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arXiv:2406.00487v1 [cs.LG] 01 Jun 2024

Optimistic Rates for Learning from Label Proportions

Gene Li thanks: Part of this work was done while GL was an intern at Google Research. Affiliation: TTIC    Lin Chen Affiliation: Google Research    Adel Javanmard Affiliation: Google Research Affiliation: University of Southern California    Vahab Mirrokni Affiliation: Google Research
Abstract

We consider a weakly supervised learning problem called Learning from Label Proportions (LLP), where examples are grouped into “bags” and only the average label within each bag is revealed to the learner. We study various learning rules for LLP that achieve PAC learning guarantees for classification loss. We establish that the classical Empirical Proportional Risk Minimization (EPRM) learning rule (Yu et al., 2014) achieves fast rates under realizability, but EPRM and similar proportion matching learning rules can fail in the agnostic setting. We also show that (1) a debiased proportional square loss, as well as (2) a recently proposed EasyLLP learning rule (Busa-Fekete et al., 2023) both achieve “optimistic rates” (Panchenko, 2002); in both the realizable and agnostic settings, their sample complexity is optimal (up to log factors) in terms of ε,δ\varepsilon,\delta, and VC dimension.

1 Introduction

We study Learning from Label Proportions (LLP), which is a framework for weakly supervised learning. In the standard supervised learning framework, the learner has access to a dataset of nn i.i.d. labeled examples {(xi,yi)}i=1n(𝒳×𝒴)n\{(x_{i},y_{i})\}_{i=1}^{n}\in(\mathcal{X}\times\mathcal{Y})^{n}, and the goal is to learn an accurate predictor f^:𝒳𝒴\widehat{f}:\mathcal{X}\mapsto\mathcal{Y}. In the LLP framework, the learner does not get access to the true labels {yi}i=1n\{y_{i}\}_{i=1}^{n}; instead, training data is organized into “bags” which contain multiple unlabeled examples, and only the average (or “aggregated”) label within the bag is provided to the learner.

The LLP framework has been studied in a long line of work, dating back to Kück and de Freitas (2005); Musicant et al. (2007). LLP is motivated by practical machine learning problems where individual labels are expensive to obtain or unavailable, see, e.g., applications in high energy physics (Dery et al., 2017), election prediction (Sun et al., 2017), and RADAR image classification (Ding et al., 2017). More recently, LLP was proposed as a mechanism to provide user privacy (Diemert et al., 2022); for example, in the Apple SKAN API (Apple, 2024) and Google Chrome’s Private Aggregation API (Google, 2024), only aggregated labels are provided for ad conversion reporting.

Problem Formulation.

Let 𝒳\mathcal{X} represent the instance space and 𝒴={0,1}\mathcal{Y}=\{0,1\} denote the label space. In LLP, we are given nn bags of examples {(Bi,αi)}i=1n\{(B_{i},\alpha_{i})\}_{i=1}^{n} where each bag Bi={xi,j}j=1kB_{i}=\{x_{i,j}\}_{j=1}^{k} contains kk instances and αi=1kj=1kyi,j\alpha_{i}=\frac{1}{k}\sum_{j=1}^{k}y_{i,j} is the average label within the bag. We assume that each instance (xi,j,yi,j)𝒟(x_{i,j},y_{i,j})\sim\mathcal{D} is independently and identically distributed (i.i.d.) according to an unknown distribution 𝒟\mathcal{D}. We also use (B,α)𝒟(B,\alpha)\sim\mathcal{D} to indicate that a bag is drawn from 𝒟\mathcal{D}. Unlike supervised learning, in LLP the learner does not get individual labels yi,jy_{i,j}, but only the aggregated label αi\alpha_{i} for a collection of kk random instances.

We study the PAC learning objective of finding an accurate instance-level predictor from label proportion data that competes with the best predictor in specified function class 𝒴𝒳\mathcal{F}\subseteq\mathcal{Y}^{\mathcal{X}}. Given parameters ε,δ(0,1)\varepsilon,\delta\in(0,1), we seek a predictor f^:𝒳𝒴\widehat{f}:\mathcal{X}\to\mathcal{Y} that satisfies the following condition with probability at least 1δ1-\delta:

(f^)inff(f)+ε,where(f)𝔼(x,y)𝒟[01(y,f(x))],\displaystyle\mathcal{L}(\widehat{f})\leq\inf_{f\in\mathcal{F}}\mathcal{L}(f)+\varepsilon,\quad\text{where}\quad\mathcal{L}(f)\coloneqq\En_{(x,y)\sim\mathcal{D}}[\ell^{01}(y,f(x))], (1)

and 01:𝒴×𝒴{0,1}\ell^{01}:\mathcal{Y}\times\mathcal{Y}\to\{0,1\} is the classification loss, defined as 01(y,y^)=𝟙{yy^}\ell^{01}(y,\widehat{y})=\mathbbm{1}\left\{y\neq\widehat{y}\right\}. When the bag size k=1k=1, this becomes the classic binary classification setup, for which it is known precisely that the VC dimension of \mathcal{F} characterizes the sample complexity of learning.

The fundamental question in LLP is to establish the sample complexity, i.e., the number of bags nn required to guarantee Eq. (1), in terms of the VC dimension of \mathcal{F}, bag size kk, and accuracy parameters ε,δ\varepsilon,\delta. Our paper investigates several proposed learning rules designed to directly minimize classification loss. We establish generalization bounds for these learning rules under both the realizable setting (where inff(f)=0\inf_{f\in\mathcal{F}}\mathcal{L}(f)=0 and fast 1/n1/n rates are possible) and the agnostic setting (where inff(f)\inf_{f\in\mathcal{F}}\mathcal{L}(f) can be arbitrary, and one gets slow 1/n1/\sqrt{n} rates). Specifically, we adopt the optimistic rates framework (Panchenko, 2002; Srebro et al., 2010) which uses localized uniform convergence bounds to show generalization guarantees that interpolate between the realizable and agnostic setting.

Notation.

We denote the marginal label proportion p=[y=1]p=\Pr[y=1]. Whenever the function class \mathcal{F} is clear from the context, we denote d=VC()d=\mathrm{VC}(\mathcal{F}). We adopt standard big-oh notation and use O~()\widetilde{O}(\cdot) to hide poly(logk,logn)\mathrm{poly}(\log k,\log n) dependencies in our bounds.

1.1 Our Contributions

We obtain the following results on LLP for the classification objective (1).

Success and Failure of Empirical Proportional Risk Minimization (Section 2):

We study the Empirical Proportional Risk Minimization (EPRM) learning rule (Yu et al., 2014), which is a natural extension of Empirical Risk Minimization (ERM) to the LLP setting. Concretely we prove that under realizability, f^EPRMargminf1ni=1n𝟙{1kj=1kf(xi,j)αt}\widehat{f}_{\mathrm{EPRM}}\coloneqq\argmin_{f\in\mathcal{F}}\tfrac{1}{n}\sum_{i=1}^{n}\mathbbm{1}\{\tfrac{1}{k}\sum_{j=1}^{k}f(x_{i,j})\neq\alpha_{t}\} achieves the sample complexity guarantee

n=O(dlogklog(1/ε)+log(1/δ)ε).\displaystyle n=O\left(\frac{d\log k\cdot\log(1/\varepsilon)+\log(1/\delta)}{\varepsilon}\right).

However, in the agnostic setting we show that EPRM cannot attain polynomial (in kk) sample complexity, and similar “folklore” learning rules based on minimization of proportional square or log losses even fail to return a predictor with constant suboptimality.

Optimistic Rates for Debiased Square Loss (Section 3):

To address the failure of proportional loss learning rules in the agnostic setting, we consider a simple debiased variant of the proportional square loss. We show that the debiased square loss learning rule f^DSQ\widehat{f}_{\mathrm{DSQ}} achieves the optimistic rate:

(f^DSQ)L+O~(k2(d+log(1/δ))n+Lk2(d+log(1/δ))n),\displaystyle\mathcal{L}(\widehat{f}_{\mathrm{DSQ}})\leq L^{\star}+\widetilde{O}\left(\frac{k^{2}\left(d+\log(1/\delta)\right)}{n}+\sqrt{\frac{L^{\star}\cdot k^{2}\left(d+\log(1/\delta)\right)}{n}}\right),

where we denote L=minf(f)L^{\star}=\min_{f\in\mathcal{F}}\mathcal{L}(f). Here, observe that under realizability (with L=0L^{\star}=0), f^DSQ\widehat{f}_{\mathrm{DSQ}} enjoys a fast 1/n1/n rate, while in the agnostic setting we recover the 1/n1/\sqrt{n} rate, both of which are optimal (up to log factors) in terms of dd, log(1/δ)\log(1/\delta), and nn.

Optimistic Rates for EasyLLP (Section 4):

We study an alternative approach called EasyLLP, which was recently proposed by Busa-Fekete et al. (2023). Specialized to the classification setting, they show a 1/n1/\sqrt{n} rate (hiding dependence on dd, kk, and δ\delta). We improve upon their result to show an optimistic rate similar to the one achieved by the debiased square loss, showing that EasyLLP can indeed adapt to realizability. Our analysis reveals a curious phenomenon: in the realizable setting, EasyLLP exhibits a separation between loss estimation (which is necessarily Ω(1/n)\Omega(1/\sqrt{n}) even for the optimal predictor ff^{\star}) and learning (which is O~(1/n)\widetilde{O}(1/n) by the optimistic rates guarantee).

Lower Bounds (Section 5):

We investigate the optimal dependence on the bag size kk, since our bounds are tight (up to log factors) in the other parameters. A trivial lower bound of n=Ω(1/k)n=\Omega(1/k) follows because LLP with nn bags is only harder than supervised learning with nknk bags. It turns out this cannot be improved in general for all \mathcal{F}, but we give an explicit example of a function class \mathcal{F} for which the minimax sample complexity has larger dependence on kk.

Experiments (Section 6):

We empirically evaluate gradient-based versions of the learning rules considered herein on binary classification versions of MNIST and CIFAR10 studied in Busa-Fekete et al. (2023) for a wide range of NN architectures. We find that proportion matching and the debiased square loss perform the best; furthermore, we demonstrate that the debiased square loss enjoys faster optimization than proportion matching, as early in training the debiased square loss is a better estimate of the true instance-level loss.

1.2 Related Work

Learning from Label Proportions.

The problem of LLP has been studied in a long line of work (Chen et al., 2006; Musicant et al., 2007; Kück and de Freitas, 2005; Quadrianto et al., 2008). Most works either assume some kind of distributional assumptions on bag/label generation (Kück and de Freitas, 2005; Quadrianto et al., 2008; Patrini et al., 2014; Scott and Zhang, 2020; Zhang et al., 2022), construct bags adaptively (Chen et al., 2023; Javanmard et al., 2024b), or study approaches to minimize a surrogate loss (Rueping, 2010; Yu et al., 2013; Qi et al., 2016; Shi et al., 2019; Dulac-Arnold et al., 2019; Javanmard et al., 2024a). On the computational side, Saket (2021); Saket (2022) shows that even in the realizable setting learning linear thresholds for LLP is NP-hard and study SDP relaxations for this task. Thus, the aforementioned papers are not directly relevant to the goals of this work in providing distribution-free statistical guarantees on classification loss.

Several works provide guarantees on the instance-level classification loss. Yu et al. (2014) introduce the EPRM learning rule. While their main focus is proving guarantees on the proportional risk, they show how to translate these to instance-level guarantees when the bags are “pure”—meaning the label proportions αt\alpha_{t} are close to 00 or 11. They also give numerical bounds which indicate that EPRM achieves instance-level guarantees under realizability. Chen et al. (2023) introduce a similar debiased square loss learning rule and prove a O(1/n)O(1/\sqrt{n}) rate in the agnostic setting; while Busa-Fekete et al. (2023) introduce the EasyLLP framework and prove a O(1/n)O(1/\sqrt{n}) rate.

Optimistic Rates.

Optimistic rates date back to seminal work of Vapnik and Chervonenkis (2015) and were expanded upon by Bousquet (2002); Koltchinskii and Panchenko (2000); Bartlett et al. (2005) using the technique of localized Rademacher complexities. Optimistic rates have been studied in various other contexts such as optimization with smooth losses (Srebro et al., 2010), multi-task learning (Yousefi et al., 2018; Watkins et al., 2023), vector-valued learning (Reeve and Kaban, 2020), and overparameterized regression (Zhou et al., 2021; Zhou et al., 2022; Zhou et al., 2023).

2 Empirical Proportional Risk Minimization

The most direct approach to finding a good predictor is the so-called empirical proportional risk minimization (EPRM) approach, which simply returns a predictor that matches the most label proportions (Yu et al., 2014). Concretely, we can consider the learning rule

f^EPRMargminf1ni=1n𝟙{1kj=1kf(xi,j)αi}.\displaystyle\widehat{f}_{\mathrm{EPRM}}\coloneqq\argmin_{f\in\mathcal{F}}\frac{1}{n}\sum_{i=1}^{n}\mathbbm{1}\left\{\frac{1}{k}\sum_{j=1}^{k}f(x_{i,j})\neq\alpha_{i}\right\}. (2)

Observe that in the setting of k=1k=1, EPRM recovers the classical empirical risk minimizer (ERM).

More generally, one can also define learning rules which minimize some other bag loss :[0,1]×[0,1]\ell:[0,1]\times[0,1]\to\mathbb{R} between the predicted label proportions α^=1kif(xi)\widehat{\alpha}=\frac{1}{k}\sum_{i}f(x_{i}) and the true label proportions α\alpha, i.e., the square loss SQ(α,α^)=(αα^)2\ell_{\mathrm{SQ}}(\alpha,\widehat{\alpha})=(\alpha-\widehat{\alpha})^{2} or the log loss LOG(α,α^)=αlogα^(1α)log(1α^)\ell_{\mathrm{LOG}}(\alpha,\widehat{\alpha})=-\alpha\log\widehat{\alpha}-(1-\alpha)\log(1-\widehat{\alpha}). In the literature, these are also called EPRM or proportion matching learning rules, and they are a “folklore” approach which, in conjunction with gradient-based methods, are competitive in practice (Busa-Fekete et al., 2023). To disambiguate, we exlusively refer to the learning rule (2) as EPRM and call the more general class of these learning rules as proportion matching.

In this section, we provide theoretical results which substantiate conventional wisdom surrounding EPRM. First, we show that under realizability, the EPRM learning rule attains fast rates for classification. However, in the agnostic setting, we illustrate that proportion matching can be ill-behaved, as we demonstrate an example for which minimizing the proportional risk gives no guarantees on the instance-level classification performance.

2.1 Fast Rates for EPRM Under Realizability

We show the following generalization guaranteee for EPRM under realizability.

Theorem 1.

Let \mathcal{F} be a symmetric function class, i.e. if ff\in\mathcal{F}, then 1f1-f\in\mathcal{F}. Let bag size k11k\geq 11, ε(0,1/(4k))\varepsilon\in\left(0,1/(4k)\right), and δ(0,1)\delta\in(0,1). As long as n=O(dlogklog(1/ε)+log(1/δ)ε)n=O(\tfrac{d\log k\cdot\log(1/\varepsilon)+\log(1/\delta)}{\varepsilon}), for any realizable distribution 𝒟\mathcal{D}, with probability at least 1δ1-\delta over the draw of the sample, (f^EPRM)ε\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})\leq\varepsilon.

Previous works have suggested that EPRM (or more generally, proportion matching learning rules) can succeed under realizability: Yu et al. (2014) give numerical evidence to show that a guarantee on bag proportions can be translated to a guarantee on the instance-level classification error, and Busa-Fekete et al. (2023) show that under some conditions on the loss function, minimizers for a population proportion matching loss are also minimizers for the instance-level loss. However, to the best of our knowledge, Theorem 1 is the first result which provides a concrete generalization bound for bounded VC classes for the EPRM learning rule.

A few comments about Theorem 1 are in order.

  • The assumption that \mathcal{F} is symmetric is mild and due to technical reasons. Note that any nonsymmetric \mathcal{F} can be enlarged to a symmetric one with VC dimension at most 2d+12d+1.

  • The bag size assumption is technically required, and we conjecture it can be removed. It is a mild assumption since if one has bags of size k10k\leq 10, then one can preprocess the dataset to combine sets of bags to form a larger bag of size at least 11, and then compute f^EPRM\widehat{f}_{\mathrm{EPRM}} on the preprocessed dataset. This achieves the same guarantee (albeit with smaller range of ε\varepsilon).

Proof of Theorem 1.

The proof has three steps. First, we show via standard uniform convergence arguments that the predictor f^EPRM\widehat{f}_{\mathrm{EPRM}} must have small population proportional matching error. Next, we relate the proportion matching error to the classification error to show that f^EPRM\widehat{f}_{\mathrm{EPRM}} must have classification error (f^EPRM)[ε,1ε]\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})\notin[\varepsilon,1-\varepsilon]. Finally, we show that f^EPRM\widehat{f}_{\mathrm{EPRM}} must have classification error (f^EPRM)ε\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})\leq\varepsilon, as otherwise we would have selected the predictor 1f^EPRM1-\widehat{f}_{\mathrm{EPRM}}.

Step 1.

We rewrite our problem as a binary classification problem and apply standard uniform convergence guarantees. Define the function class 𝒢:𝒳n×[0,1]{0,1}\mathcal{G}:\mathcal{X}^{n}\times[0,1]\to\{0,1\} as

𝒢={gf:(B,α)𝟙{1kj=1kf(xj)α}:f}.\displaystyle\mathcal{G}=\left\{g_{f}:(B,\alpha)\mapsto\mathbbm{1}\left\{\frac{1}{k}\sum_{j=1}^{k}f(x_{j})\neq\alpha\right\}:f\in\mathcal{F}\right\}.

We claim that VC(𝒢)O(dlogk)\mathrm{VC}(\mathcal{G})\leq O(d\log k). To show this, consider any X={(B1,α1),,(Bm,αm)}X=\{(B_{1},\alpha_{1}),\cdots,(B_{m},\alpha_{m})\}, and define the projection of 𝒢\mathcal{G} onto XX as 𝒢X{(gf(B1,α1),,gf(Bm,αm)):f}\mathcal{G}_{X}\coloneqq\{(g_{f}(B_{1},\alpha_{1}),\cdots,g_{f}(B_{m},\alpha_{m})):f\in\mathcal{F}\}. It suffices to show that when m=O(dlogk)m=O(d\log k), the set of labellings for XX is of size |𝒢X|<2m\lvert\mathcal{G}_{X}\rvert<2^{m}. Observe that the labelling of XX is determined by the labellings of \mathcal{F} on the mkmk points, so by Sauer’s lemma |𝒢X|(emk/d)d\lvert\mathcal{G}_{X}\rvert\leq(emk/d)^{d}. Therefore, when m=O(dlogk)m=O(d\log k) we have |𝒢X|<2m\lvert\mathcal{G}_{X}\rvert<2^{m}.

The EPRM can be written as f^EPRMargminf1ni=1n01(0,gf(Bi,αi))\widehat{f}_{\mathrm{EPRM}}\coloneqq\argmin_{f\in\mathcal{F}}\frac{1}{n}\sum_{i=1}^{n}\ell^{01}(0,g_{f}(B_{i},\alpha_{i})). Applying the uniform convergence guarantee for VC classes (Shalev-Shwartz and Ben-David, 2014, e.g.,), we see that as long as n=O(dlogklog(1/ε)+log(1/δ)ε)n=O(\frac{d\log k\cdot\log(1/\varepsilon)+\log(1/\delta)}{\varepsilon}), with probability at least 1δ1-\delta we have

(B,α)𝒟[1kj=1kf^EPRM(xj)α]ε.\displaystyle\Pr_{(B,\alpha)\sim\mathcal{D}}\bigg[\frac{1}{k}\sum_{j=1}^{k}\widehat{f}_{\mathrm{EPRM}}(x_{j})\neq\alpha\bigg]\leq\varepsilon. (3)

Henceforth, we will condition on the event in Eq. (3) holding.

Step 2.

Now we show that a proportional risk guarantee of the form Eq. (3) translates to a guarantee on the instance-level loss ()\mathcal{L}(\cdot). Let ff^{\star}\in\mathcal{F} be the optimal predictor that achieves (f)=0\mathcal{L}(f^{\star})=0, and for any ff\in\mathcal{F} define dis(f,f){x𝒳:f(x)f(x)}\mathrm{dis}(f,f^{\star})\coloneqq\left\{x\in\mathcal{X}:f(x)\neq f^{\star}(x)\right\} to be the disagreement region on which ff and ff^{\star} disagree. By definition, (f^EPRM)=x𝒟[xdis(f^EPRM,f)]\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})=\Pr_{x\sim\mathcal{D}}[x\in\mathrm{dis}(\widehat{f}_{\mathrm{EPRM}},f^{\star})]. We will show that (f^EPRM)[ε,1ε]\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})\notin[\varepsilon,1-\varepsilon].

To do so, we bound the probability that f^EPRM\widehat{f}_{\mathrm{EPRM}} does not match the proportion on a freshly sampled bag. We already have an upper bound on this from Eq. (3). Now we compute a lower bound.

(B,α)𝒟[1kj=1kf^EPRM(xj)α]\displaystyle\Pr_{(B,\alpha)\sim\mathcal{D}}\bigg[\frac{1}{k}\sum_{j=1}^{k}\widehat{f}_{\mathrm{EPRM}}(x_{j})\neq\alpha\bigg] (B,α)𝒟[j=1k𝟙{xjdis(f^EPRM,f)} is odd]\displaystyle\geq\Pr_{(B,\alpha)\sim\mathcal{D}}\bigg[\sum_{j=1}^{k}\mathbbm{1}\left\{x_{j}\in\mathrm{dis}(\widehat{f}_{\mathrm{EPRM}},f^{\star})\right\}\text{ is odd}\bigg]
=1212(12(f^EPRM))k.\displaystyle=\frac{1}{2}-\frac{1}{2}\left(1-2\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})\right)^{k}. (4)

The first inequality follows because if the bag contains an odd number of points in the disagreement set, then it is impossible for f^EPRM\widehat{f}_{\mathrm{EPRM}} and ff^{\star} to have the same proportional label.

For sake of contradiction, suppose that (f^EPRM)[ε,1ε]\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})\in[\varepsilon,1-\varepsilon]. Then we know 1212(12(f^EPRM))k1212(12ε)k\tfrac{1}{2}-\tfrac{1}{2}(1-2\mathcal{L}(\widehat{f}_{\mathrm{EPRM}}))^{k}\geq\tfrac{1}{2}-\tfrac{1}{2}(1-2\varepsilon)^{k}. However, if ε(0,1/2)\varepsilon\in(0,1/2), we arrive at a contradiction, since ε(0,1/2)\varepsilon\in(0,1/2) implies that for any bag size k2k\geq 2, we have 12ε>(12ε)k1-2\varepsilon>(1-2\varepsilon)^{k}, which implies that ε<1212(12ε)k\varepsilon<\frac{1}{2}-\frac{1}{2}\left(1-2\varepsilon\right)^{k}, so Eqs. (3) and (4) cannot simultaneously hold. Therefore we must have (f^EPRM)[ε,1ε]\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})\notin[\varepsilon,1-\varepsilon].

Step 3.

Now we establish that (f^EPRM)[1ε,1]\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})\notin[1-\varepsilon,1]. We claim the following: for any near optimal predictor f~ε{f:(f)ε}\widetilde{f}\in\mathcal{F}_{\varepsilon}\coloneqq\{f\in\mathcal{F}:\mathcal{L}(f)\leq\varepsilon\}, we must have

1ni=1n𝟙{1kj=1kf~(xi,j)αi}<1ni=1n𝟙{1kj=1k1f~(xi,j)αi}.\displaystyle\frac{1}{n}\sum_{i=1}^{n}\mathbbm{1}\left\{\frac{1}{k}\sum_{j=1}^{k}\widetilde{f}(x_{i,j})\neq\alpha_{i}\right\}<\frac{1}{n}\sum_{i=1}^{n}\mathbbm{1}\left\{\frac{1}{k}\sum_{j=1}^{k}1-\widetilde{f}(x_{i,j})\neq\alpha_{i}\right\}. (5)

Call this event (f~)\mathcal{E}(\widetilde{f}). From here, the result that (f^EPRM)[1ε,1]\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})\notin[1-\varepsilon,1] follows because if (f^EPRM)[1ε,1]\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})\in[1-\varepsilon,1], then it could not have been the EPRM, as the predictor 1f^EPRM1-\widehat{f}_{\mathrm{EPRM}} has strictly better empirical proportional risk and also lies in the class \mathcal{F} by the symmetric assumption.

We now prove Eq. (5). Consider any predictor f~ε\widetilde{f}\in\mathcal{F}_{\varepsilon}. Define the indicator variable Zi{0,1}Z_{i}\in\{0,1\} as Zi=𝟙{1kj=1kf~(xi,j)=αi}𝟙{1kj=1k1f~(xi,j)αi}Z_{i}=\mathbbm{1}\{\frac{1}{k}\sum_{j=1}^{k}\widetilde{f}(x_{i,j})=\alpha_{i}\}\cdot\mathbbm{1}\{\frac{1}{k}\sum_{j=1}^{k}1-\widetilde{f}(x_{i,j})\neq\alpha_{i}\}. We see that {1niZi>1/2}(f~)\left\{\frac{1}{n}\sum_{i}Z_{i}>1/2\right\}\subseteq\mathcal{E}(\widetilde{f}). We bound the expectation of ZiZ_{i} as:

𝔼[Zi](B,α)𝒟[xjB:xjdis(f~,f) and α12].\displaystyle\En[Z_{i}]\geq\Pr_{(B,\alpha)\sim\mathcal{D}}\left[\forall x_{j}\in B:x_{j}\notin\mathrm{dis}(\widetilde{f},f^{\star})\text{ and }\alpha\neq\frac{1}{2}\right].

To lower bound this, we can bound the two events separately. First we have

(B,α)𝒟[xjB:xjdis(f~,f)](1ε)k34,\displaystyle\Pr_{(B,\alpha)\sim\mathcal{D}}\left[\forall x_{j}\in B:x_{j}\notin\mathrm{dis}(\widetilde{f},f^{\star})\right]\geq(1-\varepsilon)^{k}\geq\frac{3}{4},

where the last inequality is true whenever ε1/(4k)\varepsilon\leq 1/(4k). In addition,

(B,α)𝒟[α1/2]=1(B,α)𝒟[α=1/2]12k(kk/2)113k/2+1/2,\displaystyle\Pr_{(B,\alpha)\sim\mathcal{D}}[\alpha\neq 1/2]=1-\Pr_{(B,\alpha)\sim\mathcal{D}}[\alpha=1/2]\geq 1-2^{-k}{k\choose k/2}\geq 1-\frac{1}{\sqrt{3k/2+1/2}},

by Stirling’s approximation. Using the law of total probability we get

1\displaystyle 1 (B,α)𝒟[xjB:xjdis(f^,f) or α12]\displaystyle\geq\Pr_{(B,\alpha)\sim\mathcal{D}}\Big[\forall x_{j}\in B:x_{j}\notin\mathrm{dis}(\widehat{f},f^{\star})\text{ or }\alpha\neq\frac{1}{2}\Big]
=(B,α)𝒟[xjB:xjdis(f^,f)]+(B,α)𝒟[α12]\displaystyle=\Pr_{(B,\alpha)\sim\mathcal{D}}\Big[\forall x_{j}\in B:x_{j}\notin\mathrm{dis}(\widehat{f},f^{\star})\Big]+\Pr_{(B,\alpha)\sim\mathcal{D}}\Big[\alpha\neq\frac{1}{2}\Big]
B[xjB:xjdis(f^,f) and α12]\displaystyle\quad\quad\quad-\Pr_{B}\Big[\forall x_{j}\in B:x_{j}\notin\mathrm{dis}(\widehat{f},f^{\star})\text{ and }\alpha\neq\frac{1}{2}\Big]
34+113k/2+1/2(B,α)𝒟[xjB:xjdis(f^,f) and α12],\displaystyle\geq\frac{3}{4}+1-\frac{1}{\sqrt{3k/2+1/2}}-\Pr_{(B,\alpha)\sim\mathcal{D}}\Big[\forall x_{j}\in B:x_{j}\notin\mathrm{dis}(\widehat{f},f^{\star})\text{ and }\alpha\neq\frac{1}{2}\Big],

so therefore

(B,α)𝒟[xjB:xjdis(f^,f) and α12]3413k/2+1/2.\displaystyle\Pr_{(B,\alpha)\sim\mathcal{D}}\Big[\forall x_{j}\in B:x_{j}\notin\mathrm{dis}(\widehat{f},f^{\star})\text{ and }\alpha\neq\frac{1}{2}\Big]\geq\frac{3}{4}-\frac{1}{\sqrt{3k/2+1/2}}.

Whenever k11k\geq 11 the RHS is at least 0.5070.507. As a consequence by Hoeffding’s inequality, we have [(f~)c][1niZi1/2]exp(2n0.072)\Pr[\mathcal{E}(\widetilde{f})^{c}]\leq\Pr[\tfrac{1}{n}\sum_{i}Z_{i}\leq 1/2]\leq\exp\left(-2n\cdot 0.07^{2}\right). By union bound, we have [fε:(f~)c]Γ(nk)exp(2n0.072)\Pr[\exists f\in\mathcal{F}_{\varepsilon}:\mathcal{E}(\widetilde{f})^{c}]\leq\Gamma_{\mathcal{F}}(nk)\cdot\exp\left(-2n\cdot 0.07^{2}\right), where Γ:\Gamma_{\mathcal{F}}:\mathbb{N}\to\mathbb{N} is the growth function for \mathcal{F}. Setting the RHS to δ\delta and using Sauer’s lemma we get that as long as n=O(dlogk+log(1/δ))n=O\left(d\log k+\log(1/\delta)\right), the event (f~)\mathcal{E}(\widetilde{f}) holds for all f~ε\widetilde{f}\in\mathcal{F}_{\varepsilon}.

Putting it together.

Therefore, with probability at least 12δ1-2\delta, (f^EPRM)ε\mathcal{L}(\widehat{f}_{\mathrm{EPRM}})\leq\varepsilon as long as ε1/(4k)\varepsilon\leq 1/(4k) and

n=O(dlogklog(1/ε)+log(1/δ)ε).\displaystyle n=O\left(\frac{d\log k\cdot\log(1/\varepsilon)+\log(1/\delta)}{\varepsilon}\right).

After rescaling δ\delta, this concludes the proof of Theorem 1. ∎

2.2 Proportion Matching Fails in Agnostic Setting

In the agnostic setting, we illustrate how proportion matching can perform quite poorly.

Example: EPRM may require Ω(2k)\Omega(2^{k}) sample complexity.

Fix any ε(0,1/2)\varepsilon\in(0,1/2) and consider the input space 𝒳={x}\mathcal{X}=\{x\}, with 𝒟\mathcal{D} given by (x,1)(x,1) with probability 1/2+ε1/2+\varepsilon and (x,0)(x,0) with probability 1/2ε1/2-\varepsilon. Let \mathcal{F} consist of two functions f0(x)=0f_{0}(x)=0 and f1(x)=1f_{1}(x)=1. The optimal predictor within the class \mathcal{F} is f1f_{1}. However, observe that the loss estimates for proportion matching are L^(f0)11ni=1n𝟙{αi=0}\widehat{L}(f_{0})\coloneqq 1-\frac{1}{n}\sum_{i=1}^{n}\mathbbm{1}\{\alpha_{i}=0\} and L^(f1)11ni=1n𝟙{αi=1}\widehat{L}(f_{1})\coloneqq 1-\frac{1}{n}\sum_{i=1}^{n}\mathbbm{1}\{\alpha_{i}=1\}. For small ε\varepsilon, unless the number of bags is exponential in kk, with constant probability we do not see any “pure” bags (with αi=0\alpha_{i}=0 or αi=1\alpha_{i}=1), so we have no way of distinguishing which of f0f_{0} and f1f_{1} achieves smaller loss.

Proportion matching can fail.

One may object that the previous failure mode is due to the fact that we are using a noncontinuous measure of discrepancy between the predicted and true label proportion, and such issues can be resolved if we minimize a continuous measure of discrepancy. We show that this does not help, as proportion matching approaches can return a predictor with constant suboptimality. This is because in the agnostic setting, the predictor which matches the bag-level proportions may not be the optimal instance-level predictor.

Consider the learning rule that minimizes the proportional square loss:

f^SQargminfL^SQ(f)=1ni=1n(1kj=1kf(xi,j)αi)2.\displaystyle\widehat{f}_{\mathrm{SQ}}\coloneqq\argmin_{f\in\mathcal{F}}\widehat{L}_{\mathrm{SQ}}(f)=\frac{1}{n}\sum_{i=1}^{n}\bigg(\frac{1}{k}\sum_{j=1}^{k}f(x_{i,j})-\alpha_{i}\bigg)^{2}. (6)

We show that in general, minimizing the proportional square loss can fail in the agnostic setting.

Proposition 1.

There exists a \mathcal{F} with VC()=1\mathrm{VC}(\mathcal{F})=1 and distribution 𝒟\mathcal{D} such that for any δ(0,1)\delta\in(0,1), bag size k7k\geq 7, and sample size n=Ω(log(1/δ))n=\Omega(\log(1/\delta)), with probability at least 1δ1-\delta, the learning rule f^SQ\widehat{f}_{\mathrm{SQ}} is 1/31/3-suboptimal.

The proportional square loss learning rule, as well as the proportional log loss learning rule

f^LOGargminf1ni=1nαilog(1kj=1kf(xi,j))(1αi)log(11kj=1kf(xi,j))\displaystyle\widehat{f}_{\mathrm{LOG}}\coloneqq\argmin_{f\in\mathcal{F}}\frac{1}{n}\sum_{i=1}^{n}-\alpha_{i}\cdot\log\bigg(\frac{1}{k}\sum_{j=1}^{k}f(x_{i,j})\bigg)-(1-\alpha_{i})\cdot\log\bigg(1-\frac{1}{k}\sum_{j=1}^{k}f(x_{i,j})\bigg) (7)

are regarded as folklore learning rules, and they were evaluated in the context of gradient-based learning (Busa-Fekete et al., 2023). Busa-Fekete et al. show that while gradient-based minimization of either the proportional square or log loss performs well in practice, it can fail in synthetic experimental settings. Proposition 1 demonstrates a simple theoretical failure mode for f^SQ\widehat{f}_{\mathrm{SQ}}. For completeness, in Appendix A we provide a similar result for the failure of f^LOG\widehat{f}_{\mathrm{LOG}} on the same construction, but note that even in the standard classification setting (k=1k=1) it is well known that minimizing surrogate losses like the log loss do not necessarily give guarantees on the classification error in the agnostic setting (Ben-David et al., 2012). Lastly, we remark that Appendix A of Scott and Zhang (2020) shows an example of similar flavor that in the limit as the bag size kk\to\infty, proportion matching learning rules can suffer constant suboptimality.

Proof of Proposition 1.

Let 𝒳={x(1),x(2)}\mathcal{X}=\{x^{(1)},x^{(2)}\}. Let ={f1,f2}\mathcal{F}=\{f_{1},f_{2}\} where f1(x)=𝟙{x=x(1)}f_{1}(x)=\mathbbm{1}\{x=x^{(1)}\} and f2(x)=𝟙{x=x(2)}f_{2}(x)=\mathbbm{1}\{x=x^{(2)}\}. The distribution 𝒟\mathcal{D} is (x,y)Unif({(x(1),1),(x(1),0),(x(2),1)})(x,y)\sim\mathrm{Unif}(\{(x^{(1)},1),(x^{(1)},0),(x^{(2)},1)\}). We can calculate that (f1)=2/3\mathcal{L}(f_{1})=2/3 and (f2)=1/3.\mathcal{L}(f_{2})=1/3. However we also have 𝔼f1=2/3\En f_{1}=2/3 while 𝔼f2=1/3\En f_{2}=1/3, and p=2/3p=2/3. While f1f_{1} in expectation matches the marginal label proportion, it is actually 1/31/3-suboptimal compared to f2f_{2}.

We compute the expectations of the bag-level losses for f1f_{1} and f2f_{2}. For f1f_{1} we have

𝔼[(1kjf1(xj)α)2]=1k2𝔼[(jf1(xj)yi)2]=1k2𝔼[j(f1(xj)yj)2]=23k.\displaystyle\En\bigg[\Big(\frac{1}{k}\sum_{j}f_{1}(x_{j})-\alpha\Big)^{2}\bigg]=\frac{1}{k^{2}}\En\bigg[\Big(\sum_{j}f_{1}(x_{j})-y_{i}\Big)^{2}\bigg]=\frac{1}{k^{2}}\En\bigg[\sum_{j}\left(f_{1}(x_{j})-y_{j}\right)^{2}\bigg]=\frac{2}{3k}.

For f2f_{2} we have

𝔼[(1kjf2(xj)α)2]\displaystyle\En\bigg[\Big(\frac{1}{k}\sum_{j}f_{2}(x_{j})-\alpha\Big)^{2}\bigg] =1k2𝔼[j(f2(xj)yj)2]+1k2𝔼[jj(f2(xj)yj)(f2(xj)yj)]\displaystyle=\frac{1}{k^{2}}\En\bigg[\sum_{j}\left(f_{2}(x_{j})-y_{j}\right)^{2}\bigg]+\frac{1}{k^{2}}\En\bigg[\sum_{j\neq j^{\prime}}\left(f_{2}(x_{j})-y_{j}\right)\left(f_{2}(x_{j^{\prime}})-y_{j^{\prime}}\right)\bigg]
=13k+k19k=k+29k.\displaystyle=\frac{1}{3k}+\frac{k-1}{9k}=\frac{k+2}{9k}.

Fix any k7k\geq 7. Then the expectation of the bag-level loss of f1f_{1} is at most 2/212/21 while the expectation of f2f_{2} is at least 1/91/9. By Hoeffding’s inequality, we have with probability at least 1δ1-\delta, that ^(f1)2/21+2log(1/δ)/n\widehat{\mathcal{L}}(f_{1})\leq 2/21+\sqrt{2\log(1/\delta)/n} and ^(f2)1/92log(1/δ)/n\widehat{\mathcal{L}}(f_{2})\geq 1/9-\sqrt{2\log(1/\delta)/n}. So as long as n=Ω(log(1/δ))n=\Omega(\log(1/\delta)), the proportional square loss learning rule will return the wrong predictor. ∎

3 Debiased Square Loss

In this section, we show that a simple debiasing of the square loss L^SQ()\widehat{L}_{\mathrm{SQ}}(\cdot) results in a learning rule that achieves optimal rates in both the realizable and the agnostic settings. Computing the expectation of the square loss, for any predictor ff,

𝔼[L^SQ(f)]\displaystyle\En\left[\widehat{L}_{\mathrm{SQ}}(f)\right] =1k2𝔼(B,α)𝒟[j(f(xj)yj)2+jj(f(xj)yj)(f(xj)yj)]\displaystyle=\frac{1}{k^{2}}\cdot\En_{(B,\alpha)\sim\mathcal{D}}\bigg[\sum_{j}(f(x_{j})-y_{j})^{2}+\sum_{j\neq j^{\prime}}(f(x_{j})-y_{j})(f(x_{j^{\prime}})-y_{j^{\prime}})\bigg]
=(i)1k(f)+k1k𝔼(x,y),(x,y)𝒟[(f(x)y)(f(x)y)]\displaystyle\overset{(i)}{=}\frac{1}{k}\cdot\mathcal{L}(f)+\frac{k-1}{k}\cdot\En_{(x,y),(x^{\prime},y^{\prime})\sim\mathcal{D}}\Big[(f(x)-y)(f(x^{\prime})-y^{\prime})\Big]
=(ii)1k(f)+k1k(𝔼fp)2.\displaystyle\overset{(ii)}{=}\frac{1}{k}\cdot\mathcal{L}(f)+\frac{k-1}{k}\cdot\left(\En f-p\right)^{2}. (8)

Equality (i)(i) uses the fact that for any jj, (f(xj)yj)2=𝟙{f(xj)yj}(f(x_{j})-y_{j})^{2}=\mathbbm{1}\left\{f(x_{j})\neq y_{j}\right\}. Equality (ii)(ii) uses independence of the samples.

Rearranging Eq. (8), we can see that the quantity kL^SQ(f)(k1)(𝔼fp)2k\cdot\widehat{L}_{\mathrm{SQ}}(f)-(k-1)\cdot\left(\En f-p\right)^{2} is an unbiased estimate of (f)\mathcal{L}(f). Of course, the caveat is that we do not have access to the values 𝔼f\En f and pp, but we can replace them with their empirical counterparts, giving us the debiased square loss learning rule:

f^DSQargminfL^DSQ(f)=1ni=1nk(1kj=1kf(xi,j)αi)2(k1)(𝔼^fp^)2,\displaystyle\widehat{f}_{\mathrm{DSQ}}\coloneqq\argmin_{f\in\mathcal{F}}\penalty\ \widehat{L}_{\mathrm{DSQ}}(f)=\frac{1}{n}\sum_{i=1}^{n}k\cdot\bigg(\frac{1}{k}\sum_{j=1}^{k}f(x_{i,j})-\alpha_{i}\bigg)^{2}-(k-1)\left(\widehat{\En}f-\widehat{p}\right)^{2},

where 𝔼^f=1nki=1nj=1kf(xi,j)\widehat{\En}f=\frac{1}{nk}\sum_{i=1}^{n}\sum_{j=1}^{k}f(x_{i,j}) and p^=1ni=1nαi\widehat{p}=\frac{1}{n}\sum_{i=1}^{n}\alpha_{i} are empirical estimates of 𝔼f\En f and pp respectively. Strictly speaking, L^DSQ(f)\widehat{L}_{\mathrm{DSQ}}(f) is not an unbiased estimate of (f)\mathcal{L}(f), but if nn is large enough, the second term approximates (𝔼fp)2(\En f-p)^{2} closely.

A similar debiasing idea for the proportional square loss was proposed by Chen et al. (2023), where in their Theorem 2 they prove the 1/n1/\sqrt{n} rate in the agnostic setting; due to difference in the bag generation assumption (they consider generating bags by resampling a dataset without replacement), our debiasing term takes a slightly different form.

3.1 Main Result: Optimistic Rates for Debiased Square Loss

Theorem 2 (Sample Complexity Bound for f^DSQ\widehat{f}_{\mathrm{DSQ}}).

Let δ(0,1)\delta\in(0,1). Fix any distribution 𝒟\mathcal{D} and any function class \mathcal{F}, and let L=inff(f)L^{\star}=\inf_{f\in\mathcal{F}}\mathcal{L}(f). With probability at least 1δ1-\delta, we have

(f^DSQ)L+O~(k2(d+log(1/δ))n+Lk2(d+log(1/δ))n).\displaystyle\mathcal{L}(\widehat{f}_{\mathrm{DSQ}})\leq L^{\star}+\widetilde{O}\left(\frac{k^{2}\left(d+\log(1/\delta)\right)}{n}+\sqrt{\frac{L^{\star}\cdot k^{2}\left(d+\log(1/\delta)\right)}{n}}\right).

Theorem 2 shows that under realizability (with L=0L^{\star}=0), f^DSQ\widehat{f}_{\mathrm{DSQ}} enjoys a fast 1/n1/n rate, while in the agnostic setting it achieves the 1/n1/\sqrt{n} rate, both of which are optimal (up to log factors) in terms of dd, log(1/δ)\log(1/\delta), and nn. The dependence on kk is certainly loose, as under realizability, f^DSQ\widehat{f}_{\mathrm{DSQ}} and f^EPRM\widehat{f}_{\mathrm{EPRM}} are identical learning rules, and Theorem 1 only has a logk\log k dependence. We leave sharpening the dependence on kk to future work.

The proof of Theorem 2 can be found in Appendix C. At a high level, we separately show uniform convergence bounds for both the square loss and the bias correction term to their expectations, and then we combine the guarantees to get a optimistic rate bound for L^DSQ()\widehat{L}_{\mathrm{DSQ}}(\cdot), which is an unbiased estimate of (f)\mathcal{L}(f), giving us the final guarantee.

Revisiting what happens if we minimize L^SQ\widehat{L}_{\mathrm{SQ}}.

Our analysis provides an answer to the question raised by Busa-Fekete et al. (2023) on understanding when proportion matching is consistent. Suppose the learner minimizes the square loss of Eq. (6), L^SQ(f)=1ni=1nk(1kj=1kf(xi,j)αi)2\widehat{L}_{\mathrm{SQ}}(f)=\frac{1}{n}\sum_{i=1}^{n}k\cdot(\frac{1}{k}\sum_{j=1}^{k}f(x_{i,j})-\alpha_{i})^{2}. (For sake of comparison, we multiply the loss by a factor of kk.) One can show an optimistic-rate style guarantee for f^SQ\widehat{f}_{\mathrm{SQ}}, albeit one that is weaker than Theorem 2. Let LSQ()L_{\mathrm{SQ}}(\cdot) denote the expectation of L^SQ\widehat{L}_{\mathrm{SQ}}. In the proof of Theorem 2, we get the following guarantee on f^SQ\widehat{f}_{\mathrm{SQ}}:

LSQ(f^SQ)inff{LSQ(f)+O~(kd+k2log(1/δ)n+LSQ(f)k(d+log(1/δ))n)}.\displaystyle L_{\mathrm{SQ}}(\widehat{f}_{\mathrm{SQ}})\leq\inf_{f\in\mathcal{F}}\bigg\{L_{\mathrm{SQ}}(f)+\widetilde{O}\bigg(\frac{kd+k^{2}\log(1/\delta)}{n}+\sqrt{\frac{L_{\mathrm{SQ}}(f)\cdot k(d+\log(1/\delta))}{n}}\bigg)\bigg\}.

This bound essentially replaces the instance-level classification error (f)\mathcal{L}(f) with the (larger) LSQ(f)L_{\mathrm{SQ}}(f) in Theorem 2. Applying the substitution LSQ(f)=(f)+B(f)L_{\mathrm{SQ}}(f)=\mathcal{L}(f)+B(f), where B(f)(k1)(𝔼fp)2B(f)\coloneqq(k-1)\left(\En f-p\right)^{2}, we see that this bound implies

(f^SQ)inff{(f)+B(f)+O~(kd+k2log(1/δ)n+((f)+B(f))k(d+log(1/δ))n)}.\displaystyle\mathcal{L}(\widehat{f}_{\mathrm{SQ}})\leq\inf_{f\in\mathcal{F}}\bigg\{\mathcal{L}(f)+B(f)+\widetilde{O}\bigg(\tfrac{kd+k^{2}\log(1/\delta)}{n}+\sqrt{\tfrac{\left(\mathcal{L}(f)+B(f)\right)\cdot k(d+\log(1/\delta))}{n}}\bigg)\bigg\}. (9)

When is Eq. (9) a useful bound? Under realizability, we have (f)=0\mathcal{L}(f^{\star})=0 and B(f)=0B(f^{\star})=0, and since any minimizer f^EPRM\widehat{f}_{\mathrm{EPRM}} is also a minimizer of L^SQ()\widehat{L}_{\mathrm{SQ}}(\cdot) and vice versa, Eq. (9) recovers the guarantee of Theorem 1, albeit with worse dependence on kk. More generally, in the agnostic setting, as long as it is possible to achieve a good trade-off between (f)\mathcal{L}(f) and B(f)B(f), we expect f^SQ\widehat{f}_{\mathrm{SQ}} to generalize well. This reasoning suggests why proportion matching learning rules perform well in practice (Busa-Fekete et al., 2023), despite the negative result of Proposition 1. In the modern over-parameterized regime, where the function class at hand (i.e., neural networks) are nearly realizable for the data distribution, Eq. (9) delivers strong guarantees.

Lastly, observe that there is no contradiction between the guarantee of Eq. (9) and the lower bound in Proposition 1: in Proposition 1, there is no predictor for which the two terms (f)\mathcal{L}(f) and B(f)B(f) are both small, as the suboptimal predictor has larger classification loss but satisfies B(f)=0B(f)=0.

4 EasyLLP Learning Rule

Recently, Busa-Fekete et al. (2023) proposed EasyLLP, a general recipe for constructing unbiased estimators of any instance-level loss function ins:𝒴×\ell_{\mathrm{ins}}:\mathcal{Y}\times\mathbb{R}\to\mathbb{R}. Given predictor ff\in\mathcal{F} and bag (B,α)(B,\alpha), the EasyLLP loss estimate of ins\ell_{\mathrm{ins}} can be written as

EZ(f,(B,α))\displaystyle\ell_{\mathrm{EZ}}\left(f,(B,\alpha)\right) 1kj=1k(k(αp)+p)ins(1,f(xj))+(k(pα)+(1p))ins(0,f(xj))\displaystyle\coloneqq\frac{1}{k}\sum_{j=1}^{k}\left(k(\alpha-p)+p\right)\ell_{\mathrm{ins}}\left(1,f(x_{j})\right)+\left(k(p-\alpha)+(1-p)\right)\ell_{\mathrm{ins}}\left(0,f(x_{j})\right) (10)

Proposition 4.2 of Busa-Fekete et al. (2023) shows that for any loss ins\ell_{\mathrm{ins}}, EZ(f,(B,α))\ell_{\mathrm{EZ}}\left(f,(B,\alpha)\right) is an unbiased estimate of the population loss, e.g., 𝔼(B,α)𝒟[EZ(f,(B,α))]=𝔼(x,y)𝒟[ins(y,f(x))]\En_{(B,\alpha)\sim\mathcal{D}}[\ell_{\mathrm{EZ}}(f,(B,\alpha))]=\En_{(x,y)\sim\mathcal{D}}[\ell_{\mathrm{ins}}(y,f(x))]. In this work, we consider EasyLLP instantiated with the classification loss ins=01\ell_{\mathrm{ins}}=\ell^{01}. The EasyLLP estimate takes the following form:

EZ(f,(B,α))\displaystyle\ell_{\mathrm{EZ}}\left(f,(B,\alpha)\right) =(k(αp)+p)(11kj=1kf(xj))+(k(pα)+(1p))(1kj=1kf(xj)).\displaystyle=\left(k(\alpha-p)+p\right)\big(1-\frac{1}{k}\sum_{j=1}^{k}f(x_{j})\big)+\left(k(p-\alpha)+(1-p)\right)\big(\frac{1}{k}\sum_{j=1}^{k}f(x_{j})\big). (11)

The EasyLLP learning rule f^EZargminfL^EZ(f)=1ni=1nEZ(f,(Bi,αi))\widehat{f}_{\mathrm{EZ}}\coloneqq\argmin_{f\in\mathcal{F}}\widehat{L}_{\mathrm{EZ}}(f)=\frac{1}{n}\sum_{i=1}^{n}\ell_{\mathrm{EZ}}(f,(B_{i},\alpha_{i})) attains the following guarantee (Busa-Fekete et al., 2023, Theorem 5.2 of): with probability at least 1δ1-\delta,

(f^EZ)inff(f)+O~(d+klog(1/δ)n).\displaystyle\mathcal{L}(\widehat{f}_{\mathrm{EZ}})\leq\inf_{f\in\mathcal{F}}\mathcal{L}(f)+\widetilde{O}\bigg(\sqrt{\frac{d+k\log(1/\delta)}{n}}\bigg). (12)

However, the question of whether EasyLLP can adapt to realizability to achieve fast rates remained open. The standard observation which enables fast rates under realizability is that loss estimates for (nearly) optimal predictors converge to their expectations at a rate of O(1/n)O(1/n). As a consequence, in supervised learning, one can show that any predictor f^\widehat{f} which achieves training error 0 (i.e., any ERM) has generalization error at most (f^)O~(d/n)\mathcal{L}(\widehat{f})\leq\widetilde{O}(d/n).

In contrast, the EasyLLP loss estimate does not satisfy this property. Even for the optimal predictor ff^{\star}, the EasyLLP loss estimate of Eq. (11) is a random quantity, since

EZ(f,(B,α))\displaystyle\ell_{\mathrm{EZ}}\left(f^{\star},(B,\alpha)\right) =(k(αp)+p)(1α)+(k(pα)+(1p))α\displaystyle=\left(k(\alpha-p)+p\right)\cdot\left(1-\alpha\right)+\left(k(p-\alpha)+(1-p)\right)\cdot\alpha

takes values depending on the bag label proportion α\alpha which itself is distributed as 1/kBin(k,p)1/k\cdot\mathrm{Bin}(k,p). As the following proposition shows, the loss estimate only concentrates at a Θ(1/n)\Theta(1/\sqrt{n}) rate.

Proposition 2.

There exists a realizable distribution 𝒟\mathcal{D} such that for any ε(0,1)\varepsilon\in(0,1), the EasyLLP loss estimate of ff^{\star} with bag size k=2k=2 requires Θ(1/ε2)\Theta(1/\varepsilon^{2}) samples in order for L^EZ(f)ε\widehat{L}_{\mathrm{EZ}}(f^{\star})\leq\varepsilon with constant probability.

Proof.

Consider the setting 𝒳={x0,x1}\mathcal{X}=\{x_{0},x_{1}\} with 𝒟\mathcal{D} that returns (x0,0)(x_{0},0) with probability 1/21/2 and (x1,1)(x_{1},1) with probability 1/21/2. Let f(x)=𝟙{x=x1}f^{\star}(x)=\mathbbm{1}\left\{x=x_{1}\right\} be the optimal predictor achieving (f)=0\mathcal{L}(f^{\star})=0. For any bag (B,α)(B,\alpha) the EasyLLP estimate of the loss can be written as

EZ(f,(B,α))\displaystyle\ell_{\mathrm{EZ}}(f^{\star},(B,\alpha)) =(k(α1/2)+1/2)(1α)+(k(1/2α)+1/2)α\displaystyle=\left(k(\alpha-1/2)+1/2\right)\cdot(1-\alpha)+\left(k(1/2-\alpha)+1/2\right)\cdot\alpha
=2kα2+2αkk/2+1/2.\displaystyle=-2k\alpha^{2}+2\alpha k-k/2+1/2.

For bag size k=2k=2, we have EZ(f,(B,α))1/2Rad(1/2)\ell_{\mathrm{EZ}}(f^{\star},(B,\alpha))\sim 1/2\cdot\mathrm{Rad}(1/2). From here, we can apply standard anti-concentration bounds for sums of Rademacher random variables. We let xj=EZ(f,(Bi,αi))x_{j}=\ell_{\mathrm{EZ}}(f^{\star},(B_{i},\alpha_{i})), and we can observe that by Paley-Zygmund (Theorem 7) that

[1ni=1nxj0.1n]=12[|1ni=1nxj|0.1n]Ω(1).\displaystyle\Pr\bigg[\frac{1}{n}\sum_{i=1}^{n}x_{j}\geq\frac{0.1}{\sqrt{n}}\bigg]=\frac{1}{2}\Pr\bigg[\Big|\frac{1}{n}\sum_{i=1}^{n}x_{j}\Big|\geq\frac{0.1}{\sqrt{n}}\bigg]\geq\Omega(1).

Thus, if n=c/ε2n=c/\varepsilon^{2} for sufficiently small c>0c>0 then with constant probability we have L^EZ(f)ε\widehat{L}_{\mathrm{EZ}}(f^{\star})\geq\varepsilon. On the flip side, we know that by Hoeffding’s inequality, n=O(1/ε2)n=O(1/\varepsilon^{2}) suffices for L^EZ(f)ε\widehat{L}_{\mathrm{EZ}}(f^{\star})\leq\varepsilon with constant probability. This proves the proposition. ∎

4.1 Main Result: Optimistic Rates for EasyLLP

Despite the fact that the EasyLLP loss estimates only concentrate to their expectations at a rate of O(1/n)O(1/\sqrt{n}), it is possible improve upon the guarantee in Eq. (12) and show that the EasyLLP learning rule instantiated with the classification loss attains optimistic rates.

Theorem 3 (Sample Complexity Bound for f^EZ\widehat{f}_{\mathrm{EZ}}).

Let δ(0,1)\delta\in(0,1). Fix any distribution 𝒟\mathcal{D} and function class \mathcal{F}, and let Linff(f)L^{\star}\coloneqq\inf_{f\in\mathcal{F}}\mathcal{L}(f). With probability at least 1δ1-\delta we have

(f^EZ)L+O~(k2(d+log(1/δ))n+Lk2(d+log(1/δ))n).\displaystyle\mathcal{L}(\widehat{f}_{\mathrm{EZ}})\leq L^{\star}+\widetilde{O}\bigg(\frac{k^{2}(d+\log(1/\delta))}{n}+\sqrt{\frac{L^{\star}\cdot k^{2}\left(d+\log(1/\delta)\right)}{n}}\bigg).

The bound for EasyLLP is order-wise identical to that shown for the debiased square loss (Theorem 2). In the agnostic setting, the guarantee in Theorem 3 is worse in terms of dependence on kk compared to Eq. (12); we leave sharpening the dependence on kk to future work. In the realizable setting, Proposition 2 and Theorem 3 together show that there is a separation between the rate of estimation (which is necessarily Ω(1/n)\Omega(1/\sqrt{n})) and the rate of learning (which is O~(1/n)\widetilde{O}(1/n)).

We sketch the proof ideas for Theorem 3, and we defer the full proof to Appendix D.

Proof Sketch. There is no hope for us to prove Theorem 3 through the usual route of showing uniform convergence bound like the following:

for allf,|(f)L^EZ(f)|O~(k2(d+log(1/δ))n+(f)k2(d+log(1/δ))n),\displaystyle\text{for all}\penalty\ f\in\mathcal{F},\lvert\mathcal{L}(f)-\widehat{L}_{\mathrm{EZ}}(f)\rvert\leq\widetilde{O}\bigg(\frac{k^{2}\left(d+\log(1/\delta)\right)}{n}+\sqrt{\frac{\mathcal{L}(f)\cdot k^{2}\left(d+\log(1/\delta)\right)}{n}}\bigg),

since the previous display directly contradicts Proposition 2 for f=argminf(f)f^{\star}=\argmin_{f\in\mathcal{F}}\mathcal{L}(f).

Instead, we make the critical observation that the offset empirical losses concentrate at the optimistic rate. Even though the learner does not know the identity of ff^{\star}, it is still true that minimizing L^EZ()\widehat{L}_{\mathrm{EZ}}(\cdot) is the same as minimizing the offset loss Γ^(,f)L^EZ()L^EZ(f)\widehat{\Gamma}(\cdot,f^{\star})\coloneqq\widehat{L}_{\mathrm{EZ}}(\cdot)-\widehat{L}_{\mathrm{EZ}}(f^{\star}), so we can equivalently think of the learning rule as minimizing Γ^(,f)\widehat{\Gamma}(\cdot,f^{\star}). The following lemma shows that one can prove an optimistic rate for the offset empirical losses. In the lemma, we use Γ(f,f)(f)(f)\Gamma(f,f^{\star})\coloneqq\mathcal{L}(f)-\mathcal{L}(f^{\star}) to denote the expected difference in classification error.

Lemma 1.

Let f=argminf(f)f^{\star}=\argmin_{f\in\mathcal{F}}\mathcal{L}(f). Then with probability at least 1δ1-\delta we have for all ff\in\mathcal{F}

|Γ^(f,f)Γ(f,f)|O~(kd+k2log(1/δ)n+(f)k2(d+log(1/δ))n).\displaystyle\left\lvert\widehat{\Gamma}(f,f^{\star})-\Gamma(f,f^{\star})\right\rvert\leq\widetilde{O}\bigg(\frac{kd+k^{2}\log(1/\delta)}{n}+\sqrt{\frac{\mathcal{L}(f)\cdot k^{2}\left(d+\log(1/\delta)\right)}{n}}\bigg).

Lemma 1 says that while the EasyLLP empirical estimate L^EZ\widehat{L}_{\mathrm{EZ}} is only O(1/n)O(1/\sqrt{n}) close to the true classification error, the estimate of the difference in classification error with that of ff^{\star} is more accurate, as much of the fluctuations in the bag estimates L^EZ()\widehat{L}_{\mathrm{EZ}}(\cdot) gets canceled out by subtraction.

In light of Lemma 1, Theorem 3 follows from standard approach of translating uniform convergence bounds to guarantees on the returned predictor f^EZ\widehat{f}_{\mathrm{EZ}}, see Appendix D.3 for more details.

Estimating the Marginal Label Proportion.

The EasyLLP learning rule requires knowledge of the marginal label proportion p=[y=1]p=\Pr[y=1]. We demonstrate that the optimistic rate guarantee in Theorem 3 can be attained without knowledge of pp, instead splitting the dataset into half to estimate p^\widehat{p} from a separate dataset (this was claimed without proof by Busa-Fekete et al. (2023)).

Corollary 1 (Sample Complexity Bound for f^EZ\widehat{f}_{\mathrm{EZ}} with Sample Splitting).

Let δ(0,1)\delta\in(0,1). Fix any distribution 𝒟\mathcal{D} and function class \mathcal{F}, and let Linff(f)L^{\star}\coloneqq\inf_{f\in\mathcal{F}}\mathcal{L}(f). Then with probability at least 1δ1-\delta the EasyLLP learning rule with sample splitting satisfies

(f^EZ)\displaystyle\mathcal{L}(\widehat{f}_{\mathrm{EZ}}) L+O~(k2(d+log(1/δ))n+Lk2(d+log(1/δ))n).\displaystyle\leq L^{\star}+\widetilde{O}\bigg(\frac{k^{2}(d+\log(1/\delta))}{n}+\sqrt{\frac{L^{\star}\cdot k^{2}\left(d+\log(1/\delta)\right)}{n}}\bigg).

The details of the sample splitting procedure and the proof of Corollary 1 are shown in Appendix D.4. In contrast to the debiased square loss learning rule, where we could estimate square loss and bias terms from the same dataset, here our analysis requires a separate dataset in order to estimate p^\widehat{p}. However, we conjecture that sample splitting is not required for the guarantee in Corollary 1.

5 Lower Bounds

The sample complexity bounds we prove in Theorem 1, 2, and 3 are optimal (up to log factors) in terms of the dependence on dd, nn, and log(1/δ)\log(1/\delta). However, the question remains of resolving the optimal dependence on the bag size kk.

For every function class \mathcal{F}, we have the trivial lower bound on the minimax sample complexity of n=Ω(dlog(1/δ)/(kε))n=\Omega\left(d\log(1/\delta)/(k\varepsilon)\right) for the realizable setting and n=Ω(dlog(1/δ)/(kε2))n=\Omega\left(d\log(1/\delta)/(k\varepsilon^{2})\right) for the agnostic setting, since LLP with nn bags is only harder than supervised learning with nknk examples. In general, the 1/k1/k dependence in the lower bound cannot be improved, as there are function classes for which it is tight. For example, consider the function class ={f0,f1}\mathcal{F}=\{f_{0},f_{1}\} where fi(x)=if_{i}(x)=i for i{0,1}i\in\{0,1\}. For this class \mathcal{F}, observing the label proportion α\alpha allows us to compute the average instance-level classification loss over the bag for both f0f_{0} and f1f_{1}. Therefore, LLP with nn bags is no harder than supervised learning with nknk labeled examples, so \mathcal{F} can be PAC learned with O~(1/(kε))\widetilde{O}(1/(k\varepsilon)) samples in the realizable setting and O~(1/(kε2))\widetilde{O}(1/(k\varepsilon^{2})) in the agnostic setting.

For specific function classes \mathcal{F}, it is possible to improve the 1/k1/k lower bound.

Theorem 4.

For any d3d\geq 3, there exists a function class \mathcal{F} with VC()=d\mathrm{VC}(\mathcal{F})=d such that any learning rule for LLP that PAC learns \mathcal{F} for bag size kO(2d/logd)k\leq O(2^{d}/\log d) with ε1/16\varepsilon\leq 1/16, and δ1/15\delta\leq 1/15 requires Ω(dlogk)\Omega\left(\tfrac{d}{\log k}\right) samples in the realizable setting and Ω(max(dlogk,dkε2))\Omega\left(\max\left(\tfrac{d}{\log k},\tfrac{d}{\sqrt{k}\varepsilon^{2}}\right)\right) samples in the agnostic setting.

The proof of Theorem 4 can be found in Appendix E. In the realizable setting, Theorem 4 gives a stronger lower bound when the accuracy parameter ε\varepsilon is a constant; however it is open to show a lower bound which dominates the trivial one of Ω(d/(kε))\Omega(d/(k\varepsilon)) for ε0\varepsilon\to 0. In the agnostic setting, the lower bound in Theorem 4 dominates the trivial one for all ε1/16\varepsilon\leq 1/16.

6 Experiments

In this section we empirically evaluate the performance of the learning rules discussed herein and present results which illustrate the differences of the learning rules in practical implementations.

Learning Rule Implementations.

Since minimizing the 0-1 loss is computationally intractable, we consider algorithmic variants of LLP learning rules which minimize a surrogate loss using minibatch stochastic gradient descent (SGD). Fix a parameterized function class ={fθ:θΘ}[0,1]𝒳\mathcal{F}=\{f_{\theta}:\theta\in\Theta\}\subseteq[0,1]^{\mathcal{X}}. For EasyLLP, we minimize the loss (10) with ins:𝒴×\ell_{\mathrm{ins}}:\mathcal{Y}\times\mathbb{R}\to\mathbb{R} instantiated to be the square loss EZ.Sq or the log loss EZ.Log. Directly minimizing the debiased square loss via SGD is not straightforward: for every gradient update, the second term in the loss requires computing 𝔼^fθ\widehat{\En}f_{\theta} (i.e., the average prediction of fθf_{\theta} over the entire dataset). We approximate it with an exponential moving average v^\widehat{v} with parameter β(0,1)\beta\in(0,1). Given a minibatch S={(B1,α1),,(B,α)}S=\{(B_{1},\alpha_{1}),\cdots,(B_{\ell},\alpha_{\ell})\}, we perform the updates

v^+\displaystyle\widehat{v}^{+} =βv^+(1β)1ki=1j=1kfθ(xi,j),\displaystyle=\beta\cdot\widehat{v}+(1-\beta)\cdot\frac{1}{\ell k}\sum_{i=1}^{\ell}\sum_{j=1}^{k}f_{\theta}(x_{i,j}), (13)
θ+\displaystyle\theta^{+} =θηθ(1i=1k(1kj=1kfθ(xi,j)αi)2(k1)(v^+p^)2).\displaystyle=\theta-\eta\cdot\nabla_{\theta}\left(\frac{1}{\ell}\sum_{i=1}^{\ell}k\cdot\bigg(\frac{1}{k}\sum_{j=1}^{k}f_{\theta}(x_{i,j})-\alpha_{i}\bigg)^{2}-(k-1)\left(\widehat{v}^{+}-\widehat{p}\right)^{2}\right). (14)

When performing backpropagation for the gradients on the debiasing term (k1)(v^+p^)2(k-1)\left(\widehat{v}^{+}-\widehat{p}\right)^{2}, the backpropagation does not go through v^\widehat{v}, only the summation over fθ(xi,j)f_{\theta}(x_{i,j}). We refer to this as DebiasedSq. For DebiasedSq, EZ.Log, and EZ.Sq, the label proportion p^\widehat{p} is estimated as the average of the training set labels. Lastly, we also consider algorithms that minimize the proportion matching square loss (6) and log loss (7), which we refer to as PM.Sq and PM.Log respectively.

6.1 Comparison of LLP Learning Rules

The goal of this experiment is to understand how these learning rules perform under various bag sizes and function classes.

Experimental Setup.

We adopt a similar experiment setup as in (Busa-Fekete et al., 2023). We run LLP algorithms on the MNIST odd vs. even task and the CIFAR10 animal vs. machine task (binary classification versions of the MNIST and CIFAR10). We consider 5 architectures: a linear model, small two layer NN (with 100 hidden units), a large two layer NN (with 1000 hidden units), a small CNN, and a large CNN. We experiment with bag sizes k{10,100,1000}k\in\{10,100,1000\}. For each dataset, model, bag size, and algorithm, we run 10 trials with different random seeds. To select the learning rate, we report the best average achieved test error for learning rates in {0.01,0.005,0.001,0.0005,0.0001}\{0.01,0.005,0.001,0.0005,0.0001\}. In all of our experiments we use the Adam optimizer (Kingma and Ba, 2014) with minibatches of size 10001000 and train for 100 epochs. For DebiasedSq, we use the approximate version in Eqs. (13)-(14) with β=0.99\beta=0.99. Further details can be found in Appendix G.1.

(a) MNIST Odd vs. Even
(b) CIFAR Animal vs. Machine
Figure 1: Training curves of various algorithms for LLP, using the large CNN architecture and bag size k=100k=100. One standard deviation confidence bands are plotted.
(a) Loss Estimates Tracking
(b) EZ.Sq Loss Estimates
(c) DebiasedSq Loss Estimates
Figure 2: Left: Loss estimates throughout training. We run a single trial of EZ.Sq and DebiasedSq on MNIST Odd vs. Even using the large CNN architecture, bag size k=10k=10, and optimally chosen learning rate. Using the test set, we plot the averaged true square loss 1ntesti=1ntest1kj=1k(fθ(xi,j)yi,j)2\tfrac{1}{n_{\mathrm{test}}}\sum_{i=1}^{n_{\mathrm{test}}}\tfrac{1}{k}\sum_{j=1}^{k}(f_{\theta}(x_{i,j})-y_{i,j})^{2} vs. the estimated square loss 1ntesti=1ntest^est(Bi,αi)\tfrac{1}{n_{\mathrm{test}}}\sum_{i=1}^{n_{\mathrm{test}}}\widehat{\ell}_{\mathrm{est}}(B_{i},\alpha_{i}), where ^est\widehat{\ell}_{\mathrm{est}} is either the EZ.Sq/DebiasedSq loss estimate. Middle and Right: Histogram of true per-bag square losses {1kj=1k(fθ(xi,j)yi,j)2}i=1ntest\{\tfrac{1}{k}\sum_{j=1}^{k}(f_{\theta}(x_{i,j})-y_{i,j})^{2}\}_{i=1}^{n_{\mathrm{test}}} and per-bag loss estimates {^est(Bi,αi)}i=1ntest\{\widehat{\ell}_{\mathrm{est}}(B_{i},\alpha_{i})\}_{i=1}^{n_{\mathrm{test}}} for EZ.Sq/DebiasedSq loss estimates at epoch 10.

Discussion of Results.

In Figure 1 as well as Table 1 and 2 (in the Appendix), we see that the best algorithm is a tossup between DebiasedSq, PM.Sq, and PM.Log.11 1 The results in (Busa-Fekete et al., 2023) suggest that EasyLLP and the proportion matching baselines have similar performance; however, note that they only train for 20 epochs. We find that EasyLLP is competitive in early stages of training but eventually DebiasedSq, PM.Sq, and PM.Log outperform it. In light of our discussion in Section 3.1, this may be not that surprising, since the model classes we work with are expressive enough to minimize both the proportion matching loss and the bias, so we do not have the failure mode shown in Proposition 1. Furthermore, in Figure 1 we see that both versions of EasyLLP exhibits overfitting as training progresses, thus necessitating early stopping; this was also observed in (Busa-Fekete et al., 2023). We observe that DebiasedSq, PM.Sq, and PM.Log also sometimes exhibits overfitting (see Figure 1(b)), but to a much less degree.

One hypothesis for why DebiasedSq performs better than EZ.Sq is that is that the debiased square loss is a more accurate estimate of the true loss than the EasyLLP loss (see Proposition 2 and the discussion before it). In Figure 2, we compare how well the estimated losses track the true square loss on the test set. Although we plot a single trial, we found that the behavior was similar across different random seeds. In Figure 2(a), the DebiasedSq loss closely tracks the true square loss, but EZ.Sq is consistently an overestimate of the true square loss. (Interestingly, we observe that the “shape” of the loss curve is still preserved). An explanation for this phenomenon can be found in Figure 2(b) and Figure 2(c), where we plot the histogram of ground-truth per-bag square losses as well as the corresponding per-bag square loss estimates in the test set. The histogram of per-bag loss estimates for DebiasedSq is quite similar to the histogram of ground-truth per-bag losses. On the other hand for EZ.Sq, the histogram of per-bag loss estimates has large variance and a heavy left tail. Since the randomness in the loss estimates is due to the bag generation procedure, it is likely that the bags with highly negative EZ.Sq loss estimates (i.e., taking values <2<-2) were not generated, so the sample mean overestimates the true square loss.

(a) MNIST Odd vs. Even
(b) CIFAR Animal vs. Machine
Figure 3: Training curves for PM.Sq and DebiasedSq with various β\beta for bag size k=10k=10 on the small CNN architecture. We use a fixed learning rate of 0.0010.001 and run full-batch GD for 1000 epochs. Each line is an average over 10 trials with different random seeds.

6.2 Benefits of Debiasing for Optimization

The previous results suggest that the performance of PM.Sq and DebiasedSq is similar, and that there might not be a need for debiasing in practice, as the failure mode in Proposition 1 is arguably pathological. On the contrary, we present evidence that (accurate) debiasing can improve optimization. In Figure 3, we compare the performance of PM.Sq and DebiasedSq with varying β\beta for full-batch GD. (We also investigated large batch SGD, but we found this effect was more striking with larger batches, so we only report the full-batch GD results.) Here, we can compute 𝔼^fθ\widehat{\En}f_{\theta} in every epoch at no additional cost, so the debiasing term can be exactly computed for every gradient update.

We find that in early stages of training, debiasing results in faster optimization. For CIFAR Animal vs. Machine, GD on PM.Sq seems to get stuck at a bad local minimum near initialization and only escapes after \sim25 epochs, but GD on DebiasedSq is able to escape the local minimum much faster. Early in training, the model fθf_{\theta} does not fit the label proportion p^\widehat{p} well, so the debiasing term B(fθ)B(f_{\theta}) is large, and therefore the PM.Sq loss will overestimate the true instance-level square loss. On the contrary, DebiasedSq uses more accurate estimates of the instance-level square loss, which is why we see benefits of debiasing. Later in training, as B(fθ)0B(f_{\theta})\to 0, the PM.Sq/DebiasedSq loss estimates are more similar, explaining why the different lines seem to converge to similar test errors.

7 Discussion

Our work studies various learning rules for minimizing classification loss in the LLP framework. We show that EPRM attains fast rates under realizability, but EPRM and other proportion matching approaches can fail in the agnostic setting. For the debiased square loss and EasyLLP learning rules, we prove optimistic rate sample complexity bounds which are optimal (up to log factors) in terms of the dependence on dd, nn, and log(1/δ)\log(1/\delta) in both the realizable and agnostic settings. In addition, we investigate the optimal dependence on kk from the lower bound side. We also compare the empirical performance of gradient-based versions of these learning rules and demonstrate the benefits of debiasing for optimization.

For clarity of exposition, we focus on binary classification, but we note that both the debiased square loss and EasyLLP learning rules can be extended to the multi-label multi-class setting (as also observed in Chen et al. (2023); Busa-Fekete et al. (2023)) by using one-hot encoding to write the label as a binary vector. In this way the learning task decouples into multiple binary classification tasks, and it would be straightforward to extend our analysis to this more general setting.

Our work leaves open several future directions. The most immediate one is to resolve the optimal dependence on kk. On the upper bound side, we believe that the dependence on kk can be improved in our analysis, and leave this to future work. Our optimistic rate results (Theorem 2 and 3) are stated for a more general setting, and their proofs do not use the combinatorial structure of the LLP problem in the way that the proof of Theorem 1 does, thus hinting at a source of looseness. On the lower bound side, there is room to improve upon the construction in Theorem 4. Other directions for future work include understanding the debiased square loss and EasyLLP learning rules in a more unified manner, designing a debiased variant of log loss for LLP, and studying the role of optimization in gradient-based algorithms for LLP.

Acknowledgements

We thank Robert Istvan Busa-Fekete, Travis Dick, Claudio Gentile, Nathan Srebro, and Han Shao for helpful discussions. Adel Javanmard is supported in part by the NSF Award DMS-2311024 and the Sloan fellowship in Mathematics.

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Appendix A Failure of Proportional Log Loss in Agnostic Setting

Proposition 3.

There exists a \mathcal{F} with VC()=1\mathrm{VC}(\mathcal{F})=1 and distribution 𝒟\mathcal{D} such that as long as nn, kk, and δ\delta satisfy the relationship k18log(2n/δ)k\geq 18\log(2n/\delta), with probability at least 1δ1-\delta, the learning rule f^LOG\widehat{f}_{\mathrm{LOG}} is 1/31/3-suboptimal.

Therefore, unless nn is exponentially large in the bag size kk, the proportional log loss learning rule will return a suboptimal predictor.

Proof.

We will use the same construction as in the proof of Proposition 1. For every bag (Bi,αi)(B_{i},\alpha_{i}) we will calculate the difference in proportional log loss for f1f_{1} and f2f_{2}. For any i[n]i\in[n], let βi1kj=1k𝟙{xi,j=x(1)}\beta_{i}\coloneqq\frac{1}{k}\sum_{j=1}^{k}\mathbbm{1}\left\{x_{i,j}=x^{(1)}\right\}. Then we have

LOG(f1,(Bi,αi))\displaystyle\ell_{\mathrm{LOG}}(f_{1},(B_{i},\alpha_{i})) =αilogβi(1αi)log(1βi)\displaystyle=-\alpha_{i}\log\beta_{i}-(1-\alpha_{i})\log(1-\beta_{i})
LOG(f2,(Bi,αi))\displaystyle\ell_{\mathrm{LOG}}(f_{2},(B_{i},\alpha_{i})) =αilog(1βi)(1αi)logβi.\displaystyle=-\alpha_{i}\log(1-\beta_{i})-(1-\alpha_{i})\log\beta_{i}.

Therefore the difference between the two losses is

LOG(f2,(Bi,αi))LOG(f1,(Bi,αi))=(2αi1)logβi1βi.\displaystyle\ell_{\mathrm{LOG}}(f_{2},(B_{i},\alpha_{i}))-\ell_{\mathrm{LOG}}(f_{1},(B_{i},\alpha_{i}))=(2\alpha_{i}-1)\log\frac{\beta_{i}}{1-\beta_{i}}.

We now show that with high probability, for all i[n]i\in[n] we have LOG(f2,(Bi,αi))LOG(f1,(Bi,αi))0\ell_{\mathrm{LOG}}(f_{2},(B_{i},\alpha_{i}))-\ell_{\mathrm{LOG}}(f_{1},(B_{i},\alpha_{i}))\geq 0. By Hoeffding’s inequality (Theorem 6) and union bound we have

[i[n]:αi<1/2 or βi<1/2]2nexp(k18).\displaystyle\Pr[\exists i\in[n]:\alpha_{i}<1/2\text{ or }\beta_{i}<1/2]\leq 2n\exp\left(-\frac{k}{18}\right).

Thus as long as k18log(2n/δ)k\geq 18\log(2n/\delta), with probability at least 1δ1-\delta we have LOG(f2,(Bi,αi))LOG(f1,(Bi,αi))0\ell_{\mathrm{LOG}}(f_{2},(B_{i},\alpha_{i}))-\ell_{\mathrm{LOG}}(f_{1},(B_{i},\alpha_{i}))\geq 0 for all i[n]i\in[n]. This implies that f^LOG=f1\widehat{f}_{\mathrm{LOG}}=f_{1}, the predictor which is 1/31/3-suboptimal. ∎

Appendix B Technical Background for Optimistic Rates

B.1 Uniform Convergence via Local Rademacher Complexity

In this section, we establish technical several results on uniform convergence using local Rademacher complexities. Recall the worst-case Rademacher for a function class 𝒢𝒳\mathcal{G}\subseteq\mathbb{R}^{\mathcal{X}} for any nn\in\mathbb{N}

n(𝒢)=supx1,,xn𝒳n𝔼σ[supf𝒢1ni=1nσif(xj)],\displaystyle\mathfrak{R}_{n}(\mathcal{G})=\sup_{x_{1},\cdots,x_{n}\in\mathcal{X}^{n}}\penalty\ \En_{\sigma}\left[\sup_{f\in\mathcal{G}}\frac{1}{n}\sum_{i=1}^{n}\sigma_{i}f(x_{j})\right],

where {σi}i=1n\left\{\sigma_{i}\right\}_{i=1}^{n} are i.i.d. Rademacher random variables. For any gg, we denote 𝔼[g]\En[g] to denote its expectation and 𝔼^n[g]\widehat{\En}_{n}[g] to denote the empirical average over a sample {xi}i=1n\{x_{i}\}_{i=1}^{n} where xix_{i} are i.i.d. drawn.

Lemma 2 (Modified Version of Lemma 6.2 in Bousquet (2002)).

Let 𝒢𝒳\mathcal{G}\subseteq\mathbb{R}^{\mathcal{X}} be a class of functions such that gb\|g\|_{\infty}\leq b for all g𝒢g\in\mathcal{G} and let (𝒢k)n\left(\mathcal{G}_{k}\right)_{n\in\mathbb{N}} be a sequence of subsets of 𝒢\mathcal{G} such that supg𝒢k𝔼[g2]A+Bγk\sup_{g\in\mathcal{G}_{k}}\En[g^{2}]\leq A+B\gamma_{k}, where γk=b/2k\gamma_{k}=b/2^{k} and A,B>0A,B>0 are constants.

Then for all δ(0,1)\delta\in(0,1) with probability at least 1δ1-\delta, for all k0k\geq 0 and g𝒢kg\in\mathcal{G}_{k}:

|𝔼[g]𝔼^n[g]|6n(𝒢k)+2(A+Bγk)(log1δ+cloglogbγk)n+6b(log1δ+cloglogbγk)n.\displaystyle\lvert\En[g]-\widehat{\En}_{n}[g]\rvert\leq 6\mathfrak{R}_{n}(\mathcal{G}_{k})+\sqrt{\frac{2(A+B\gamma_{k})\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}}+\frac{6b\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}.

Here, c>0c>0 is an absolute constant.

Lemma 2 is a trivial modification of Lemma 6.2 in Bousquet (2002), so we omit the proof details.

Definition 1 (Sub-Root Function).

The function ϕ:0\phi:\mathbb{R}_{\geq 0}\to\mathbb{R} is said to be a sub-root function if ϕ\phi is non-negative, non-decreasing, not identically zero, and ϕ(r)/r\phi(r)/\sqrt{r} is non-increasing.

Assumption 1 (Variance-Expectation Bound).

The function class 𝒢𝒳\mathcal{G}\subseteq\mathbb{R}^{\mathcal{X}} satisfies the following properties: (1) 𝔼[g]0\En[g]\geq 0 for all g𝒢g\in\mathcal{G} and (2) there exists constants A,B0A,B\geq 0 such that for all g𝒢g\in\mathcal{G}, 𝔼[g2]A+B𝔼[g]\En[g^{2}]\leq A+B\En[g].

Assumption 1 is a slight generalization of the variance-expectation bound stated in Assumption 1.4 from Bousquet (2002); the difference is that we allow an additional constant offset A0A\geq 0.

Theorem 5 (Modified Version of Theorem 6.2 in Bousquet (2002)).

Let 𝒢\mathcal{G} be a class of functions such that for all g𝒢g\in\mathcal{G}, gb\lVert g\rVert_{\infty}\leq b and 𝒢\mathcal{G} satisfies Assumption 1 with parameters (A,B)(A,B).

Let ϕn\phi_{n} be a sub-root function such that

𝔼σ[supg:𝔼^n[g2]r1ni=1nσig(xi)]ϕn(r).\displaystyle\En_{\sigma}\left[\sup_{g:\widehat{\En}_{n}[g^{2}]\leq r}\frac{1}{n}\sum_{i=1}^{n}\sigma_{i}g(x_{i})\right]\leq\phi_{n}(r).

Define rnr_{n}^{\star} to be the largest solution of ϕn(r)=r\phi_{n}(r)=r. For any δ>0\delta>0, we have with probability at least 1δ1-\delta for all g𝒢g\in\mathcal{G}:

|𝔼[g]𝔼^n[g]|C(brn+rn(A+B𝔼[g])+B𝔼[g](log1δ+cloglogn)n+r0).\displaystyle\lvert\En[g]-\widehat{\En}_{n}[g]\rvert\leq C\left(br^{\star}_{n}+\sqrt{r^{\star}_{n}(A+B\En[g])}+\sqrt{\frac{B\En[g]\left(\log\frac{1}{\delta}+c\log\log n\right)}{n}}+r_{0}\right).

where C>0C>0 is an absolute numerical constant and r02b2A(log1δ+cloglogn)n+22b2(log1δ+cloglogn)nr_{0}\coloneqq\sqrt{\tfrac{2b^{2}A\left(\log\tfrac{1}{\delta}+c\log\log n\right)}{n}}+\tfrac{22b^{2}\left(\log\tfrac{1}{\delta}+c\log\log n\right)}{n}.

Proof.

For all kk\in\mathbb{N} define γk=b/2k\gamma_{k}=b/2^{k}. We define 𝒢k{g𝒢:γk+1<𝔼[g]γk}\mathcal{G}_{k}\coloneqq\{g\in\mathcal{G}:\gamma_{k+1}<\En[g]\leq\gamma_{k}\}, so that 𝒢=k0𝒢k\mathcal{G}=\cup_{k\geq 0}\mathcal{G}_{k}.

By Lemma 2 and Assumption 1, we have with probability at least 1δ1-\delta for all k0k\geq 0 and g𝒢kg\in\mathcal{G}_{k}:

|𝔼^n[g]𝔼[g]|8n(𝒢k)+2(A+Bγk)(log1δ+cloglogbγk)n+20b(log1δ+cloglogbγk)n.\displaystyle\lvert\widehat{\En}_{n}[g]-\En[g]\rvert\leq 8\mathfrak{R}_{n}(\mathcal{G}_{k})+\sqrt{\frac{2(A+B\gamma_{k})\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}}+\frac{20b\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}. (15)

In addition, we can apply Lemma 2 to the squares of g𝒢kg\in\mathcal{G}_{k} to get that with probability at least 1δ1-\delta for all k0k\geq 0 and every g𝒢kg\in\mathcal{G}_{k}:

|𝔼^n[g2]𝔼[g2]|\displaystyle\lvert\widehat{\En}_{n}[g^{2}]-\En[g^{2}]\rvert 8n(𝒢k2)+2b2(A+Bγk)(log1δ+cloglogbγk)n+20b2(log1δ+cloglogbγk)n\displaystyle\leq 8\mathfrak{R}_{n}(\mathcal{G}_{k}^{2})+\sqrt{\frac{2b^{2}(A+B\gamma_{k})\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}}+\frac{20b^{2}\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}
16bn(𝒢k)+2b2(A+Bγk)(log1δ+cloglogbγk)n+20b2(log1δ+cloglogbγk)n,\displaystyle\leq 16b\mathfrak{R}_{n}(\mathcal{G}_{k})+\sqrt{\frac{2b^{2}(A+B\gamma_{k})\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}}+\frac{20b^{2}\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}, (16)

where the last inequality uses the fact that xx2x\mapsto x^{2} is 2b2b-Lipschitz and centered at 0.

Now we condition on Eq. (15) and (16), which happens with probability at least 12δ1-2\delta. By Eq. (16) and Assumption 1, for all g𝒢kg\in\mathcal{G}_{k},

𝔼^n[g2](A+Bγk)+16bn(𝒢k)+2b2(A+Bγk)(log1δ+cloglogbγk)n+20b2(log1δ+cloglogbγk)n.\displaystyle\widehat{\En}_{n}[g^{2}]\leq\left(A+B\gamma_{k}\right)+16b\mathfrak{R}_{n}(\mathcal{G}_{k})+\sqrt{\frac{2b^{2}(A+B\gamma_{k})\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}}+\frac{20b^{2}\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}.

Define the RHS of the previous display to be UkU_{k}. By definition of ϕn\phi_{n}, we know that

n(𝒢k)supx1,,xn𝔼σ[supg:𝔼^n[g2]Uk1ni=1nσig(xi)]ϕn(Uk),\displaystyle\mathfrak{R}_{n}(\mathcal{G}_{k})\leq\sup_{x_{1},\cdots,x_{n}}\En\sigma\left[\sup_{g:\widehat{\En}_{n}[g^{2}]\leq U_{k}}\frac{1}{n}\sum_{i=1}^{n}\sigma_{i}g(x_{i})\right]\leq\phi_{n}(U_{k}),

so therefore

Uk(A+Bγk)+16bϕn(Uk)+2b2(A+Bγk)(log1δ+cloglogbγk)n+20b2(log1δ+cloglogbγk)n.\displaystyle U_{k}\leq(A+B\gamma_{k})+16b\phi_{n}(U_{k})+\sqrt{\frac{2b^{2}(A+B\gamma_{k})\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}}+\frac{20b^{2}\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}.

Let us define k0k_{0} to be the largest value such that γk0+1bn\gamma_{k_{0}+1}\geq\frac{b}{n}. For all kk0k\leq k_{0}, we have cloglogbγkcloglognc\log\log\frac{b}{\gamma_{k}}\leq c\log\log n. Therefore

Uk\displaystyle U_{k} A+Bγk+16bϕn(Uk)+2b2(A+Bγk)(log1δ+cloglogn)n+20b2(log1δ+cloglogn)n\displaystyle\leq A+B\gamma_{k}+16b\phi_{n}(U_{k})+\sqrt{\frac{2b^{2}(A+B\gamma_{k})\left(\log\frac{1}{\delta}+c\log\log n\right)}{n}}+\frac{20b^{2}\left(\log\frac{1}{\delta}+c\log\log n\right)}{n}
A+2Bγk+16bϕn(Uk)+2b2A(log1δ+cloglogn)n+22b2(log1δ+cloglogn)nr0.\displaystyle\leq A+2B\gamma_{k}+16b\phi_{n}(U_{k})+\underbrace{\sqrt{\frac{2b^{2}A\left(\log\frac{1}{\delta}+c\log\log n\right)}{n}}+\frac{22b^{2}\left(\log\frac{1}{\delta}+c\log\log n\right)}{n}}_{\eqqcolon r_{0}}.

If UkrnU_{k}\geq r^{\star}_{n}, then by definition of the sub-root function ϕn(Uk)/Ukϕn(rn)/rn=rn\phi_{n}(U_{k})/\sqrt{U_{k}}\leq\phi_{n}(r^{\star}_{n})/\sqrt{r^{\star}_{n}}=\sqrt{r^{\star}_{n}}, so

Uk16bUkrn+A+2Bγk+r0C(b2rn+A+Bγk+r0)rn(γk)\displaystyle U_{k}\leq 16b\sqrt{U_{k}r^{\star}_{n}}+A+2B\gamma_{k}+r_{0}\leq C\left(b^{2}r^{\star}_{n}+A+B\gamma_{k}+r_{0}\right)\eqqcolon r_{n}(\gamma_{k})

for some absolute constant C>0C>0. The last inequality holds by Fact 1. In addition if Uk<rnU_{k}<r^{\star}_{n}, the conclusion of the previous display trivially holds.

By Eq. (15), we also know that for any g𝒢kg\in\mathcal{G}_{k}

𝔼[g]\displaystyle\En[g] 𝔼^n[g]+8n(𝒢k)+2Bγk(log1δ+cloglogbγk)n+r0\displaystyle\leq\widehat{\En}_{n}[g]+8\mathfrak{R}_{n}(\mathcal{G}_{k})+\sqrt{\frac{2B\gamma_{k}\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}}+r_{0}
𝔼^n[g]+8ϕn(rn(γk))+2Bγk(log1δ+cloglogbγk)n+r0.\displaystyle\leq\widehat{\En}_{n}[g]+8\phi_{n}\left(r_{n}(\gamma_{k})\right)+\sqrt{\frac{2B\gamma_{k}\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}}+r_{0}.

By definition of 𝒢k\mathcal{G}_{k} and γk\gamma_{k}, we know that γk2𝔼[g]\gamma_{k}\leq 2\En[g] so that

𝔼[g]\displaystyle\En[g] 𝔼^n[g]+8ϕn(rn(2𝔼[g]))+4B𝔼[g](log1δ+cloglogbγk)n+r0\displaystyle\leq\widehat{\En}_{n}[g]+8\cdot\phi_{n}\left(r_{n}(2\En[g])\right)+\sqrt{\frac{4B\En[g]\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}}+r_{0}
𝔼^n[g]+8rnC(b2rn+A+B𝔼[g]+r0)+4B𝔼[g](log1δ+cloglogbγk)n+r0\displaystyle\leq\widehat{\En}_{n}[g]+8\sqrt{r^{\star}_{n}}\cdot\sqrt{C\left(b^{2}r^{\star}_{n}+A+B\En[g]+r_{0}\right)}+\sqrt{\frac{4B\En[g]\left(\log\frac{1}{\delta}+c\log\log\frac{b}{\gamma_{k}}\right)}{n}}+r_{0}
𝔼^n[g]+C(brn+rn(A+B𝔼[g]))+4B𝔼[g](log1δ+cloglogn)n+2r0,\displaystyle\leq\widehat{\En}_{n}[g]+C^{\prime}\left(br^{\star}_{n}+\sqrt{r^{\star}_{n}(A+B\En[g])}\right)+\sqrt{\frac{4B\En[g]\left(\log\frac{1}{\delta}+c\log\log n\right)}{n}}+2r_{0},

for some absolute numerical constant C>0C^{\prime}>0. When kk0k\geq k_{0}, we have γkbn\gamma_{k}\leq\frac{b}{n}, so the previous display also trivially holds.

Lastly, we can repeat the same argument with the function class 𝒢={g:g𝒢}\mathcal{G}^{\prime}=\left\{-g:g\in\mathcal{G}\right\} to get the two-sided bound. This concludes the proof of Theorem 5. ∎

B.2 Calculating the Complexity Radius for LLP Losses

Now we present results which allow us to calculate the complexity radius rnr^{\star}_{n} for function classes with bounded VC dimension.

Lemma 3 (Refined Dudley’s Inequality, Lemma A.1 from Srebro et al. (2010)).

For any function class :𝒳\mathcal{F}:\mathcal{X}\to\mathbb{R},

n()infη>0{4η+12ηsupf𝔼^n[f2]log𝒩2(,ε,n)n𝑑ε}.\displaystyle\mathfrak{R}_{n}(\mathcal{F})\leq\inf_{\eta>0}\left\{4\eta+12\int_{\eta}^{\sup_{f\in\mathcal{F}}\sqrt{\widehat{\En}_{n}[f^{2}]}}\sqrt{\frac{\log\mathcal{N}_{2}(\mathcal{F},\varepsilon,n)}{n}}\penalty\ d\varepsilon\right\}.
Lemma 4 (Theorem 2.14 in Mendelson (2003)).

Let {0,1}𝒳\mathcal{F}\subseteq\{0,1\}^{\mathcal{X}} with VC dimension dd. Then

log𝒩2(,ε,n)dlog(4e2log2e2ε)+(2d)log(1ε).\displaystyle\log\mathcal{N}_{2}(\mathcal{F},\varepsilon,n)\leq d\log\left(4e^{2}\log\frac{2e^{2}}{\varepsilon}\right)+(2d)\cdot\log\left(\frac{1}{\varepsilon}\right).

We next present a lemma which allows us to calculate the local Rademacher complexity for various loss functions for LLP. Fix any 𝒴𝒳\mathcal{F}\subseteq\mathcal{Y}^{\mathcal{X}} with VC dimension dd. Consider any loss :[0,1]×[0,1]\ell:[0,1]\times[0,1]\to\mathbb{R}. Define the constrained loss class

(r){(B,α)(1kj=1kf(xj),α):f,𝔼^n[(1kj=1kf(xj),α)2]r}.\displaystyle\mathcal{L}_{\ell}(r)\coloneqq\left\{\left(B,\alpha\right)\mapsto\ell\left(\frac{1}{k}\sum_{j=1}^{k}f(x_{j}),\alpha\right):f\in\mathcal{F},\widehat{\En}_{n}\left[\ell\left(\frac{1}{k}\sum_{j=1}^{k}f(x_{j}),\alpha\right)^{2}\right]\leq r\right\}.

We also let \mathcal{L}_{\ell} denote the unrestricted loss class which contains all bag-level losses for ff\in\mathcal{F}.

Lemma 5.

Let :[0,1]×[0,1]\ell:[0,1]\times[0,1]\to\mathbb{R} be any λ\lambda-Lipschitz (in the first argument) bag loss. For any n>dn>d and any r>0r>0, we have

n((r))Crdlog(λnr)n\displaystyle\mathfrak{R}_{n}(\mathcal{L}_{\ell}(r))\leq C\sqrt{\frac{rd\log\left(\frac{\lambda n}{r}\right)}{n}}

for some absolute numerical constant C>0C>0. Furthermore, if we denote the sub-root function ϕn(r)Crdlog(λnr)n\phi_{n}(r)\coloneqq C\sqrt{\frac{rd\log(\frac{\lambda n}{r})}{n}}, and let rnr^{\star}_{n} be the largest number such that ϕn(r)=r\phi_{n}(r)=r, we have

rnO(dlog(λn)n).\displaystyle r^{\star}_{n}\leq O\left(\frac{d\log(\lambda n)}{n}\right).
Proof.

We use Lemma 3 applied to (r)\mathcal{L}_{\ell}(r). This gives

n((r))infη>0{4η+12ηrlog𝒩2((r),ε,n)n𝑑ε}.\displaystyle\mathfrak{R}_{n}(\mathcal{L}_{\ell}(r))\leq\inf_{\eta>0}\left\{4\eta+12\int_{\eta}^{\sqrt{r}}\sqrt{\frac{\log\mathcal{N}_{2}(\mathcal{L}_{\ell}(r),\varepsilon,n)}{n}}\penalty\ d\varepsilon\right\}. (17)

From here we bound the covering numbers of the loss class in terms of the function class as

log𝒩2((r),ε,n)log𝒩2(,ε,n)log𝒩2(,ε/λ,nk).\displaystyle\log\mathcal{N}_{2}(\mathcal{L}_{\ell}(r),\varepsilon,n)\leq\log\mathcal{N}_{2}(\mathcal{L}_{\ell},\varepsilon,n)\leq\log\mathcal{N}_{2}(\mathcal{F},\varepsilon/\lambda,nk).

The last inequality follows because for any f,fεf,f_{\varepsilon} and (B1,α1),,(Bn,αn)(B_{1},\alpha_{1}),\cdots,(B_{n},\alpha_{n}):

1ni=1n((1kj=1kf(xi,j),α)(1kj=1kfε(xi,j),α))2\displaystyle\hskip-20.00003pt\sqrt{\frac{1}{n}\sum_{i=1}^{n}\left(\ell\left(\frac{1}{k}\sum_{j=1}^{k}f(x_{i,j}),\alpha\right)-\ell\left(\frac{1}{k}\sum_{j=1}^{k}f_{\varepsilon}(x_{i,j}),\alpha\right)\right)^{2}}
λ1ni=1n(1kif(xi,j)fε(xi,j))2λ1nki=1nj=1k(f(xi,j)fε(xi,j))2,\displaystyle\leq\lambda\cdot\sqrt{\frac{1}{n}\sum_{i=1}^{n}\left(\frac{1}{k}\sum_{i}f(x_{i,j})-f_{\varepsilon}(x_{i,j})\right)^{2}}\leq\lambda\sqrt{\frac{1}{nk}\sum_{i=1}^{n}\sum_{j=1}^{k}\left(f(x_{i,j})-f_{\varepsilon}(x_{i,j})\right)^{2}},

where we use the λ\lambda-Lipschitz property of \ell as well as Jensen’s inequality. Thus, an empirical 2\ell_{2} cover of \mathcal{F} at scale ε/λ\varepsilon/\lambda for nknk points implies an empirical 2\ell_{2} cover of \mathcal{L} at scale ε\varepsilon for nn points.

From here, we can apply the covering number bound of Lemma 4 to get

log𝒩2(,ε/λ,nk)dlog(4e2log2e2λε)+(2d)log(λε)Cdlog(λε),\displaystyle\log\mathcal{N}_{2}(\mathcal{F},\varepsilon/\lambda,nk)\leq d\log\left(4e^{2}\log\frac{2e^{2}\lambda}{\varepsilon}\right)+(2d)\cdot\log\left(\frac{\lambda}{\varepsilon}\right)\leq Cd\log\left(\frac{\lambda}{\varepsilon}\right),

for some absolute numerical constant C>0C>0.

Now we can plug the covering number bound back into Eq. (17) to get

n((r))infη>0{4η+12Cηrdlogλεn𝑑ε}.\displaystyle\mathfrak{R}_{n}(\mathcal{L}_{\ell}(r))\leq\inf_{\eta>0}\left\{4\eta+12\sqrt{C}\int_{\eta}^{\sqrt{r}}\sqrt{\frac{d\log\frac{\lambda}{\varepsilon}}{n}}\penalty\ d\varepsilon\right\}.

Choosing η=Θ(rd/n)\eta=\Theta(\sqrt{rd/n}) gives

n((r))Crdlog(λnr)n,\displaystyle\mathfrak{R}_{n}(\mathcal{L}_{\ell}(r))\leq C^{\prime}\sqrt{\frac{rd\log\left(\frac{\lambda n}{r}\right)}{n}},

for some absolute numerical constant C>0C^{\prime}>0. For the proof of the second part, it is easy to see that the solution to the equation Crndlog(λnrn)n=rC^{\prime}\sqrt{\tfrac{r^{\star}_{n}d\log(\frac{\lambda n}{r^{\star}_{n}})}{n}}=r must satisfy rnC′′dlog(λn)nr^{\star}_{n}\leq C^{\prime\prime}\cdot\tfrac{d\log(\lambda n)}{n} for some absolute constant C′′>0C^{\prime\prime}>0. This concludes the proof of Lemma 5. ∎

Appendix C Proof of Theorem 2

C.1 Notation and Preliminaries

We define several quantities which will be used in the proof. The loss function L^DSQ(f)\widehat{L}_{\mathrm{DSQ}}(f) is the difference between a proportion matching term and a debiasing term. For a given bag (B,α)(B,\alpha) and function ff, we let SQ(f,(B,α))k(1kj=1kf(xj)α)2\ell_{\mathrm{SQ}}(f,(B,\alpha))\coloneqq k\cdot(\frac{1}{k}\sum_{j=1}^{k}f(x_{j})-\alpha)^{2} and let

L^SQ(f)1ni=1nSQ(f,(Bi,αi)),andLSQ(f)𝔼B[SQ(f,(B,α))].\displaystyle\widehat{L}_{\mathrm{SQ}}(f)\coloneqq\frac{1}{n}\sum_{i=1}^{n}\ell_{\mathrm{SQ}}(f,(B_{i},\alpha_{i})),\quad\text{and}\quad L_{\mathrm{SQ}}(f)\coloneqq\En_{B}\left[\ell_{\mathrm{SQ}}(f,(B,\alpha))\right].

In addition, we let

B^(f)(k1)(𝔼^fp^)2,andB(f)(k1)(𝔼fp)2.\displaystyle\widehat{B}(f)\coloneqq(k-1)\cdot(\widehat{\En}f-\widehat{p})^{2},\quad\text{and}\quad B(f)\coloneqq(k-1)\cdot(\En f-p)^{2}.

Written in this notation, we have L^DSQ(f)L^SQ(f)B^(f)\widehat{L}_{\mathrm{DSQ}}(f)\coloneqq\widehat{L}_{\mathrm{SQ}}(f)-\widehat{B}(f). In addition, recall that we showed in Section 3 that for any predictor ff we have (f)=LSQ(f)B(f)\mathcal{L}(f)=L_{\mathrm{SQ}}(f)-B(f).

We also establish several elementary facts about B^(f)\widehat{B}(f), L^SQ\widehat{L}_{\mathrm{SQ}}, and L^DSQ\widehat{L}_{\mathrm{DSQ}}.

Lemma 6.

The following statements are true for any predictor ff and dataset {(Bi,αi)}i=1n\left\{(B_{i},\alpha_{i})\right\}_{i=1}^{n}.

  1. 1.

    B^(f)k1kL^SQ(f)\widehat{B}(f)\leq\frac{k-1}{k}\cdot\widehat{L}_{\mathrm{SQ}}(f).

  2. 2.

    L^DSQ(f)0\widehat{L}_{\mathrm{DSQ}}(f)\geq 0.

  3. 3.

    L^SQ(f)kL^DSQ(f)\widehat{L}_{\mathrm{SQ}}(f)\leq k\cdot\widehat{L}_{\mathrm{DSQ}}(f).

Proof.

Fix any predictor ff. For the first statement, using Jensen’s inequality we can compute that

B^(f)\displaystyle\widehat{B}(f) =(k1)(1ni=1n1kj=1kf(xi,j)αi)2\displaystyle=(k-1)\cdot\left(\frac{1}{n}\sum_{i=1}^{n}\frac{1}{k}\sum_{j=1}^{k}f(x_{i,j})-\alpha_{i}\right)^{2}
(k1)1ni=1n(1kj=1kf(xi,j)αi)2=k1kL^SQ(f).\displaystyle\leq(k-1)\cdot\frac{1}{n}\sum_{i=1}^{n}\left(\frac{1}{k}\sum_{j=1}^{k}f(x_{i,j})-\alpha_{i}\right)^{2}=\frac{k-1}{k}\widehat{L}_{\mathrm{SQ}}(f).

The second statement is a simple consequence of the first statement. For the third statement we have

L^SQ(f)=L^DSQ(f)+B^(f)L^DSQ(f)+k1kL^SQ(f),\displaystyle\widehat{L}_{\mathrm{SQ}}(f)=\widehat{L}_{\mathrm{DSQ}}(f)+\widehat{B}(f)\leq\widehat{L}_{\mathrm{DSQ}}(f)+\frac{k-1}{k}\cdot\widehat{L}_{\mathrm{SQ}}(f),

and rearranging yields the statement. This proves Lemma 6. ∎

C.2 Showing Optimistic Rates

To prove Theorem 2, we separately prove optimistic rates for the square loss and debiasing terms, and then combine the guarantees. Consider the function class

𝒢{(B,α)k(1kj=1kf(xj)α)2:f}.\displaystyle\mathcal{G}\coloneqq\left\{(B,\alpha)\mapsto k\cdot\left(\frac{1}{k}\sum_{j=1}^{k}f(x_{j})-\alpha\right)^{2}:f\in\mathcal{F}\right\}.

For any function g𝒢g\in\mathcal{G}, we know that gk\|g\|\leq k and that gg is kk-Lipschitz in the first argument; furthermore, by nonnegativity, 𝒢\mathcal{G} satisfies Assumption 1 with parameters A=0A=0, B=kB=k. Thus we apply Theorem 5 to 𝒢\mathcal{G} to get that with probability at least 1δ1-\delta for all g𝒢g\in\mathcal{G},

|𝔼[g]𝔼^n[g]|C(krn+rn(k𝔼[g])+k𝔼[g](log1δ+cloglogn)n+k2(log1δ+cloglogn)n).\displaystyle\lvert\En[g]-\widehat{\En}_{n}[g]\rvert\leq C\left(kr^{\star}_{n}+\sqrt{r^{\star}_{n}(k\En[g])}+\sqrt{\frac{k\En[g]\left(\log\frac{1}{\delta}+c\log\log n\right)}{n}}+\tfrac{k^{2}\left(\log\tfrac{1}{\delta}+c\log\log n\right)}{n}\right).

where rnr^{\star}_{n} is the critical radius and C>0C>0 is an absolute numerical constant. Applying Lemma 5, we can also calculate the critical radius as

rnO(dlog(kn)n),\displaystyle r^{\star}_{n}\leq O\left(\frac{d\log(kn)}{n}\right),

so therefore we have with probability at least 1δ1-\delta for any ff\in\mathcal{F}:

|L^SQ(f)LSQ(f)|O~(kd+k2log1δn+LSQ(f)k(d+log1δ)n).\displaystyle\left\lvert\widehat{L}_{\mathrm{SQ}}(f)-L_{\mathrm{SQ}}(f)\right\rvert\leq\widetilde{O}\left(\frac{kd+k^{2}\log\tfrac{1}{\delta}}{n}+\sqrt{\frac{L_{\mathrm{SQ}}(f)\cdot k\left(d+\log\frac{1}{\delta}\right)}{n}}\right). (18)

Next, we show a uniform convergence bound which relates B^(f)\widehat{B}(f) to B(f)B(f). Consider the partition

𝒢+\displaystyle\mathcal{G}^{+} {(x,y)f(x)y:f,𝔼fp},\displaystyle\coloneqq\left\{(x,y)\mapsto f(x)-y:f\in\mathcal{F},\En f\geq p\right\},
𝒢\displaystyle\mathcal{G}^{-} {(x,y)yf(x):f,𝔼f<p}.\displaystyle\coloneqq\left\{(x,y)\mapsto y-f(x):f\in\mathcal{F},\En f<p\right\}.

The function class 𝒢+\mathcal{G}^{+} is 11-Lipschitz, bounded in [1,+1][-1,+1], and satisfies Assumption 1 with A=0A=0, B=1B=1. Applying Theorem 5 to 𝒢+\mathcal{G}^{+} and using Lemma 5 we get that with probability at least 1δ1-\delta for any ff\in\mathcal{F} such that 𝔼fp\En f\geq p:

||𝔼^fp^||𝔼fp||O~((𝔼fp)(d+log1δ)nk+d+log1δnk).\displaystyle\left\lvert\lvert\widehat{\En}f-\widehat{p}\rvert-\lvert\En f-p\rvert\right\rvert\leq\widetilde{O}\left(\sqrt{\frac{(\En f-p)\left(d+\log\frac{1}{\delta}\right)}{nk}}+\frac{d+\log\frac{1}{\delta}}{nk}\right). (19)

This implies

(𝔼^fp^)2\displaystyle(\widehat{\En}f-\widehat{p})^{2} (𝔼fp)2+O~((𝔼fp)3(d+log1δ)nk+d+log1δnk)\displaystyle\leq(\En f-p)^{2}+\widetilde{O}\left(\sqrt{\frac{(\En f-p)^{3}\left(d+\log\frac{1}{\delta}\right)}{nk}}+\frac{d+\log\frac{1}{\delta}}{nk}\right)
B^(f)\displaystyle\Rightarrow\quad\widehat{B}(f) B(f)+O~(B(f)(d+log1δ)n+d+log1δn).\displaystyle\leq B(f)+\widetilde{O}\left(\sqrt{\frac{B(f)\left(d+\log\frac{1}{\delta}\right)}{n}}+\frac{d+\log\frac{1}{\delta}}{n}\right). (20)

Likewise, for the reverse inequality we can also get that

(𝔼fp)2\displaystyle(\En f-p)^{2} (𝔼^fp^)2+O~(|𝔼^fp^|(𝔼fp)(d+log1δ)nk+d+log1δnk)\displaystyle\leq(\widehat{\En}f-\widehat{p})^{2}+\widetilde{O}\left(\lvert\widehat{\En}f-\widehat{p}\rvert\cdot\sqrt{\frac{(\En f-p)\left(d+\log\frac{1}{\delta}\right)}{nk}}+\frac{d+\log\frac{1}{\delta}}{nk}\right)
(𝔼^fp^)2+O~((𝔼fp)3(d+log1δ)nk+d+log1δnk)\displaystyle\leq(\widehat{\En}f-\widehat{p})^{2}+\widetilde{O}\left(\sqrt{\frac{(\En f-p)^{3}\left(d+\log\frac{1}{\delta}\right)}{nk}}+\frac{d+\log\frac{1}{\delta}}{nk}\right)
B(f)\displaystyle\Rightarrow\quad B(f) B^(f)+O~(B(f)(d+log1δ)n+d+log1δn),\displaystyle\leq\widehat{B}(f)+\widetilde{O}\left(\sqrt{\frac{B(f)\left(d+\log\frac{1}{\delta}\right)}{n}}+\frac{d+\log\frac{1}{\delta}}{n}\right), (21)

where the second inequality uses the bound in Eq. (19).

Combining Eq. (20) and (21), we get the two-sided bound for all ff\in\mathcal{F} such that 𝔼fp\En f\geq p:

|B(f)B^(f)|O~(B(f)(d+log1δ)n+d+log1δn)\displaystyle\lvert B(f)-\widehat{B}(f)\rvert\leq\widetilde{O}\left(\sqrt{\frac{B(f)\left(d+\log\frac{1}{\delta}\right)}{n}}+\frac{d+\log\frac{1}{\delta}}{n}\right) (22)

Following the same approach, we can show that the Eq. (22) also holds for any ff\in\mathcal{F} such that 𝔼f<p\En f<p, as the function class 𝒢\mathcal{G}^{-} also is 1-Lipschitz, bounded in [1,+1][-1,+1] and satisfies Assumption 1 with A=0,B=1A=0,B=1. Therefore we conclude that with probability at least 1δ1-\delta, Eq. (22) holds uniformly for all ff\in\mathcal{F}.

Now we use Eqs. (18) and (22) to get

(f^DSQ)\displaystyle\mathcal{L}(\widehat{f}_{\mathrm{DSQ}}) =LSQ(f^DSQ)B(f^DSQ)\displaystyle=L_{\mathrm{SQ}}(\widehat{f}_{\mathrm{DSQ}})-B(\widehat{f}_{\mathrm{DSQ}})
L^SQ(f^DSQ)B^(f^DSQ)\displaystyle\leq\widehat{L}_{\mathrm{SQ}}(\widehat{f}_{\mathrm{DSQ}})-\widehat{B}(\widehat{f}_{\mathrm{DSQ}})
+O~(kd+k2log1δn+LSQ(f^DSQ)k(d+log1δ)n+B(f^DSQ)(d+log1δ)n)\displaystyle\quad\quad\quad+\widetilde{O}\left(\frac{kd+k^{2}\log\tfrac{1}{\delta}}{n}+\sqrt{\frac{L_{\mathrm{SQ}}(\widehat{f}_{\mathrm{DSQ}})\cdot k\left(d+\log\frac{1}{\delta}\right)}{n}}+\sqrt{\frac{B(\widehat{f}_{\mathrm{DSQ}})\left(d+\log\frac{1}{\delta}\right)}{n}}\right)
L^SQ(f^DSQ)B^(f^DSQ)+O~(kd+k2log1δn+(f^DSQ)k2(d+log1δ)n)\displaystyle\leq\widehat{L}_{\mathrm{SQ}}(\widehat{f}_{\mathrm{DSQ}})-\widehat{B}(\widehat{f}_{\mathrm{DSQ}})+\widetilde{O}\left(\frac{kd+k^{2}\log\tfrac{1}{\delta}}{n}+\sqrt{\frac{\mathcal{L}(\widehat{f}_{\mathrm{DSQ}})\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right)
L^SQ(f)B^(f)+O~(kd+k2log1δn+(f^DSQ)k2(d+log1δ)n)\displaystyle\leq\widehat{L}_{\mathrm{SQ}}(f^{\star})-\widehat{B}(f^{\star})+\widetilde{O}\left(\frac{kd+k^{2}\log\tfrac{1}{\delta}}{n}+\sqrt{\frac{\mathcal{L}(\widehat{f}_{\mathrm{DSQ}})\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right)
(f)+O~(kd+k2log1δn+(f^DSQ)k2(d+log1δ)n).\displaystyle\leq\mathcal{L}(f^{\star})+\widetilde{O}\left(\frac{kd+k^{2}\log\tfrac{1}{\delta}}{n}+\sqrt{\frac{\mathcal{L}(\widehat{f}_{\mathrm{DSQ}})\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right).

The third inequality uses the fact that for any predictor ff, we have kLSQ(f)+B(f)=k(f)+(k+1)B(f)(k2+k1)(f)kL_{\mathrm{SQ}}(f)+B(f)=k\mathcal{L}(f)+(k+1)B(f)\leq(k^{2}+k-1)\mathcal{L}(f). The fourth and fifth inequalities uses the optimality of f^DSQ\widehat{f}_{\mathrm{DSQ}} and ff^{\star} for the empirical and population minimization problems respectively.

Finally, using the inequality Fact 1 we get that

(f^DSQ)\displaystyle\mathcal{L}(\widehat{f}_{\mathrm{DSQ}}) (f)+O~(kd+k2log1δn+(f^DSQ)k2(d+log1δ)n)\displaystyle\leq\mathcal{L}(f^{\star})+\widetilde{O}\left(\frac{kd+k^{2}\log\tfrac{1}{\delta}}{n}+\sqrt{\frac{\mathcal{L}(\widehat{f}_{\mathrm{DSQ}})\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right)
(f)+O~(k2(d+log1δ)n+(f)+kd+k2log1δnk2(d+log1δ)n)\displaystyle\leq\mathcal{L}(f^{\star})+\widetilde{O}\left(\frac{k^{2}\left(d+\log\tfrac{1}{\delta}\right)}{n}+\sqrt{\mathcal{L}(f^{\star})+\frac{kd+k^{2}\log\tfrac{1}{\delta}}{n}}\cdot\sqrt{\frac{k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right)
=(f)+O~(k2(d+log1δ)n+(f)k2(d+log1δ)n)\displaystyle=\mathcal{L}(f^{\star})+\widetilde{O}\left(\frac{k^{2}\left(d+\log\tfrac{1}{\delta}\right)}{n}+\sqrt{\frac{\mathcal{L}(f^{\star})\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right)

This concludes the proof of Theorem 2.

Appendix D Proofs for Section 4

D.1 Offset Loss Class

To analyze the performance of f^EZ\widehat{f}_{\mathrm{EZ}}, we consider an offset version of the EasyLLP loss estimate. Specifically, let finff(f)f^{\star}\coloneqq\inf_{f\in\mathcal{F}}\mathcal{L}(f). We define the offset loss

Γ^(f,f)L^EZ(f)L^EZ(f)=1ni=1n(k(2αi2p)+(2p1))(1kj=1kf(xi,j)f(xi,j))\displaystyle\widehat{\Gamma}(f,f^{\star})\coloneqq\widehat{L}_{\mathrm{EZ}}(f)-\widehat{L}_{\mathrm{EZ}}(f^{\star})=\frac{1}{n}\sum_{i=1}^{n}\left(k(2\alpha_{i}-2p)+(2p-1)\right)\cdot\left(\frac{1}{k}\sum_{j=1}^{k}f^{\star}(x_{i,j})-f(x_{i,j})\right)

Moreover, we use Γ(f,f)\Gamma(f,f^{\star}) to denote its expectation, and we have 𝔼[Γ^(f,f)]=(f)(f)\En[\widehat{\Gamma}(f,f^{\star})]=\mathcal{L}(f)-\mathcal{L}(f^{\star}). Clearly, minimizing the original EasyLLP loss is equivalent to minimizing Γ^(f,f)\widehat{\Gamma}(f,f^{\star}).

We also define the (empirical) second moment of the Γ\Gamma function as

Γ^2(f,f)1ni=1n(k(2αi2p)+(2p1))2(1kj=1kf(xi,j)f(xi,j))2,\displaystyle\widehat{\Gamma}^{2}(f,f^{\star})\coloneqq\frac{1}{n}\sum_{i=1}^{n}\left(k(2\alpha_{i}-2p)+(2p-1)\right)^{2}\cdot\left(\frac{1}{k}\sum_{j=1}^{k}f^{\star}(x_{i,j})-f(x_{i,j})\right)^{2},

and use Γ2(f,f)\Gamma^{2}(f,f^{\star}) to denote 𝔼[Γ^2(f,f)]\En\left[\widehat{\Gamma}^{2}(f,f^{\star})\right].

We show that the offset loss class

𝒢{(B,α)(k(2α2p)+(2p1))(1kj=1kf(xj)f(xj)):f}\displaystyle\mathcal{G}\coloneqq\left\{(B,\alpha)\mapsto\left(k(2\alpha-2p)+(2p-1)\right)\cdot\left(\frac{1}{k}\sum_{j=1}^{k}f^{\star}(x_{j})-f(x_{j})\right):f\in\mathcal{F}\right\}

satisfies Assumption 1.

Lemma 7.

The function class 𝒢\mathcal{G} satisfies Assumption 1 with A=8k2(f)A=8k^{2}\mathcal{L}(f^{\star}) and B=4k2B=4k^{2}.

Proof.

First we observe that because f=argminf(f)f^{\star}=\argmin_{f\in\mathcal{F}}\mathcal{L}(f), we have Γ(f,f)0\Gamma(f,f^{\star})\geq 0 for all ff\in\mathcal{F}. For the variance bound we can compute that

Γ2(f,f)\displaystyle\Gamma^{2}(f,f^{\star}) 4𝔼[(j=1kf(xj)f(xj))2]4k2𝔼[(f(x)f(x))2]\displaystyle\leq 4\cdot\En\left[\left(\sum_{j=1}^{k}f^{\star}(x_{j})-f(x_{j})\right)^{2}\right]\leq 4k^{2}\En\left[\left(f^{\star}(x)-f(x)\right)^{2}\right]
=4k2𝔼[(f(x)y+yf(x))2]4k2((f)+(f))\displaystyle=4k^{2}\En\left[\left(f^{\star}(x)-y+y-f(x)\right)^{2}\right]\leq 4k^{2}\left(\mathcal{L}(f)+\mathcal{L}(f^{\star})\right)
=8k2(f)+4k2Γ(f,f).\displaystyle=8k^{2}\mathcal{L}(f^{\star})+4k^{2}\Gamma(f,f^{\star}).

The first inequality uses the fact that (k(2α2p)+(2p1))[2k+1,2k1]\left(k(2\alpha-2p)+(2p-1)\right)\in[-2k+1,2k-1]. The second inequality uses the independence of the {xj}\{x_{j}\} as well as Jensen’s inequality. The third inequality uses the fact that (f(x)y)2=𝟙{f(x)y}(f(x)-y)^{2}=\mathbbm{1}\left\{f(x)\neq y\right\}, and that the cross terms satisfy (f(x)y)(yf(x))0(f^{\star}(x)-y)(y-f(x))\leq 0.

This concludes the proof of Lemma 7. ∎

D.2 Uniform Convergence for Offset Loss Class

Now we are ready to prove a uniform convergence bound for the function class 𝒢\mathcal{G}.

Lemma 8 (Lemma 1, restated).

Let f=argminf(f)f^{\star}=\argmin_{f\in\mathcal{F}}\mathcal{L}(f). Then with probability at least 1δ1-\delta we have for all ff\in\mathcal{F}

|Γ^(f,f)Γ(f,f)|O~(kd+k2log1δn+(f)k2(d+log1δ)n).\displaystyle\left\lvert\widehat{\Gamma}(f,f^{\star})-\Gamma(f,f^{\star})\right\rvert\leq\widetilde{O}\left(\frac{kd+k^{2}\log\frac{1}{\delta}}{n}+\sqrt{\frac{\mathcal{L}(f)\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right).
Proof.

By Lemma 7, we know that the function class 𝒢\mathcal{G} satisfies Assumption 1 with A=8k2(f)A=8k^{2}\mathcal{L}(f^{\star}) and B=4k2B=4k^{2}. We also know that gk\|g\|_{\infty}\leq k for all g𝒢g\in\mathcal{G}. Therefore, we can apply Theorem 5 to get that with probability at least 1δ1-\delta for all ff\in\mathcal{F},

|Γ^(f,f)Γ(f,f)|\displaystyle\hskip-10.00002pt\left\lvert\widehat{\Gamma}(f,f^{\star})-\Gamma(f,f^{\star})\right\rvert
C(krn+rn2k2(f)+k2(f)(log1δ+cloglogn)n+k2(log1δ+cloglogn)n).\displaystyle\leq C\left(kr^{\star}_{n}+\sqrt{r^{\star}_{n}\cdot 2k^{2}\mathcal{L}(f)}+\sqrt{\frac{k^{2}\mathcal{L}(f)\left(\log\frac{1}{\delta}+c\log\log n\right)}{n}}+\tfrac{k^{2}\left(\log\tfrac{1}{\delta}+c\log\log n\right)}{n}\right). (23)

where rnr^{\star}_{n} is the critical radius and C>0C>0 is an absolute numerical constant.

Since the function (B,α)(k(2α2p)+(2p1))(1kj=1kf(xj)f(xj))(B,\alpha)\mapsto\left(k(2\alpha-2p)+(2p-1)\right)\cdot\left(\frac{1}{k}\sum_{j=1}^{k}f^{\star}(x_{j})-f(x_{j})\right) is 2k2k-Lipschitz, we have by Lemma 5 that

rn=O(dlog(kn)n).\displaystyle r^{\star}_{n}=O\left(\frac{d\log(kn)}{n}\right).

Plugging this into Eq. (23) we get the conclusion of Lemma 1. ∎

D.3 Proof of Theorem 3

Now we will prove the final generalization bound for EasyLLP. Using Lemma 1, we get that

Γ(f^EZ,f)\displaystyle\Gamma(\widehat{f}_{\mathrm{EZ}},f^{\star}) Γ^(f^EZ,f)+O~(kd+k2log1δn+(f^EZ)k2(d+log1δ)n)\displaystyle\leq\widehat{\Gamma}(\widehat{f}_{\mathrm{EZ}},f^{\star})+\widetilde{O}\left(\frac{kd+k^{2}\log\frac{1}{\delta}}{n}+\sqrt{\frac{\mathcal{L}(\widehat{f}_{\mathrm{EZ}})\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right)
=Γ^(f^EZ,f)+O~(kd+k2log1δn+(Γ(f^EZ,f)+(f))k2(d+log1δ)n)\displaystyle=\widehat{\Gamma}(\widehat{f}_{\mathrm{EZ}},f^{\star})+\widetilde{O}\left(\frac{kd+k^{2}\log\frac{1}{\delta}}{n}+\sqrt{\frac{\left(\Gamma(\widehat{f}_{\mathrm{EZ}},f^{\star})+\mathcal{L}(f^{\star})\right)\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right)

In fact, we know that since ff^{\star}\in\mathcal{F}, we must have Γ^(f^EZ,f)Γ^(f,f)=0\widehat{\Gamma}(\widehat{f}_{\mathrm{EZ}},f^{\star})\leq\widehat{\Gamma}(f^{\star},f^{\star})=0, so using Fact 1 we can further upper bound this as

Γ(f^EZ,f)\displaystyle\Gamma(\widehat{f}_{\mathrm{EZ}},f^{\star}) O~(k2(d+log1δ)n+(f)k2(d+log1δ)n).\displaystyle\leq\widetilde{O}\left(\frac{k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}+\sqrt{\frac{\mathcal{L}(f^{\star})\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right).

Plugging in the definition of Γ\Gamma proves the bound.

D.4 Proof of Corollary 1

We describe the sample splitting version of the EasyLLP learning rule, which allows us to use an estimate p^\widehat{p} instead of the true marginal label proportion pp. We assume that we are given a dataset of size 2n2n, denoted S={(Bi,αi)}i=12nS=\{(B_{i},\alpha_{i})\}_{i=1}^{2n}. We split it randomly into two equally-sized parts SS and SS^{\prime}. In the proof, we will use i[n]i\in[n] to index bags in SS and i[n]i^{\prime}\in[n] to index bags in SS^{\prime}.

  1. 1.

    Using SS, estimate marginal label proportion p^=1ni=1nαi\widehat{p}=\frac{1}{n}\sum_{i=1}^{n}\alpha_{i}.

  2. 2.

    Return f^EZargminf1ni=1nEZ(f,(Bi,αi))\widehat{f}_{\mathrm{EZ}}\coloneqq\argmin_{f\in\mathcal{F}}\frac{1}{n}\sum_{i^{\prime}=1}^{n}\ell_{\mathrm{EZ}}\left(f,(B_{i^{\prime}},\alpha_{i^{\prime}})\right) using SS^{\prime}, where EZ(,)\ell_{\mathrm{EZ}}(\cdot,\cdot) is defined with the plug-in estimate p^\widehat{p} instead of pp:

    EZ(f,(B,α))=(k(αp^)+p^)(11kj=1kf(xj))+(k(p^α)+(1p^))(1kj=1kf(xj)).\displaystyle\ell_{\mathrm{EZ}}\left(f,(B,\alpha)\right)=\left(k(\alpha-\widehat{p})+\widehat{p}\right)\cdot\left(1-\frac{1}{k}\sum_{j=1}^{k}f(x_{j})\right)+\left(k(\widehat{p}-\alpha)+(1-\widehat{p})\right)\cdot\left(\frac{1}{k}\sum_{j=1}^{k}f(x_{j})\right).

For any q[0,1]q\in[0,1] let us define the quantity

Γ^q(f,f)1ni=1n(k(2αi2q)+(2q1))(1kj=1kf(xi,j)f(xi,j)),\displaystyle\widehat{\Gamma}_{q}(f,f^{\star})\coloneqq\frac{1}{n}\sum_{i^{\prime}=1}^{n}\left(k(2\alpha_{i^{\prime}}-2q)+(2q-1)\right)\cdot\left(\frac{1}{k}\sum_{j=1}^{k}f^{\star}(x_{i^{\prime},j})-f(x_{i^{\prime},j})\right),

which represents the empirical offset loss estimated on SS^{\prime} if we plugged in the value qq for the marginal label proportion.

By Lemma 1 we know that for any ff\in\mathcal{F}

Γp(f,f)\displaystyle\Gamma_{p}(f,f^{\star}) Γ^p(f,f)+O~(kd+k2log1δn+(f)k2(d+log1δ)n).\displaystyle\leq\widehat{\Gamma}_{p}(f,f^{\star})+\widetilde{O}\left(\frac{kd+k^{2}\log\frac{1}{\delta}}{n}+\sqrt{\frac{\mathcal{L}(f)\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right).

From here, we need to relate Γ^p(f,f)\widehat{\Gamma}_{p}(f,f^{\star}) to Γ^p^(f,f)\widehat{\Gamma}_{\widehat{p}}(f,f^{\star}). We can bound the difference as:

|Γ^p(f,f)Γ^p^(f,f)|(2k2)|p^p||1ni=1n1kj=1kf(xi,j)f(xi,j)Ξ(f)|.\displaystyle\left\lvert\widehat{\Gamma}_{p}(f,f^{\star})-\widehat{\Gamma}_{\widehat{p}}(f,f^{\star})\right\rvert\leq(2k-2)\cdot\left\lvert\widehat{p}-p\right\rvert\cdot\Big\lvert\underbrace{\frac{1}{n}\sum_{i^{\prime}=1}^{n}\frac{1}{k}\sum_{j=1}^{k}f^{\star}(x_{i^{\prime},j})-f(x_{i^{\prime},j})}_{\eqqcolon\Xi(f)}\Big\rvert.

By Hoeffding’s inequality (Theorem 6), with probability at least 1δ1-\delta over SS, we have |p^p|2log(2/δ)nk\left\lvert\widehat{p}-p\right\rvert\leq\sqrt{\frac{2\log(2/\delta)}{nk}}. Henceforth we condition on this event holding over SS.

Now we prove a uniform convergence bound on |Ξ()|\lvert\Xi(\cdot)\rvert for all ff\in\mathcal{F}. For any ff\in\mathcal{F} we write that

|Ξ(f)|\displaystyle\lvert\Xi(f)\rvert =|1ni=1n1kj=1kf(xi,j)f(xi,j)|1nki=1nj=1k(f(xi,j)f(xi,j))2\displaystyle=\left\lvert\frac{1}{n}\sum_{i^{\prime}=1}^{n}\frac{1}{k}\sum_{j=1}^{k}f^{\star}(x_{i^{\prime},j})-f(x_{i^{\prime},j})\right\rvert\leq\sqrt{\frac{1}{nk}}\cdot\sqrt{\sum_{i^{\prime}=1}^{n}\sum_{j=1}^{k}\left(f^{\star}(x_{i^{\prime},j})-f(x_{i^{\prime},j})\right)^{2}}
1nki=1nj=1k(f(xi,j)yi,j)2+(yi,jf(xi,j))2=L^(f)+L^(f).\displaystyle\leq\sqrt{\frac{1}{nk}}\cdot\sqrt{\sum_{i^{\prime}=1}^{n}\sum_{j=1}^{k}\left(f^{\star}(x_{i^{\prime},j})-y_{i^{\prime},j}\right)^{2}+(y_{i^{\prime},j}-f(x_{i^{\prime},j}))^{2}}=\sqrt{\widehat{L}(f^{\star})+\widehat{L}(f)}.

Here, we use L^()\widehat{L}(\cdot) to denote the empirical classification loss on SS^{\prime}. The first inequality follows by Cauchy-Schwarz. The second inequality uses the fact that for a,b,c{0,1}a,b,c\in\{0,1\} we have (ab)(bc)0(a-b)(b-c)\leq 0. The last equality follows because (f(x)y)2=𝟙{f(x)y}(f(x)-y)^{2}=\mathbbm{1}\left\{f(x)\neq y\right\}. Now we use the standard uniform convergence guarantee: with probability at least 1δ1-\delta over SS^{\prime}, for any ff\in\mathcal{F}:

|(f)L^(f)|O~((f)+d+log1δnk).\displaystyle\lvert\mathcal{L}(f)-\widehat{L}(f)\rvert\leq\widetilde{O}\left(\mathcal{L}(f)+\frac{d+\log\frac{1}{\delta}}{nk}\right).

So therefore with probability at least 1δ1-\delta over SS^{\prime} we have for all ff\in\mathcal{F}:

|Ξ(f)|L^(f)+L^(f)O~((f)+d+log1δnk).\displaystyle\lvert\Xi(f)\rvert\leq\sqrt{\widehat{L}(f^{\star})+\widehat{L}(f)}\leq\widetilde{O}\left(\sqrt{\mathcal{L}(f)+\frac{d+\log\frac{1}{\delta}}{nk}}\right).

Thus with probability at least 12δ1-2\delta over the draws of S1S_{1} and S2S_{2}, we have for all ff\in\mathcal{F}:

|Γ^p(f,f)Γ^p^(f,f)|O~(d+log1δn+(f)klog1δn)\displaystyle\left\lvert\widehat{\Gamma}_{p}(f,f^{\star})-\widehat{\Gamma}_{\widehat{p}}(f,f^{\star})\right\rvert\leq\widetilde{O}\left(\frac{d+\log\frac{1}{\delta}}{n}+\sqrt{\frac{\mathcal{L}(f)\cdot k\log\frac{1}{\delta}}{n}}\right) (24)

Now we are ready to prove the final guarantee. Similar to the proof of Theorem 3 we compute that

Γp(f^EZ,f)\displaystyle\Gamma_{p}(\widehat{f}_{\mathrm{EZ}},f^{\star}) Γ^p(f^EZ,f)+O~(kd+k2log1δn+(f^EZ)k2(d+log1δ)n)\displaystyle\leq\widehat{\Gamma}_{p}(\widehat{f}_{\mathrm{EZ}},f^{\star})+\widetilde{O}\left(\frac{kd+k^{2}\log\frac{1}{\delta}}{n}+\sqrt{\frac{\mathcal{L}(\widehat{f}_{\mathrm{EZ}})\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right)
Γ^p^(f^EZ,f)+O~(kd+k2log1δn+(f^EZ)k2(d+log1δ)n)\displaystyle\leq\widehat{\Gamma}_{\widehat{p}}(\widehat{f}_{\mathrm{EZ}},f^{\star})+\widetilde{O}\left(\frac{kd+k^{2}\log\frac{1}{\delta}}{n}+\sqrt{\frac{\mathcal{L}(\widehat{f}_{\mathrm{EZ}})\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right)
O~(kd+k2log1δn+(f^EZ)k2(d+log1δ)n),\displaystyle\leq\widetilde{O}\left(\frac{kd+k^{2}\log\frac{1}{\delta}}{n}+\sqrt{\frac{\mathcal{L}(\widehat{f}_{\mathrm{EZ}})\cdot k^{2}\left(d+\log\frac{1}{\delta}\right)}{n}}\right),

where the second line uses Eq. (24) and the last line follows from the fact that Γ^p^(f^EZ,f)Γ^p^(f,f)=0\widehat{\Gamma}_{\widehat{p}}(\widehat{f}_{\mathrm{EZ}},f^{\star})\leq\widehat{\Gamma}_{\widehat{p}}(f^{\star},f^{\star})=0. From here, the proof concludes similarly as the proof of Theorem 3 in Appendix D.3 by using Fact 1 and plugging in the definition of Γp(,f)\Gamma_{p}(\cdot,f^{\star}).

Appendix E Proof of Theorem 4

First, we describe the construction, then separately prove the lower bound for both the realizable and agnostic settings.

E.1 Construction

We define the instance space 𝒳={0,1}2d\mathcal{X}=\{0,1\}^{2^{d}} and let ={fi:i[2d]}\mathcal{F}=\{f_{i}:i\in[2^{d}]\} where the function fif_{i} is defined as fi(x)=x[i]f_{i}(x)=x[i]. It is clear that VC()d\mathrm{VC}(\mathcal{F})\leq d. Now we define a family of distributions 𝒟i\mathcal{D}_{i} for i[2d]i\in[2^{d}]. For some parameter choice γ[0,1/2]\gamma\in[0,1/2], each 𝒟i\mathcal{D}_{i} is defined as follows: the example xUnif({0,1}2d)x\sim\mathrm{Unif}(\{0,1\}^{2^{d}}) and y=fi(x)y=f_{i}(x) with probability 1/2+γ1/2+\gamma, y=1fi(x)y=1-f_{i}(x) with probability 1/2γ1/2-\gamma. For any predictor ff the classification loss for distribution 𝒟i\mathcal{D}_{i} can be written as

𝒟i(f)inff𝒟i(f)=2γ𝔼xUnif({0,1}2d)[𝟙{f(x)x[i]}].\displaystyle\mathcal{L}_{\mathcal{D}_{i}}(f)-\inf_{f^{\prime}\in\mathcal{F}}\mathcal{L}_{\mathcal{D}_{i}}(f^{\prime})=2\gamma\cdot\En_{x\sim\mathrm{Unif}(\{0,1\}^{2^{d}})}\left[\mathbbm{1}\left\{f(x)\neq x[i]\right\}\right].

Furthermore, for any predictor ff, as well as distributions 𝒟i\mathcal{D}_{i} and 𝒟j\mathcal{D}_{j} we have the separation condition

𝒟i(f)inff𝒟i(f)+𝒟j(f)inff𝒟j(f)\displaystyle\hskip-20.00003pt\mathcal{L}_{\mathcal{D}_{i}}(f)-\inf_{f^{\prime}\in\mathcal{F}}\mathcal{L}_{\mathcal{D}_{i}}(f^{\prime})+\mathcal{L}_{\mathcal{D}_{j}}(f)-\inf_{f^{\prime}\in\mathcal{F}}\mathcal{L}_{\mathcal{D}_{j}}(f^{\prime})
=2γ𝔼xUnif({0,1}2d)[𝟙{f(x)x[i]}+𝟙{f(x)x[j]}]γ.\displaystyle=2\gamma\cdot\En_{x\sim\mathrm{Unif}(\{0,1\}^{2^{d}})}\left[\mathbbm{1}\left\{f(x)\neq x[i]\right\}+\mathbbm{1}\left\{f(x)\neq x[j]\right\}\right]\geq\gamma. (25)

E.2 Realizable Setting

For the realizable setting result, we use the construction with γ=1/2\gamma=1/2. We claim that for any learning rule for LLP that PAC learns \mathcal{F}, there exists a distribution 𝒟i\mathcal{D}_{i} for which it requires n=Ω(d/logk)n=\Omega(d/\log k) samples in expectation. Let us define 𝒟¯\bar{\mathcal{D}} to be the averaged distribution where one first draws iUnif([2d])i\sim\mathrm{Unif}([2^{d}]) then samples the bag (B,α)𝒟i(B,\alpha)\sim\mathcal{D}_{i}. Using the separation condition of Eq. (25), we invoke Fano’s inequality (e.g., Lemma 3 of Yu (1997)) to get

inff^supi[2d]𝔼𝒟i[𝒟i(f^)]14(1n12di=12dKL(𝒟i𝒟¯)+log2d)=14(1nKL(𝒟1𝒟¯)+log2d),\displaystyle\inf_{\widehat{f}}\sup_{i\in[2^{d}]}\En_{\mathcal{D}_{i}}\left[\mathcal{L}_{\mathcal{D}_{i}}\left(\widehat{f}\right)\right]\geq\frac{1}{4}\left(1-\frac{n\cdot\frac{1}{2^{d}}\sum_{i=1}^{2^{d}}\mathrm{KL}\left(\mathcal{D}_{i}\|\bar{\mathcal{D}}\right)+\log 2}{d}\right)=\frac{1}{4}\left(1-\frac{n\cdot\mathrm{KL}\left(\mathcal{D}_{1}\|\bar{\mathcal{D}}\right)+\log 2}{d}\right), (26)

where the equality follows by symmetry of the distributions 𝒟i\mathcal{D}_{i}.

From here we need to estimate KL(𝒟1𝒟¯)\mathrm{KL}\left(\mathcal{D}_{1}\|\bar{\mathcal{D}}\right). By chain rule for KL divergence, we see that

KL(𝒟1𝒟¯)\displaystyle\mathrm{KL}\left(\mathcal{D}_{1}\|\bar{\mathcal{D}}\right) =KL(𝒟1[B]𝒟¯[B])+𝔼B𝒟1[KL(𝒟1[α|B]𝒟¯[α|B])]\displaystyle=\mathrm{KL}\left(\Pr_{\mathcal{D}_{1}}[B]\|\Pr_{\bar{\mathcal{D}}}[B]\right)+\En_{B\sim\mathcal{D}_{1}}\left[\mathrm{KL}\left(\Pr_{\mathcal{D}_{1}}[\alpha\penalty\ |\penalty\ B]\|\Pr_{\bar{\mathcal{D}}}[\alpha\penalty\ |\penalty\ B]\right)\right]
=𝔼B𝒟1[KL(𝒟1[α|B]𝒟¯[α|B])]\displaystyle=\En_{B\sim\mathcal{D}_{1}}\left[\mathrm{KL}\left(\Pr_{\mathcal{D}_{1}}[\alpha\penalty\ |\penalty\ B]\|\Pr_{\bar{\mathcal{D}}}[\alpha\penalty\ |\penalty\ B]\right)\right]
=𝔼BUnif({0,1}2d)[KL(𝒟1[α|B]𝒟¯[α|B])],\displaystyle=\En_{B\sim\mathrm{Unif}(\{0,1\}^{2^{d}})}\left[\mathrm{KL}\left(\Pr_{\mathcal{D}_{1}}[\alpha\penalty\ |\penalty\ B]\|\Pr_{\bar{\mathcal{D}}}[\alpha\penalty\ |\penalty\ B]\right)\right],

since all of the 𝒟i\mathcal{D}_{i} have the same marginal over 𝒳\mathcal{X}. Fix a bag B={x1,,xk}B=\{x_{1},\cdots,x_{k}\}, and let us define the vector z=1kj=1kxj[0,1]2dz=\frac{1}{k}\sum_{j=1}^{k}x_{j}\in[0,1]^{2^{d}}. We calculate that

KL(𝒟1[α|B]𝒟¯[α|B])\displaystyle\mathrm{KL}\left(\Pr_{\mathcal{D}_{1}}[\alpha\penalty\ |\penalty\ B]\|\Pr_{\bar{\mathcal{D}}}[\alpha\penalty\ |\penalty\ B]\right) =α{0,1k,,1}α𝒟1[α|B]logα𝒟1[α|B]α𝒟¯[α|B]\displaystyle=\sum_{\alpha\in\{0,\tfrac{1}{k},\cdots,1\}}\Pr_{\alpha\sim\mathcal{D}_{1}}\left[\alpha\penalty\ |\penalty\ B\right]\cdot\log\frac{\Pr_{\alpha\sim\mathcal{D}_{1}}\left[\alpha\penalty\ |\penalty\ B\right]}{\Pr_{\alpha\sim\bar{\mathcal{D}}}\left[\alpha\penalty\ |\penalty\ B\right]}
=log1α𝒟¯[α=z[1]|B]\displaystyle=\log\frac{1}{\Pr_{\alpha\sim\bar{\mathcal{D}}}\left[\alpha=z[1]\penalty\ |\penalty\ B\right]}
=log112d+12di>1α𝒟i[α=z[1]|B]\displaystyle=\log\frac{1}{\frac{1}{2^{d}}+\frac{1}{2^{d}}\sum_{i>1}\Pr_{\alpha\sim\mathcal{D}_{i}}\left[\alpha=z[1]\penalty\ |\penalty\ B\right]}
=log112d+12di>1𝟙{z[i]=z[1]}\displaystyle=\log\frac{1}{\frac{1}{2^{d}}+\frac{1}{2^{d}}\sum_{i>1}\mathbbm{1}\left\{z[i]=z[1]\right\}}
min{d,log2di>1𝟙{z[i]=z[1]}}.\displaystyle\leq\min\left\{d,\log\frac{2^{d}}{\sum_{i>1}\mathbbm{1}\left\{z[i]=z[1]\right\}}\right\}.

The second line follows from the fact that once we fix BB, the value of α=z[1]\alpha=z[1] is deterministic under 𝒟1\mathcal{D}_{1}. Putting it together we get that, we get

KL(𝒟1𝒟¯)𝔼BUnif({0,1}2d)[min{d,log2di>1𝟙{z[i]=z[1]}}].\displaystyle\mathrm{KL}\left(\mathcal{D}_{1}\|\bar{\mathcal{D}}\right)\leq\En_{B\sim\mathrm{Unif}(\{0,1\}^{2^{d}})}\left[\min\left\{d,\log\frac{2^{d}}{\sum_{i>1}\mathbbm{1}\left\{z[i]=z[1]\right\}}\right\}\right].

Observe that kz[i]k\cdot z[i] is distributed as independent Bin(k,1/2)\mathrm{Bin}(k,1/2) variables for all i[2d]i\in[2^{d}]. Using Lemma 9, we have for all i>1i>1 that B[z[i]=z[1]]1/(k+1)\Pr_{B}[z[i]=z[1]]\geq 1/(k+1), since the random variable z[i]z[i] takes at most k+1k+1 values. Thus, applying Chernoff bounds we have for any δ(0,1)\delta\in(0,1),

B[i>1𝟙{z[i]=z[1]}(1δ)2d1k+1]1exp(δ222d1k+1).\displaystyle\Pr_{B}\left[\sum_{i>1}\mathbbm{1}\left\{z[i]=z[1]\right\}\geq\left(1-\delta\right)\frac{2^{d}-1}{k+1}\right]\geq 1-\exp\left(-\frac{\delta^{2}}{2}\cdot\frac{2^{d}-1}{k+1}\right).

Let us pick δ=2(k+1)logd/(2d1)\delta=\sqrt{2(k+1)\log d/(2^{d}-1)}; by our assumption on kk we have δ1/2\delta\leq 1/2. This guarantees that the above event, call it \mathcal{E}, happens with probability at least 11/d1-1/d. Therefore, we get

KL(𝒟1𝒟¯)d1d+log2d2(k+1)2d12+log(k+1).\displaystyle\mathrm{KL}\left(\mathcal{D}_{1}\|\bar{\mathcal{D}}\right)\leq d\cdot\frac{1}{d}+\log\frac{2^{d}\cdot 2(k+1)}{2^{d}-1}\leq 2+\log(k+1). (27)

Therefore, plugging in Eq. (27) into Eq. (26) we see that

inff^supi[2d]𝔼𝒟i[𝒟i(f^)]14(1n(2+log(k+1))+log2d).\displaystyle\inf_{\widehat{f}}\sup_{i\in[2^{d}]}\En_{\mathcal{D}_{i}}\left[\mathcal{L}_{\mathcal{D}_{i}}\left(\widehat{f}\right)\right]\geq\frac{1}{4}\left(1-\frac{n\left(2+\log(k+1)\right)+\log 2}{d}\right).

For n=Cd/(log(k+1))n=C^{\prime}d/(\log(k+1)) where C>0C^{\prime}>0 is a sufficiently small constant we have inff^supi[2d]𝔼𝒟i[𝒟i(f^)]18\inf_{\widehat{f}}\sup_{i\in[2^{d}]}\En_{\mathcal{D}_{i}}\left[\mathcal{L}_{\mathcal{D}_{i}}\left(\widehat{f}\right)\right]\geq\frac{1}{8}. Fix any learning rule f^\widehat{f}, and let 𝒟i\mathcal{D}_{i^{\star}} be the distribution which witnesses the supremum. We have

18𝔼𝒟i[𝒟i(f^)]𝒟i[𝒟i(f^)>116]+116𝒟i[𝒟i(f^)116]\displaystyle\frac{1}{8}\leq\En_{\mathcal{D}_{i^{\star}}}\left[\mathcal{L}_{\mathcal{D}_{i^{\star}}}\left(\widehat{f}\right)\right]\leq\Pr_{\mathcal{D}_{i^{\star}}}\left[\mathcal{L}_{\mathcal{D}_{i^{\star}}}\left(\widehat{f}\right)>\frac{1}{16}\right]+\frac{1}{16}\cdot\Pr_{\mathcal{D}_{i^{\star}}}\left[\mathcal{L}_{\mathcal{D}_{i^{\star}}}\left(\widehat{f}\right)\leq\frac{1}{16}\right]

which implies that 𝒟i[𝒟i(f^)>116]1/15\Pr_{\mathcal{D}_{i^{\star}}}\left[\mathcal{L}_{\mathcal{D}_{i^{\star}}}\left(\widehat{f}\right)>\frac{1}{16}\right]\geq 1/15. We conclude that any learning rule which PAC learns \mathcal{F} with parameters (ε,δ)=(1/16,1/15)(\varepsilon,\delta)=(1/16,1/15) requires n=Ω(d/logk)n=\Omega(d/\log k) samples.

E.3 Agnostic Setting

For the agnostic setting result we use the construction with γ=ε\gamma=\varepsilon. Again, by Fano’s inequality we get

inff^supi[2d]𝔼𝒟i[𝒟i(f^)inff𝒟i(f)]\displaystyle\inf_{\widehat{f}}\sup_{i\in[2^{d}]}\En_{\mathcal{D}_{i}}\left[\mathcal{L}_{\mathcal{D}_{i}}\left(\widehat{f}\right)-\inf_{f^{\prime}\in\mathcal{F}}\mathcal{L}_{\mathcal{D}_{i}}(f^{\prime})\right] ε2(1n12di=12dKL(𝒟i𝒟¯)+log2d)\displaystyle\geq\frac{\varepsilon}{2}\left(1-\frac{n\cdot\frac{1}{2^{d}}\sum_{i=1}^{2^{d}}\mathrm{KL}\left(\mathcal{D}_{i}\|\bar{\mathcal{D}}\right)+\log 2}{d}\right)
=ε2(1nKL(𝒟1𝒟¯)+log2d),\displaystyle=\frac{\varepsilon}{2}\left(1-\frac{n\cdot\mathrm{KL}\left(\mathcal{D}_{1}\|\bar{\mathcal{D}}\right)+\log 2}{d}\right), (28)

where the equality follows by symmetry of the distributions 𝒟i\mathcal{D}_{i}.

From here we need to estimate KL(𝒟1𝒟¯)\mathrm{KL}\left(\mathcal{D}_{1}\|\bar{\mathcal{D}}\right). By chain rule for KL divergence, we see that

KL(𝒟1𝒟¯)\displaystyle\mathrm{KL}\left(\mathcal{D}_{1}\|\bar{\mathcal{D}}\right) =𝔼BUnif({0,1}2d)[KL(𝒟1[α|B]𝒟¯[α|B])].\displaystyle=\En_{B\sim\mathrm{Unif}(\{0,1\}^{2^{d}})}\left[\mathrm{KL}\left(\Pr_{\mathcal{D}_{1}}[\alpha\penalty\ |\penalty\ B]\|\Pr_{\bar{\mathcal{D}}}[\alpha\penalty\ |\penalty\ B]\right)\right].

From here we can bound this in two ways. The first way is to use the data-processing inequality for KL divergence. Observe that in distribution 𝒟i\mathcal{D}_{i}, the label proportion α\alpha is distributed as Bin(kz[i],1/2+ε)+Bin(kkz[i],1/2ε)\mathrm{Bin}(kz[i],1/2+\varepsilon)+\mathrm{Bin}(k-kz[i],1/2-\varepsilon). Therefore, by the data processing inequality

KL(𝒟1[α|B]𝒟¯[α|B])\displaystyle\mathrm{KL}\left(\Pr_{\mathcal{D}_{1}}[\alpha\penalty\ |\penalty\ B]\|\Pr_{\bar{\mathcal{D}}}[\alpha\penalty\ |\penalty\ B]\right) KL(𝒟1[αclean|B]𝒟¯[αclean|B])\displaystyle\leq\mathrm{KL}\left(\Pr_{\mathcal{D}_{1}}[\alpha_{\mathrm{clean}}\penalty\ |\penalty\ B]\|\Pr_{\bar{\mathcal{D}}}[\alpha_{\mathrm{clean}}\penalty\ |\penalty\ B]\right)
min{d,log2di>1𝟙{z[i]=z[1]}},\displaystyle\leq\min\left\{d,\log\frac{2^{d}}{\sum_{i>1}\mathbbm{1}\left\{z[i]=z[1]\right\}}\right\},

where αclean=z[i]\alpha_{\mathrm{clean}}=z[i] in distribution 𝒟i\mathcal{D}_{i}. From here, the proof proceeds similarly as in the realizable setting. We get the lower bound of Ω(d/logk)\Omega(d/\log k), with no dependence on ε\varepsilon.

Alternatively, we can directly calculate the bound on the KL divergence:

KL(𝒟1[α|B]𝒟¯[α|B])\displaystyle\hskip-20.00003pt\mathrm{KL}\left(\Pr_{\mathcal{D}_{1}}[\alpha\penalty\ |\penalty\ B]\|\Pr_{\bar{\mathcal{D}}}[\alpha\penalty\ |\penalty\ B]\right)
12di=12dKL(𝒟1[α|B]𝒟i[α|B])\displaystyle\leq\frac{1}{2^{d}}\sum_{i=1}^{2^{d}}\mathrm{KL}\left(\Pr_{\mathcal{D}_{1}}[\alpha\penalty\ |\penalty\ B]\|\Pr_{\mathcal{D}_{i}}[\alpha\penalty\ |\penalty\ B]\right)
=2d12dKL(𝒟1[α|B]𝒟2[α|B])\displaystyle=\frac{2^{d}-1}{2^{d}}\cdot\mathrm{KL}\left(\Pr_{\mathcal{D}_{1}}[\alpha\penalty\ |\penalty\ B]\|\Pr_{\mathcal{D}_{2}}[\alpha\penalty\ |\penalty\ B]\right)
max{kz[1]kz[2],0}kl(1/2+ε1/2ε)+max{kz[2]kz[1],0}kl(1/2ε1/2+ε)\displaystyle\leq\max\{kz[1]-kz[2],0\}\cdot\mathrm{kl}(1/2+\varepsilon\|1/2-\varepsilon)+\max\{kz[2]-kz[1],0\}\cdot\mathrm{kl}(1/2-\varepsilon\|1/2+\varepsilon)
|z[1]z[2]|O(kε2).\displaystyle\leq\lvert z[1]-z[2]\rvert\cdot O(k\varepsilon^{2}).

The first line uses the convexity of KL. The second line uses the symmetry of the distributions 𝒟i\mathcal{D}_{i}. The third line follows because under 𝒟i\mathcal{D}_{i}, kαBin(kz[i],1/2+ε)+Bin(kkz[i],1/2ε)k\alpha\sim\mathrm{Bin}(kz[i],1/2+\varepsilon)+\mathrm{Bin}(k-kz[i],1/2-\varepsilon), then applying chain rule for KL divergence. The last line applies the bound on the KL divergence between two Bernoulli random variables.

Now we investigate the expected difference between z[1]z[1] and z[2]z[2]. Note that both of these are independently distributed as Bin(k,1/2)/k\mathrm{Bin}(k,1/2)/k. By Hoeffding’s inequality, we see that for any i[2d]i\in[2^{d}],

B[|z[i]12|log(2k)2k]1k,\displaystyle\Pr_{B}\left[\left\lvert z[i]-\frac{1}{2}\right\rvert\geq\sqrt{\frac{\log(2k)}{2k}}\right]\leq\frac{1}{k},

so by union bound, with probability at least 12/k1-2/k we have |z[1]z[2]|2log(2k)k\lvert z[1]-z[2]\rvert\leq\sqrt{\frac{2\log(2k)}{k}}. Using this, we can compute the bound that

KL(𝒟1𝒟¯)\displaystyle\mathrm{KL}\left(\mathcal{D}_{1}\|\bar{\mathcal{D}}\right) O(kε2)𝔼B[|z[1]z[2]|]\displaystyle\leq O(k\varepsilon^{2})\cdot\En_{B}\left[\lvert z[1]-z[2]\rvert\right]
O(kε2)(2k+2log(2k)k)\displaystyle\leq O(k\varepsilon^{2})\cdot\left(\frac{2}{k}+\sqrt{\frac{2\log(2k)}{k}}\right)
O(ε2)(2+2klog(2k)).\displaystyle\leq O(\varepsilon^{2})\cdot\left(2+\sqrt{2k\log(2k)}\right).

Plugging the previous display into Eq. (28), we see that if n=Cd/(kε2)n=C^{\prime}d/(\sqrt{k}\varepsilon^{2}) where C>0C^{\prime}>0 is a sufficiently small constant we have inff^supi[2d]𝔼𝒟i[𝒟i(f^)inff𝒟i(f)]ε4\inf_{\widehat{f}}\sup_{i\in[2^{d}]}\En_{\mathcal{D}_{i}}\left[\mathcal{L}_{\mathcal{D}_{i}}\left(\widehat{f}\right)-\inf_{f^{\prime}\in\mathcal{F}}\mathcal{L}_{\mathcal{D}_{i}}(f^{\prime})\right]\geq\frac{\varepsilon}{4}. As with the realizable setting proof, we can translate this to a lower bound for PAC learning; we omit the details.

Appendix F Technical Lemmas

Fact 1.

For any A,B,C0A,B,C\geq 0 if AB+CAA\leq B+C\sqrt{A} then AB+C2+BC2B+2C2A\leq B+C^{2}+\sqrt{B}C\leq 2B+2C^{2}.

Theorem 6 (Hoeffding’s Inequality).

Let Z1,,ZnZ_{1},\cdots,Z_{n} be independent bounded random variables with Zi[a,b]Z_{i}\in[a,b] for all i[n]i\in[n]. Then

[|1nt=1nZi𝔼[Zi]|t]2exp(2nt2(ba)2).\displaystyle\Pr\left[\left\lvert\frac{1}{n}\sum_{t=1}^{n}Z_{i}-\En[Z_{i}]\right\rvert\geq t\right]\leq 2\exp\left(\frac{-2nt^{2}}{(b-a)^{2}}\right).
Theorem 7 (Paley-Zygmund).

Let Z0Z\geq 0 be a random variable with finite variance. For any θ[0,1]\theta\in[0,1],

[Zθ𝔼Z](1θ)2𝔼[Z]2𝔼[Z2].\displaystyle\Pr\left[Z\geq\theta\cdot\En Z\right]\geq(1-\theta)^{2}\frac{\En[Z]^{2}}{\En[Z^{2}]}.
Lemma 9.

Let XX and YY be two i.i.d. discrete random variables which are supported on a set of size kk. Then [X=Y]1/k\Pr\left[X=Y\right]\geq 1/k.

Proof.

Let Ω\Omega be the support. The probability that X=YX=Y can be calculated as

[X=Y]=xΩ[X=x]21k(xΩ[X=x])2=1k.\displaystyle\Pr\left[X=Y\right]=\sum_{x\in\Omega}\Pr[X=x]^{2}\geq\frac{1}{k}\cdot\left(\sum_{x\in\Omega}\Pr[X=x]\right)^{2}=\frac{1}{k}.

The inequality uses the fact that v1kv2\|v\|_{1}\leq\sqrt{k}\|v\|_{2} for any vkv\in\mathbb{R}^{k}. ∎

Appendix G Experimental Details

G.1 Implementation Details

Our code can be found on GitHub at https://github.com/GXLI97/llp_experiments. All experiments were run on an NVIDIA RTX A6000 GPU using Tensorflow and Keras.

We elaborate on the architectures used in our experiments.

  • The linear model has a single dense output layer with 1 unit and sigmoid activation.

  • The small two layer NN comprises of a dense layer with 100 units and ReLU activation, followed by a dense output layer with 1 unit and sigmoid activation.

  • The large two layer NN comprises of a dense layer with 1000 units and ReLU activation, followed by a dense output layer with 1 unit and sigmoid activation.

  • The small CNN is the same architecture in (Busa-Fekete et al., 2023):

    • Convolutional layer with 32 kernels of size 3×33\times 3 and ReLU activation.

    • Max pooling layer with pool 2×22\times 2.

    • Convolutional layer with 64 kernels of size 3×33\times 3 and ReLU activation.

    • Max pooling layer with pool size 2×22\times 2.

    • Flatten layer.

    • Dropout layer with drop rate 0.50.5.

    • Dense output layer with 1 unit and sigmoid activation.

  • The large CNN is the same architecture in (Busa-Fekete et al., 2023):

    • Convolutional layer with 32 kernels of size 3×33\times 3 and ReLU activation.

    • Convolutional layer with 32 kernels of size 3×33\times 3 and ReLU activation.

    • Max pooling layer with pool 2×22\times 2.

    • Dropout layer with drop rate 0.250.25.

    • Convolutional layer with 64 kernels of size 3×33\times 3 and ReLU activation.

    • Convolutional layer with 64 kernels of size 3×33\times 3 and ReLU activation.

    • Max pooling layer with pool 2×22\times 2.

    • Dropout layer with drop rate 0.250.25.

    • Flatten layer.

    • Dense layer with 512 units and ReLU activation.

    • Dropout layer with drop rate 0.50.5.

    • Dense output layer with 1 unit and sigmoid activation.

G.2 Results Summary

EZ.Log EZ.Sq PM.Log PM.Sq DebiasedSq
k=10k=10 Linear 10.9 10.2 9.47 9.51 9.50
Two Layer Small 5.32 3.93 1.56 1.56 1.56
Two Layer Large 4.56 3.50 1.28 1.18 1.19
CNN Small 3.89 2.47 1.06 1.08 1.02
CNN Large 3.54 1.58 0.417 0.416 0.422
k=100k=100 Linear 14.2 12.9 11.8 11.8 11.8
Two Layer Small 8.72 8.33 3.77 3.84 3.80
Two Layer Large 9.29 8.50 3.21 2.80 2.80
CNN Small 6.89 5.82 2.11 2.12 2.08
CNN Large 5.86 4.63 0.826 0.774 0.841
k=1000k=1000 Linear 21.3 21.5 19.9 19.9 19.9
Two Layer Small 20.5 20.5 20.1 20.2 20.2
Two Layer Large 19.6 21.0 18.2 18.5 18.4
CNN Small 16.1 16.3 14.1 14.4 14.3
CNN Large 15.3 15.2 13.1 13.7 13.7
Table 1: MNIST Test Error (%) for LLP Algorithms. Best error is reported in bold.
EZ.Log EZ.Sq PM.Log PM.Sq DebiasedSq
k=10k=10 Linear 19.2 18.7 18.4 18.4 18.4
Two Layer Small 16.3 15.8 14.6 14.7 14.7
Two Layer Large 18.0 16.5 13.2 13.4 13.4
CNN Small 10.5 10.2 8.30 8.09 8.23
CNN Large 10.5 10.2 8.19 8.11 8.18
k=100k=100 Linear 22.1 21.0 20.5 20.4 20.4
Two Layer Small 20.1 20.1 19.3 19.3 19.3
Two Layer Large 20.3 24.5 19.5 19.7 19.7
CNN Small 17.4 17.3 13.0 13.0 12.9
CNN Large 20.4 18.7 13.2 13.3 13.4
k=1000k=1000 Linear 24.0 45.7 28.5 28.5 28.5
Two Layer Small 37.2 49.8 36.8 36.7 36.7
Two Layer Large 27.1 54.0 35.1 37.9 37.4
CNN Small 33.5 33.6 35.7 35.8 35.6
CNN Large 36.1 36.1 35.9 36.8 36.6
Table 2: CIFAR Test Error (%) for LLP Algorithms. Best error is reported in bold.