ξ∥2=Ω(1)=Ω(∥Mz∥2)\|\xi\|_{2}=\Omega(1)=\Omega(\|\mathbf{M}z\|_{2}). With high probability, a linear classifier using direction Mw⋆\mathbf{M}w^{\star} cannot be used to classify xx correctly. Yet, this direction Mw⋆\mathbf{M}w^{\star} is still locally positively correlated with the labeling function yy, especially for well-trained, well-generalizing neural networks when the ξ\xi can be “de-noised”. (Stochastic) gradient descent, as a local update algorithm, only exams the local correlation between the update direction and the labeling function, and it does not exam whether this direction can be used in the final result. Thus, this dense direction Mw⋆\mathbf{M}w^{\star} will be accumulated step by step, leading to a non-robust part of each of the features during clean training. In fact, even if we use wi=Miw_{i}=\mathbf{M}_{i} as initialization as opposed to random initialization, continuing clean training will still accumulate these small but dense mixtures. We illustrate this in Figure 9.

Conclusion

In this paper, we made a first step towards understanding how, in principle, the features in a neural network are learned during the training process, and why after clean training, these provably well-generalizing features are still provably non-robust. Our main conclusion is that during the clean training process using (stochastic) gradient descent, neural network will accumulate, in all features, some “dense mixture directions” that have low correlations with any natural input, but are extremely vulnerable to (dense) adversarial perturbations. During adversarial training, such dense mixtures are purified to make the model more robust. Our results suggest that the non-robustness of clean training is mainly due to two reasons:

the inductive bias of (stochastic) gradient descent, and

the “sparse coding” structure of the data.

Both reasons are necessary in some sense. First, a robust model is also a global minimizer of the clean training objective (at least in our setting); but even with proper regularization and infinite training examples to avoid over-fitting, gradient descent still has inductive bias towards finding a non-robust model. Second, it is easy to come up with data sets— such as linear-classifier labels over well-conditioned mixture-of-Gaussians like inputs— where clean training using gradient descent directly achieves the best robust accuracy. Thus, to understand the non-robustness of neural networks, we more or less have to take into account the gradient descent algorithm and the structure of the inputs.

Indeed, our step is still very provisional. We immediately see a plethora of extensions from our work. First of all, natural images have much richer structures than sparsity; hence, those “non-robust mixtures” accumulated by clean training might also carry structural properties other than density. Moreover, we would like to extend our work to the clean and robust training of multi-layer neural networks, possibly with hierarchical feature purification processes. (Our experiments in Figure 10 have confirmed on such hierarchical feature purification phenomenon.) Indeed, understanding the whole picture of adversarial examples and adversarial training might require a complete understanding of deep learning.

Experiment Details

We perform experiments using three standard architectures, AlexNet, ResNet-16, and ResNet-34 with basic blocks, and tested on the CIFAR-10 dataset.We used the implementations from https://github.com/bearpaw/pytorch-classification. We used their default random crop and random flip as data augmentation.

We discover that learning rate 0.10.1 for good for ResNet and 0.020.02 is good for AlexNet; while weight decay 0.00010.0001 is good for ResNet and 0.00050.0005 is good for AlexNet (this was also recommended by the git repo authors). We use standard SGD with 0.9 momentum as the training algorithm. During adversarial training, we have implemented:

In Table 1, we present robust/clean accuracies against such attackers after vanilla clean training / vanilla adversarial training. We emphasize here that, in practice, one can also first perform clean training and then apply adversarial traing using the clean-trained weights as initialization (like we have theoretically studied in this paper). This does not affect the overall performance of both robust and clean accuracies.

Visualizing the first layer of any trained architecture is trivial: for instance, for AlexNet, the weight tensor of the first layer is 3×11×113\times 11\times 11 which gives the RGB color of 11×1111\times 11 patches (and this was precisely what we presented in Figure 1). However, such visualization can be less meaningful for ResNet because the tensors are of dimension 3×3×33\times 3\times 3.

Visualizing the features presented by deeper convolutional layers is an active research area, dating back at least to . Perhaps the most naive approach is to start from a randomly initialize image (of size 3×32×323\times 32\times 32), then take a specific neuron nn at some layer, and repeatedly take its gradient with respect to the image. If we keep adding this gradient to the input image, then ideally this gives us the image which “excites” nn the most. Unfortunately, it is a common knowledge in this area that this naive approach does not lead to “visually meaningful” images as we go (even slightly) deeper into a network (see e.g. the left column of Figure 10).

In existing literature, researchers have tried to various ways to resolve this issue (see e.g. an extensive survey by Olah et al. and the references therein). At a high level, some penalizes the image to remove high-frequency noise ; some searches for images that can still excite the given neuron after jittering ; and some searches only in the space of “real data” by building a model (e.g. using GAN) to capture the prior .

We observe that, if the model is robustly trained, then one can directly apply the naive approach to visualize features of the deep layers, and the resulting images can be “visually very meaningful.” See Figure 10.This should not be surprising given that the “jittering” technique is known to work in practice on visualizing clean models. Our theory in fact explains this phenomenon: the dense mixtures accumulated during clean training are extremely harmful to the visualization effect, since they are “visually meaningless.” After robust training, such dense mixtures are removed so the visualization starts to align better with human concepts.

Throughout this paper we stick to this naive approach for visualizing features of deep layers.Specifically, starting from a random input image, we take 2000 gradient steps to update the image so that the given neuron at a specific layer is excited the most. We added a weight decay factor to incentivize the image to go to RGB (128,128,128) — except in Figure 6 we incentivize the image to go to RGB (0,0,0).

2 Feature Purification at Deeper Layers

(stage 1) perform TT epochs of adversarial training;

3 Sparse Reconstruction of Input Data and of Adversarial Perturbation

Perhaps more importantly, our theory suggests that for clean-trained models, adversarial perturbations (we refer to as clean_delta) have “dense mixtures”; while for robust-trained models, adversarial perturbations (we refer to as robust_delta) are “more pure.” This was visually illustrated in Figure 7. Now, to better quantify this observation, we compare how sparse clean_delta and robust_delta can be reconstructed from robust features. See the second row of Figure 12.In fact, we have also re-scaled the perturbations so that they have similar mean and standard deviations comparing to real input images. This allows one to also compare the two rows of Figure 12. From this experiment, we confirm that in practice, adversarial perturbations on robust models are more “pure” and closer to real input images.

We point out when comparing how sparse clean_delta and robust_delta can be reconstructed from robust features, we did not cheat. For instance, in principle clean_delta may not lie in the span of robust features and if so, it cannot be (sparsely) reconstructed from them. In our experiments (namely, the second row of Figure 12), we noticed that clean_delta almost lies in the span of robust features (with regression error <0.00005<0.00005 for AlexNet and <10−9<10^{-9} for ResNet).

4 Comparing Different Attackers

We also demonstrate in Figure 13 that feature purification occurs against several different attackers.

5 Feature Purification is a Low-Rank Update

Recall from Theorem 5.2 and illustrated in Section 6.2 that the non-robustness of neurons in a clean-trained model, only comes from a common dense mixture direction Mw⋆\mathbf{M}w^{\star}. This suggests, during the robust (e.g., adversarial) training, we do not need to re-train all the parameters; it suffices to search only for a hidden mixture direction. We use experiments to support this finding.

Before we do so, please note we derived this “common dense direction” theory using a two-layered, binary classification setting. When there are multiple classes, one would expect there to be more dense mixture directions. In such a case, “low-rank update” is a more suitable choice.Specifically, consider a Conv2D unit of k×kk\times k kernel size from nn in-channels to mm out-channels, its weight matrix (tensor) W\mathbf{W} is of dimension m×n×k×km\times n\times k\times k. When performing a rank-rr update to it, we can construct two other Conv2D units, A\mathbf{A} of dimension r×n×k×kr\times n\times k\times k, and B\mathbf{B} of dimension m×r×1×1m\times r\times 1\times 1. Now, during adversarial training, we perform low-rank update W+BA\mathbf{W}+\mathbf{B}\mathbf{A} by fixing W\mathbf{W} to be the clean-train parameters, and only letting A,B\mathbf{A},\mathbf{B} be trainable. For simplicity, we use zero initialization for A\mathbf{A} and Gaussian initialization for B\mathbf{B}, to ensure a smooth transition between clean and adversarial training. As for parameters, for this simple illustration we did not tune much, and simply set 0.050.05 for learning rate and 5e-5 for weight decay. Both of them are just slightly smaller than the commonly used choices (for training ResNet) as we now have much fewer number of trainable parameters.

Now on CIFAR-10, we compare clean training, (traditional) adversarial training, as well as first conducting clean training and then performing an adversarially trained, low-rank update on all the convolutional parameters. We use the (pre-activation) ResNet-28 architecture as well as its widened versions ResNet-28-3/5/10 . Recall ResNet-28-xx has three groups of basic convolutional layers, each of 16x,32x16x,32x and 64x64x channels respectively.

We use “rank rr” to denote a rank-rr update to all of the convolutional parameters.

We use “rank r∗r^{*}” to denote rank-rr, 2r2r, 4r4r updates respectively to the convolutional parameters in the three groups.

We present our findings in Figure 14. For example, for the ResNet-28-10 model:

Using the clean-trained model weights alone, the robust accuracies are zeros.

Now, on top of such clean-trained weights, barely training a rank-1 (or rank 1∗1^{*}) update for each convolutional matrix, or equivalently only about 1160\frac{1}{160} (less than 1 percent) of the parameters, one can already recover more than 80% of the maximum robust accuracy.

Similarly, in the “rank 8∗8^{*}” case we train only about 5%=81605\%=\frac{8}{160} of the parameters, one can recover more than 90% of the maximum robust accuracy.

We believe this preliminary experiment can be useful in supporting our theory, and may be of independent interests for other applications. (Indeed, in a follow-up work we found an application of such low-rank update to language model fine-tuning .)

We give a quick overview of the structure of our appendix sections.

In Section A, we warm up the readers by calculating the gradient of the objective, and demonstrating that polynomially many samples are sufficient for the training.

In Section B, we formally introduce Sj,pot(t)\mathcal{S}_{j,pot}^{(t)}, the set of “potentially lucky neurons” and Sj,sure(t)\mathcal{S}_{j,sure}^{(t)}, the set of “surely lucky neurons” at iteration tt. In particular, we shall emphasize on how those notions evolve as tt increases.

In Section C, we formally prove how “lucky neurons” continue to be lucky, and more importantly, for every neuron ii that is lucky in direction jj, why it grows faster than other unlucky directions j′j^{\prime}, and how much faster. Specifically, Theorem C.1 corresponds to the initial “lottery-winning” phase where the accuracy remains around 50%50\%; and Theorem C.2 corresponds to the later phase where large signals become even larger and eventually most neurons become “pure + dense mix” of the form (6.1). This is the most difficult section of this paper.

In Section D, we prove that why clean training gives good clean (testing) accuracy. It is based on the structural theorem given by Theorem C.2, and requires some non-trivial manipulations of probability theory results (such as introducing a high-probability, Bernstein form of the McDiarmid’s inequality).

In Section E, we prove that why the model obtained from clean training is non-robust. It formally shows how the “dense mixtures” become accumulated step by step during clean training.

In Section G, we prove lower bounds for the neural tangent kernel model given by two-layer networks.

In Section H, we give missing details of some probability theory lemmas.

Appendix A Notations and Warmups

We find it perhaps a good exercise to do some simple calculations to warmup the readers with our notations, before going into the proofs.

Global Assumptions. Throughout the proof,

We choose m=d1+c0m=d^{1+c_{0}} for a very small constant c0∈(0,1)c_{0}\in(0,1).

(One should think of c0=0.0001c_{0}=0.0001 for a simple reading. Our proof generalizers to larger m=\poly(d)m=\poly(d) since having more neurons does not hurt performance, but we ignore the analysis so as to provide the simplest notations.)

We choose λ=log⁡log⁡log⁡dd\lambda=\frac{\log\log\log d}{d} for simplicity.

(The purpose of log⁡log⁡log⁡d\log\log\log d factor is to simplify notations, and it can be tightened to constant.)

Whenever we write “for random xx”, “for random zz” or “for random ξ\xi”, we mean that the come from the distributions introduced in Section 2 with x=Mz+ξx=\mathbf{M}z+\xi.

Suppose Z={x(1),…,x(N)}\mathcal{Z}=\{x^{(1)},\dots,x^{(N)}\} are i.i.d. samples from D\mathcal{D} and y(i)=y(x(i))y^{(i)}=y(x^{(i)}), and suppose N≥\poly(d)N\geq\poly(d) for some sufficiently large polynomial. Let f=ftf=f_{t} and suppose b(t)≤\poly(d)b^{(t)}\leq\poly(d) and σρ≥1\poly(d)\sigma_{\rho}\geq\frac{1}{\poly(d)}. Then, for every w1,…,wNw_{1},\dots,w_{N} that may depend on the randomness of Z\mathcal{Z} and satisfies ∥wi∥≤\poly(d)\|w_{i}\|\leq\poly(d), it satisfies

In addition, suppose for every i∈[N]i\in[N], we have an i.i.d. random sample ρ(i)∼N(0,σρ2I)\rho^{(i)}\sim\mathcal{N}(0,\sigma_{\rho}^{2}\mathbf{I}) that is independent of Z\mathcal{Z} and ww. Then, with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over ρ\rho, we have

The proof of the second part can be done by trivial Hoeffding bounds. ∎

Appendix B Neuron Structure and Initialization Properties

We consider m=d1+c0m=d^{1+c_{0}} for a very small constant c0∈(0,1)c_{0}\in(0,1), and consider constants c1>c2c_{1}>c_{2} to be chosen shortly. Let us define a few notations to characterize each neuron’s behavior.

Recall wi(t)w_{i}^{(t)} is the weight for the ii-th neuron at iteration tt. We shall choose a parameter σw(t)\sigma_{w}^{(t)} at each iteration t≥0t\geq 0 and define the following notions. Consider any dimension j∈[d]j\in[d].

Let Sj,sure(t)⊆[m]\mathcal{S}_{j,sure}^{(t)}\subseteq[m] be those neurons i∈[m]i\in[m] satisfying

⟨wi(t),Mj⟩2≥(c1+c2)(σw(t))2log⁡d\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle^{2}\geq(c_{1}+c_{2})(\sigma_{w}^{(t)})^{2}\log d,

⟨wi(t),Mj′⟩2<(c1−c2)(σw(t))2log⁡d\langle w_{i}^{(t)},\mathbf{M}_{j^{\prime}}\rangle^{2}<(c_{1}-c_{2})(\sigma_{w}^{(t)})^{2}\log d for every j′≠jj^{\prime}\neq j,

sign(⟨wi(t),Mj⟩)=sign(wj⋆)\mathsf{sign}(\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle)=\mathsf{sign}(w^{\star}_{j}).

Let Sj,pot(t)⊆[m]\mathcal{S}_{j,pot}^{(t)}\subseteq[m] be those neurons i∈[m]i\in[m] satisfying

⟨wi(t),Mj⟩2≥(c1−c2)(σw(t))2log⁡d\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle^{2}\geq(c_{1}-c_{2})(\sigma_{w}^{(t)})^{2}\log d

Let Sept(t)⊆[m]\mathcal{S}_{ept}^{(t)}\subseteq[m] be the set of neurons i∈[m]i\in[m] satisfying

∥wi(t)∥22≤2(σw(t))2d\|w_{i}^{(t)}\|_{2}^{2}\leq 2(\sigma_{w}^{(t)})^{2}d

⟨wi(t),Mj⟩2≥(c1−c2)(σw(t))2log⁡d\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle^{2}\geq(c_{1}-c_{2})(\sigma_{w}^{(t)})^{2}\log d for at most O(1)O(1) many j∈[d]j\in[d].

⟨wi(t),Mj⟩2≥2(σw(t))2log⁡d\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle^{2}\geq 2(\sigma_{w}^{(t)})^{2}\sqrt{\log d} for at most 2−log⁡dd2^{-\sqrt{\log d}}d many j∈[d]j\in[d].

∣⟨wi(t),Mj⟩∣≤σw(t)log⁡d|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\leq\frac{\sigma_{w}^{(t)}}{\log d} for at least Ω(dlog⁡d)\Omega(\frac{d}{\log d}) many j∈[d]j\in[d].

Suppose each wi(0)∼N(0,σ02I)w_{i}^{(0)}\sim\mathcal{N}(0,\sigma_{0}^{2}\mathbf{I}) and suppose σw(0)=σ0\sigma_{w}^{(0)}=\sigma_{0}. For every constants c0∈(0,1)c_{0}\in(0,1) and γ∈(0,0.1)\gamma\in(0,0.1), by choosing c1=2+2(1−γ)c0c_{1}=2+2(1-\gamma)c_{0} and c2=γc0c_{2}=\gamma c_{0}, we have with probability ≥1−o(1/d3)\geq 1-o(1/d^{3}) over the random initialization, for all j∈[d]j\in[d]:

Recall b(t)b^{(t)} is the bias at iteration tt, and let us introduce more notions.

Let Sept+(t)⊆[m]\mathcal{S}_{ept+}^{(t)}\subseteq[m] be the set of neurons i∈[m]i\in[m] satisfying

∥wi(t)∥22≤(σw(t))2dlog⁡2d\|w_{i}^{(t)}\|_{2}^{2}\leq\frac{(\sigma_{w}^{(t)})^{2}d}{\log^{2}d},

∣⟨wi(t),Mj⟩∣≥σw(t)log⁡d\left|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle\right|\geq\frac{\sigma_{w}^{(t)}}{\log d} for at most O(1)O(1) many j∈[d]j\in[d].

Let Sept++(t)⊆[m]\mathcal{S}_{ept++}^{(t)}\subseteq[m] be the set of neurons i∈[m]i\in[m] satisfying

Let Sj,pot+(t)⊆[m]\mathcal{S}_{j,pot+}^{(t)}\subseteq[m] be the set of neurons i∈[m]i\in[m] satisfying

∣⟨wi(t),Mj⟩∣≥kdβb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\geq\frac{k}{d\beta}b^{(t)}.

Let Sj,sure+(t)⊆[m]\mathcal{S}_{j,sure+}^{(t)}\subseteq[m] be the set of neurons i∈[m]i\in[m] satisfying

⟨wi(t),Mj⟩2≥4k(b(t))2\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle^{2}\geq 4k(b^{(t)})^{2},

sign(⟨wi(t),Mj⟩)=sign(wj⋆)\mathsf{sign}(\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle)=\mathsf{sign}(w^{\star}_{j}).

Note that we do not have good properties on Sept+(t)\mathcal{S}_{ept+}^{(t)}, Sept++(t)\mathcal{S}_{ept++}^{(t)}, Sj,pot+(t)\mathcal{S}_{j,pot+}^{(t)} or Sj,sure+(t)\mathcal{S}_{j,sure+}^{(t)} at initialization t=0t=0; however, they will gradually begin to satisfy certain properties as the training process goes. See Section C for details.

Recall if gg is standard Gaussian, then for every t>0t>0,

Therefore, for every i∈[m]i\in[m] and j∈[d]j\in[d],

p1=Pr⁡[⟨wi(0),Mj⟩2≥(c1+c2)σ02log⁡d]=Θ(1log⁡d)⋅1d(c1+c2)/2=Θ(1log⁡d)⋅1d⋅d(1−γ/2)c0p_{1}=\operatornamewithlimits{\mathbf{Pr}}[\langle w_{i}^{(0)},\mathbf{M}_{j}\rangle^{2}\geq(c_{1}+c_{2})\sigma_{0}^{2}\log d]=\Theta(\frac{1}{\log d})\cdot\frac{1}{d^{(c_{1}+c_{2})/2}}=\Theta(\frac{1}{\sqrt{\log d}})\cdot\frac{1}{d\cdot d^{(1-\gamma/2)c_{0}}}

p2=Pr⁡[⟨wi(0),Mj⟩2≥(c1−c2)σ02log⁡d]=Θ(1log⁡d)⋅1d(c1−c2)/2=Θ(1log⁡d)⋅1d⋅d(1−3γ/2)c0p_{2}=\operatornamewithlimits{\mathbf{Pr}}[\langle w_{i}^{(0)},\mathbf{M}_{j}\rangle^{2}\geq(c_{1}-c_{2})\sigma_{0}^{2}\log d]=\Theta(\frac{1}{\log d})\cdot\frac{1}{d^{(c_{1}-c_{2})/2}}=\Theta(\frac{1}{\sqrt{\log d}})\cdot\frac{1}{d\cdot d^{(1-3\gamma/2)c_{0}}}

We first lower bound ∣Sj,sure(0)∣|\mathcal{S}_{j,sure}^{(0)}|. For every i∈[m]i\in[m], with probability at least p1/2⋅(1−p2)d−1≥Ω(1log⁡d)⋅dγ2c0mp_{1}/2\cdot(1-p_{2})^{d-1}\geq\Omega(\frac{1}{\sqrt{\log d}})\cdot\frac{d^{\frac{\gamma}{2}c_{0}}}{m} it satisfies

By concentration with respect to all mm choices of i∈[m]i\in[m], we know with probability at least 1−o(1d3)1-o(\frac{1}{d^{3}}) it satisfies ∣Sj,sure(0)∣=Ω(dγ4c0)|\mathcal{S}_{j,sure}^{(0)}|=\Omega\left(d^{\frac{\gamma}{4}c_{0}}\right).

We next upper bound ∣Sj,pot(0)∣|\mathcal{S}_{j,pot}^{(0)}|. For every i∈[m]i\in[m], with probability at most p2<O(1log⁡d)⋅d3γ2c0mp_{2}<O(\frac{1}{\sqrt{\log d}})\cdot\frac{d^{\frac{3\gamma}{2}c_{0}}}{m} it satisfies

By concentration with respect to all mm choices of ii, we know with probability at least 1−o(1d3)1-o(\frac{1}{d^{3}}) it satisfies ∣Sj,pot(0)∣=O(d2γc0)|\mathcal{S}_{j,pot}^{(0)}|=O(d^{2\gamma c_{0}}).

As for Sept(0)\mathcal{S}_{ept}^{(0)}, we first note that for every i∈[m]i\in[m], by chi-square distribution’s tail bound, with probability at least 1−o(1/d3)1-o(1/d^{3}) it satisfies ∥wi(0)∥22∈[σ02d2,2σ02d]\|w_{i}^{(0)}\|_{2}^{2}\in\left[\frac{\sigma_{0}^{2}d}{2},2\sigma_{0}^{2}d\right].

For every i∈[m]i\in[m], the probability of existing q=20/c0q=20/c_{0} different

is at most dq⋅(p2)q≤d−q⋅c02≤1d4md^{q}\cdot(p_{2})^{q}\leq d^{-q\cdot\frac{c_{0}}{2}}\leq\frac{1}{d^{4}m}. Union bounding over all possible i∈[m]i\in[m] gives the proof that, with probability at least 1−1/d41-1/d^{4}, for all but at most q=O(1)q=O(1) values of j∈[d]j\in[d], it satisfies ⟨wi(0),Mj⟩2<(c1−c2)(σw(t))2log⁡d\langle w_{i}^{(0)},\mathbf{M}_{j}\rangle^{2}<(c_{1}-c_{2})(\sigma_{w}^{(t)})^{2}\log d.

For every i∈[m]i\in[m] and j∈[d]j\in[d], with probability at least 1−e−log⁡d1-e^{-\sqrt{\log d}} it satisfies ⟨wi(0),Mj⟩2≤2σ02log⁡d\langle w_{i}^{(0)},\mathbf{M}_{j}\rangle^{2}\leq 2\sigma_{0}^{2}\sqrt{\log d}. Therefore, with probability at least 1−o(1/d3)1-o(1/d^{3}), there are ≥d(1−2−log⁡d)\geq d(1-2^{-\sqrt{\log d}}) indices j∈[d]j\in[d] satisfying ⟨wi(0),Mj⟩2≤2σ02log⁡d\langle w_{i}^{(0)},\mathbf{M}_{j}\rangle^{2}\leq 2\sigma_{0}^{2}\sqrt{\log d}.

For every i∈[m]i\in[m] and j∈[d]j\in[d], with probability at least 140000log⁡d\frac{1}{40000\sqrt{\log d}} it satisfies ∣⟨wi(0),Mj⟩∣≤σ010000log⁡d|\langle w_{i}^{(0)},\mathbf{M}_{j}\rangle|\leq\frac{\sigma_{0}}{10000\sqrt{\log d}}. Therefore, with probability at least 1−o(1/d3)1-o(1/d^{3}), there are 1100000log⁡d\frac{1}{100000\sqrt{\log d}} indices j∈[d]j\in[d] satisfying ∣⟨wi(0),Mj⟩∣≤σ010000log⁡d|\langle w_{i}^{(0)},\mathbf{M}_{j}\rangle|\leq\frac{\sigma_{0}}{10000\sqrt{\log d}}. ∎

Appendix C Neuron Structure Change During Training

For analysis purpose, we consider two phases during training. In Phase I, the neurons have moved so little so that the accuracy remains 50%50\% for binary classification; however, some neurons shall start to win lottery and form “singleton” structures. We summarize this as the following theorem.

Sj,sure(0)⊆Sj,sure(t)\mathcal{S}_{j,sure}^{(0)}\subseteq\mathcal{S}_{j,sure}^{(t)} for every j∈[d]j\in[d].

Sj,pot(0)⊇Sj,pot(t)\mathcal{S}_{j,pot}^{(0)}\supseteq\mathcal{S}_{j,pot}^{(t)} for every j∈[d]j\in[d].

Sept++(t)=[m]\mathcal{S}_{ept++}^{(t)}=[m] and Sj,pot(0)⊇Sj,pot+(t)\mathcal{S}_{j,pot}^{(0)}\supseteq\mathcal{S}_{j,pot+}^{(t)} for every j∈[d]j\in[d] at this iteration t=Tbt=T_{\mathsf{b}}.

(Recall according to Definition B.1 we have Sj,sure(t)⊆Sj,pot(t)\mathcal{S}_{j,sure}^{(t)}\subseteq\mathcal{S}_{j,pot}^{(t)}.)

In Phase II, the neurons start to move much more so that the network output becomes more meaningful; in phase II, the “singleton” neurons become even more singleton.

In the same setting as Theorem C.1, with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)}, the following holds for all t\in\big{[}T_{\mathsf{b}},\,d^{O(\log d)}/\eta\big{]}.

Sept++(t)=Sept+(t)=[m]\mathcal{S}_{ept++}^{(t)}=\mathcal{S}_{ept+}^{(t)}=[m].

Sj,sure(0)⊆Sj,sure(t)\mathcal{S}_{j,sure}^{(0)}\subseteq\mathcal{S}_{j,sure}^{(t)} for every j∈[d]j\in[d].

Sj,pot(0)⊇Sj,pot+(t)⊇Sj,pot(t)\mathcal{S}_{j,pot}^{(0)}\supseteq\mathcal{S}_{j,pot+}^{(t)}\supseteq\mathcal{S}_{j,pot}^{(t)} for every j∈[d]j\in[d].

(Recall according to Definition B.3 we have Sj,sure+(t)⊆Sj,pot(t)\mathcal{S}_{j,sure+}^{(t)}\subseteq\mathcal{S}_{j,pot}^{(t)}.)

Theorem C.2 immediately implies the first claim of Theorem 5.1 and the first claim of Theorem 5.2, after plugging in the definitions of those neuron structure sets introduced in Section B. For instance, we can write

We make several observations, when t≥Tet\geq T_{\mathsf{e}}:

Sept(t)=[m]\mathcal{S}_{ept}^{(t)}=[m] implies the cardinality of {j∈[d] ⁣:i∈Sj,pot(0)}\{j\in[d]\colon i\in\mathcal{S}_{j,pot}^{(0)}\} is ≤O(1)\leq O(1), so we can define it as Ni\mathcal{N}_{i}.

For every i∉Sj,pot(0)i\not\in\mathcal{S}_{j,pot}^{(0)}, it also satisfies i∉Sj,pot+(t)i\not\in\mathcal{S}_{j,pot+}^{(t)}, so we have ∣⟨wi(t),Mj⟩∣<kdβb(t)≤kΞ22d|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|<\frac{k}{d\beta}b^{(t)}\leq\frac{k\Xi_{2}^{2}}{d} (using b(t)≤βΞ22b^{(t)}\leq\beta\Xi_{2}^{2} from Definition C.12).

For every i∈Sj,sure+(t)⊆Sj,pot(0)i\in\mathcal{S}_{j,sure+}^{(t)}\subseteq\mathcal{S}_{j,pot}^{(0)}, we have ⟨wi(t),Mj⟩⋅sign(wi∗)≥2kb(t)>1Ξ28\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle\cdot\mathsf{sign}(w_{i}^{*})\geq 2\sqrt{k}b^{(t)}>\frac{1}{\Xi_{2}^{8}} (using b(t)=βΞ22b^{(t)}=\beta\Xi_{2}^{2} from Definition C.12 and our choice of β\beta).

They together imply the first claim of Theorem 5.2. One can similarly derive the first claim of Theorem 5.1.

We present a lemma to bound the size of the pre-activation signal.

For every t≥0t\geq 0, every i∈[m]i\in[m], every λ≥0\lambda\geq 0, every j∈[d]j\in[d]:

Let F\mathcal{F} be the event where there exists j′∈[d]j^{\prime}\in[d] with zj′≠0z_{j^{\prime}}\not=0 and ⟨wi(t),Mj′⟩2≥Ω((σw(t))2log⁡d)\langle w_{i}^{(t)},\mathbf{M}_{j^{\prime}}\rangle^{2}\geq\Omega((\sigma_{w}^{(t)})^{2}\log d). Again, by the definition of Sept(t)\mathcal{S}_{ept}^{(t)}, we know that

Thus, when neither E\mathcal{E} or F\mathcal{F} happens, we have for every j′∈[d]j^{\prime}\in[d]:

Apply Bernstein concentration bound we complete the proof that

Finally, for the ξ\xi part, let us recall ⟨wi(t),ξ⟩\langle w_{i}^{(t)},\xi\rangle variable with variance at most O(∥wi(t)∥2σx2d)≤O((σw(t))2)O(\frac{\|w_{i}^{(t)}\|^{2}\sigma_{x}^{2}}{d})\leq O((\sigma_{w}^{(t)})^{2}) and each ∣⟨wi(t),Mj⟩⟨Mj,ξ⟩∣≤σw(t)log⁡2d|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle\langle\mathbf{M}_{j},\xi\rangle|\leq\frac{\sigma_{w}^{(t)}}{\log^{2}d} w.h.p. Using Bernstein concentration of random variables, we finish the proof. ∎

Let F\mathcal{F} be the event where there exists j′∈[d]j^{\prime}\in[d] with zj′≠0z_{j^{\prime}}\not=0 and ∣⟨wi(t),Mj′⟩∣≥Ω(σw(t)log⁡d)|\langle w_{i}^{(t)},\mathbf{M}_{j^{\prime}}\rangle|\geq\Omega(\frac{\sigma_{w}^{(t)}}{\log d}). By the definition of Sept+(t)\mathcal{S}_{ept+}^{(t)}, we know that

When F\mathcal{F} does not happen, we have for every j′∈[d]j^{\prime}\in[d]: ∣⟨wi(t),Mj′zj′⟩∣≤O(σw(t)log⁡d)|\langle w_{i}^{(t)},\mathbf{M}_{j^{\prime}}z_{j^{\prime}}\rangle|\leq O(\frac{\sigma_{w}^{(t)}}{\log d}) and at the same time

Apply Bernstein concentration bound we complete the proof that

Finally, for the ξ\xi part, let us recall ⟨wi(t),ξ⟩\langle w_{i}^{(t)},\xi\rangle is a random variable with variance at most O(∥wi(t)∥2σx2d)≤O((σw(t))2log⁡2d)O(\frac{\|w_{i}^{(t)}\|^{2}\sigma_{x}^{2}}{d})\leq O(\frac{(\sigma_{w}^{(t)})^{2}}{\log^{2}d}). Using the Bernstein concentration bound, we finish the proof. ∎

C.2 Auxiliary Lemma 2: A Critical Lemma for Gradient Bound

In this section we present a critical lemma that shall be used multiple times to bound the gradient in many of the following sections. Recall c2c1∈(0,0.1)\frac{c_{2}}{c_{1}}\in(0,0.1) is a constant from Lemma B.2.

For every α>0\alpha>0, suppose ρ∼N(0,σρ2)\rho\sim\mathcal{N}(0,\sigma_{\rho}^{2}), define quantity

it always satisfies ∣Δ∣≤O(Vσρ+L)|\Delta|\leq O\left(\frac{\sqrt{V}}{\sigma_{\rho}}+L\right)

Furthermore, suppose we can write S1=(S1′,S1′′)S_{1}=(S_{1}^{\prime},S_{1}^{\prime\prime}) and S2=S2′+S2′′S_{2}=S_{2}^{\prime}+S_{2}^{\prime\prime} for (S1′,S2′)(S_{1}^{\prime},S_{2}^{\prime}) and (S1′′,S2′′)(S_{1}^{\prime\prime},S_{2}^{\prime\prime}) being independent (although S1′,S2′S_{1}^{\prime},S_{2}^{\prime} may be dependent, and S1′′,S2′′S_{1}^{\prime\prime},S_{2}^{\prime\prime} may be dependent). Then, we have

if α≤b(1−c22c1)\alpha\leq b(1-\frac{c_{2}}{2c_{1}}), then ∣Δ∣≤(e−Ω(b2/σρ2)+Γ)(min⁡{1,O(ασρ)}+Ly)+Γy|\Delta|\leq\left(e^{-\Omega\left(b^{2}/\sigma_{\rho}^{2}\right)}+\Gamma\right)\left(\min\{1,O(\frac{\alpha}{\sigma_{\rho}})\}+L_{y}\right)+\Gamma_{y}

Γ:=Pr⁡[∣S2∣≥c210c1⋅b]\Gamma:=\operatornamewithlimits{\mathbf{Pr}}\left[|S_{2}|\geq\frac{c_{2}}{10c_{1}}\cdot b\right] and Γy:=Pr⁡[∣S2′′∣≥c210c1⋅b]\Gamma_{y}:=\operatornamewithlimits{\mathbf{Pr}}\left[|S_{2}^{\prime\prime}|\geq\frac{c_{2}}{10c_{1}}\cdot b\right]

We first focus on Y(1,S1)\mathds1α+S2+ρ≥bY(1,S_{1})\mathds{1}_{\alpha+S_{2}+\rho\geq b}, and write

Focusing on the term Y(0,S1)\mathds1α+S2+ρ≥bY(0,S_{1})\mathds{1}_{\alpha+S_{2}+\rho\geq b}, by the symmetric properties of YY and (S1,S2)(S_{1},S_{2}), we have

so we can go back to (C.1) (and repeating for Y(−1,S1)Y(-1,S_{1})) to derive that

This proves Lemma lem:criticalb. Finally, when α≤b(1−c22c1)\alpha\leq b(1-\frac{c_{2}}{2c_{1}}), we can bound Δ\Delta differently

To bound the first term in (C.4) we consider two cases.:

when ∣S2∣≤b4|S_{2}|\leq\frac{b}{4}, we have Pr⁡ρ[ρ∈[b−S2−α,b−S2+α]]≤min⁡{1,ασρ}e−Ω(b2/σρ2)\operatornamewithlimits{\mathbf{Pr}}_{\rho}[\rho\in[b-S_{2}-\alpha,b-S_{2}+\alpha]]\leq\min\{1,\frac{\alpha}{\sigma_{\rho}}\}e^{-\Omega\left(b^{2}/\sigma_{\rho}^{2}\right)};

when ∣S2∣≥b4|S_{2}|\geq\frac{b}{4} (happening w.p. ≤Γ\leq\Gamma), we have \operatornamewithlimits{\mathbf{Pr}}_{\rho}[\rho\in[b-S_{2}-\alpha,b-S_{2}+\alpha]]\leq\min\{1,O\big{(}\frac{\alpha}{\sigma_{\rho}}\big{)}\}.

To bound the second term in (C.4), first recall S1=(S1′,S1′′)S_{1}=(S_{1}^{\prime},S_{1}^{\prime\prime}) and S2=S2′+S2′′S_{2}=S_{2}^{\prime}+S_{2}^{\prime\prime}, so we can write

To bound the first term in (C.6), we can take expectation over S1′′S_{1}^{\prime\prime} and use the bound LyL_{y} to derive

but since α≤(1−c22c1)⋅b\alpha\leq(1-\frac{c_{2}}{2c_{1}})\cdot b and S2′=S2−S2′′S_{2}^{\prime}=S_{2}-S_{2}^{\prime\prime}, we can further bound

Putting (C.5) and (C.7) back to (C.4), we conclude the when α≤b(1−c22c1)\alpha\leq b(1-\frac{c_{2}}{2c_{1}}), we have

C.3 Phase I: Winning lottery tickets near initialization

In Phase I.1, we pick b(t)=c1σw(t)log⁡db^{(t)}=\sqrt{c_{1}}\sigma_{w}^{(t)}\sqrt{\log d} and σρ(t)=σw(t)(log⁡log⁡log⁡d)3\sigma_{\rho}^{(t)}=\sigma_{w}^{(t)}(\log\log\log d)^{3}

We grow b(t+1)=b(t)+Cηdb^{(t+1)}=b^{(t)}+\frac{C\eta}{d} for Ta=Θ(dσ0log⁡2.5dη)T_{\mathsf{a}}=\Theta\left(\frac{d\sigma_{0}\log^{2.5}d}{\eta}\right) iterations.

In Phase I.2, we pick b(t)=c1σw(t)log⁡db^{(t)}=\sqrt{c_{1}}\sigma_{w}^{(t)}\sqrt{\log d} and σρ(t)=σw(t)⋅(log⁡log⁡log⁡d)3log⁡d\sigma_{\rho}^{(t)}=\sigma_{w}^{(t)}\cdot\frac{(\log\log\log d)^{3}}{\sqrt{\log d}}

We grow b(t+1)=b(t)+Cηdb^{(t+1)}=b^{(t)}+\frac{C\eta}{d} for Tb=Θ(d2σ0kη)T_{\mathsf{b}}=\Theta\left(\frac{d^{2}\sigma_{0}}{k\eta}\right) iterations.

Recall x=∑jMjzj+ξx=\sum_{j}\mathbf{M}_{j}z_{j}+\xi. Recall also c2c1∈(0,0.1)\frac{c_{2}}{c_{1}}\in(0,0.1) is a constant from Lemma B.2.

We define Γt\Gamma_{t} to be any value such that

\operatornamewithlimits{\mathbf{Pr}}_{x}\left[\big{|}\big{\langle}w_{i}^{(t)},\sum_{j^{\prime}\not=j}\mathbf{M}_{j^{\prime}}z_{j^{\prime}}+\xi\big{\rangle}\big{|}\geq\frac{c_{2}}{10c_{1}}b^{(t)}\right]\leq\Gamma_{t} for every i∈[m]i\in[m] and j∈[d]j\in[d];

\operatornamewithlimits{\mathbf{Pr}}_{x}\left[\big{|}\big{\langle}w_{i}^{(t)},x\big{\rangle}\big{|}\geq\frac{c_{2}}{10c_{1}}b^{(t)}\right]\leq\Gamma_{t} for every i∈[m]i\in[m]

Pr⁡x[∣ρi∣≥c210c1b(t)]≤Γt\operatornamewithlimits{\mathbf{Pr}}_{x}\left[|\rho_{i}|\geq\frac{c_{2}}{10c_{1}}b^{(t)}\right]\leq\Gamma_{t} for every i∈[m]i\in[m]

We define Γt,y\Gamma_{t,y} to be any value such that

for every i∈[m]i\in[m] and j∈[d]j\in[d], there exists Λ⊆[d]∖{j}\Lambda\subseteq[d]\setminus\{j\} with ∣Λ∣≥Ω(dlog⁡d)|\Lambda|\geq\Omega(\frac{d}{\sqrt{\log d}}) satisfying

Pr⁡x[∣⟨wi(t),∑j′∈ΛMj′zj′⟩∣≥c210c1b(t)]≤Γt,y\operatornamewithlimits{\mathbf{Pr}}_{x}\left[\left|\left\langle w_{i}^{(t)},\sum_{j^{\prime}\in\Lambda}\mathbf{M}_{j^{\prime}}z_{j^{\prime}}\right\rangle\right|\geq\frac{c_{2}}{10c_{1}}b^{(t)}\right]\leq\Gamma_{t,y}

If we are in Phase I.1 and Sept(t)=[m]\mathcal{S}_{ept}^{(t)}=[m], then we can choose Γt=e−Ω(log⁡1/4d)\Gamma_{t}=e^{-\Omega(\log^{1/4}d)} and Γt,y=1d10\Gamma_{t,y}=\frac{1}{d^{10}}.

If we are in Phase I.2 and Sept+(t)=[m]\mathcal{S}_{ept+}^{(t)}=[m], then we can choose Γt=O(kd)\Gamma_{t}=O(\frac{k}{d}) and Γt,y=1d10\Gamma_{t,y}=\frac{1}{d^{10}}.

Recall b(t)=Θ(σw(t)log⁡d)b^{(t)}=\Theta(\sigma_{w}^{(t)}\sqrt{\log d}). Applying Lemma lem:geo:0a and Lemma lem:geo:0b we immediately have

If Sept(t)=[m]\mathcal{S}_{ept}^{(t)}=[m], then Pr⁡x[∣⟨wi(t),∑j′≠jMj′zj′+ξ⟩∣≥c210c1b(t)]≤e−Ω(log⁡1/4d)\operatornamewithlimits{\mathbf{Pr}}_{x}\left[\left|\left\langle w_{i}^{(t)},\sum_{j^{\prime}\not=j}\mathbf{M}_{j^{\prime}}z_{j^{\prime}}+\xi\right\rangle\right|\geq\frac{c_{2}}{10c_{1}}b^{(t)}\right]\leq e^{-\Omega(\log^{1/4}d)};

If Sept+(t)=[m]\mathcal{S}_{ept+}^{(t)}=[m], then Pr⁡x[∣⟨wi(t),∑j′≠jMj′zj′+ξ⟩∣≥c210c1b(t)]≤O(kd)\operatornamewithlimits{\mathbf{Pr}}_{x}\left[\left|\left\langle w_{i}^{(t)},\sum_{j^{\prime}\not=j}\mathbf{M}_{j^{\prime}}z_{j^{\prime}}+\xi\right\rangle\right|\geq\frac{c_{2}}{10c_{1}}b^{(t)}\right]\leq O\left(\frac{k}{d}\right).

Now, recall x=∑j′∈[d]Mj′zj′+ξx=\sum_{j^{\prime}\in[d]}\mathbf{M}_{j^{\prime}}z_{j^{\prime}}+\xi so it differs from ∑j′≠jMj′zj′+ξ\sum_{j^{\prime}\not=j}\mathbf{M}_{j^{\prime}}z_{j^{\prime}}+\xi only by one term. Therefore, we have the same bound on Pr⁡x[∣⟨wi(t),x⟩∣≥c210c1b(t)]\operatornamewithlimits{\mathbf{Pr}}_{x}\left[\left|\left\langle w_{i}^{(t)},x\right\rangle\right|\geq\frac{c_{2}}{10c_{1}}b^{(t)}\right] by modifying the statements of Lemma lem:geo:0a and Lemma lem:geo:0b (without changing the proofs) to include this missing term.

If Sept(t)=[m]\mathcal{S}_{ept}^{(t)}=[m], Pr⁡x,ρ[∣⟨wi(t),x⟩∣≥c210c1b(t)]≤e−Ω(log⁡1/4d)\operatornamewithlimits{\mathbf{Pr}}_{x,\rho}\left[\left|\left\langle w_{i}^{(t)},x\right\rangle\right|\geq\frac{c_{2}}{10c_{1}}b^{(t)}\right]\leq e^{-\Omega(\log^{1/4}d)}

If Sept+(t)=[m]\mathcal{S}_{ept+}^{(t)}=[m], Pr⁡x,ρ[∣⟨wi(t),x⟩∣≥c210c1b(t)]≤O(kd)\operatornamewithlimits{\mathbf{Pr}}_{x,\rho}\left[\left|\left\langle w_{i}^{(t)},x\right\rangle\right|\geq\frac{c_{2}}{10c_{1}}b^{(t)}\right]\leq O\left(\frac{k}{d}\right)

At the same time, using ρi∼N(0,(σρ(t))2)\rho_{i}\sim\mathcal{N}(0,(\sigma_{\rho}^{(t)})^{2}), we also have

In Phase I.1, because σρ(t)=Θ(σw(t)(log⁡log⁡log⁡d)3log⁡d)b(t)\sigma_{\rho}^{(t)}=\Theta(\frac{\sigma_{w}^{(t)}(\log\log\log d)^{3}}{\sqrt{\log d}})b^{(t)}, we have Pr⁡ρ[∣ρi∣≥c210c1b(t)]≪e−Ω(log⁡1/4d)\operatornamewithlimits{\mathbf{Pr}}_{\rho}\left[\left|\rho_{i}\right|\geq\frac{c_{2}}{10c_{1}}b^{(t)}\right]\ll e^{-\Omega(\log^{1/4}d)}

In Phase I.2, because σρ(t)=Θ(σw(t)(log⁡log⁡log⁡d)3log⁡d)b(t)\sigma_{\rho}^{(t)}=\Theta(\frac{\sigma_{w}^{(t)}(\log\log\log d)^{3}}{\log d})b^{(t)}, we have Pr⁡ρ[∣ρi∣≥c210c1b(t)]≪O(kd)\operatornamewithlimits{\mathbf{Pr}}_{\rho}\left[\left|\rho_{i}\right|\geq\frac{c_{2}}{10c_{1}}b^{(t)}\right]\ll O\left(\frac{k}{d}\right)

C.3.2 Growth Lemmas

Our first lemma here shall be used to (lower) bound how ⟨wi(t),Mj⟩\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle (i.e., the weight with respect to neuron ii in direction Mj\mathbf{M}_{j}) grows for those i∈Sj,surei\in\mathcal{S}_{j,sure}.

Suppose we (1) either are in Phase I.1 with Sept(t)=[m]\mathcal{S}_{ept}^{(t)}=[m], (2) or are in Phase I.2 with Sept+(t)=[m]\mathcal{S}_{ept+}^{(t)}=[m]. Then, for every j∈[d]j\in[d], every i∈Sj,sure(t)i\in\mathcal{S}_{j,sure}^{(t)}, as long as ∣⟨wi(t),Mj⟩∣=O(b(t)log⁡log⁡log⁡d)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|=O(b^{(t)}\log\log\log d), the following holds:

Recall that i∈Sj,sure(t)i\in\mathcal{S}_{j,sure}^{(t)} means sign(⟨wi(t),Mj⟩)=sign(wj⋆)\mathsf{sign}(\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle)=\mathsf{sign}(w^{\star}_{j}). Without loss of generality, let us assume sign(⟨wi(t),Mj⟩)=sign(wj⋆)=1\mathsf{sign}(\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle)=\mathsf{sign}(w^{\star}_{j})=1.

First consider the case when ∣zj∣=1|z_{j}|=1. Since j∈Sj,sure(t)j\in\mathcal{S}_{j,sure}^{(t)}, we have ⟨wi(t),Mj⟩≥b(t)1+c2c1\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle\geq b^{(t)}\sqrt{1+\frac{c_{2}}{c_{1}}} so applying Lemma C.7,

For all other non-zero value ∣zj∣=s>0|z_{j}|=s>0, we have s≥1ks\geq\frac{1}{\sqrt{k}} and wish to apply Lemma C.5 to bound

In Phase I.1, to apply Lemma C.5, we choose parameters as follows:

Y=yY=y, S1=∑j′≠jwj′⋆zj′S_{1}=\sum_{j^{\prime}\not=j}w_{j^{\prime}}^{\star}z_{j^{\prime}}, S2=⟨wi(t),∑j′≠jMj′zj′+ξ⟩S_{2}=\left\langle w_{i}^{(t)},\sum_{j^{\prime}\not=j}\mathbf{M}_{j^{\prime}}z_{j^{\prime}}+\xi\right\rangle, α=⟨wi(t),Mj⟩⋅s>0\alpha=\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle\cdot s>0, ρ=ρi\rho=\rho_{i},

let Λ\Lambda be the subset defined in Lemma C.7, then we can let S1′′=(zj)j∈ΛS_{1}^{\prime\prime}=(z_{j})_{j\in\Lambda} and S2′′=⟨wi(t),∑j′∈ΛMj′zj′⟩S_{2}^{\prime\prime}=\left\langle w_{i}^{(t)},\sum_{j^{\prime}\in\Lambda}\mathbf{M}_{j^{\prime}}z_{j^{\prime}}\right\rangle

we have Γy=Γt,y=1d10\Gamma_{y}=\Gamma_{t,y}=\frac{1}{d^{10}} (from Lemma C.7) and

where inequality ① uses Lemma lem:sba and ∣Λ∣≥Ω(dlog⁡d)|\Lambda|\geq\Omega(\frac{d}{\log d}) from Lemma C.7.

Δs≥−σw(t)σρ(t)≥−O(1)(log⁡log⁡log⁡d)3\Delta_{s}\geq-\frac{\sigma_{w}^{(t)}}{\sigma_{\rho}^{(t)}}\geq-\frac{O(1)}{(\log\log\log d)^{3}} and ∣Δs∣≤σw(t)σρ(t)+s≤O(1)(log⁡log⁡log⁡d)3+s|\Delta_{s}|\leq\frac{\sigma_{w}^{(t)}}{\sigma_{\rho}^{(t)}}+s\leq\frac{O(1)}{(\log\log\log d)^{3}}+s when s=Ω(1log⁡log⁡log⁡d)s=\Omega\left(\frac{1}{\log\log\log d}\right)

∣Δs∣≤(e−Ω(b2/σρ2)+Γt)(O(ασρ)+Ly)+Γt,y≤e−Ω(log⁡1/4d)⋅s|\Delta_{s}|\leq\left(e^{-\Omega\left(b^{2}/\sigma_{\rho}^{2}\right)}+\Gamma_{t}\right)\left(O(\frac{\alpha}{\sigma_{\rho}})+L_{y}\right)+\Gamma_{t,y}\leq e^{-\Omega(\log^{1/4}d)}\cdot s when s=O(1log⁡log⁡log⁡d)s=O\left(\frac{1}{\log\log\log d}\right) (which implies α<b(t)4\alpha<\frac{b^{(t)}}{4})

In Phase I.2, the analysis is similar with different parameters: in particular,

Δs≥−σw(t)σρ(t)log⁡d≥−O(1)(log⁡log⁡log⁡d)3\Delta_{s}\geq-\frac{\sigma_{w}^{(t)}}{\sigma_{\rho}^{(t)}\sqrt{\log d}}\geq-\frac{O(1)}{(\log\log\log d)^{3}} and ∣Δs∣≤σw(t)σρ(t)log⁡d+s≤O(1)(log⁡log⁡log⁡d)3+s|\Delta_{s}|\leq\frac{\sigma_{w}^{(t)}}{\sigma_{\rho}^{(t)}\sqrt{\log d}}+s\leq\frac{O(1)}{(\log\log\log d)^{3}}+s when s=Ω(1log⁡log⁡log⁡d)s=\Omega\left(\frac{1}{\log\log\log d}\right)

∣Δs∣≤(e−Ω(b2/σρ2)+Γt)(O(ασρ)+Ly)+Γt,y≤O(kslog⁡dd)|\Delta_{s}|\leq\left(e^{-\Omega\left(b^{2}/\sigma_{\rho}^{2}\right)}+\Gamma_{t}\right)\left(O(\frac{\alpha}{\sigma_{\rho}})+L_{y}\right)+\Gamma_{t,y}\leq O(\frac{ks\log d}{d}) when s=O(1log⁡log⁡log⁡d)s=O\left(\frac{1}{\log\log\log d}\right) (which implies α<b(t)4\alpha<\frac{b^{(t)}}{4}).

(This uses Γt=O(k/d)\Gamma_{t}=O(k/d) and ασρ≤o(slog⁡d)\frac{\alpha}{\sigma_{\rho}}\leq o(s\log d).)

Taking expectation over zjz_{j} as before, and using k<d(1−c0)/2k<d^{(1-c_{0})/2} finishes the proof. ∎

Our next lemma shall be used to upper bound how ⟨wi(t),Mj⟩\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle can grown for every i∈[m]i\in[m].

Suppose we (1) either are in Phase I.1 with Sept(t)=[m]\mathcal{S}_{ept}^{(t)}=[m], (2) or are in Phase I.2 with Sept+(t)=[m]\mathcal{S}_{ept+}^{(t)}=[m]. Then, for every j∈[d]j\in[d], every i∈[m]i\in[m], the following holds:

Proof is analogous to that of Lemma C.8, and the reason we no longer need the requirement ∣⟨wi(t),Mj⟩∣=O(b(t)log⁡log⁡log⁡d)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|=O(b^{(t)}\log\log\log d) is because, when invoking Lemma C.5, it suffices for us to apply Lemma lem:criticala for every non-zero values of zz (as opposed to only those z=\Omega\big{(}\frac{1}{\log\log\log d}\big{)}) which no longer requires α≤b\alpha\leq b. ∎

Our next lemma shall be used to upper bound how ⟨wi(t),Mj⟩\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle can grown for every i∈[m]∖Sj,pot(t)i\in[m]\setminus\mathcal{S}_{j,pot}^{(t)}.

Suppose we (1) either are in Phase I.1 with Sept(t)=[m]\mathcal{S}_{ept}^{(t)}=[m], (2) or are in Phase I.2 with Sept+(t)=[m]\mathcal{S}_{ept+}^{(t)}=[m]. Then, for every j∈[d]j\in[d], every i∈[m]∖Sj,pot(t)i\in[m]\setminus\mathcal{S}_{j,pot}^{(t)}, the following holds:

where Γt\Gamma_{t} is given from Lemma C.7.

Suppose ∣zj∣=s|z_{j}|=s and without loss of generality ⟨wi(t),Mj⟩≥0\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle\geq 0. We choose α=⟨wi(t),Mj⟩⋅s\alpha=\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle\cdot s as before. Then, we have α≤c1−c2σw(t))log⁡d=b(t)1−c2c1≤b(t)(1−c22c1)\alpha\leq\sqrt{c_{1}-c_{2}}\sigma_{w}^{(t)})\sqrt{\log d}=b^{(t)}\sqrt{1-\frac{c_{2}}{c_{1}}}\leq b^{(t)}(1-\frac{c_{2}}{2c_{1}}) always holds.

Therefore, using the same notation as the proof of Lemma C.8, we always have the bound

Plugging in the parameters we finish the proof. ∎

Our final lemma shall be used to upper bound how ⟨wi(t),Mj⟩\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle can grown with respect to the noise ξ\xi in the input.

For every i∈[m]i\in[m], every j∈[d]j\in[d], the following holds:

We can define α=∣⟨Mj,ξ⟩⟩∣\alpha=|\langle\mathbf{M}_{j},\xi\rangle\rangle| and study

when α≤b(t)/4\alpha\leq b^{(t)}/4, ∣Δs∣≤∣⟨wi(t),Mj⟩∣⟨Mj,ξ⟩∣∥Mj∥2σρ(t)⋅(e−Ω(b2/σρ2)+Γt)|\Delta_{s}|\leq\frac{|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\langle\mathbf{M}_{j},\xi\rangle|}{\|\mathbf{M}_{j}\|_{2}\sigma_{\rho}^{(t)}}\cdot\left(e^{-\Omega\left(b^{2}/\sigma_{\rho}^{2}\right)}+\Gamma_{t}\right);

when α>b(t)/4\alpha>b^{(t)}/4 (which happens with exponentially small prob.), ∣Δs∣≤O(σw(t)σρ)|\Delta_{s}|\leq O(\frac{\sigma_{w}^{(t)}}{\sigma_{\rho}}).

C.3.3 Proof of Theorem C.1

Suppose in Lemma C.8 the hidden constant is 20C20C for the lower bound, that is,

Let us prove by induction with respect to tt. Suppose the properties all hold at t=0t=0. Recall from Fact A.1, for iteration tt, for every neuron i∈[m]i\in[m],

Together, we have a clean formulation for our gradient update rule:

We now prove each statement separately (and note our proofs apply both to Phase I.1 and I.2).

For every i∉Sj,pot(t)i\not\in\mathcal{S}_{j,pot}^{(t)}, by substituting Lemma C.10 and Lemma C.11 into (C.9), we have

so we also have ⟨wi(t+1),Mj⟩<c1−c2σw(t+1)log⁡d\langle w_{i}^{(t+1)},\mathbf{M}_{j}\rangle<\sqrt{c_{1}-c_{2}}\sigma_{w}^{(t+1)}\sqrt{\log d} and thus i∉Sj,pot(t+1)i\not\in\mathcal{S}_{j,pot}^{(t+1)}.

For every i∈Sj,sure(t)i\in\mathcal{S}_{j,sure}^{(t)}, suppose wlog ⟨wi(t),Mj⟩\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle is positive. Then, either ⟨wi(t),Mj⟩>Ω(b(t)log⁡log⁡log⁡d)\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle>\Omega(b^{(t)}\log\log\log d) in such a case we still have ⟨wi(t+1),Mj⟩≥c1+c2σw(t+1)log⁡d\langle w_{i}^{(t+1)},\mathbf{M}_{j}\rangle\geq\sqrt{c_{1}+c_{2}}\sigma_{w}^{(t+1)}\sqrt{\log d}. Otherwise, if ⟨wi(t),Mj⟩≤Ω(b(t)log⁡log⁡log⁡d)\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle\leq\Omega(b^{(t)}\log\log\log d) then by substituting Lemma C.8 and Lemma C.11 into (C.9), we have (using σ0≤1\poly(d)\sigma_{0}\leq\frac{1}{\poly(d)} and λ≤log⁡dd\lambda\leq\frac{\log d}{d})

so by induction we also have ⟨wi(t+1),Mj⟩≥c1+c2σw(t+1)log⁡d\langle w_{i}^{(t+1)},\mathbf{M}_{j}\rangle\geq\sqrt{c_{1}+c_{2}}\sigma_{w}^{(t+1)}\sqrt{\log d}. Combining this with Sj,pot(0)⊇Sj,pot(t+1)\mathcal{S}_{j,pot}^{(0)}\supseteq\mathcal{S}_{j,pot}^{(t+1)}, we conclude that i∈Sj,sure(t+1)i\in\mathcal{S}_{j,sure}^{(t+1)}.

To check Sept(t+1)=[m]\mathcal{S}_{ept}^{(t+1)}=[m], we need to verify four things:

⟨wi(t),Mj⟩2≥(c1−c2)(σw(t))2log⁡d\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle^{2}\geq(c_{1}-c_{2})(\sigma_{w}^{(t)})^{2}\log d for at most O(1)O(1) many j∈[d]j\in[d].

This is so because Sept(0)=[m]\mathcal{S}_{ept}^{(0)}=[m] and Sj,pot(0)⊇Sj,pot(t+1)\mathcal{S}_{j,pot}^{(0)}\supseteq\mathcal{S}_{j,pot}^{(t+1)}.

⟨wi(t+1),Mj⟩2≥2(σw(t+1))2log⁡d\langle w_{i}^{(t+1)},\mathbf{M}_{j}\rangle^{2}\geq 2(\sigma_{w}^{(t+1)})^{2}\sqrt{\log d} for at most 2−log⁡dd2^{-\sqrt{\log d}}d many j∈[d]j\in[d].

This can be derived from (C.10) in the same way.

∣⟨wi(t+1),Mj⟩∣≤σw(t+1)log⁡d|\langle w_{i}^{(t+1)},\mathbf{M}_{j}\rangle|\leq\frac{\sigma_{w}^{(t+1)}}{\log d} for at least Ω(dlog⁡d)\Omega(\frac{d}{\log d}) many j∈[d]j\in[d].

This can be derived from (C.10) in the same way.

∥wi(t)∥22≤2(σw(t))2d\|w_{i}^{(t)}\|_{2}^{2}\leq 2(\sigma_{w}^{(t)})^{2}d

For every i∈[m]i\in[m], suppose wlog ⟨wi(t),Mj⟩\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle is positive. Then, by substituting Lemma C.9 and Lemma C.11 into (C.9), we have

Applying this formula for t+1t+1 times, we derive that

and therefore applying this together with (C.10),

(Above, inequality ① uses that there are at most O(1)O(1) indices j∈[d]j\in[d] such that i∈Sj,pot(0)i\in\mathcal{S}_{j,pot}^{(0)}.)

Finally, to check Sept+(t+1)=[m]\mathcal{S}_{ept+}^{(t+1)}=[m] for t≥Ta=Θ(dσ0log⁡2.5dη)t\geq T_{\mathsf{a}}=\Theta(\frac{d\sigma_{0}\log^{2.5}d}{\eta}), we first derive that

For every i∉Sj,pot(t+1)i\not\in\mathcal{S}_{j,pot}^{(t+1)}, (C.10) gives

In particular, this together with Sept(0)=[m]\mathcal{S}_{ept}^{(0)}=[m] ensures that for every i∈[m]i\in[m], ∣⟨wi(t+1),Mj⟩∣≥σw(t+1)log⁡d\left|\langle w_{i}^{(t+1)},\mathbf{M}_{j}\rangle\right|\geq\frac{\sigma_{w}^{(t+1)}}{\log d} for at most O(1)O(1) many j∈[d]j\in[d].

For any i∈Sj,pot(0)i\in\mathcal{S}_{j,pot}^{(0)}, using (C.11) we have

Using this together with the previous item, as well as ∣Sj,pot(0)∣≤O(1)|\mathcal{S}_{j,pot}^{(0)}|\leq O(1), we have ∥wi(t+1)∥2≤O((σw(t+1))2dlog⁡3d)\|w_{i}^{(t+1)}\|^{2}\leq O(\frac{(\sigma_{w}^{(t+1)})^{2}d}{\log^{3}d}).

Putting them together we have Sept+(t+1)=[m]\mathcal{S}_{ept+}^{(t+1)}=[m] for every t≥Tat\geq T_{\mathsf{a}}.

After t=Tbt=T_{\mathsf{b}} iterations, we have σw(t)=Θ(σ0+ηdlog⁡d⋅Tb)\sigma_{w}^{(t)}=\Theta(\sigma_{0}+\frac{\eta}{d\sqrt{\log d}}\cdot T_{\mathsf{b}}), for every i∉Sj,pot(0)i\not\in\mathcal{S}_{j,pot}^{(0)}, by Lemma C.10 and Lemma C.11

Combining this with (C.11), we immediately have

This implies Sept++(t)=[m]\mathcal{S}_{ept++}^{(t)}=[m] and Sj,pot(0)⊇Sj,pot+(t)\mathcal{S}_{j,pot}^{(0)}\supseteq\mathcal{S}_{j,pot+}^{(t)} at this iteration tt.

C.4 Phase II: Signal Growth After Winning Lottery

In phase II we make the following parameter choices.

In Phase II, we pick b(t)=c1σw(t)log⁡db^{(t)}=\sqrt{c_{1}}\sigma_{w}^{(t)}\sqrt{\log d} and σρ(t)=σw(t)⋅(log⁡log⁡log⁡d)3log⁡d\sigma_{\rho}^{(t)}=\sigma_{w}^{(t)}\cdot\frac{(\log\log\log d)^{3}}{\sqrt{\log d}}.

We grow b(t+1)=b(t)+Cηdb^{(t+1)}=b^{(t)}+\frac{C\eta}{d} as before (the same as phase I.2 in Definition C.6) for each iteration, but stop growing b(t)b^{(t)} when it reaches a threshold b(t)=βΞ22b^{(t)}=\beta\Xi_{2}^{2}.

We first introduce a notation on a (high-probability) version of the coordinate Lipscthiz continuity.

At every iteration tt, for every j∈[d]j\in[d], we define Lt,j>e−Ω(log⁡2d)L_{t,j}>e^{-\Omega(\log^{2}d)} to be the smallest value such that w.p. at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over the choice of {zj′}j′≠j\{z_{j^{\prime}}\}_{j^{\prime}\not=j} and ξ\xi, for every z∈z\in and z=(z1,⋯ ,zj−1,z,zj+1,⋯ ,zd)z=(z_{1},\cdots,z_{j-1},z,z_{j+1},\cdots,z_{d}), z′=(z1,⋯ ,zj−1,0,zj+1,⋯ ,zd)z^{\prime}=(z_{1},\cdots,z_{j-1},0,z_{j+1},\cdots,z_{d}), x=Mz+ξx=\mathbf{M}z+\xi and x′=Mz′+ξx^{\prime}=\mathbf{M}z^{\prime}+\xi:

In this subsection, we provide new growth lemmas Lemma C.14, Lemma C.15, Lemma C.16, Lemma C.17 that are specific to Phase II, to replace the user of the old growth lemmas Lemma C.8, Lemma C.9, Lemma C.10, Lemma C.11 from Phase I.

Suppose we Sept+(t)=Sept++(t)=[m]\mathcal{S}_{ept+}^{(t)}=\mathcal{S}_{ept++}^{(t)}=[m]. Then, for every j∈[d]j\in[d], every i∈Sj,sure(t)i\in\mathcal{S}_{j,sure}^{(t)}, the following holds:

First, without loss of generality, assuming that sign(⟨wi(t),Mj⟩)=sign(wj⋆)=1\mathsf{sign}(\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle)=\mathsf{sign}(w^{\star}_{j})=1. Let us define z′=(z1,⋯ ,zj−1,0,zj+1,⋯ ,zd)z^{\prime}=(z_{1},\cdots,z_{j-1},0,z_{j+1},\cdots,z_{d}) and x′=Mz′+ξx^{\prime}=\mathbf{M}z^{\prime}+\xi. Define

Now, since es1+es\frac{e^{s}}{1+e^{s}} is an O(1)O(1)-Lipschitz function in ss, we know that w.p. at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)}

Let us first focus on the case that ∣zj∣=1|z_{j}|=1. As before, since j∈Sj,sure(t)j\in\mathcal{S}_{j,sure}^{(t)}, we have ⟨wi(t),Mj⟩≥b(t)1+c2c1\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle\geq b^{(t)}\sqrt{1+\frac{c_{2}}{c_{1}}} so applying Lemma C.7,

Now recall y(zj,z)=sign(wj⋆zj+⟨w⋆,z⟩)y(z_{j},z)=\mathsf{sign}(w^{\star}_{j}z_{j}+\langle w^{\star},z\rangle).

Combining this with (C.13), and using Pr⁡[∣zj∣=1]≥Ω(1/d)\operatornamewithlimits{\mathbf{Pr}}[|z_{j}|=1]\geq\Omega(1/d) finishes the proof. ∎

Suppose we Sept+(t)=Sept++(t)=[m]\mathcal{S}_{ept+}^{(t)}=\mathcal{S}_{ept++}^{(t)}=[m]. Then, for every j∈[d]j\in[d], every i∈[m]i\in[m], the following holds:

Suppose we Sept+(t)=Sept++(t)=[m]\mathcal{S}_{ept+}^{(t)}=\mathcal{S}_{ept++}^{(t)}=[m]. Then, for every j∈[d]j\in[d], every i∉Sj,pot(t)i\notin\mathcal{S}_{j,pot}^{(t)} and i∈[m]i\in[m], the following holds:

In the same notation as the proof of Lemma C.14, we have

where the last inequality uses Lemma C.7 and the fact i∉Sj,pot(t)i\not\in\mathcal{S}_{j,pot}^{(t)} (which, as before, implies if we choose α=⟨wi(t),Mj⟩⋅z\alpha=\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle\cdot z then \alpha^{2}\leq(c_{1}-c_{2})(\sigma_{w}^{(t)})^{2}\log d\leq\big{(}\frac{b^{(t)}}{4}\big{)}^{2}).

Since α<b(t)4\alpha<\frac{b^{(t)}}{4}, Lemma C.5 tells us

Combining this with (C.13), and using Pr⁡[∣zj∣=1]≥Ω(1/d)\operatornamewithlimits{\mathbf{Pr}}[|z_{j}|=1]\geq\Omega(1/d) finishes the proof. ∎

Finally, we derive a more fine-grind bound for the noise:

Suppose we Sept+(t)=Sept++(t)=[m]\mathcal{S}_{ept+}^{(t)}=\mathcal{S}_{ept++}^{(t)}=[m]. Then, for every j∈[d]j\in[d],

suppose also Sj,pot(0)⊇Sj,pot+(t)\mathcal{S}_{j,pot}^{(0)}\supseteq\mathcal{S}_{j,pot+}^{(t)} for every j∈[d]j\in[d], then

We can first decompose the noise ξ\xi into

Let us define xj′=Mz+ξj′x^{\prime}_{j}=\mathbf{M}z+\xi^{\prime}_{j}.

On one hand we have with probability at least 1−Γt1-\Gamma_{t}, ∣⟨wi(t),xj′⟩∣≤b(t)10|\langle w_{i}^{(t)},x^{\prime}_{j}\rangle|\leq\frac{b^{(t)}}{10} (using a variant of Lemma C.7). Using the randomness of ⟨ξ,Mj⟩\langle\xi,\mathbf{M}_{j}\rangle and ρi\rho_{i} we also have with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} it satisfies ∣⟨wi(t),Mj⟩⋅⟨Mj,ξ⟩∣+∣ρi∣≤b(t)10|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle\cdot\langle\mathbf{M}_{j},\xi\rangle|+|\rho_{i}|\leq\frac{b^{(t)}}{10}. Therefore, with probability at least 1−Γt−e−Ω(log⁡2d)1-\Gamma_{t}-e^{-\Omega(\log^{2}d)}, we have \mathds1⟨wi(t),x⟩+ρi≥b(t)=\mathds1⟨wi(t),xj′⟩+ρi≥b(t)\mathds{1}_{\langle w_{i}^{(t)},x\rangle+\rho_{i}\geq b^{(t)}}=\mathds{1}_{\langle w_{i}^{(t)},x^{\prime}_{j}\rangle+\rho_{i}\geq b^{(t)}}.

Otherwise, in the event that ∣⟨wi(t),xj′⟩∣≥b(t)10|\langle w_{i}^{(t)},x^{\prime}_{j}\rangle|\geq\frac{b^{(t)}}{10}, using the randomness of ρi\rho_{i}, we have that

Using the coordinate Lipscthizness, we also have

Next, we want to prove Lemma lem:noise2b. We have: denote

Since w.p. at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)}, ∣⟨wi′(t),ξ⟩∣≤∣b(t)∣10|\langle w_{i^{\prime}}^{(t)},\xi\rangle|\leq\frac{|b^{(t)}|}{10}, in this case, we know that

Next, similar to the (C.14), we also have

Combining (C.16) and (C.17) we finish the proof of Lemma lem:noise2b.

C.4.2 Growth Coupling

We also have the following lemma which says, essentially, that all those neurons i∈[m]i\in[m] satisfying ∣⟨wi(t),Mj⟩∣≥2kb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\geq 2\sqrt{k}b^{(t)} for the same jj, grows roughly in the same direction that is independent of ii.

Suppose at iteration tt, Sept+(t)=[m]\mathcal{S}^{(t)}_{ept+}=[m]. Then, for every j∈[d]j\in[d], every i∈[m]i\in[m] such that ∣⟨wi(t),Mj⟩∣≥2kb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\geq 2\sqrt{k}b^{(t)}, we have:

We first focus on the case when ⟨wi(t),Mj⟩∣≥2kb(t)\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\geq 2\sqrt{k}b^{(t)} is positive, and the reverse case is analogous. Conditional on ∣zj∣=s>0|z_{j}|=s>0, we know that s≥1ks\geq\frac{1}{\sqrt{k}}. Thus, when ∣⟨wi(t),Mj⟩∣≥2kb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\geq 2\sqrt{k}b^{(t)}, ∣⟨wi(t),Mj⟩s∣≥2b(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle s|\geq 2b^{(t)}. Now, using Sept+(t)=[m]\mathcal{S}^{(t)}_{ept+}=[m] and Lemma C.7, we can conclude that

when zj>0z_{j}>0, Pr⁡[⟨wi(t),x⟩+ρi≥b(t)∣zj=s]≥1−O(kd)\operatornamewithlimits{\mathbf{Pr}}[\langle w_{i}^{(t)},x\rangle+\rho_{i}\geq b^{(t)}\mid z_{j}=s]\geq 1-O\left(\frac{k}{d}\right);

when zj<0z_{j}<0, Pr⁡[⟨wi(t),x⟩+ρi≥b(t)∣zj=−s]≤O(kd)\operatornamewithlimits{\mathbf{Pr}}[\langle w_{i}^{(t)},x\rangle+\rho_{i}\geq b^{(t)}\mid z_{j}=-s]\leq O\left(\frac{k}{d}\right).

C.4.3 Activation Probabilities

Suppose Sept+(t)=[m]\mathcal{S}_{ept+}^{(t)}=[m] and Sj,pot(0)⊇Sj,pot+(t)\mathcal{S}_{j,pot}^{(0)}\supseteq\mathcal{S}_{j,pot+}^{(t)} for every j∈[d]j\in[d]. Then, with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)},

∣{i∈[m] s.t. ∣⟨wi(t),x⟩∣≥b(t)10}∣≤O(kΞ2).\left|\left\{i\in[m]\,s.t.\,|\langle w_{i}^{(t)},x\rangle|\geq\frac{b^{(t)}}{10}\right\}\right|\leq O(k\Xi_{2})\enspace.

∣⟨wi(t),∑j∈[d] ⁣:i∉Sj,pot+(t)Mjzj+ξ⟩∣≤b(t)10\left|\left\langle w_{i}^{(t)},\sum_{j\in[d]\colon i\not\in\mathcal{S}_{j,pot+}^{(t)}}\mathbf{M}_{j}z_{j}+\xi\right\rangle\right|\leq\frac{b^{(t)}}{10} for every i∈[m]i\in[m].

For every i∈[m]i\in[m] and j∈[d]j\in[d] with i∉Sj,pot+(t)i\not\in\mathcal{S}_{j,pot+}^{(t)}, we have ∣⟨wi(t),Mj⟩∣≤kdβb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\leq\frac{k}{d\beta}b^{(t)}. Therefore, by Bernstein’s inequality (similar to Lemma lem:geo:0b), we know with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)}, for every i∈[m]i\in[m],

With probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} it satisfies ∑j∈[d]\mathds1zj≠0≤O(k)\sum_{j\in[d]}\mathds{1}_{z_{j}\neq 0}\leq O(k) (since each zj≠0z_{j}\neq 0 with probability at most O(kd)O(\frac{k}{d})). Therefore, denoting by Λ=⋃j∈[d] ⁣:zj≠0Sj,pot+(t)\Lambda=\bigcup_{j\in[d]\colon z_{j}\neq 0}\mathcal{S}_{j,pot+}^{(t)}, we have ∣Λ∣≤O(kΞ2)|\Lambda|\leq O(k\Xi_{2}) (since every ∣Sj,pot+(t)∣≤Ξ2|\mathcal{S}_{j,pot+}^{(t)}|\leq\Xi_{2}). Now, for any i∈[m]∖Λi\in[m]\setminus\Lambda, inequality (C.18) immediately gives

Therefore, the number of i∈[m]i\in[m] satisfying ∣⟨wi(t),x⟩∣≥b(t)10\left|\left\langle w_{i}^{(t)},x\right\rangle\right|\geq\frac{b^{(t)}}{10} cannot be more than O(kΞ2)O(k\Xi_{2}). ∎

C.4.4 Coordinate Lipscthizness Bound

For every j∈[d]j\in[d], let us define γj(t)=∑i∈Sj,pot+(t)∣⟨wi(t),Mj⟩∣\gamma_{j}^{(t)}=\sum_{i\in\mathcal{S}_{j,pot+}^{(t)}}|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|. Then, suppose Sept+(t)=[m]\mathcal{S}_{ept+}^{(t)}=[m] and suppose Sj,pot(0)⊇Sj,pot+(t)\mathcal{S}_{j,pot}^{(0)}\supseteq\mathcal{S}_{j,pot+}^{(t)}, we have

By Lemma C.19 and the randomness of ρi\rho_{i}, we know with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)}, the number of activate neurons i∉Sj,pot+(t)i\not\in\mathcal{S}_{j,pot+}^{(t)}—meaning ⟨wi(t),x⟩+ρi≥0.9bi(t)\langle w_{i}^{(t)},x\rangle+\rho_{i}\geq 0.9b_{i}^{(t)} or −⟨wi(t),x⟩+ρi≥0.9bi(t)-\langle w_{i}^{(t)},x\rangle+\rho_{i}\geq 0.9b_{i}^{(t)}— is at most O(kΞ2)O(k\Xi_{2}). On the other hand, when i∉Sj,pot+(t)i\notin\mathcal{S}_{j,pot+}^{(t)}, we know that

Therefore, together, the total contribution from these active neurons with i∉Sj,pot+(t)i\notin\mathcal{S}_{j,pot+}^{(t)} is at most kΞ22d⋅O(kΞ2)<O(k2Ξ23d)<O(1Ξ23)\frac{k\Xi_{2}^{2}}{d}\cdot O(k\Xi_{2})<O(\frac{k^{2}\Xi_{2}^{3}}{d})<O\left(\frac{1}{\Xi_{2}^{3}}\right). This completes the proof. ∎

C.4.5 Regularization

Following the same argument as (C.9) from phase I, we know at any iteration tt, as long as Sept+(t)=[m]\mathcal{S}_{ept+}^{(t)}=[m],

In this and the next subsection, we shall repeatedly apply growth lemmas to (C.19). Before doing so, let us note σρ(t)=o(b(t)log⁡d)≤o(βΞ22log⁡d)\sigma_{\rho}^{(t)}=o(b^{(t)}\log d)\leq o(\beta\Xi_{2}^{2}\log d), so using our parameter choice of β\beta and using k≤d1−c0k\leq d^{1-c_{0}},

This means, when applying the aforementioned growth lemmas Lemma C.14, Lemma C.15, Lemma C.16, the additional terms kσρ(t)log⁡dd\frac{\sqrt{k}\sigma_{\rho}^{(t)}\log d}{d} and kβd3/2\frac{\sqrt{k}}{\beta d^{3/2}} are negligible.

We also have the following regularity lemma:

For every T≤dO(log⁡d)/ηT\leq d^{O(\log d)}/\eta, suppose Sept+(t)=Sept++(t)=[m]\mathcal{S}_{ept+}^{(t)}=\mathcal{S}_{ept++}^{(t)}=[m] and Sj,pot(0)⊇Sj,pot+(t)\mathcal{S}_{j,pot}^{(0)}\supseteq\mathcal{S}_{j,pot+}^{(t)} hold for every t≤Tt\leq T and j∈[d]j\in[d]. Then, we have for every t≤Tt\leq T, with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)},

By substituting Lemma C.15, Lemma lem:noise2a and (C.20) into (C.19), we have for every j∈[d]j\in[d]:

Summing up over all i∈Sj,pot(0)i\in\mathcal{S}_{j,pot}^{(0)}, and using Cauchy-Schwarz inequality together with ∣Sj,pot(0)∣≤Ξ2|\mathcal{S}_{j,pot}^{(0)}|\leq\Xi_{2}, we have

Combining this with Lt,j≤∑i∈Sj,pot(0)∣⟨wi(t),Mj⟩∣+O(1Ξ23)L_{t,j}\leq\sum_{i\in\mathcal{S}_{j,pot}^{(0)}}|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|+O(\frac{1}{\Xi_{2}^{3}}) from Lemma C.20 and our choice λ≥1d\lambda\geq\frac{1}{d}, we have (for every j∈[d]j\in[d] and t≤Tt\leq T),

This also implies Lt,j≤O(Ξ22)L_{t,j}\leq O(\Xi_{2}^{2}) as well as

Finally, for the objective value, we wish use Lt,j≤O(Ξ22)L_{t,j}\leq O(\Xi_{2}^{2}) and apply a high-probability Bernstein variant of the McDiarmid’s inequality (see Lemma H.3).

Specifically, consider random z,ξ,ρz,\xi,\rho. For notation simplicity, let us write ξ=∑j∈[d]Mjξj\xi=\sum_{j\in[d]}\mathbf{M}_{j}\xi_{j} for i.i.d. random ξj∼N(0,σx2d)\xi_{j}\sim\mathcal{N}(0,\frac{\sigma_{x}^{2}}{d}).

Now, for every j∈[m]j\in[m], suppose we change zjz_{j} to zj′z^{\prime}_{j} and ξj\xi_{j} to ξj′\xi^{\prime}_{j} with the same distribution. Then, with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)},

This implies with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)},

Therefore, we can apply Lemma H.3 to derive that with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)},

We also prove this Lemma, which gives a lower bound on the loss:

In every iteration tt, define Lmax⁡:=max⁡j∈[d]{Lt,j}L_{\max}:=\max_{j\in[d]}\{L_{t,j}\} and suppose Lmax⁡≤O(Ξ22)L_{\max}\leq O(\Xi_{2}^{2}). Then we have:

Let \alpha\in\big{[}\frac{1}{(\Xi_{2})^{5}},1\big{]} be a fixed value to be chosen later, and S0⊆[d]\mathcal{S}_{0}\subseteq[d] be an arbitrary subset of size ∣S0∣=αd|\mathcal{S}_{0}|=\alpha d. Consider a randomly sampled vector zz and let x=Mz+ξx=\mathbf{M}z+\xi be the corresponding input. We construct another z′z^{\prime} that is generated from the following process

Let Sre,z⊆S0\mathcal{S}_{re,z}\subseteq\mathcal{S}_{0} be the set consisting of all i∈S0i\in\mathcal{S}_{0} with ∣zi∣=Θ(1k)|z_{i}|=\Theta\left(\frac{1}{\sqrt{k}}\right).

For all i∉Sre,zi\notin\mathcal{S}_{re,z}, pick zi′=ziz^{\prime}_{i}=z_{i}.

For all i∈Sre,zi\in\mathcal{S}_{re,z}, pick zi′=ziz^{\prime}_{i}=z_{i} or zi′=−ziz^{\prime}_{i}=-z_{i} each with probability 0.50.5, independently at random.

Obviously, z′z^{\prime} has the same distribution as zz. Now, let us define x′=Mz′+ξx^{\prime}=\mathbf{M}z^{\prime}+\xi, y′=sign(⟨w⋆,z′⟩)y^{\prime}=\mathsf{sign}(\langle w^{\star},z^{\prime}\rangle).

Since ∣S0∣=αd|\mathcal{S}_{0}|=\alpha d, recalling the distribution property that Pr⁡[∣zi∣=Θ(1k)]=Ω(kd)\operatornamewithlimits{\mathbf{Pr}}\left[|z_{i}|=\Theta\left(\frac{1}{\sqrt{k}}\right)\right]=\Omega\left(\frac{k}{d}\right), we know with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over the choice of zz, ∣Sre,z∣=Θ(αk)|\mathcal{S}_{re,z}|=\Theta(\alpha k). We call this event E1(z)\mathcal{E}_{1}(z).

Let us denote by bi=zi′zi∈{−1,1}b_{i}=\frac{z^{\prime}_{i}}{z_{i}}\in\{-1,1\} for every i∈Sre,zi\in\mathcal{S}_{re,z}. We can therefore write ft(w(t);x′,ρ)=f(z,b,ξ,ρ)f_{t}(w^{(t)};x^{\prime},\rho)=f(z,b,\xi,\rho) to emphasize that the randomness comes from z,b,ξ,ρz,b,\xi,\rho. Using the definition of coordinate Lipscthizness, we know with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over z,ξ,ρz,\xi,\rho, it satisfies

Let E2(z,ξ,ρ)\mathcal{E}_{2}(z,\xi,\rho) denote the event where the above statement holds.

Now, conditioning on E1(z)\mathcal{E}_{1}(z) and E2(z,ξ,ρ)\mathcal{E}_{2}(z,\xi,\rho) both hold, we can apply standard MiDiarmid’s inequality (see Lemma H.2) over the randomness of bb, and derive that with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over bb,

Let E3(b∣∣z,ξ,ρ)\mathcal{E}_{3}(b||z,\xi,\rho) denote the (conditional) event where the above statement holds.

In sum, by combining E1,E2,E3\mathcal{E}_{1},\mathcal{E}_{2},\mathcal{E}_{3}, we know with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over z,z′,ξ,ρz,z^{\prime},\xi,\rho, it satisfies

As a simple corollary, if we generate another copy z′′z^{\prime\prime} in the same way as z′z^{\prime}, and denote by x′′=Mz′′+ξx^{\prime\prime}=\mathbf{M}z^{\prime\prime}+\xi, then with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over z,z′,z′′,ξ,ρz,z^{\prime},z^{\prime\prime},\xi,\rho, it satisfies

Now, let us denote by y′=sign(⟨w⋆,z′⟩)y^{\prime}=\mathsf{sign}(\langle w^{\star},z^{\prime}\rangle) and y′′=sign(⟨w⋆,z′′⟩)y^{\prime\prime}=\mathsf{sign}(\langle w^{\star},z^{\prime\prime}\rangle) and compare them. Let us write

Thus, we have y′=sign(A+B)y^{\prime}=\mathsf{sign}(A+B) and y′′=sign(A+C)y^{\prime\prime}=\mathsf{sign}(A+C).

First using a minor variant of Lemma lem:sbb, we have To be precise, we can do so since we still have at least (1−α)d≥d2(1-\alpha)d\geq\frac{d}{2} coordinates.

Denote this event by E4(z)\mathcal{E}_{4}(z).

Next, conditioning on any fixed zz which satisfies E1(z)\mathcal{E}_{1}(z) and E4(z)\mathcal{E}_{4}(z), we know that BB and CC become independent, each controlled by ∣Sre,z∣=Θ(αk)|\mathcal{S}_{re,z}|=\Theta(\alpha k) random Bernoulli variables. Therefore, we can apply a Wasserstein distance version of the central limit theorem (that can be derived from , full statement see [6, Appendix A.2]) to derive that, for a Gaussian variable g∼(0,V2)g\sim(0,V^{2}) where V2=∑j∈Sre,z(zj)2=Θ(α)V^{2}=\sum_{j\in\mathcal{S}_{re,z}}(z_{j})^{2}=\Theta(\alpha), the Wasserstein distance:

This means with probability at least Ω(1)\Omega(1), it satisfies B∈[0,α]B\in[0,\sqrt{\alpha}] and C≤−5αC\leq-5\sqrt{\alpha}.

To sum up, we know with probability at least Ω(α)\Omega(\sqrt{\alpha}), it satisfies A,B∈[0,α]A,B\in[0,\sqrt{\alpha}] and C≤−5αC\leq-5\sqrt{\alpha}. This means y′≠y′′y^{\prime}\neq y^{\prime\prime}, or in symbols,

Finally, conditioning on both (C.21) and (C.22) happen, we know that

C.4.6 Proof of Theorem C.2

We first prove that for every t≥Tbt\geq T_{\mathsf{b}},

Note from the definitions the relationship Sj,pot(t)⊆Sj,pot+(t)\mathcal{S}_{j,pot}^{(t)}\subseteq\mathcal{S}_{j,pot+}^{(t)} always holds, so we only need to prove the second inclusion.

Suppose (C.23) holds until iteration tt. Then, for every i∉Sj,pot(0)i\not\in\mathcal{S}_{j,pot}^{(0)}, let us apply Lemma C.16, Lemma lem:noise2a together with (C.20) and Lt,j≤O(Ξ22)L_{t,j}\leq O(\Xi_{2}^{2}) (using Lemma C.21) to (C.19). We get

Therefore, for those tt that are sufficiently large so that b(t+1)=βΞ22b^{(t+1)}=\beta\Xi_{2}^{2}, we have (using λ≥1d\lambda\geq\frac{1}{d})

and for those tt that are still small so that b(t+1)=Θ(η(t+1)d)b^{(t+1)}=\Theta(\frac{\eta(t+1)}{d}), we have

Together, this means i∉Sj,pot+(t+1)i\not\in\mathcal{S}_{j,pot+}^{(t+1)} so (C.23) holds for all t≥Tbt\geq T_{\mathsf{b}} and T≤dO(log⁡d)/ηT\leq d^{O(\log d)}/\eta.

Phase II.1. We will construct a threshold TeT_{\mathsf{e}} and prove inductively for all t∈[Tb,Te]t\in[T_{\mathsf{b}},T_{\mathsf{e}}]. Initially at t=Tbt=T_{\mathsf{b}}, by Lemma C.20 we have Lt,j=o(1)L_{t,j}=o(1). As long as Lt,j=o(1)L_{t,j}=o(1) holds for all j∈[d]j\in[d], we have

for every i∈[m]i\in[m], substituting Lemma C.15, Lemma lem:noise2a and (C.20) into (C.19),

for every i∉Sj,pot(t)i\not\in\mathcal{S}_{j,pot}^{(t)}, substituting Lemma C.16, Lemma lem:noise2a and (C.20) into (C.19),

Since for each ii, the number of jj satisfying i∈Sj,pot(t)i\in\mathcal{S}_{j,pot}^{(t)} is at most O(1)O(1) (using Sj,pot(t)⊆Sj,pot(0)\mathcal{S}_{j,pot}^{(t)}\subseteq\mathcal{S}_{j,pot}^{(0)} and Sept(0)=[m]\mathcal{S}_{ept}^{(0)}=[m]), we have

These bounds together mean several things:

Lt,j=o(1)L_{t,j}=o(1) for all j∈[d]j\in[d] and t∈[Tb,Te]t\in[T_{\mathsf{b}},T_{\mathsf{e}}] with Te=Θ(dηΞ2log⁡d)T_{\mathsf{e}}=\Theta\left(\frac{d}{\eta\Xi_{2}\log d}\right).

Indeed, (C.24) gives ∣⟨wi(t),Mj⟩∣≤O(1Ξ2log⁡d)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\leq O(\frac{1}{\Xi_{2}\log d}), but the number of jj satisfying i∈Sj,pot(t)i\in\mathcal{S}_{j,pot}^{(t)} is at most O(1)O(1). So we can apply Lemma C.20 to get Lt,j=o(1)L_{t,j}=o(1).

Sept++(t)=[m]\mathcal{S}_{ept++}^{(t)}=[m] for all t∈[Tb,Te]t\in[T_{\mathsf{b}},T_{\mathsf{e}}].

for those tt that are small so that σw(t)=Θ(ηdlog⁡dt)\sigma_{w}^{(t)}=\Theta(\frac{\eta}{d\sqrt{\log d}}t), we have (C.24) implies ∥wi(t)∥≤O(log⁡d⋅σw(t))≪σw(t)β\|w_{i}^{(t)}\|\leq O(\sqrt{\log d}\cdot\sigma_{w}^{(t)})\ll\frac{\sigma_{w}^{(t)}}{\beta}; and

for those tt that are large so that σw(t)=Θ(βΞ22log⁡d)\sigma_{w}^{(t)}=\Theta(\frac{\beta\Xi_{2}^{2}}{\sqrt{\log d}}), we have (C.24) implies ∥wi(t)∥≤O(1Ξ2log⁡d)≪σw(t)β\|w_{i}^{(t)}\|\leq O(\frac{1}{\Xi_{2}\log d})\ll\frac{\sigma_{w}^{(t)}}{\beta}.

Together we have i∈Sept++(t)i\in\mathcal{S}_{ept++}^{(t)}.

Sept+(t)=[m]\mathcal{S}_{ept+}^{(t)}=[m] for all t∈[Tb,Te]t\in[T_{\mathsf{b}},T_{\mathsf{e}}].

This is a direct corollary of Sept++(t)=[m]\mathcal{S}_{ept++}^{(t)}=[m] together with the property that the number of jj satisfying i∈Sj,pot+(t)i\in\mathcal{S}_{j,pot+}^{(t)} is at most O(1)O(1).

Next, let us consider any j∈[d]j\in[d] with i∈Sj,sure(0)i\in\mathcal{S}_{j,sure}^{(0)}. At any iteration t∈[Tb,Te]t\in[T_{\mathsf{b}},T_{\mathsf{e}}], substituting Lemma C.14, Lemma lem:noise2a, (C.24), and (C.20) into (C.19),

The value ∣⟨wi(t),Mj⟩∣|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle| keeps increasing as tt increases, until it reaches Θ(ηd⋅1ηλ)=Θ(1dλ)\Theta(\frac{\eta}{d}\cdot\frac{1}{\eta\lambda})=\Theta(\frac{1}{d\lambda}) and at that point it may decrease but will not fall below Θ(1dλ)\Theta(\frac{1}{d\lambda}). This ensures i∈Sj,sure(t)i\in\mathcal{S}_{j,sure}^{(t)}.

At t=Tet=T_{\mathsf{e}}, we must have i∈Sj,sure+(t)i\in\mathcal{S}_{j,sure+}^{(t)} because

To sum up, at iteration t=Tet=T_{\mathsf{e}}, we have

for i∈Sj,sure(0)i\in\mathcal{S}_{j,sure}^{(0)}, ∣⟨wi(t),Mj⟩∣≥Ω(1Ξ2log⁡d)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\geq\Omega\left(\frac{1}{\Xi_{2}\log d}\right);

for i∈Sj,pot(0)i\in\mathcal{S}_{j,pot}^{(0)}, ∣⟨wi(t),Mj⟩∣≤O(1Ξ2log⁡d)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\leq O(\frac{1}{\Xi_{2}\log d})

for i∉Sj,pot(0)i\not\in\mathcal{S}_{j,pot}^{(0)}, ∣⟨wi(t),Mj⟩∣≤O(kdΞ2)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\leq O(\frac{k}{d\Xi_{2}})

Phase II.2. We first make a quick observation that

Sept+(t)=Sept++(t)=[m]\mathcal{S}_{ept+}^{(t)}=\mathcal{S}_{ept++}^{(t)}=[m] for all t≥Tet\geq T_{\mathsf{e}}.

Indeed, from iteration t=Tet=T_{\mathsf{e}} on, we have b(t)=βΞ22b^{(t)}=\beta\Xi_{2}^{2}. Using Lemma C.21 we have for every i∈[m]i\in[m], ∥wi(t)∥≤O(Ξ22)≤σw(t)β\|w_{i}^{(t)}\|\leq O(\Xi_{2}^{2})\leq\frac{\sigma_{w}^{(t)}}{\beta}. Thus, Sept++(t)=[m]\mathcal{S}_{ept++}^{(t)}=[m] holds for all t≥Tet\geq T_{\mathsf{e}}. As for Sept+(t)=[m]\mathcal{S}_{ept+}^{(t)}=[m], it is a simple corollary of Sept++(t)=[m]\mathcal{S}_{ept++}^{(t)}=[m] together with the property that the number of jj satisfying i∈Sj,pot+(t)i\in\mathcal{S}_{j,pot+}^{(t)} is at most O(1)O(1).

Next, we claim for every i∈Sj,sure+(Te)i\in\mathcal{S}_{j,sure+}^{(T_{\mathsf{e}})} and every t≥Tet\geq T_{\mathsf{e}}, it must hold that

for some sufficiently large constant C′>1C^{\prime}>1. We prove by induction. Suppose (C.25) holds for tt and we consider t+1t+1. By the definition of i∈Sj,sure+(t)i\in\mathcal{S}_{j,sure+}^{(t)}, we know ∣⟨wi(t),Mj⟩∣≥2kb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\geq 2\sqrt{k}b^{(t)}. Now, consider every other i′∈[m]∖{i}i^{\prime}\in[m]\setminus\{i\}

if ∣⟨wi′(t),Mj⟩∣<2C′∣⟨wi(t),Mj⟩∣|\langle w_{i^{\prime}}^{(t)},\mathbf{M}_{j}\rangle|<2C^{\prime}|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|, then after one iteration we still have ∣⟨wi′(t+1),Mj⟩∣<C′∣⟨wi(t+1),Mj⟩∣|\langle w_{i^{\prime}}^{(t+1)},\mathbf{M}_{j}\rangle|<C^{\prime}|\langle w_{i}^{(t+1)},\mathbf{M}_{j}\rangle|.

if ∣⟨wi′(t),Mj⟩∣>2C′∣⟨wi(t),Mj⟩∣|\langle w_{i^{\prime}}^{(t)},\mathbf{M}_{j}\rangle|>2C^{\prime}|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|, then we have

Therefore, applying Lemma C.18 and Lemma lem:noise2a (for ii and i′i^{\prime}), and using β≤1k\beta\leq\frac{1}{\sqrt{k}}, we have

Taking the difference and using (C.26), we have

thus we continue to have ∣⟨wi′(t+1),Mj⟩∣≤C′∣⟨wi(t+1),Mj⟩∣|\langle w_{i^{\prime}}^{(t+1)},\mathbf{M}_{j}\rangle|\leq C^{\prime}|\langle w_{i}^{(t+1)},\mathbf{M}_{j}\rangle|.

Putting these together we show that the first half of (C.25) holds at t+1t+1.

As for why i∈Sj,sure+(t+1)i\in\mathcal{S}_{j,sure+}^{(t+1)}, we consider two cases.

If ∣⟨wi(t),Mj⟩∣≥4kb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\geq 4\sqrt{k}b^{(t)}, then in one iteration we should still have ∣⟨wi(t+1),Mj⟩∣≥2kb(t)=2kb(t+1)|\langle w_{i}^{(t+1)},\mathbf{M}_{j}\rangle|\geq 2\sqrt{k}b^{(t)}=2\sqrt{k}b^{(t+1)}.

If ∣⟨wi(t),Mj⟩∣≤4kb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\leq 4\sqrt{k}b^{(t)}, then by the first half of (C.25) together with Lemma C.20, we know the Lipscthizness Lt,j≤O(kb(t)⋅Ξ2)+O(1Ξ23)≤o(1)L_{t,j}\leq O(\sqrt{k}b^{(t)}\cdot\Xi_{2})+O\left(\frac{1}{\Xi_{2}^{3}}\right)\leq o(1). In this case, we also have (see (C.26)) ∥wi(t)∥2≤O(k(b(t))2)=o(1)\|w_{i}^{(t)}\|^{2}\leq O(k(b^{(t)})^{2})=o(1). Applying Lemma C.14 and Lemma lem:noise2a again we have

Putting both cases together we have i∈Sj,sure+(t+1)i\in\mathcal{S}_{j,sure+}^{(t+1)} so the second half of (C.25) holds at t+1t+1.

Appendix D Clean Accuracy Convergence Analysis

In this section we show the upper bound on how the clean training of a two-layer neural network can learn the labeling function from NN training samples {xi,yi}i=1N\{x_{i},y_{i}\}_{i=1}^{N} up to small generalization error.

Our convergence analysis will rely on the following (what we call) coupling function which is the first-order approximation of the neural network.

At every iteration tt, we define a linear function in μ\mu

and it equals the output of the real network at point μ=w(t)\mu=w^{(t)} both on zero and first order:

In the analysis, we shall also identify a special choice μ⋆\mu^{\star} defined as follows.

Recall S1,sure(0),…,Sd,sure(0)⊆[m]\mathcal{S}_{1,sure}^{(0)},\dots,\mathcal{S}_{d,sure}^{(0)}\subseteq[m] are disjoint, so we construct μ1⋆,…,μm⋆\mu^{\star}_{1},\dots,\mu^{\star}_{m} by

Above, α=o(1)\alpha=o(1) is a parameter to be chosen later. One can easily check (using Lemma B.2) that

∑i∈[m]∥μi⋆∥2≤O(α2Ξ1d)\sum_{i\in[m]}\|\mu^{\star}_{i}\|^{2}\leq O(\frac{\alpha^{2}}{\Xi_{1}}d) and ∑i∈[m]∥μi⋆∥3≤O(α3Ξ12d)\sum_{i\in[m]}\|\mu^{\star}_{i}\|^{3}\leq O(\frac{\alpha^{3}}{\Xi_{1}^{2}}d)

More interestingly, our so-constructed μ⋆\mu^{\star} satisfies (to be proved in Section D.2)

Suppose Sept+(t)=Sept++(t)=[m]\mathcal{S}_{ept+}^{(t)}=\mathcal{S}_{ept++}^{(t)}=[m], Sj,pot(0)⊇Sj,pot+(t)\mathcal{S}_{j,pot}^{(0)}\supseteq\mathcal{S}_{j,pot+}^{(t)} and Sj,sure(0)⊆Sj,sure+(t)\mathcal{S}_{j,sure}^{(0)}\subseteq\mathcal{S}_{j,sure+}^{(t)} for every j∈[d]j\in[d]. Then,

with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over x,ρx,\rho, gt(μ⋆;x,ρ)=α⟨w⋆,z⟩±O(1Ξ22)g_{t}(\mu^{\star};x,\rho)=\alpha\langle w^{\star},z\rangle\pm O(\frac{1}{\Xi_{2}^{2}})

We are now ready to prove Theorem D.1. Since wi(t+1)=wi(t)−η∇wiObj~t(w(t))w^{(t+1)}_{i}=w^{(t)}_{i}-\eta\nabla_{w_{i}}\widetilde{\mathbf{Obj}}_{t}(w^{(t)}), we have the identity

which is a convex function in μ\mu because gt(μ;x,ρ)g_{t}(\mu;x,\rho) is linear in μ\mu. We have, for every t≥Tet\geq T_{\mathsf{e}},

Above, ① uses the definition of gtg_{t}, ② uses Lemma lem:g-coupling:baseb (and Theorem C.2 for the prerequisite for Lemma lem:g-coupling:baseb), and ③ uses Claim D.5 for the bound on ∥μi⋆∥2\|\mu_{i}^{\star}\|^{2} and ∥μi⋆∥3\|\mu_{i}^{\star}\|^{3}.

Therefore, after telescoping for t=Te,Te+1,…,Te+T−1t=T_{\mathsf{e}},T_{\mathsf{e}}+1,\dots,T_{\mathsf{e}}+T-1, and using η≤1\poly(d)\eta\leq\frac{1}{\poly(d)}, we have

and this finishes the proof. ■\blacksquare

D.2 Proof of Claim D.6: Main Coupling

The proof of Lemma lem:g-coupling:basea comes from Claim D.7 and Claim D.8 below. In the two claims, we split gt(μ⋆;x)=gt,1+gt,4g_{t}(\mu^{\star};x)=g_{t,1}+g_{t,4} into two terms, and bound them separately. Define

Pr⁡x,ρ[gt,1(μ⋆;x,ρ)=α⟨w⋆,z⟩]≥1−e−Ω(log⁡2d)\operatornamewithlimits{\mathbf{Pr}}_{x,\rho}[g_{t,1}(\mu^{\star};x,\rho)=\alpha\langle w^{\star},z\rangle]\geq 1-e^{-\Omega(\log^{2}d)}

Recall for each i∈Sj,sure(0)i\in\mathcal{S}_{j,sure}^{(0)},

it satisfies i∈Sj,sure+(t)i\in\mathcal{S}_{j,sure+}^{(t)} so ∣⟨wi(t),Mj⟩∣≥2kb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\geq 2\sqrt{k}b^{(t)};

it also implies i∉Sj′,pot+(t)i\not\in\mathcal{S}_{j^{\prime},pot+}^{(t)} for any j′≠jj^{\prime}\neq j, so ∣⟨wi(t),Mj′⟩∣≤kdβb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j^{\prime}}\rangle|\leq\frac{k}{d\beta}b^{(t)};

recall ρi∼N(0,(σρ(t))2)\rho_{i}\sim\mathcal{N}(0,(\sigma_{\rho}^{(t)})^{2}) for σρ(t)=Θ(b(t)⋅(log⁡log⁡log⁡d)3log⁡d)\sigma_{\rho}^{(t)}=\Theta(b^{(t)}\cdot\frac{(\log\log\log d)^{3}}{\log d}).

recall ⟨wi(t),ξ⟩\langle w_{i}^{(t)},\xi\rangle is a variable with variance at most O(∥wi(t)∥2σx2d)O(\frac{\|w_{i}^{(t)}\|^{2}\sigma_{x}^{2}}{d}) for σx=O(1)\sigma_{x}=O(1).

Applying Lemma C.19, we know with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} it satisfies

and when this happens it satisfies, whenever zj≠0z_{j}\neq 0,

Summing up over all i∈Sj,sure(0)i\in\mathcal{S}_{j,sure}^{(0)} and j∈[d]j\in[d], we have with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over x,ρx,\rho: gt,1(μ⋆;x)=α⟨w⋆,z⟩g_{t,1}(\mu^{\star};x)=\alpha\langle w^{\star},z\rangle. ∎

Pr⁡x,ρ[∣gt,4(x,ρ)∣≤O(1Ξ22)]≥1−e−Ω(log⁡2d)\operatornamewithlimits{\mathbf{Pr}}_{x,\rho}[|g_{t,4}(x,\rho)|\leq O(\frac{1}{\Xi_{2}^{2}})]\geq 1-e^{-\Omega(\log^{2}d)}

Let us write ξ=∑j∈[d]Mjξj\xi=\sum_{j\in[d]}\mathbf{M}_{j}\xi_{j} where each ξj\xi_{j} is i.i.d. Let us write

We note that gt,4(x,ρ)g_{t,4}(x,\rho) is a random variable that depends on independent variables

so we also want to write it as gt,4(z,ξ,ρ)g_{t,4}(z,\xi,\rho) and gt,4,i(z,ξ,ρi)g_{t,4,i}(z,\xi,\rho_{i}).

We can without loss of generality assume as if ∣ρi∣≤b(t)10|\rho_{i}|\leq\frac{b^{(t)}}{10} and ∣ξj∣≤b(t)Ξ210:=B|\xi_{j}|\leq\frac{b^{(t)}}{\Xi_{2}^{10}}:=B always hold, both of which happen with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)}. In the rest of the proof we condition on this happens. By symmetry we have

We wish to apply a high-probability version of the McDiarmid’s inequality (see Lemma H.3) to bound gt,4g_{t,4}. In order to do so, we need to check the sensitivity of gt,4(x,ρ)g_{t,4}(x,\rho) regarding every random variable.

For every zjz_{j}, suppose we perturb it to an arbitrary zj′∈z^{\prime}_{j}\in. We also write z′=(z−j,zj′)z^{\prime}=(z_{-j},z^{\prime}_{j}) and x′=Mz′+ξx^{\prime}=\mathbf{M}z^{\prime}+\xi.

Now, for every i∈Sj,pot+(t)i\in\mathcal{S}_{j,pot+}^{(t)}, we have the naive bound

and there are at most ∣Sj,pot+(t)∣≤Ξ2|\mathcal{S}_{j,pot+}^{(t)}|\leq\Xi_{2} such neurons ii.

For every i∉Sj,pot+(t)i\not\in\mathcal{S}_{j,pot+}^{(t)}, we have ∣⟨wi(t),Mj⟩∣≤kdβb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\leq\frac{k}{d\beta}b^{(t)}. Define event

When event Ei\mathcal{E}_{i} does not happen, we have \mathds1⟨wi(t),x⟩+ρi≥b(t)=\mathds1⟨wi(t),x′⟩+ρi≥b(t)\mathds{1}_{\langle w_{i}^{(t)},x\rangle+\rho_{i}\geq b^{(t)}}=\mathds{1}_{\langle w_{i}^{(t)},x^{\prime}\rangle+\rho_{i}\geq b^{(t)}}, and thus

When Ei\mathcal{E}_{i} happens, using the randomness of ρi\rho_{i}, we have

Note with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)}, the number of i∈[m]i\in[m] with Ei\mathcal{E}_{i} holds is at most O(kΞ2)O(k\Xi_{2}) (using Lemma C.19). Therefore, by applying Chernoff bound, we know

This means two things that both hold with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over z−j,ξ,ρz_{-j},\xi,\rho:

For all zj,zj′z_{j},z^{\prime}_{j}, ∣gt,4(x,ρ)−gt,4(x′,ρ)∣≤O(b(t))⋅(klog⁡ddβ⋅kΞ2+Ξ2)≤O(kb(t))<o(1Ξ22)|g_{t,4}(x,\rho)-g_{t,4}(x^{\prime},\rho)|\leq O(b^{(t)})\cdot\left(\frac{k\log d}{d\beta}\cdot k\Xi_{2}+\Xi_{2}\right)\leq O(\sqrt{k}b^{(t)})<o(\frac{1}{\Xi_{2}^{2}})

For every ξj\xi_{j}, suppose we perturb it to ξj′∈[−B,B]\xi_{j}^{\prime}\in[-B,B]. We write ξ′=ξ+Mj(ξj′−ξj)\xi^{\prime}=\xi+\mathbf{M}_{j}(\xi^{\prime}_{j}-\xi_{j}) and x′=Mz+ξ′x^{\prime}=\mathbf{M}z+\xi^{\prime}.

Now, for every i∈Sj,pot+(t)i\in\mathcal{S}_{j,pot+}^{(t)}, with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} we have ∣⟨wi(t),∑j′≠jMj′ξj′⟩∣≤b(t)10|\langle w_{i}^{(t)},\sum_{j^{\prime}\neq j}\mathbf{M}_{j^{\prime}}\xi_{j^{\prime}}\rangle|\leq\frac{b^{(t)}}{10}. Therefore, if it also happens that ∣⟨wi(t),Mz⟩∣≤b(t)10|\langle w_{i}^{(t)},\mathbf{M}z\rangle|\leq\frac{b^{(t)}}{10}, then gt,4,i(z,ξ,ρi)=gt,4,i(z,ξ′,ρi)g_{t,4,i}(z,\xi,\rho_{i})=g_{t,4,i}(z,\xi^{\prime},\rho_{i}). In other words, we have

Summing up over i∈Sj,pot+(t)i\in\mathcal{S}_{j,pot+}^{(t)}, and taking expectation in zz, we have

where the last inequality uses a variant of Lemma C.7 and ∣Sj,pot+(t)∣≤Ξ2|\mathcal{S}_{j,pot+}^{(t)}|\leq\Xi_{2}.

For every i∉Sj,pot+(t)i\not\in\mathcal{S}_{j,pot+}^{(t)}, we have ∣⟨wi(t),Mj⟩∣≤kdβb(t)|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\leq\frac{k}{d\beta}b^{(t)}. Define event

When event Ei\mathcal{E}_{i} does not happen, we have \mathds1⟨wi(t),x⟩+ρi≥b(t)=\mathds1⟨wi(t),x′⟩+ρi≥b(t)\mathds{1}_{\langle w_{i}^{(t)},x\rangle+\rho_{i}\geq b^{(t)}}=\mathds{1}_{\langle w_{i}^{(t)},x^{\prime}\rangle+\rho_{i}\geq b^{(t)}}, and thus

When Ei\mathcal{E}_{i} happens, using the randomness of ρi\rho_{i}, we have

Note with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)}, the number of i∈[m]i\in[m] with Ei\mathcal{E}_{i} holds is at most O(kΞ2)O(k\Xi_{2}) (using a minor variant of Lemma C.19). Therefore, by applying Chernoff bound, we know with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over z,ξ−j,ρz,\xi_{-j},\rho

Taking expectation over zz, we have with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over ξ−j,ρ\xi_{-j},\rho:

Putting the two cases together, we have with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over ξ−j,ρ\xi_{-j},\rho:

This means two things that both hold with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over ξ−j,ρ\xi_{-j},\rho:

We are now ready to apply the high-probability version of the McDiarmid’s inequality (see Lemma H.3). We apply it twice. In the first time, we use the perturbation on zz to derive that, with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over z,ξ,ρz,\xi,\rho:

In the second time, we use the perturbation on ξ\xi to derive that, with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} over ξ,ρ\xi,\rho,

This finishes the proof of Lemma lem:g-coupling:basea. We are only left to prove Lemma lem:g-coupling:baseb.

By Lipscthiz continuity of the log⁡(1+e−x)\log(1+e^{-x}) function, we know with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)},

Taking expectation (and using the exponential tail) we have

Note if we take expectation over zz, we have

where ① uses Lemma lem:sba. This finishes the proof of Lemma lem:g-coupling:baseb. ■\blacksquare

Appendix E Why Clean Training is Non-Robust

The proof of Theorem E.1 relies on the following main lemma (to be proved in Section E.1). It says that towards the end of clean training, neurons wi(t)w_{i}^{(t)} have a (small) common direction in Mw⋆\mathbf{M}w^{\star}.

With the help of Lemma E.2, one can calculate that by perturbing input in this direction −y⋅Mw⋆-y\cdot\mathbf{M}w^{\star}, the output label of the network can change dramatically. This is the proof of Theorem E.1 and details can be found in Section E.2.

Before proving Lemma E.2, let us first present Claim E.3.

Applying Lemma lem:noise2b and using y⟨w∗,z⟩=∣⟨w∗,z⟩∣y\langle w^{*},z\rangle=|\langle w^{*},z\rangle| and ∥wi(t)∥≤O(Ξ22)\|w_{i}^{(t)}\|\leq O(\Xi_{2}^{2}) (see Lemma C.21), we have

Using Sj,sure(0)⊆Sj,sure+(t)\mathcal{S}_{j,sure}^{(0)}\subseteq\mathcal{S}_{j,sure+}^{(t)}, and a similar analysis to Lemma C.19, we know with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} it satisfies ∑i∈S\mathds1⟨wi(s),x⟩+ρi≥b(s)≥Ω(k)\sum_{i\in\mathcal{S}}\mathds{1}_{\langle w_{i}^{(s)},x\rangle+\rho_{i}\geq b^{(s)}}\geq\Omega(k). Therefore, the above inequality gives

Now, using small ball probability Lemma lem:sba we have

where the last inequality ① uses Lemma C.21 and Lemma C.22.

Therefore, using ∣S∣≤dΞ2|\mathcal{S}|\leq d\Xi_{2}, we have

so we conclude for every t≥Ω(1ηλΞ22)t\geq\Omega(\frac{1}{\eta\lambda\Xi_{2}^{2}}) it satisfies

E.2 Proof of Theorem E.1

Therefore, setting δ=δ0Mw⋆\delta=\delta_{0}\mathbf{M}w^{\star} for some δ0∈(0,βd)\delta_{0}\in(0,\frac{\beta}{\sqrt{d}}), and using ∥wi(t)∥2≤Ξ22\|w_{i}^{(t)}\|_{2}\leq\Xi_{2}^{2} (since Sept++(t)=[m]\mathcal{S}_{ept++}^{(t)}=[m]), we have ∣⟨wi(t),δ⟩∣≤δ0Ξ22d≤b(t)|\langle w_{i}^{(t)},\delta\rangle|\leq\delta_{0}\Xi_{2}^{2}\sqrt{d}\leq b^{(t)}. Using this, we can sum up over all i∈Si\in\mathcal{S}:

where ji∈[d]j_{i}\in[d] is the unique index such that i∈Sji,sure(0)i\in\mathcal{S}_{j_{i},sure}^{(0)}. We can rewrite the decrement

Using ∣∑i∈Sj,sure(0)⟨wi(t),Mw⋆⟩∣≤Ξ22d∣Sj,sure(0)∣≤Ξ23d|\sum_{i\in\mathcal{S}_{j,sure}^{(0)}}\langle w_{i}^{(t)},\mathbf{M}w^{\star}\rangle|\leq\Xi_{2}^{2}\sqrt{d}|\mathcal{S}_{j,sure}^{(0)}|\leq\Xi_{2}^{3}\sqrt{d} and \big{(}\mathds{1}_{w^{\star}_{j}z_{j}>0}+\mathds{1}_{w^{\star}_{j}z_{j}<0}\big{)}=1 with probability Θ(kd)\Theta(\frac{k}{d}), we can apply Bernstein’s inequality and derive

Also using \big{(}\mathds{1}_{w^{\star}_{j}z_{j}>0}+\mathds{1}_{w^{\star}_{j}z_{j}<0}\big{)}=1 with probability Θ(kd)\Theta(\frac{k}{d}), we can derive using Lemma E.2 that

Combining the above equations and using k>d(1−c0)/3k>d^{(1-c_{0})/3} , we have with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)},

For the remainder terms, we using ∣⟨wi(t),δ⟩∣≤δ0Ξ22d≤b(t)/2|\langle w_{i}^{(t)},\delta\rangle|\leq\delta_{0}\Xi_{2}^{2}\sqrt{d}\leq b^{(t)}/2, we have

Using Lemma E.2 and Lemma C.19 we have with probability at least 1−eΩ(log⁡2d)1-e^{\Omega(\log^{2}d)},

Putting together the bounds for ♣\clubsuit and ♠\spadesuit we have

In other words, choosing δ0=Ξ210k2\delta_{0}=\frac{\Xi_{2}^{10}}{k^{2}}, then combining with ∣ft(x)∣≤O(Ξ22log⁡d)|f_{t}(x)|\leq O(\Xi_{2}^{2}\log d) from Lemma C.21, we immediately have ft(x−δ)<0f_{t}(x-\delta)<0.

Using an analogous proof, one can also show that ft(x+δ)>0f_{t}(x+\delta)>0. Therefore, if we choose a perturb direction −δ0yMw∗=−yδ-\delta_{0}y\mathbf{M}w^{*}=-y\delta, we have

This means the robust accuracy is below e−Ω(log⁡2d)e^{-\Omega(\log^{2}d)}. Finally, using ∥Mw∗∥2≤O(d)\|\mathbf{M}w^{*}\|_{2}\leq O(\sqrt{d}) and ∥Mw∗∥∞≤O(∑j∈[d]∥Mj∥∞)=O(∥M∥1)\|\mathbf{M}w^{*}\|_{\infty}\leq O(\sum_{j\in[d]}\|\mathbf{M}_{j}\|_{\infty})=O(\|\mathbf{M}\|_{1}) finishes the proof.

Note that a similar proof as above also shows

Appendix F Robust Training Through Local Feature Purification

Suppose we run clean training for Tf≥Ω(dΞ26η)T_{\mathsf{f}}\geq\Omega(\frac{d\Xi_{2}^{6}}{\eta}) iterations following Theorem D.1. From this iteration on, let us perform TT more steps of robust training.

During the robust training phase, let us consider an arbitrary (norm-bounded) adversarial perturbation algorithm AA. Recall from Definition 4.2 that, given the current network ff (which includes hidden weights {wi}\{w_{i}\}, output weights {ai}\{a_{i}\}, bias {bi}\{b_{i}\} and smoothing parameter σρ\sigma_{\rho}), an input xx, a label yy, and some internal random string rr, the perturbation algorithm AA outputs a vector satisfying

for some t∈[Tf,Tf+Tg]t\in[T_{\mathsf{f}},T_{\mathsf{f}}+T_{\mathsf{g}}]. Then,

Consider for instance σx=0\sigma_{x}=0, c0=0.00001c_{0}=0.00001, and sufficiently large d>1d>1.

for some t∈[Tf,Tf+Tg]t\in[T_{\mathsf{f}},T_{\mathsf{f}}+T_{\mathsf{g}}]. Then,

With additional efforts, one can also prove that Theorem F.1 and Theorem F.4 holds with high probability for all TT in the range T=Θ(Tg)T=\Theta(T_{\mathsf{g}}). We do not prove it here since it is not beyond the scope of this paper.

We first note some simple structural properties that are corollaries of Theorem C.2.

At iteration t=Tft=T_{\mathsf{f}}, for every neuron i∈[m]i\in[m], we can write

where Si⊆{j∈[d]∣i∈Sj,pot(0)}\mathcal{S}_{i}\subseteq\{j\in[d]\mid i\in\mathcal{S}_{j,pot}^{(0)}\} with ∣Si∣=O(1)|\mathcal{S}_{i}|=O(1), ∣αi,j∣≤O(Ξ22)|\alpha_{i,j}|\leq O(\Xi_{2}^{2}) and max⁡j∈[d]{∣⟨ui,Mj⟩∣}=kΞ22d\max_{j\in[d]}\{|\langle u_{i},\mathbf{M}_{j}\rangle|\}=\frac{k\Xi_{2}^{2}}{d}.

We can let αi,j=⟨wi(t),Mj⟩\alpha_{i,j}=\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle and let uiu_{i} be the remaining part. We have ∣Si∣≤O(1)|\mathcal{S}_{i}|\leq O(1) because Sept(0)=[m]\mathcal{S}_{ept}^{(0)}=[m]. We have ∣αi,j∣=∣⟨wi(t),Mj⟩∣≤∥wi(t)∥≤O(Ξ22)|\alpha_{i,j}|=|\langle w_{i}^{(t)},\mathbf{M}_{j}\rangle|\leq\|w_{i}^{(t)}\|\leq O(\Xi_{2}^{2}). We also have

We next introduce an important notation that shall be used throughout the proofs of this section.

F.2 Robust Coupling

At every iteration tt, recalling wi(t)=gi+vi(t)w_{i}^{(t)}=g_{i}+v_{i}^{(t)}, we define a linear function in μ\mu

and it equals the output of the real network at point μ=v(t)\mu=v^{(t)} both on its zero and first order:

We shall show in this section that, recalling w(t)=g+v(t)w^{(t)}=g+v^{(t)}, then

It is perhaps worth nothing that the “closeness” of the above terms depend on two things,

One is regarding how small ∑i∈[m]∥vi(t)∥22\sum_{i\in[m]}\|v_{i}^{(t)}\|_{2}^{2} is, and this shall later be automatically guaranteed via implicit regularization of first-order methods.

The other is regarding how small ∥vi(t)∥2\|v_{i}^{(t)}\|_{2} or ∥vi(t)∥1\|v_{i}^{(t)}\|_{1} is for every individual neuron i∈[m]i\in[m]. This is a bit non-trivial to prove, and we shall spend the entire Section F.3 to deal with this.

As a corollary, in the event of r≤O(kΞ22d)r\leq O\left(\frac{k\Xi_{2}^{2}}{\sqrt{d}}\right) and r′≤1r^{\prime}\leq 1 and using m=d1+c0m=d^{1+c_{0}}, we have

Let us abbreviate the notations by setting vi=vi(t)v_{i}=v_{i}^{(t)} and b=b(t)b=b^{(t)}.

To upper bound ∣gt(v(t);x+δ,x,ρ)−ft(w(t);x+δ,ρ)∣|g_{t}(v^{(t)};x+\delta,x,\rho)-f_{t}(w^{(t)};x+\delta,\rho)| it suffices to upper bound ∣V1−V2∣|V_{1}-V_{2}| for

(and one also needs to take into account the reverse part, whose proof is analogous).

We first make some calculations. Using the definition of gig_{i}, we have Pr⁡x[⟨gi,x⟩≥∣b∣/10]≤O(kd)\operatornamewithlimits{\mathbf{Pr}}_{x}[\langle g_{i},x\rangle\geq|b|/10]\leq O\left(\frac{k}{d}\right) for every i∈[m]i\in[m]. Thus, we can easily calculate that Here, the spectral norm bound of ∑i∈[m]gigi⊤\sum_{i\in[m]}g_{i}g_{i}^{\top} holds for the following reason. Each gig_{i} is a sparse vector supported only on ∣Si∣=O(1)|\mathcal{S}_{i}|=O(1) coordinates, and thus gigi⊤⪯Dig_{i}g_{i}^{\top}\preceq\mathbf{D}_{i} holds for a diagonal matrix Di\mathbf{D}_{i} that where [Di]j,j=∥gi∥2≤O(Ξ24)[\mathbf{D}_{i}]_{j,j}=\|g_{i}\|^{2}\leq O(\Xi_{2}^{4}) for j∈Sij\in\mathcal{S}_{i} and [Di]j,j=0[\mathbf{D}_{i}]_{j,j}=0 otherwise. Now, using the fact that ∣Sj,pot(0)∣≤Ξ2|\mathcal{S}_{j,pot}^{(0)}|\leq\Xi_{2}, we immediately have that D1+⋯+Dm⪯O(Ξ25)⋅Id×d\mathbf{D}_{1}+\cdots+\mathbf{D}_{m}\preceq O(\Xi_{2}^{5})\cdot\mathbf{I}_{d\times d}.

Case 1, ∣⟨vi,x⟩∣≤b10|\langle v_{i},x\rangle|\leq\frac{b}{10} and ∣⟨gi,x⟩∣≤b10|\langle g_{i},x\rangle|\leq\frac{b}{10} both happen. In this case, it must satisfy ∣⟨gi+vi,δ⟩∣≤(∥gi∥+∥vi∥)⋅τ≤O(Ξ22+r′)⋅τ≤b10|\langle g_{i}+v_{i},\delta\rangle|\leq(\|g_{i}\|+\|v_{i}\|)\cdot\tau\leq O(\Xi_{2}^{2}+r^{\prime})\cdot\tau\leq\frac{b}{10}. . Also, with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)}, it satisfies ∣ρi∣≤b10|\rho_{i}|\leq\frac{b}{10}. To sum up, with high probability we have

Case 2, either ∣⟨vi,x⟩∣>b10|\langle v_{i},x\rangle|>\frac{b}{10} or ∣⟨gi,x⟩∣>b10|\langle g_{i},x\rangle|>\frac{b}{10}. In this case, to satisfy \mathds1⟨gi+vi,x+δ⟩+ρi≥b≠\mathds1⟨gi+vi,x⟩+ρi≥b\mathds{1}_{\langle g_{i}+v_{i},x+\delta\rangle+\rho_{i}\geq b}\not=\mathds{1}_{\langle g_{i}+v_{i},x\rangle+\rho_{i}\geq b}, one must have ∣⟨gi+vi,x+δ⟩−b+ρi∣≤∣⟨vi,δ⟩∣+∣⟨gi,δ⟩∣|\langle g_{i}+v_{i},x+\delta\rangle-b+\rho_{i}|\leq|\langle v_{i},\delta\rangle|+|\langle g_{i},\delta\rangle|. Also, using the randomness of ρi\rho_{i}, we have

As a corollary, in the event of r≤O(kΞ22d)r\leq O\left(\frac{k\Xi_{2}^{2}}{\sqrt{d}}\right) and using m=d1+c0m=d^{1+c_{0}}, we have

To upper bound ∣gt(0;x+δ,x,ρ)−ft(v(Tf),x)∣|g_{t}(0;x+\delta,x,\rho)-f_{t}(v^{(T_{\mathsf{f}})},x)| it suffices to upper bound ∣V3−V4∣|V_{3}-V_{4}| for

(and one also needs to take into account the reverse part, whose proof is analogous).

Let us define s=∑i∈[m]\mathds1⟨gi+vi,x⟩+ρi≥bs=\sum_{i\in[m]}\mathds{1}_{\langle g_{i}+v_{i},x\rangle+\rho_{i}\geq b}. By the properties that (1) gig_{i} is only supported on Si\mathcal{S}_{i} with ∣Si∣≤O(1)|\mathcal{S}_{i}|\leq O(1), (2) for each j∈[d]j\in[d] at most Ξ2\Xi_{2} of the gig_{i} are supported on ii, and (3) ∥gi∥2≤O(Ξ22)\|g_{i}\|_{2}\leq O(\Xi_{2}^{2}), we can obtain

Using a similar analysis to (F.4), we have

Above, inequality ① is due to a similar analysis as (F.6), and inequality ② is because ∣ρi∣≤b/4|\rho_{i}|\leq b/4 with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)}. Next, let us recall ⟨ui,Mj⟩≤rd\langle u_{i},\mathbf{M}_{j}\rangle\leq\frac{r}{\sqrt{d}} and thus, by Bernstein’s inequality, with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)},

Combining the bounds on ∣V6−V4∣|V_{6}-V_{4}|, ∣V3−V5∣|V_{3}-V_{5}|, and ∣V5−V6∣|V_{5}-V_{6}| finishes the proof. ∎

As a corollary, in the event of r≤O(kΞ22d)r\leq O\left(\frac{k\Xi_{2}^{2}}{\sqrt{d}}\right) and r′≤O(kΞ22⋅∥M∥∞)r^{\prime}\leq O(k\Xi_{2}^{2}\cdot\|\mathbf{M}\|_{\infty}) and using m=d1+c0m=d^{1+c_{0}}, we have

The proof is analogous to Lemma F.11 so we only highly the differences. In fact, we only need to change (F.1), (F.2) and (F.3) with the following calculations.

Putting those into the rest of the proof (to replace (F.1), (F.2) and (F.3)) finishes the proof. ∎

As a corollary, in the event of r≤O(kΞ22d)r\leq O\left(\frac{k\Xi_{2}^{2}}{\sqrt{d}}\right) and using m=d1+c0m=d^{1+c_{0}}, we have

The proof is analogous to Lemma F.12 so we only highly the differences. Recall we have defined

Let us define s=∑i∈[m]\mathds1⟨gi+vi,x⟩+ρi≥bs=\sum_{i\in[m]}\mathds{1}_{\langle g_{i}+v_{i},x\rangle+\rho_{i}\geq b}. By the properties that (1) gig_{i} is only supported on Si\mathcal{S}_{i} with ∣Si∣≤O(1)|\mathcal{S}_{i}|\leq O(1), (2) for each j∈[d]j\in[d] at most Ξ2\Xi_{2} of the gig_{i} are supported on ii, and (3) ∥gi∥2≤O(Ξ22)\|g_{i}\|_{2}\leq O(\Xi_{2}^{2}), we can obtain

F.3 Individual Neuron Growth Lemma

As a corollary, suppose we run robust training from iteration TfT_{\mathsf{f}} to Tf+TT_{\mathsf{f}}+T with Tη≤o(db)T\eta\leq o(db), τ≤1klog⁡d\tau\leq\frac{1}{\sqrt{k}\log d} and σx≤o(d2b2(Tη)2klog⁡d)\sigma_{x}\leq o(\frac{d^{2}b^{2}}{(T\eta)^{2}\sqrt{k}\log d}), then

First of all we can reuse the analysis of (F.4) and derive that

Using the property of gig_{i} we have \mathds1⟨gi,x⟩≥b/10≤∑j′∈Si\mathds1zj′≠0\mathds{1}_{\langle g_{i},x\rangle\geq b/10}\leq\sum_{j^{\prime}\in\mathcal{S}_{i}}\mathds{1}_{z_{j^{\prime}}\neq 0} for ∣Si∣≤O(1)|\mathcal{S}_{i}|\leq O(1) and therefore

Putting (F.7), (F.8), (F.9), (F.10) these together, we have

Now, suppose we run robust training for t=Tf,Tf+1,…,Tf+T−1t=T_{\mathsf{f}},T_{\mathsf{f}}+1,\dots,T_{\mathsf{f}}+T-1 and suppose for all of them we have ∥vi(T)∥2≤r′\|v_{i}^{(T)}\|_{2}\leq r^{\prime} satisfied. Then, using the gradient update formula (see e.g. (C.19))

This means, in order to show ∥vi(Tf+T)∥2≤r′\|v_{i}^{(T_{\mathsf{f}}+T)}\|_{2}\leq r^{\prime} we can choose any r′>0r^{\prime}>0 satisfying

Using the assumption of Tη≤o(db)T\eta\leq o(db) (which also implies (Tη)2≤o(d2.5b2k)(T\eta)^{2}\leq o(\frac{d^{2.5}b^{2}}{k})), τ≤1klog⁡d\tau\leq\frac{1}{\sqrt{k}\log d} (which also implies τ≤o(d2b2(Tη)2klog⁡d)\tau\leq o(\frac{d^{2}b^{2}}{(T\eta)^{2}\sqrt{k}\log d})), and σx≤o(d2b2(Tη)2klog⁡d)\sigma_{x}\leq o(\frac{d^{2}b^{2}}{(T\eta)^{2}\sqrt{k}\log d}), we can choose

As a corollary, suppose we run robust training from iteration TfT_{\mathsf{f}} to Tf+TT_{\mathsf{f}}+T with Tη≤db2∥M∥∞2kΞ23T\eta\leq\frac{db^{2}}{\|\mathbf{M}\|_{\infty}^{2}k\Xi_{2}^{3}} and \tau\leq o\big{(}\frac{b^{2}}{T\eta\cdot k\Xi_{2}^{2}\|\mathbf{M}\|_{\infty}}\big{)}, then

Similar to the proof of Lemma F.15, and using ∥vi∥2≤∥vi∥1≤r′\|v_{i}\|_{2}\leq\|v_{i}\|_{1}\leq r^{\prime}, we have

Since with probability at least 1−e−Ω(log⁡d)1-e^{-\Omega(\log d)} it satisfies ∣⟨x,u⟩∣=O(max⁡j∈[d]∥Mj∥1log⁡d)|\langle x,u\rangle|=O(\max_{j\in[d]}\|\mathbf{M}_{j}\|_{1}\log d), we can conclude that ∥h∥1=O(κ∥M∥∞log⁡d)\|h\|_{1}=O(\kappa\|\mathbf{M}\|_{\infty}\log d). Together we have

Now, suppose we run robust training for t=Tf,Tf+1,…,Tf+T−1t=T_{\mathsf{f}},T_{\mathsf{f}}+1,\dots,T_{\mathsf{f}}+T-1 and suppose for all of them we have ∥vi(T)∥1≤r′\|v_{i}^{(T)}\|_{1}\leq r^{\prime} satisfied. Then, using the gradient update formula (see e.g. (C.19))

Recalling ∣⟨vi(Tf),Mj⟩∣≤kΞ22d|\langle v_{i}^{(T_{\mathsf{f}})},\mathbf{M}_{j}\rangle|\leq\frac{k\Xi_{2}^{2}}{d} from (F.8), we have

This means, to prove that ∥vi(Tf)∥1≤r′\left\|v_{i}^{(T_{\mathsf{f}})}\right\|_{1}\leq r^{\prime}, we can choose any r′r^{\prime} satisfying

and using the assumption of Tη≤db2∥M∥∞2kΞ23T\eta\leq\frac{db^{2}}{\|\mathbf{M}\|_{\infty}^{2}k\Xi_{2}^{3}} (which implies Tη≤O(d)T\eta\leq O(d)), and \tau\leq o\big{(}\frac{b^{2}}{T\eta\cdot k\Xi_{2}^{2}\|\mathbf{M}\|_{\infty}}\big{)} (which implies τ≤1ηT\tau\leq\frac{1}{\eta T}), we can choose

F.4 Robust Convergence

We are now ready to prove the main convergence theorem (that is, Theorem F.1 and F.4) for robust learning. Let us first calculate a simple bound:

∣∑i∈[m]Reg(gi)−Reg(wi(Tf))∣≤O(kdΞ24)\left|\sum_{i\in[m]}\mathbf{Reg}(g_{i})-\mathbf{Reg}(w_{i}^{(T_{\mathsf{f}})})\right|\leq O(k\sqrt{d}\Xi_{2}^{4})

Recalling ∥gi∥2≤O(Ξ22)\|g_{i}\|_{2}\leq O(\Xi_{2}^{2}) and ∥ui∥≤O(kΞ22d)\|u_{i}\|\leq O(\frac{k\Xi_{2}^{2}}{\sqrt{d}}) from Proposition F.8, we have

Since wi(t+1)=wi(t)−η∇wiRobObj~t(w(t))w^{(t+1)}_{i}=w^{(t)}_{i}-\eta\nabla_{w_{i}}\widetilde{\mathbf{RobObj}}_{t}(w^{(t)}), we have the identity

Applying (a variant of) Lemma A.2 (which requires us to use the Lipscthiz continuity assumption on AA, see Definition 4.2), we know that by letting

Let us also define the clean objective and the pseudo objective as follows:

which is a convex function in μ\mu because gt(μ;x+δ,x,ρ)g_{t}(\mu;x+\delta,x,\rho) is linear in μ\mu.

Now, we inductively prove that at every iteration t∈[Tf,Tf+T]t\in[T_{\mathsf{f}},T_{\mathsf{f}}+T], it satisfies

In the base case t=Tft=T_{\mathsf{f}} this is obvious due to Proposition F.8. Next, suppose (F.12) and (F.13) hold at iteration tt. Using the notation wi(t)=gi+v(t)w_{i}^{(t)}=g_{i}+v^{(t)} and the Lipscthiz continuity of log⁡(1+et)\log(1+e^{t}), we have Note to apply Lemma F.11 we also need to check τ≤o(bΞ22+r′)\tau\leq o(\frac{b}{\Xi_{2}^{2}+r^{\prime}}) but this is automatically satisfied under our parameter choice τ≤1k⋅dc0\tau\leq\frac{1}{\sqrt{k}\cdot d^{c_{0}}}.

Therefore, we can bound the left hand side of (F.11) as follows:

Putting this back to (F.11) and telescoping for t=Tf,Tf+1,…,Tf+T0−1t=T_{\mathsf{f}},T_{\mathsf{f}}+1,\dots,T_{\mathsf{f}}+T_{0}-1 for any T0≤TT_{0}\leq T, we have

so (F.12) holds at iteration t=Tf+T0t=T_{\mathsf{f}}+T_{0}. We can then also apply Lemma F.15 which ensures (F.13) holds at iteration t=Tf+T0t=T_{\mathsf{f}}+T_{0}.

Finally, let us go back to (F.14) and choose T0=T=Θ(k2Ξ24mlog⁡dηd)T_{0}=T=\Theta(\frac{k^{2}\Xi_{2}^{4}m\log d}{\eta d}). It implies

Note that our final choice of TT also ensures that the pre-requisite Tη≤o(db)T\eta\leq o(db) and τ,σx≤o(d2b2(Tη)2klog⁡d)\tau,\sigma_{x}\leq o(\frac{d^{2}b^{2}}{(T\eta)^{2}\sqrt{k}\log d}) of Lemma F.15 hold. ∎

The proof is nearly identical to that of Theorem F.1. In particular, we want to inductively prove that at every iteration t∈[Tf,Tf+T]t\in[T_{\mathsf{f}},T_{\mathsf{f}}+T], it satisfies

We also need to redo the following calculations:Note to apply Lemma F.13 we also need to check τ≤o(bΞ22+r′)\tau\leq o(\frac{b}{\Xi_{2}^{2}+r^{\prime}}) but this is automatically satisfied under our parameter choice for τ\tau.

F.5 Fast Gradient Method (FGM) Robust Training

This means for at least 1−O(1log⁡d)1-O(\frac{1}{\log d}) probability mass of inputs xx, we have

For those choices of xx, using the fact that gtg_{t} is linear in δ\delta, we also have

Therefore, for all of those xx (with total mass ≥1−o(1)\geq 1-o(1)) satisfying both, we can first apply (F.17) (with δ=δ⋆\delta=\delta^{\star}) to derive

Applying (F.18) then we obtain (for any δ\delta)

This means, the output of the network ftf_{t} is robust at point xx against any perturbation δ\delta with radius τ\tau. We finish the proof of Corollary F.2. ∎

Recall from Definition 5.6 that the feature mapping of the neural tangent kernel for our two-layer network ff is

Therefore, given weights {vi}i∈[m]\{v_{i}\}_{i\in[m]}, the NTK function p(x)p(x) is given as

With probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)} it satisfies ∥ζ+δ∥∞≤τ\|\zeta+\delta\|_{\infty}\leq\tau. When this happens, we must have p(z+δ+ζ)⋅sign(⟨w⋆,z⟩)≥0p(z+\delta+\zeta)\cdot\mathsf{sign}(\langle w^{\star},z\rangle)\geq 0. ∎

One can carefully apply the Taylor expansion of the smoothed indicator function (using the randomness of ζ\zeta), to derive the following claim. (Detailed proof in Section G.4.)

Consider any NTK function p(x)p(x) with parameters ∥wi∥2≥d2\|w_{i}\|_{2}\geq\frac{\sqrt{d}}{2}, ∥wi∥∞≤log⁡2d\|w_{i}\|_{\infty}\leq\log^{2}d, ρi∼N(0,σρi2)\rho_{i}\sim\mathcal{N}(0,\sigma_{\rho_{i}}^{2}) with ∣σρi∣≤do(1)|\sigma_{\rho_{i}}|\leq d^{o(1)} and ∣bi∣≤do(1)|b_{i}|\leq d^{o(1)}. Suppose τ∈[1d1/5,1]\tau\in[\frac{1}{d^{1/5}},1], then there exists coefficients {ci,r,ci,r′,ci′′}i∈[m],r≥0\{c_{i,r},c^{\prime}_{i,r},c^{\prime\prime}_{i}\}_{i\in[m],r\geq 0} with

each ∣ci,r∣,∣ci,r′∣≤O(1)|c_{i,r}|,|c^{\prime}_{i,r}|\leq O\left(1\right),

each ∣ci,r′∣≤∣ci,r∣⋅O(d−0.1r)|c^{\prime}_{i,r}|\leq|c_{i,r}|\cdot O(d^{-0.1}r),

each ∣ci,r∣≥Ω(1d2)|c_{i,r}|\geq\Omega\left(\frac{1}{d^{2}}\right) for every odd constant r≥1r\geq 1

each ∣ci′′∣≤O(d−1/4)|c^{\prime\prime}_{i}|\leq O(d^{-1/4}).

so that, for every zz with ∥z∥1≤d1/4\|z\|_{1}\leq d^{1/4} and every δ\delta with ∥δ∥∞≤τ/2\|\delta\|_{\infty}\leq\tau/2 and MW(δ)≤τd1/4M_{W}(\delta)\leq\tau d^{1/4}, we have:

Using ∣ci,r∣≥Ω(1d2)|c_{i,r}|\geq\Omega\left(\frac{1}{d^{2}}\right) for odd constant r≥1r\geq 1, and ∣ci,r′∣≤O(d−0.1)⋅∣ci,r∣|c^{\prime}_{i,r}|\leq O(d^{-0.1})\cdot|c_{i,r}|, by applying Lemma G.4,Specifically, one should substitute \|v_{i}\|_{2}\big{(}c_{i,r}\frac{v_{i}}{\|v_{i}\|_{2}}+c^{\prime}_{i,r}\frac{w_{i}}{\|w_{i}\|_{2}}\big{)} as the new viv_{i} when applying Lemma G.4. we know that when r=3C+3r=3C+3 (say wlog. 3C+33C+3 is odd),

Also, for a parameter q=dq=\sqrt{d}, let us apply Lemma G.5 to derive

Let R≥3C+3R\geq 3C+3 be a constant to be chosen later, λmax⁡=max⁡r<R{λr}\lambda_{\max}=\max_{r<R}\{\lambda_{r}\}, and let δmax⁡\delta_{\max} be the choice of δ\delta which maximizes the value of λmax⁡\lambda_{\max}.

Consider the high probability event that MW(z)=O~(1)M_{W}(z)=\widetilde{O}(1), then using MW(δmax⁡)≤τqM_{W}(\delta_{\max})\leq\tau\sqrt{q}, we have

Next, for every s∈[1/2,1]s\in[1/2,1], let us define

On one hand, by applying Lemma G.5 twice for each rr, we know for every set of vectors z1,⋯ ,zqz_{1},\cdots,z_{q} with ∥zi∥∞≤1,MW(zi)=O~(1)\|z_{i}\|_{\infty}\leq 1,M_{W}(z_{i})=\widetilde{O}(1) and supp(zi)∩supp(zj)=∅\text{supp}(z_{i})\cap\text{supp}(z_{j})=\varnothing for i≠ji\not=j, it satisfies

This means by Markov’s inequality, for at least (1−1log⁡d)(1-\frac{1}{\log d}) fraction of the indices j∈[q]j\in[q], denoting them by Λ⊆[q]\Lambda\subseteq[q], it satisfies

On the other hand, by Claim G.7, we know that there is an s∈[1/2,1]s\in[1/2,1] such that

Without loss of generality, suppose q<R(0,s)q_{<R}(0,s) is positive and q<R(0,s)≥Ω(λmax⁡)q_{<R}(0,s)\geq\Omega\left(\lambda_{\max}\right).

Combining the two, when τ100000(C+1)≥1d\tau^{100000(C+1)}\geq\frac{1}{d}, we derive that for those j∈Λj\in\Lambda,

Thus, combining with (G.1), (G.2) and (G.3), we have for those j∈Λj\in\Lambda,

This finishes the proof of Theorem G.1 ■\blacksquare

G.2 Tensor Lower Bound

Next, for each degree-rr homogenous part of the polynomial expansion of Claim G.3, we can write it as a tensor and lower bound its Frobenius norm as follows.

We have as long as r≥3Cr\geq 3C, then w.p. ≥1−e−Ω(log⁡2d)\geq 1-e^{-\Omega(\log^{2}d)} over the randomness of {wi}i∈[m]\{w_{i}\}_{i\in[m]}, for every {vi}i∈[m]\{v_{i}\}_{i\in[m]} we have

Consider any fixed j∈[m]j\in[m], and some γ∈\gamma\in to be chosen later.

Let us define x=wj2∥wj∥2+γvj2∥vj∥2x=\frac{w_{j}}{2\|w_{j}\|_{2}}+\gamma\frac{v_{j}}{2\|v_{j}\|_{2}} which satisfies ∥x∥2≤1\|x\|_{2}\leq 1. We have

Note that for every j≠ij\not=i, with probability at least 1−e−Ω(log⁡2d)1-e^{-\Omega(\log^{2}d)},

This implies that as long as ∣γ∣≤1d|\gamma|\leq\frac{1}{\sqrt{d}},

Since the above lower bound holds for every ∣γ∣≤1d|\gamma|\leq\frac{1}{\sqrt{d}} and every j∈[d]j\in[d], we immediately know

This implies our bound on the Frobenius norm as well. ∎

G.3 Tensor Perturbation

We present the following critical lemma, which serves as the major step to prove the non-robustness of Neural Tangent Kernel:

For the first item, we can simply let δ∼N(0,τ2log⁡2d)\delta\sim\mathcal{N}\left(0,\frac{\tau^{2}}{\log^{2}d}\right). This choice of δ\delta satisfies ∥δ∥∞≤τ\|\delta\|_{\infty}\leq\tau and MW(δ)≤τqM_{W}(\delta)\leq\tau\sqrt{q} with high probability. Furthermore, by applying anti-concentration of Gaussian polynomials (see for instance [5, Lemma I.1]), we know with at least constant probability ∣T(δ⊗r)∣≥Ω(∥T∥F)τr|T(\delta^{\otimes r})|\geq\frac{\Omega(\|T\|_{F})}{\tau^{r}}. This proves the first item.

To see the second item, we first note by tensor rr-linearity and symmetry,

and therefore we only need to bound the terms on the right hand side for any fixed r′∈[r]r^{\prime}\in[r].

From these notions one can directly calculate that

On the other hand, we have ∥Zi∥∞≤τ\|Z_{i}\|_{\infty}\leq\tau and moreover, using the randomness of ξi,j\xi_{i,j}, we know w.h.p. ∣MW(Zi)∣=O~(τq)|M_{W}(Z_{i})|=\widetilde{O}(\tau\sqrt{q}) for every i∈[q]i\in[q]. Hence, by Claim G.8, we know that

Putting them together, we have ∣∑i∈[q]γiT(zi⊗r′,y⊗(r−r′))∣=O~(λτr′)\left|\sum_{i\in[q]}\gamma_{i}T(z_{i}^{\otimes r^{\prime}},y^{\otimes(r-r^{\prime})})\right|=\widetilde{O}\left(\frac{\lambda}{\tau^{r^{\prime}}}\right), and since this holds for every γi∈{−1,1}\gamma_{i}\in\{-1,1\}, we conclude that:

Putting this back to the binomial expansion finishes the proof. ∎

G.4 Smoothed ReLU Taylor Series: Proof of Claim G.3

We first note the following Taylor expansion formula for smoothed ReLU.

Let a≥0a\geq 0 be any real and ρ∼N(0,σ2)\rho\sim\mathcal{N}(0,\sigma^{2}) for σ≥a\sigma\geq a. Then, for every x∈[−a,a]x\in[-a,a],

where ∣c2i∣=Θ(1i!),∣c2i+1′∣=Θ(1(i+1)!)|c_{2i}|=\Theta\left(\frac{1}{i!}\right),|c_{2i+1}^{\prime}|=\Theta\left(\frac{1}{(i+1)!}\right)

so using Taylor expansion of e−x22σ2e^{-\frac{x^{2}}{2\sigma^{2}}} we prove the first equation. As for the second equation, we have

Using Taylor expansion and integrating once, we prove the second equation. ∎

Specifically, for each i∈[m]i\in[m], denoting by x=z+δx=z+\delta, we wish to apply Claim G.6 to

We first deal with the ♡\heartsuit part. Using Claim G.6, we have

for ∣c2r+1′∣=Θ(1(r+1)!)|c_{2r+1}^{\prime}|=\Theta\left(\frac{1}{(r+1)!}\right). Similarly, we also have

Putting them together, and using the fact that bi≪d−0.2≪σb_{i}\ll d^{-0.2}\ll\sigma, we can write

for ∣c2r′′∣≤O(1(r)!)|c^{\prime\prime}_{2r}|\leq O\left(\frac{1}{(r)!}\right) for every r≥0r\geq 0 and ∣c2r+1′′∣≥Ω(1(r+1)!)|c^{\prime\prime}_{2r+1}|\geq\Omega\left(\frac{1}{(r+1)!}\right).

Let us now focus on the ♢\diamondsuit part. Let vi∥v_{i}^{\parallel} be the part of viv_{i} that is parallel to wiw_{i}. Then obviously we have

Above, the last ① is due to σρi≤do(1)\sigma_{\rho_{i}}\leq d^{o(1)} and ∥wi∥2≥Ω(d)\|w_{i}\|_{2}\geq\Omega(\sqrt{d}).

for ∣c2r∣=Θ(1r!)|c_{2r}|=\Theta\left(\frac{1}{r!}\right). Putting them together, and doing the same thing for the symmetric part, we have

Above, using the property of bi≪d−0.2≪σb_{i}\ll d^{-0.2}\ll\sigma, equation ① holds for some ∣c2r′′′∣≤O(d−0.1r!)|c^{\prime\prime\prime}_{2r}|\leq O(\frac{d^{-0.1}}{r!}) and ∣c2r+1′′′∣≤O(d−0.1(r+1)!)|c^{\prime\prime\prime}_{2r+1}|\leq O(\frac{d^{-0.1}}{(r+1)!}).

Finally, putting the bounds for ♡\heartsuit and ♢\diamondsuit together, and using τ∥wi∥2log⁡8d≤σ≤τ∥wi∥2\frac{\tau\|w_{i}\|_{2}}{\log^{8}d}\leq\sigma\leq\tau\|w_{i}\|_{2}, we derive that

for ∣ci,r′′′′∣≤O(1)|c^{\prime\prime\prime\prime}_{i,r}|\leq O(1) for every r≥0r\geq 0, cr′′′′′≤O(d−0.1r)⋅cr′′′′c^{\prime\prime\prime\prime\prime}_{r}\leq O(d^{-0.1}r)\cdot c^{\prime\prime\prime\prime}_{r} for every r≥0r\geq 0, and ∣ci,r′′′′∣≥Ω(1d2)|c^{\prime\prime\prime\prime}_{i,r}|\geq\Omega\left(\frac{1}{d^{2}}\right) for every odd constant r≥1r\geq 1. This finishes the proof of Claim G.3. ∎

G.5 Simple Lemmas

We have the following claim relating polynomial value with its coefficients:

Conversely, by writing ∑r=0Rcr′(x−0.5)r=∑r=0Rcrxr\sum_{r=0}^{R}c_{r}^{\prime}(x-0.5)^{r}=\sum_{r=0}^{R}c_{r}x^{r}, we also have the other direction and therefore

Now, notice that ∣drdxrq(x)∣x=0∣=Θ(∣cr′∣)\left|\frac{d^{r}}{dx^{r}}q(x)\mid_{x=0}\right|=\Theta(|c_{r}^{\prime}|), so we can apply Markov brother’s inequality to derive that

Using this Claim, we also have the following claim about symmetric tensor:

λ1≤λ2\lambda_{1}\leq\lambda_{2} is obvious so let us prove the other direction. Define polynomial

The coefficient of p(s)p(s) at degree ∑r′∈[r](r+1)r′−1\sum_{r^{\prime}\in[r]}(r+1)^{r^{\prime}-1} is Θ(T(x1,x2,⋯ ,xr))\Theta\left(T(x_{1},x_{2},\cdots,x_{r})\right). Thus, applying Claim G.7 and appropriately scaling the operator, we complete the proof. ∎

Appendix H Appendix for Probability Theory

For every subset Λ⊆[d]\Lambda\subseteq[d], every rr, and every t>0t>0,

For every subset Λ⊆[d]\Lambda\subseteq[d] with ∣Λ∣≥Ω(d)|\Lambda|\geq\Omega(d), and every t>0t>0,

Recall we have Pr⁡zj[zj≠0]≥Ω(kd)\operatornamewithlimits{\mathbf{Pr}}_{z_{j}}[z_{j}\neq 0]\geq\Omega(\frac{k}{d}) for each j∈Λj\in\Lambda. Let Λ′⊆Λ\Lambda^{\prime}\subseteq\Lambda be the subset of such indices jj with non-zero zjz_{j}, so by our assumption we have ∣zj∣≥1k|z_{j}|\geq\frac{1}{\sqrt{k}} for each j∈Λ′j\in\Lambda^{\prime}. By Chernoff bound, with probability at least 1−e−Ω(∣Λ∣k/d)1-e^{-\Omega(|\Lambda|k/d)}, we know ∣Λ′∣≥Ω(kd)⋅∣Λ∣|\Lambda^{\prime}|\geq\Omega(\frac{k}{d})\cdot|\Lambda|.

Conditioning on such Λ′\Lambda^{\prime}, by the Littlewood-Offord problem (a.k.a. small ball probability theorem, or anti-concentration for sum of Bernoulli variables, see ), we know

Using the property of Gaussian variables and B=O(1)B=O(1), we have

and using the above Wasserstein distance bound, we have

H.2 McDiarmid’s Inequality and An Extension

We state the standard McDiarmid’s inequality,

Consider independent random variables x1,⋯ ,xn∈Xx_{1},\cdots,x_{n}\in{\cal X} and a mapping f:Xn→Rf:{\cal X}^{n}\rightarrow\R. If for all i∈[n]i\in[n] and for all y1,⋯ ,yn,yi′∈Xy_{1},\cdots,y_{n},y_{i}^{\prime}\in{\cal X}, the function ff satisfies

We prove a more general version of McDiarmid’s inequality,

Let w1,…,wNw_{1},\dots,w_{N} be independent random variables and f ⁣:(w1,…,wN)↦[0,B]f\colon(w_{1},\dots,w_{N})\mapsto[0,B]. Suppose it satisfies for every k∈{2,3,…,N}k\in\{2,3,\dots,N\},

with probability at least 1−p1-p over w1,…,wNw_{1},\dots,w_{N}, it satisfies

with probability at least 1−p1-p over w1,…,wk−1,wk+1,…,wNw_{1},\dots,w_{k-1},w_{k+1},\dots,w_{N}, it satisfies

For each t=1,…,N−1t=1,\dots,N-1, we have with probability at least 1−p1-\sqrt{p} over w1,…,wtw_{1},\dots,w_{t}, it satisfies

We also have with probability at least 1−p1-\sqrt{p} over w1,…,wtw_{1},\dots,w_{t}, it satisfies

We denote by K≤tK_{\leq t} the event (over w≤t=(w1,…,wt)w_{\leq t}=(w_{1},\dots,w_{t})) that the above two statements hold. We know that Pr⁡[w≤t∈K≤t]≥1−2p\operatornamewithlimits{\mathbf{Pr}}[w_{\leq t}\in K_{\leq t}]\geq 1-2\sqrt{p}. For notational simplicity, we denote by K≤NK_{\leq N} the full set over all possible (w1,…,wN)(w_{1},\dots,w_{N}).

Define random variable XtX_{t} (which depends only on w1,…,wtw_{1},\dots,w_{t}) as

For every tt and fixed w1,…,wt−1w_{1},\dots,w_{t-1}.

If (w≤1,…,w<t)∉K≤1×⋯×K<t(w_{\leq 1},\dots,w_{<t})\not\in K_{\leq 1}\times\cdots\times K_{<t}, then Xt=Xt−1=0X_{t}=X_{t-1}=0.

If (w≤1,…,w<t)∈K≤1×⋯×K<t(w_{\leq 1},\dots,w_{<t})\in K_{\leq 1}\times\cdots\times K_{<t},

If w≤t∉K≤tw_{\leq t}\not\in K_{\leq t}, then Xt−Xt−1=0−Xt−1≤0X_{t}-X_{t-1}=0-X_{t-1}\leq 0.

Recall the property w<t∈K<tw_{<t}\in K_{<t}, we know with probability at least 1−p1-\sqrt{p} over wtw_{t} and w>tw_{>t}, it satisfies

Taking expectation over wtw_{t} and w>tw_{>t}, we have

This precisely means Xt−Xt−1≤c+pBX_{t}-X_{t-1}\leq c+\sqrt{p}B.

Using the property w<t∈K<tw_{<t}\in K_{<t}, we know with probability at least 1−p1-\sqrt{p} over w>tw_{>t}, it satisfies

Taking expectation also over w>tw_{>t}, we have

In sum, we have just shown that for all choices of w1,…,wt−1w_{1},\dots,w_{t-1},

and we have XN=f(w1,…,wN)X_{N}=f(w_{1},\dots,w_{N}) with probability at least 1−2Np1-2N\sqrt{p} (and XN=0X_{N}=0 with the remaining probability). Also recalling

Let us state, for completeness’ sake, a simple one-sided Bernstein form of martingale concentration (that we do not know a good reference to it).

Suppose we have a submartingale sequence X0,X1,…,XNX_{0},X_{1},\dots,X_{N}, satisfying:

Define potential function Ψt=eη2cXt\Psi_{t}=e^{\frac{\eta}{2c}X_{t}} for some η∈(0,1)\eta\in(0,1) to be chosen later. We have

where the inequality is due to ey≤1+y+y2e^{y}\leq 1+y+y^{2} which holds for all −∞<y≤0.5-\infty<y\leq 0.5. Taking conditional expectation, we have

Choosing the optimal η∈(0,1)\eta\in(0,1) gives us bound

References