# IMPROVED CONCENTRATION FOR MEAN ESTIMATORS VIA SHRINKAGE

Antônio Catão, Lucas Resende, Paulo Orenstein

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Antônio Catão, Lucas Resende, Paulo Orenstein. IMPROVED CONCENTRATION FOR MEAN ESTIMATORS VIA SHRINKAGE. 2025. hal-05418081

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Distributed under a Creative Commons Attribution 4.0 International License
IMPROVED CONCENTRATION FOR MEAN ESTIMATORS VIA
SHRINKAGE

Antônio Catão
IMPA

antonio.catao@impa.br

Lucas Resende
ENSAE

lucas.resende@ensae.fr

Paulo Orenstein
IMPA

pauloo@impa.br

ABSTRACT

We study a class of robust mean estimators $\hat{\mu}$ obtained by adaptively shrinking the weights of sample points far from a base estimator $\hat{\kappa}$. Given a data-dependent scaling factor $\hat{\alpha}$ and a weighting function $w: [0, \infty) \to [0, 1]$, we let $\hat{\mu} = \hat{\kappa} + \frac{1}{\eta} \sum_{i=1}^{n} (X_i - \hat{\kappa})w(\hat{\alpha}|X_i - \hat{\kappa}|)$. We prove that, under mild assumptions over $w$, these estimators achieve stronger concentration bounds than the base estimate $\hat{\kappa}$, including sub-Gaussian guarantees. This framework unifies and extends several existing approaches to robust mean estimation in $\mathbb{R}$. Through numerical experiments, we show that our shrinking approach translates to faster concentration, even for small sample sizes.

**Keywords** Shrinkage · Mean Estimation · Robustness · Sub-Gaussian Concentration

# 1 Introduction

Let $P$ be an unknown distribution over $\mathbb{R}$ and let $X_{1:n} = (X_i)_{i=1}^n \sim P^{\otimes n}$ be an independent sample from $P$. Our goal is to approximate the mean $\mu := \mathbb{E}X_1$ through an estimator $\hat{\mu} = \hat{\mu}(X_1, \dots, X_n)$. In this setting, a canonical choice for $\hat{\mu}$ is the empirical mean $\bar{X} := \frac{1}{n} \sum_{i=1}^n X_i$. Indeed, this estimator possesses many appealing properties, such as consistency, asymptotic efficiency, and admissibility among large classes of distributions.

Nevertheless, as observed by Tukey [1948], the empirical mean presents important limitations when the underlying distribution deviates from the Gaussian. This motivated the study of robust mean estimation, whose aim is to provide good estimators under challenging settings. For instance, Huber [1964] developed a systematic treatment for location estimation to minimize the asymptotic variance under adversarial distributions. Typically, robust mean estimators try to improve on the empirical mean by reducing the sensitivity to unlikely but extreme points in the sample, often exploiting the idea of “outlyingness”, through order statistics or distance from a good centrality estimate. Some examples are given by the following estimators:

$$ \hat{\mu}_{TM}^{(k)} = \frac{1}{n-2k} \sum_{i=k+1}^{n-k} X_{(i)}, \quad (1) $$

$$ \hat{\mu}_{WM}^{(k)} = \frac{1}{n} \sum_{i=1}^{n} X_{(i \lor (k+1) \land (n-k))}, \quad (2) $$

$$ \hat{\mu}_{LV}^{\eta} = \hat{\kappa} + \frac{1}{n} \sum_{i=1}^{n} (X_i - \kappa)(1 - \min(\alpha^2 |X_i - \hat{\kappa}|^p, 1)), \quad (3) $$

where $X_{(i)}$ denotes the $i$-th order statistic, $\alpha$ satisfies $\sum_{i=1}^n \min(\alpha^2 |X_i - \hat{\kappa}|^2, 1) = \eta$, and $\hat{\kappa}$ is a mean-estimator. The trimmed (1) and Winsorized (2) means [Tukey and McLaughlin, 1963, Huber, 1964] exemplify the former approach, while the estimator introduced by Lee and Valiant [2022] (3) follows the latter.

Inspired by the aforementioned ideas, the present paper studies the effect of shrinking the weights of extreme points in order to minimize their impact in mean estimation. To this end, we propose a family of estimators that uses a base estimate $\hat{\kappa}$ and adds a correction term that takes into account the deviations from $\hat{\kappa}$ to the data, assigning weights for
each sample point that decrease with the distance to $\hat{\kappa}$. More specifically, let $w: [0, \infty) \to [0, 1]$ be a non-increasing function and let $\hat{\alpha} > 0$ be a data-dependent scaling factor defined as a $Z$-estimator<fnref n="1" /> via $\sum_{i=1}^n w(\hat{\alpha}|X_i - \hat{\kappa}|) = n - \eta$, where $\eta \in (0, n)$ is a parameter. We then define estimators of the form

$$ \hat{\mu}(\hat{\kappa}; X_{1:n}) := \hat{\kappa} + \frac{1}{n} \sum_{i=1}^{n} (X_i - \hat{\kappa})w(\hat{\alpha}|X_i - \hat{\kappa}|). \quad (4) $$

Many robust estimators can be written in this form for suitable choices of $w$, such as (3), as well as variants of (1) and (2) (see Section 5). The main objective of this work is to present non-asymptotic high-probability bounds for the error of $\hat{\mu}(\hat{\kappa}; X_{1:n})$, under mild assumptions on $w$. In particular, we show that its concentration often improves upon that of $\hat{\kappa}$ and attains optimal rates when the error of $\hat{\kappa}$ is $O(1)$ with high probability. In Section 1.1.3 we also present other desirable properties.

# 1.1 Main results

## 1.1.1 Optimal concentration via shrinkage

We prove concentration for $\hat{\mu}$ under adversarial contamination; we assume that an underlying sample $X_{1:n} \sim P^{\otimes n}$ is adversarially contaminated with contamination level $\varepsilon \in [0, 1]$, resulting in a corrupted sample $X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon) := \{Y_{1:n} : \#\{i \in \{1, \dots, n\} : X_i \neq Y_i\} \leq n\varepsilon\}$. The statistician has access only to the corrupted sample $X_{1:n}^{\varepsilon}$ and an independent estimate $\hat{\kappa}$ (possibly computed in an independently contaminated sample). We let $\hat{\mu}^{\varepsilon} = \hat{\mu}(\hat{\kappa}; X_{1:n}^{\varepsilon})$ as in (4). The concentration rates of $\hat{\mu}^{\varepsilon}$ depends upon the concentration of $\hat{\kappa}$. We make this dependence explicit through the following definition: for $\delta \in (0, 1)$, define

$$ R_{\hat{\kappa}}(\delta) := \inf \{r > 0 : \mathbb{P}[|\hat{\kappa} - \mu| > r] \leq \delta\}. \quad (5) $$

We then have the following informal version of the main result, presented in Section 3, which holds under mild assumptions on $w$ that are fully delineated in Section 2:

**Theorem 1.1 (Concentration under adversarial contamination - informal statement).** Assume $\eta = \ln \frac{1}{\delta} + \frac{3}{2}n\varepsilon$ and that $\hat{\mu}^{\varepsilon} = \hat{\mu}(\hat{\kappa}; X_{1:n}^{\varepsilon})$ is computed on the contaminated sample $X_{1:n}^{\varepsilon}$, with $\hat{\mu}$ given by (4) and $\hat{\kappa}$ computed on an independent (possibly corrupted) sample. Further assume that $R_{\hat{\kappa}}(\delta) \leq C_{\hat{\kappa}}\nu_p$ for all $p > 1$. For $\delta \in (0, 1)$, $n \in \mathbb{N}$ large enough, for all atom-free distributions $P$, with probability at least $1 - 4\delta$, for all possible contaminated samples $X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon)$,

$$ |\hat{\mu}^{\varepsilon} - \mu| \leq C_{w, \hat{\kappa}, \varepsilon} \left[ \inf_{1<p \leq 2} \nu_p \left( \frac{1}{n} \ln \frac{1}{\delta} \right)^{1-\frac{1}{p}} + \inf_{p>1} \nu_p \varepsilon^{1-\frac{1}{p}} \right], \quad (6) $$

where $\nu_p = \mathbb{E}[|X - \mu|^p]^{\frac{1}{p}}$, and $C_{w, \hat{\kappa}}$ depends only on $w$ and $C_{\hat{\kappa}}$.

As we will show in Section 3, the sample median, hereafter denoted by $M$, satisfies the bound $R_M(\delta) \leq \frac{2\nu_p}{1/2-\varepsilon}$ for $\delta > e^{-c(1-2\varepsilon)m}$ and all $p > 1$ when computed on a $\varepsilon$-contaminated sample of size $m$. The necessity of the terms inside the brackets in (6) follows from [Devroye et al., 2016, Theorem 3.1] and [Minsker, 2019, Lemma 5.4], respectively. Thus, for a wide class of reweighting schemes $w$, the associated shrinkage estimator achieves optimal concentration bounds. It is worth noting that this result holds even when the population mean of $P$ is far from its population median, implying that the shrinkage estimators are able to substantially improve upon the concentration rates of an inadequate mean estimator, often leading to optimal bounds.

## 1.1.2 Improved concentration rates

Indeed, our main result is more general; it guarantees concentration bounds without necessarily assuming that $R_{\hat{\kappa}}$ is conveniently bounded. For the sake of exposition, let us consider the uncontaminated case ($\varepsilon = 0$) and assume that the underlying distribution $P$ has finite variance. Another statement that may be extracted from the main result Theorem 3.1 is described as follows:

**Theorem 1.2 (Main result - informal statement).** Let $\hat{\kappa}$ be independent of $X_{1:n}$, $n$ and $\delta > e^{-cn}$. Take $\eta = \ln \frac{1}{\delta}$ and $\hat{\mu} = \hat{\mu}(\hat{\kappa}; X_{1:n})$ as in (4). If $X_{1:n} \sim P^{\otimes n}$ for some atom-free distribution $P$ satisfying $\nu_2(P) := \nu_2 = \mathbb{E}[|X_1 - \mu|^2]^{\frac{1}{2}} < \infty$, then there exists a constant $C_w > 0$ depending only on the choice of $w$ such that, with probability at least $1 - 4\delta$,

$$ |\hat{\mu} - \mu| \leq C_w(\nu_2 + R_{\hat{\kappa}}(\delta))\sqrt{\frac{1}{n}\ln\frac{1}{\delta}}. $$

<footnote n="1">¹This definition suffices for continuous choices of $w$; the general definition is given in (7).</footnote>
In other words, if $\hat{\kappa}$ is such that $R_{\hat{\kappa}}(\delta) \gg \sqrt{\frac{1}{n} \ln \frac{1}{\delta}}$ for some regime of $(n, \delta)$ such that $\delta \gg e^{-cn}$, then the estimator $\hat{\mu}$ offers an improvement over $\hat{\kappa}$ in terms of concentration rates. In particular, let $\hat{\kappa}$ be the empirical average of some independent sample of size $m = \lfloor nq \rfloor$ for some constant $q \in (0, 1)$, let $\{P_k\}_{k \ge 1}$ a sequence of distributions for which Chebyshev's inequality is sharp [Catoni, 2012], and take $\delta = \frac{1}{16m}$ such that $\delta \gg e^{-cn}$. One has that $R_{\bar{X}}(\delta) = \Theta(1)$, implying that the shrinkage estimator improves upon the base estimate in terms of concentration rates, which is often the case even for already optimally robust estimators such as the trimmed mean (see computational experiments in Section 6).

Contrastingly with respect to the result presented in Theorem 1.1, the bound given in Theorem 1.2 is still informative if $R_{\hat{\kappa}}(\delta) \gg 1$: in the setup of the previous paragraph, if we instead take $m = n$ and $\delta = cn^{-\frac{5}{4}}$, we would have that $R_{\bar{X}} = \Theta(n^{\frac{1}{4}}) \gg 1$, while

$$C_{w,\hat{\kappa}}(\nu_2 + R_{\hat{\kappa}}(\delta))\sqrt{\frac{1}{n} \ln \frac{1}{\delta}} \le C n^{-\frac{1}{4}} \sqrt{\ln n} \ll 1.$$

### 1.1.3 Breakdown point, asymptotic efficiency and affine equivariance

Beyond optimal concentration rates under mild assumptions on $\hat{\kappa}$, our shrinkage estimator $\hat{\mu}$ also inherits other desirable properties from the base estimator. For example, if $\hat{\kappa}$ is affine-equivariant then so is $\hat{\mu}(\hat{\kappa})$. Another desirable property that $\hat{\mu}(\hat{\kappa}; X_{1:n})$ enjoys is asymptotic efficiency; fix $\eta > 0$ and denote $\hat{\mu}_n = \hat{\mu}(\hat{\kappa}_n; X_{1:n})$ for a sequence of $\hat{\kappa}_n$ independent of $X_{1:n}$. If we assume that $\nu_p < \infty$ for any $p > 2$ and that $(\hat{\kappa}_n - \mu)n^{-\frac{1}{2}} \to 0$ in probability, then $\sqrt{n}(\hat{\mu}_n - \mu) \xrightarrow{d} N(0, \nu_2^2)$. That is, the shrinkage estimator $\hat{\mu}_n$ is undistinguishable from the empirical mean in terms of asymptotic distribution. Notice that the assumption that $(\hat{\kappa}_n - \mu)n^{-\frac{1}{2}} \to 0$ is very weak: in particular it holds for any constant base estimator $\hat{\kappa}_n$.

Finally, we also show that, under mild assumptions on $w$, the breakdown point of the estimator $\hat{\mu}(\hat{\kappa})$, given by $(X_{1:n}, Y_{1:m}) \mapsto \hat{\mu}(\hat{\kappa}(Y_{1:m}); X_{1:n})$,

$$\varepsilon^*(\hat{\mu}(\hat{\kappa}), (X_{1:n}, Y_{1:m})) := \inf \left\{ \varepsilon \in [0, 1] : \sup_{\substack{X_{1:n} \in A(X_{1:n}, \varepsilon) \\ Y_{1:m} \in A(Y_{1:m}, \varepsilon)}} |\hat{\mu}(\hat{\kappa}(Y_{1:m}); X_{1:n}) - \hat{\mu}(\hat{\kappa}(Y_{1:m}^{\varepsilon}); X_{1:n})| = \infty \right\}.$$

is at least as high as that of $\hat{\kappa}$, as long as $\eta \ge n\varepsilon^*(\hat{\kappa}; Y_{1:m})$. When $\hat{\kappa}$ is taken to be the sample median of $Y_{1:m}$ and $\eta \ge n/2$, this implies that $\varepsilon^*(\hat{\mu}(\hat{\kappa}), (X_{1:n}, Y_{1:m})) = 1/2$ for every pair of samples $(X_{1:n}, Y_{1:m})$. The formal statements of these properties are given and proved in Section 4.

## 1.2 Related works

Estimators satisfying concentration bounds such as the ones given in (6) have been the focus of many recent papers in robust mean estimation. In particular, for the case $\varepsilon = 0$ and restricting the infima to $p = 2$, these estimators are called $\delta$-dependent sub-Gaussian mean estimators. Indeed, Devroye et al. [2016] showed that such guarantees are optimal in terms of probabilistic concentration. Recent efforts have established similar high-probability bounds. The first work to study whether estimators are sub-Gaussian in the above sense was Catoni [2012]. In this paper, Catoni shows that Chebyshev's inequality is sharp, implying that the empirical mean concentrates badly for certain distributions with finite variance. He then proposes a sub-Gaussian estimator that is constructed as a Z-estimator satisfying $\sum_{i=1}^n \psi(X_i - \mu) = 0$, for $\psi(x) = \phi(\beta x)$ where $\beta > 0$ is a parameter, $\phi$ is increasing and satisfies $-\ln(1 - x + x^2/2) \le \phi(x) \le \ln(1 + x + x^2/2)$. The parameter $\beta$ plays the role of a scaling factor and the bounds on $\phi$ enforce diminishing contributions from outliers. Catoni [2012] only manages to obtain sub-Gaussian deviations assuming knowledge of an upper bound $\nu \ge \sigma^2$ and setting $\beta \approx \sqrt{2/(n\nu) \ln(1/\delta)}$. With no prior knowledge of $\sigma^2$, Lepski's method [Lepskii, 1991] can be employed to select $\beta$, yielding an almost sub-Gaussian bound. In fact, Devroye et al. [2016] showed that, in order for an estimator to attain these guarantees without prior knowledge of $\sigma^2$, it needs to be $\delta$-dependent, justifying our choice of the parameter $\eta$ as a function of $\delta$ in Theorems 1.1 and 1.2.

Following Catoni [2012], several papers have shown that classical estimators are also sub-Gaussian (see the survey Lugosi and Mendelson [2019a]). For instance, Lerasle and Oliveira [2011] showed that the median-of-means [Ne-mirovskij and Yudin, 1983, Jerrum et al., 1986, Alon et al., 1996] is sub-Gaussian when $P$ has second moment; Bubeck et al. [2013] showed that (6) also holds when the infima are restricted to any single $p \in (1, 2)$. Recently, Oliveira et al. [2025] showed the general adversarially contaminated case (6) for the trimmed mean (1), and Lugosi and Mendelson [2019a] proved that the Winsorized mean (2) is sub-Gaussian assuming the existence of second moment.
In this context, Lee and Valiant [2022] seek to obtain the optimal sub-Gaussian constant $L = \sqrt{2}$ under second moment and asymptotic regimes of $(n, \delta)$ by employing the estimator (3). They do so through a reweighting scheme on top of a sub-Gaussian base estimator $\hat{\kappa}$, exploiting a data-dependent scaling factor analogous to the Z-estimator $\hat{\alpha}$. While we take heavy inspiration in this work, our framework is fundamentally more general and conceptually distinct in two key aspects. First, we make no structural assumptions on the base estimator $\hat{\kappa}$, which can be any preliminary estimate of location, and may not even be sub-Gaussian. Second, our work introduces a broad class of shrinkage functions $w(\cdot)$. This generality comes at the cost of assuming that $\hat{\kappa}$ is independent of $X_{1:n}$, which we believe to be an artifact of our proofs.

Other novel sub-Gaussian estimators have also been proposed. Mathieu [2022] studies concentration properties of mean estimators through their influence function under the adversarial contamination context. Minsker and Ndaoud [2021] introduce an estimator that splits the sample in $K$ buckets, and use the empirical deviation $\hat{\sigma}_j$ and empirical mean $\hat{\mu}_j$ of each bucket $j = 1, 2, \dots, K$ to construct the weighted average $\sum_{j=1}^K \hat{\mu}_j \hat{\sigma}_j^{-r} / \sum_{j=1}^K \hat{\sigma}_j^{-r}$ for some parameter $r \ge 1$, showing that it is robust to a weaker form of contamination. Minsker [2023] studies some modifications to the median-of-means construction to achieve better constants (see also Lecué and Lerasle [2020]).

The idea of using an initial location estimate and then reweighting the sample points to obtain a better estimate is not new. W-estimators were introduced by Mosteller and Tukey [1977] as estimators $(\sum_{i=1}^n w_i)^{-1} \sum_{i=1}^n w_i X_i$, where the weights $w_i$ are a non-increasing function of the normalized deviation $|X_i - \hat{\kappa}|$, and $\hat{\kappa}$ is chosen to be the sample median. Donoho and Gasko [1992] then generalized this idea to higher dimensions, providing breakdown point guarantees for such estimators. In fact, beyond the estimator given in (4), we also show that concentration properties hold for a related estimator $\hat{\mu}_W$, introduced in Section 2.

# 2 Preliminaries

## 2.1 Notation and setup

Define $[m] := \{1, \dots, m\}$. Let $\hat{P}_n$ denote the empirical distribution of $X_{1:n} \sim P^{\otimes n}$, and let $X \sim P$ and $Pf = E[f(X)]$, with $\|f\|_{L_p(P)} = (E[|f(X)|^p])^{1/p}$, $\|f\|_{\infty} = \sup_{x \in X} |f(x)|$. Denote $\mathcal{A}(X_{1:n}, \varepsilon)$ the set of all possible $\varepsilon$-contaminated samples $\{X_{1:n}^{\varepsilon} : \#\{i \in [n] : X_i \neq X_i^{\varepsilon}\} \le n\varepsilon\}$. Let $\mu = \mathbb{E}X_1, \nu_p = \mathbb{E}[|X_1 - \mu|^p]^{\frac{1}{p}}$.

Given $p > 1$, let $\mathcal{P}$ denote a known family of distributions in $\mathcal{P}_p := \{P : P$ has finite $p$-th moment$\}$. We have access to an adversarially contaminated sample $X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon)$ where $X_{1:n} \sim P^{\otimes n}$ for some unknown $P \in \mathcal{P}$ and we aim to estimate $\mu$. We are given a base estimator $\hat{\kappa}$, and are interested in improving it through shrinkage. As in Section 1, we consider $w : [0, \infty) \to [0, 1]$, a right-continuous non-increasing function satisfying $w(0) = 1$ and $\lim_{t \to \infty} w(t) = 0$. When $w$ is continuous, the data-dependent scaling factor $\hat{\alpha}$ may be defined as a Z-estimator. However, for the sake of generality, we will consider the following alternative definition:

$$ \hat{\alpha}(\kappa; X_{1:n}) = \hat{\alpha}(\kappa) := \inf \left\{ \alpha > 0 : \sum_{i=1}^{n} w(\alpha | X_i - \kappa|) \le n - \eta \right\}, \quad (7) $$

where $\eta$ is deemed the shrinkage level, since it determines how much of the sample will be discarded. When $w$ is continuous, the Z-estimator defined by $\frac{1}{n} \sum_{i=1}^n w(\hat{\alpha}(\hat{\kappa})|X_i - \hat{\kappa}|) = 1 - \frac{\eta}{n}$, as proposed in Section 1, is a particular case of the more general definition given by (7) above. In that sense, $w(\hat{\alpha}(\hat{\kappa})|X_i - \hat{\kappa}|)$ can be thought of as the amount by which $X_i$ will be shrunk. Our proposed estimator is then given as in (4), or, equivalently,

$$ \hat{\mu}(\hat{\kappa}; X_{1:n}) = \hat{\mu}(\hat{\kappa}) := \hat{\kappa} + \frac{1}{n} \sum_{i=1}^{n} (X_i - \hat{\kappa}) w(\hat{\alpha}(\hat{\kappa}) | X_i - \hat{\kappa}|) = \frac{\eta}{n} \hat{\kappa} + \frac{1}{n} \sum_{i=1}^{n} X_i w(\hat{\alpha}(\hat{\kappa}) | X_i - \hat{\kappa}|). \quad (8) $$

The last equality in (8) holds if $w$ is continuous. Thus, the resulting estimator can be thought of as an interpolation between $\hat{\kappa}$ (when $\eta = n$) and the empirical mean (when $\eta = 0$). For the contaminated versions of the aforementioned estimators, we denote $\hat{\alpha}^{\varepsilon}(\kappa) = \hat{\alpha}(\kappa; X_{1:n}^{\varepsilon})$, $\hat{\mu}^{\varepsilon}(\kappa) = \hat{\mu}(\kappa; X_{1:n}^{\varepsilon})$, $\hat{\alpha}^{\varepsilon} = \hat{\alpha}(\hat{\kappa}; X_{1:n}^{\varepsilon})$, and $\hat{\mu}^{\varepsilon} = \hat{\mu}(\hat{\kappa}; X_{1:n}^{\varepsilon})$.

**Remark 1.** The alternative estimator

$$ \hat{\mu}_W(\kappa; X_{1:n}) := \sum_{i=1}^{n} \frac{X_i w(\hat{\alpha}(\kappa) | X_i - \kappa|)}{\sum_{i=1}^{n} w(\hat{\alpha}(\kappa) | X_i - \kappa|)}, \quad (9) $$

where $\hat{\alpha}$ is the one given in (7), also satisfies the concentration bounds given in Theorem 3.1 as a corollary. The proof is relegated to the appendix. This estimator is a W-estimator when $\hat{\kappa}$ is chosen to be the sample median.
The function $w$ controls how we treat deviant sample points: a compactly-supported $w$ may discard a few sample points altogether, whereas a strictly positive $w$ includes some proportion of each sample point in the average of the deviations. Naturally, the performance of $\hat{\mu}^{\epsilon}$ is necessarily dependent on the quality of this base estimator, which is made clear in our concentration bounds given by Theorem 3.1. Nevertheless, we will show that this dependence is not overly strong and, therefore, that we can achieve good concentration bounds for $\hat{\mu}^{\epsilon}$ even if $\hat{\kappa}$ does not concentrate well.

## 2.2 Assumptions

To obtain concentration bounds for $\hat{\mu}^{\epsilon}$, we will assume the following four hypotheses.

**Assumption 1.** All distributions $P \in \mathcal{P}$ are atom-free.

This assumption ensures that there are no ties in the uncontaminated data (that is, $|X_i - \hat{\kappa}| \neq |X_j - \hat{\kappa}|$ for $i \neq j$ — we allow ties for the contaminated sample $X_{1:n}^{\epsilon}$). This yields tight lower bounds on $\sum_{i=1}^n w(\hat{\alpha}(\hat{\kappa})|X_i - \hat{\kappa}|)$ (see Lemma B.1) and makes $\mathbb{E}[w(\alpha|X_1|)]$ continuous in $\alpha$ (see the proof of Lemma 3.4). This assumption is relatively weak: it is not necessary if $w$ is continuous and, even if that is not the case, we can consider the distribution $P'$ of $X' = X_1 + \sigma Z$ for $Z \sim N(0, 1)$ independent of $X_1$ and $\sigma > 0$ arbitrarily small. Then, $P'$ is atom-free, with the same mean as $P$ and centered $L^p$ norm $\nu_p'$ bounded by $\nu_p + \sigma(\mathbb{E}[|Z|^p])^{1/p}$. Our results can then be applied to $P'$.

**Assumption 2.** The function $w(\cdot)$ satisfies $\sup_{t \ge 0} tw(t) < \infty$.

This hypothesis ensures that $w(t)$ decays at least as fast as $1/t$ for large $t$. Consider the function

$$m(\alpha) := \sup_{x \in \mathbb{R}} |x| w(\alpha|x|) = \alpha^{-1} \sup_{t \ge 0} tw(t), \quad (10)$$

for $\alpha > 0$. Informally, $m(\hat{\alpha}^{\epsilon})$ measures the maximum contribution that a sample point can have towards the sum in (8). Note that Assumption 2 is satisfied whenever $w$ has compact support, which is the case for some natural choices of $w$ (see Section 5). By Assumption 2, $m(\alpha) < \infty$ for all $\alpha > 0$ and by (10), $m(\alpha)$ is decreasing, so lower-bounding $\hat{\alpha}^{\epsilon}$ by a positive value with high probability implies that $\frac{1}{n} \sum_{i=1}^n (X_i - \hat{\kappa})w(\hat{\alpha}^{\epsilon}|X_i - \hat{\kappa}|)$ is a sum of bounded random variables (conditioned on a high-probability event). This will be a core tool to obtain concentration bounds for $\hat{\mu}^{\epsilon}$, relying on $m(\underline{\alpha})$ where $\underline{\alpha}$ is some lower bound of $\hat{\alpha}^{\epsilon}$.

**Assumption 3.** The function $w(\cdot)$ satisfies $w(t) \ge (1 - t^p)_+$ for all $t \ge 0$.

This hypothesis ensures $w(t)$ does not decay too quickly for small $t$. In particular, it allows us to bound $m(\underline{\alpha})$ through a lower bound on the $\underline{\alpha}$ mentioned above. Intuitively, for differentiable $w$, this hypothesis guarantees that $w'(0) = 0$ and, therefore, that the closest sample points to the base estimate contribute similarly. Observe that if this assumption holds for $q > 1$, it then holds for all $p \in (1, q]$. In particular, one may choose $w$ such that $w(t) \ge 1_{t<1}$, which implies that Assumption 3 holds for all $p > 1$.

**Assumption 4.** The base estimator $\hat{\kappa}$ is independent from the sample $X_{1:n}$.

This assumption is used to control the empirical processes that arise in our proof, and we believe it is ultimately just a proof artifact. Nevertheless, to obtain sub-Gaussian concentration bounds, we only require $|\hat{\kappa} - \mu|$ to be of order $\nu_p$. In practice, this means that for a sample of size $n$, one can allocate a small subset of observations to estimate $\hat{\kappa}$, reserving most of the data for evaluating (8). We empirically investigate several data-splitting strategies in Section 6.

# 3 Concentration of shrinkage estimators

We now state the main result of this paper in full generality, which provides concentration bounds for the shrinkage estimator $\hat{\mu}^{\epsilon}$ defined in (8). The main ingredients for the proof are discussed in Section 3.1, with details included in Appendix B.

**Theorem 3.1 (Concentration of shrinkage estimators under adversarial contamination).** Let $X_{1:n} \sim P^{\otimes n}$ with $P \in \mathcal{P}$. Let $\epsilon \in [0, \frac{1}{2})$. Under Assumptions 1 to 4, set $\eta = \ln \frac{4}{\delta} + (1 + \xi)\epsilon n$ for some $\xi > 0$. Assume $(1 + 2\xi)\epsilon + \frac{\bar{c}}{n} \ln \frac{4}{\delta} < 1$, where $\bar{c} = 16 + 4\xi^{-1}$. If $\hat{\mu}^{\epsilon} = \hat{\mu}(\hat{\kappa}; X_{1:n}^{\epsilon})$ is as in (8) and $R_{\hat{\kappa}}$ is as in (5), then, with probability at least $1 - \delta$, for all $X_{1:n}^{\epsilon} \in \mathcal{A}(X_{1:n}, \epsilon)$, $|\hat{\mu}^{\epsilon} - \mu|$ is at most

$$C_{w, \xi} \left\{ \inf_{1 < q \le 2 \wedge p} \left( \nu_q + \nu_q^{\frac{q}{2}} R_{\hat{\kappa}} \left( \frac{\delta}{4} \right)^{1-\frac{q}{2}} \right) \left( \frac{1}{n} \ln \frac{4}{\delta} \right)^{1-\frac{1}{q}} + \inf_{1 < q \le p} \left( \nu_q + R_{\hat{\kappa}} \left( \frac{\delta}{4} \right) \right) \left( \frac{1}{n} \ln \frac{4}{\delta} + \epsilon \right)^{1-\frac{1}{q}} \right\} \quad (11)$$
for $C_{w,\xi} > 0$ depending only on $w$ and $\xi$.

In particular, if there exist constants $C, \delta_{\min} > 0$ depending only on the base estimator such that $R_{\hat{\kappa}}(\frac{\delta}{4}) \le C\nu_p$ for $\delta \ge \delta_{\min}$ and $P \in \mathcal{P}$, then $\hat{\mu}^\epsilon$ attains for $\delta \ge \delta_{\min} \lor 4e^{-\frac{1-(1+2\epsilon)\epsilon}{\epsilon}}n$:

$$ \mathbb{P} \left[ \sup_{X_{1:n}^{\epsilon} \in \mathcal{A}(X_{1:n}, \epsilon)} |\hat{\mu}^{\epsilon} - \mu| > \tilde{C}_{w,\xi} \left\{ \nu_{2 \wedge p} \left( \frac{1}{n} \ln \frac{4}{\delta} \right)^{1-\frac{1}{2\wedge p}} + \nu^p \epsilon^{1-\frac{1}{p}} \right\} \right] \le \delta, \quad (12) $$

where $\tilde{C}_{w,\xi} = C_{w,\xi}(2C + 3)$.

**Remark 2.** While $R_{\hat{\kappa}} = O(1)$ is sufficient for attaining sub-Gaussian guarantees, for example, one can conclude from (11) that having moments $p > 2$ leads to an improved bound, in the sense that $\hat{\mu}^\epsilon$ might retain light tails even if $\hat{\kappa}$ has poorer concentration. For example, if $p = 4$, $\epsilon = 0$, and $\hat{\kappa}$ is such that

$$ R_{\hat{\kappa}}(\delta) = \Theta \left( \left( \frac{1}{n} \ln \frac{1}{\delta} \right)^{-1/4} \right), $$

with an increasingly poor concentration even as $\frac{1}{n} \ln \frac{1}{\delta} \to 0$, then $\hat{\mu}^\epsilon$ still retains sub-Gaussian rate tails. Notice also that if $w(t) \ge 1_{t<1}$, Assumption 3 is satisfied for every $p > 1$ so, in particular, the infimum in (11) can be taken in $[1, 2]$ for the first term and in $(1, \infty)$ for the second. It is under this scenario that Theorem 1.1 holds (taking $\xi = \frac{1}{2}$). We explore choices of $w$ satisfying this lower bound in Section 5.

Informally, Theorem 3.1 describes the error associated to the estimation of the population mean $\mu$ through the estimator $\hat{\mu}^\epsilon$. The error in (11) can be decomposed into two terms: one that depends on $R_{\hat{\kappa}}$ and one that does not. The latter term corresponds to the optimal order of mean estimation. The former term is associated with the quality of the base estimate combined with the contamination level and is significantly mitigated when $\delta \gg e^{-cn}$, since the term $(n^{-1} \ln(4\delta^{-1}))^{1-1/p}$ is small in this regime, leaving only the error associated to the contamination $\epsilon^{1-\frac{1}{p}}$. This means that for sufficiently large sample sizes, $\hat{\mu}^\epsilon$ may concentrate well around the mean $\mu$ even if $\hat{\kappa}$ does not. Remark 2 emphasizes the magnification of this effect as $P$ presents finite moments of higher order.

We now explore the consequences of Theorem 3.1, exhibiting examples of shrinkage estimators with specific choices of base estimators that yield mean-estimators with optimal concentration rates. We will consider estimates computed on an independent sample $Y_{1:m} \sim P^{\otimes m}$. To account for possible adversarial contamination (with the same level $\epsilon$) on the sample $Y_{1:m}$, we allow $\hat{\kappa}$ to take the form of $\hat{\kappa}(f(Y_{1:m}, \epsilon))$, where $f: \mathbb{R}^m \times [0, 1] \to \mathbb{R}^m$ is a Borel-measurable function such that $f(y_{1:m}, \gamma) \in \mathcal{A}(y_{1:m}, \gamma)$ for every possible sample $y_{1:m} \in \mathbb{R}^m$ and contamination level $\gamma \in [0, 1]$. Notice that $f(y_{1:m}, 0) = y_{1:m}$.

**Example 1 (Empirical mean).** Assume that $\epsilon = 0$. Let $\hat{\kappa} = \hat{\kappa}(Y_{1:m}) = \bar{Y}$. Recall that the empirical mean $\bar{Y}$ satisfies $R_{\bar{Y}}(\delta) \le \nu_2 \sqrt{\frac{1}{\delta m}}$ for $\delta \in (0, 1)$. Then, (12) holds for $p = 2$, $C = 1$, and $\delta_{\min} = \frac{4}{m}$. Notice that $\epsilon = 0$ is necessary in this case, since the empirical mean is not robust to adversarial contamination. In fact, $\sup_{X_{1:n}^{\epsilon} \in \mathcal{A}(X_{1:n}, \epsilon)} \hat{\mu}^{\epsilon}(\bar{Y}^{\epsilon}; X_{1:n}^{\epsilon}) \to \infty$ as any $Y_i^{\epsilon} \to \infty$, rendering this estimator useless under adversarial contamination.

**Proposition 3.2 (Concentration of empirical quantiles).** Let $\gamma \in (0, 1)$. Then, for every $\delta \ge 2e^{-c\gamma m}$, the empirical $\gamma$-quantile $\hat{Q}_{\gamma}(Y_{1:m})$ satisfies $R_{\hat{\kappa}}(\delta) \le C_{\gamma}\nu_p$, for every $P \in \mathcal{P}_p$, where $c_{\gamma} = \frac{((1-\gamma)\wedge\gamma)^2}{16}$ and $C_{\gamma} = \frac{2}{\gamma\wedge(1-\gamma)}$.

The proof is deferred to Appendix B.

**Example 2 (Empirical quantiles).** Let $\hat{\kappa} = \hat{\kappa}(f(Y_{1:n}, \epsilon)) = \hat{Q}_{\gamma}(f(Y_{1:n}, \epsilon))$ be an empirical quantile of a possibly contaminated version of the sample $Y_{1:n}$. It is easy to see that $\hat{Q}_{\gamma-\epsilon}(Y_{1:n}) \le \hat{Q}_{\gamma}(Y_{1:n}^{\epsilon}) \le \hat{Q}_{\gamma+\epsilon}(Y_{1:n})$ when $\gamma \in (\epsilon, 1-\epsilon)$ and $Y_{1:n}^{\epsilon} \in \mathcal{A}(Y_{1:n}, \epsilon)$. By the proof of Proposition 3.2, $R_{\hat{\kappa}}(\delta) \le \nu_p \left( \frac{2}{\gamma\wedge(1-\gamma)-\epsilon} \right)$ when $\delta \ge \exp\{-\frac{m}{16}(\gamma \wedge (1-\gamma) - \epsilon)^2\}$, for every $p > 1$. In particular, the sample median satisfies the condition $R_{\hat{\kappa}}(\delta) \le 4\nu_p(1-2\epsilon)^{-1}$ for any $\epsilon \in [0, 1/2)$. Then, if $\delta \ge 4\exp\{-\frac{m}{16}(1/2 - \epsilon)^2\} \lor 4\exp\{-\frac{1-(1+2\epsilon)\epsilon}{\epsilon}n\}$, with probability at least $1 - \delta$,

$$ \sup_{X_{1:n}^{\epsilon} \in \mathcal{A}(X_{1:n}, \epsilon)} |\hat{\mu}^{\epsilon} - \mu| \le C_{w,\xi,\epsilon} \left\{ \inf_{1<q \le 2 \wedge p} \nu_q \left( \frac{1}{n} \ln \frac{4}{\delta} \right)^{1-\frac{1}{q}} + \inf_{1<q \le p} \nu_q \epsilon^{1-\frac{1}{q}} \right\}, \quad (13) $$

for some $C_{w,\xi,\epsilon} > 0$ depending only on $w$, $\xi$, and $\epsilon$.
**Example 3 (Constant base estimate).** When $\hat{\kappa} = \lambda$ is constant-valued, $R_{\hat{\kappa}}(\delta) = |\lambda - \mu|$. In particular, if $|\lambda - \mu| \le \nu_p$ for all $p > 1$, then, for $\delta \ge 4 \exp\left\{-\frac{1-(1+2\xi)\epsilon}{\bar{c}}n\right\}$, with probability at least $1 - \delta$,

$$ \sup_{X_{1:n}^{\epsilon} \in \mathcal{A}(X_{1:n}, \epsilon)} |\hat{\mu}^{\epsilon} - \mu| \le C_{w,\epsilon} \left\{ \inf_{1<q \le 2\wedge p} \nu_q \left( \frac{1}{n} \ln \frac{4}{\delta} \right)^{1-\frac{1}{q}} + \inf_{1<q \le p} \nu_q \epsilon^{1-\frac{1}{q}} \right\}, $$

for some $C_{w,\epsilon} > 0$ depending only on $w$ and $\epsilon$.

**Example 4 (Optimal $\delta$-dependent mean-estimators).** Assume that the $\delta$-dependent base estimator $\hat{\kappa}$ is optimal in the sense that there exist $L_{\hat{\kappa}}$, $\delta_{\min}(\hat{\kappa}, m) > 0$ such that

$$ \mathbb{P} \left[ \sup_{Y_{1:m}^{\epsilon} \in \mathcal{A}(Y_{1:m}, \epsilon)} |\hat{\kappa}(Y_{1:m}^{\epsilon}) - \mu| > L_{\hat{\kappa}} \left\{ \inf_{q \in (1,2]} \nu_q \left( \frac{1}{m} \ln \left( 1 + \frac{1}{\delta} \right) \right)^{1-\frac{1}{q}} + \inf_{q>1} \nu_q \epsilon^{1-\frac{1}{q}} \right\} \right] \le \delta, $$

for all $\delta \ge \delta_{\min}(\hat{\kappa}, m)$. Then $\hat{\kappa} = \hat{\kappa}(f(Y_{1:m}, \epsilon))$ is such that

$$ R_{\hat{\kappa}}(\delta) \le L_{\hat{\kappa}} \left\{ \inf_{q \in (1,2]} \nu_q \left( \frac{1}{m} \ln \left( 1 + \frac{1}{\delta} \right) \right)^{1-\frac{1}{q}} + \inf_{q>1} \nu_q \epsilon^{1-\frac{1}{q}} \right\}, $$

and we may attain similar bounds to (13).

## 3.1 Proof of Theorem 3.1

We now state the results that are used to prove Theorem 3.1. For this section and the appendices, denote the uncontaminated shrinkage estimator based on $\hat{\kappa}$ given by $\hat{\mu}(\hat{\kappa}; X_{1:n})$ as $\hat{\mu}$. Similarly, denote $\hat{\alpha} := \hat{\alpha}(\hat{\kappa}; X_{1:n})$. The proof of Theorem 3.1 follows by showing that $\hat{\mu}$ concentrates well and that the error $|\hat{\mu}^{\epsilon} - \hat{\mu}|$ is small. Theorem 3.3 and lemmas 3.4 and 3.5 are dedicated to showing the former, while Lemma 3.6 justifies the latter. The main difficulty in proving concentration bounds for $\hat{\mu}$ is to control the dependence between the parameter $\hat{\alpha}$ and the sample $X_{1:n}$. To that end, we will first assume that, for each $\kappa \in \mathbb{R}$, $\hat{\alpha}(\kappa) := \hat{\alpha}(\kappa; X_{1:n})$ is, with high probability, in an interval $I_{\alpha}(\kappa)$ such that the shrinkage estimator's error may be properly controlled, for fixed $\kappa$. This is done by bounding $|\hat{\mu}(\kappa; X_{1:n}) - \mu|$ by $\sup_{\alpha \in I_{\alpha}(\kappa)} |s(\alpha, \kappa) - \mu|$ where

$$ s(\alpha, \kappa) := \kappa + \frac{1}{n} \sum_{i=1}^{n} (X_i - \kappa)w(\alpha|X_i - \kappa|). $$

The bias induced by a fixed choice of $\alpha$ and $\kappa$ on the weighted sum $s(\alpha, \kappa)$ is

$$ b(\alpha, \kappa) := \mathbb{E}s(\alpha, \kappa) - \mu = \mathbb{E}(\kappa + (X - \kappa)w(\alpha|X - \kappa|)) - \mu. $$

The estimator $\hat{\mu}$ is obtained taking $\kappa = \hat{\kappa}$ and $\alpha = \hat{\alpha}$. Formally, if it holds that $\hat{\alpha}(\kappa) \in I_{\alpha}(\kappa)$ with high probability for some set $I_{\alpha}(\kappa)$, then, with high probability,

$$ |s(\hat{\alpha}(\kappa), \kappa) - \mu| \le \sup_{\alpha \in I_{\alpha}(\kappa)} |s(\alpha, \kappa) - \mu| \le \sup_{\alpha \in I_{\alpha}(\kappa)} |b(\alpha, \kappa)| + \sup_{\alpha \in I_{\alpha}(\kappa)} |s(\alpha, \kappa) - \mu - b(\alpha, \kappa)|. \quad (14) $$

We may then use the concentration of $\hat{\kappa}$ in some interval $I_{\hat{\kappa}}$ along with its independence with respect to the sample $X_{1:n}$ to derive a high-probability bound on the error $|\hat{\mu} - \mu|$. The upper bound in (14) can be thought of as a bias-variance decomposition. Theorem 3.3 below shows that, assuming the existence of the appropriate set $I_{\alpha}(\kappa)$, we can control the variance term $\sup_{\alpha \in I_{\alpha}(\kappa)} |s(\alpha, \kappa) - \mu - b(\alpha, \kappa)|$ in (14) via $\|F_{\kappa}\|_{L^2(P)}^2$ and $m(\alpha)$ (see (10)). Lemma 3.4 then guarantees the existence of a useful choice of $I_{\alpha}(\kappa)$, which allows for a good bound on the bias and variance terms through Lemma 3.5.

**Theorem 3.3 (Bias-variance decomposition).** Under Assumptions 2 and 4, let $I_{\hat{\kappa}}$ be a deterministic set such that $\hat{\kappa} \in I_{\hat{\kappa}}$ with probability at least $1 - \frac{\delta}{4}$ and $I_{\alpha}(\kappa)$ be a family of sets indexed by $\kappa \in I_{\hat{\kappa}}$ such that, for every $\kappa \in I_{\hat{\kappa}}$,

$$ \mathbb{P}[\hat{\alpha}(\kappa) \in I_{\alpha}(\kappa)] \ge 1 - \frac{\delta}{2}. \quad (15) $$

Define $\mathcal{F}_{\kappa} = \{x \mapsto \kappa + (x - \kappa)w(\alpha|x - \kappa)| - \mu : \alpha \in I_{\alpha}(\kappa)\}$ and $F_{\kappa}(x) = \sup_{f \in \mathcal{F}_{\kappa}} |f(x)|$ for all $x \in \mathbb{R}$. Then, with probability at least $1 - \delta$,

$$ |\hat{\mu} - \mu| \le \sup_{\kappa \in I_{\hat{\kappa}}} \left\{ \sup_{\alpha \in I_{\alpha}(\kappa)} |b(\alpha, \kappa)| + (2K + \sqrt{2}) \|F_{\kappa}\|_{L^2(P)} \sqrt{\frac{1}{n} \ln \frac{4}{\delta}} + \sup_{\alpha \in I_{\alpha}(\kappa)} \frac{8m(\alpha)}{3n} \ln \frac{4}{\delta} \right\}. \quad (16) $$
The proof of Theorem 3.3 is achieved through a Bennett-type concentration inequality for suprema of empirical processes (Lemma A.3) applied to $\sup_{\alpha \in I_\alpha(\kappa)} |s(\alpha, \kappa) - \mu - b(\alpha, \kappa)|$ followed by an integration over the distribution of $\hat{\kappa}$ (Assumption 4). To that end, our empirical process must be indexed by uniformly bounded functions, which is guaranteed by Assumption 2, since

$$ \|s(\alpha, \kappa) - \mu - b(\alpha, \kappa)\|_{\infty} = \left\| \frac{1}{n} \sum_{i=1}^{n} (X_i - \kappa)w(\alpha|X_i - \kappa) - \mathbb{E}[(X_i - \kappa)w(\alpha|X_i - \kappa)] \right\|_{\infty} \le 2m(\alpha), $$

assuming that $\sup_{\alpha \in I_\alpha(\kappa)} m(\alpha) < \infty$. The application of Lemma A.3 inequality relies on this last bound, but it is easy to see that the inequality (16) holds trivially when $\sup_{\alpha \in I_\alpha(\kappa)} m(\alpha)$ is infinite, although we will show that, by the construction of the appropriate set $I_\alpha(\kappa)$ in Lemma 3.4 and the bounds given in Lemma 3.5, this is finite and properly controlled. A detailed proof is presented in Appendix B.

The next lemma provides the definition of the set $I_\alpha(\kappa)$ satisfying (15). Intuitively, we will construct sets $I_\alpha(\kappa) := (\underline{\alpha}(\kappa), \bar{\alpha}(\kappa))$, with $\underline{\alpha}(\kappa), \bar{\alpha}(\kappa)$ given implicitly through $w$ such that $\hat{\alpha}(\kappa) \in I_\alpha(\kappa)$ for all $\kappa \in I_{\bar{\kappa}}$ with high probability. Note that for most of the following bounds we do not require a direct bound on the values of $\hat{\alpha}(\kappa)$, since the bias and variance terms depend on $\hat{\alpha}(\kappa)$ only through $w$ (except for $m(\underline{\alpha}(\kappa))$, which is dealt with in Lemma 3.5). This allows us to attain good bounds without restrictive conditions on $w$.

**Lemma 3.4.** Under Assumption 1, there exist absolute constants $c, \bar{c} > 0$ such that $\eta = \ln \frac{4}{\delta} + (1 + \xi)\varepsilon n$ and $\underline{\alpha}(\kappa) < \bar{\alpha}(\kappa)$ are implicitly defined by

$$ \mathbb{E}w(\underline{\alpha}(\kappa)|X - \kappa|) = 1 - \frac{c \ln \frac{4}{\delta}}{n} - \frac{\xi}{7}\varepsilon \quad \text{and} \quad \mathbb{E}w(\bar{\alpha}(\kappa)|X - \kappa|) = 1 - \frac{\bar{c} \ln \frac{4}{\delta}}{n} - (1 + 2\xi)\varepsilon $$

for every $\kappa \in I_{\bar{\kappa}}$ and (15) holds with $I_\alpha(\kappa) = (\underline{\alpha}(\kappa), \bar{\alpha}(\kappa))$ whenever $\frac{\bar{c} \ln \frac{4}{\delta}}{n} + (1 + 2\xi)\varepsilon < 1$. Moreover, for all $\kappa \in I_{\bar{\kappa}}$,

$$ \mathbb{P}\left[\underline{\alpha}(\kappa) < \inf_{X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon)} \hat{\alpha}^{\varepsilon}(\kappa) \le \hat{\alpha}(\kappa) < \bar{\alpha}(\kappa)\right] \ge 1 - \frac{\delta}{2} \quad (17) $$

The inequality (17) implies (15). Accordingly, the proof of Lemma 3.4 consists in showing the former. We prove the stronger statement (17) in order to provide a good bound for the contamination error $|\hat{\mu}^{\varepsilon} - \hat{\mu}|$. To that end, we observe that $\inf_{X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon)} \hat{\alpha}^{\varepsilon}(\kappa) \le \hat{\alpha}(\kappa)$ holds by definition and so the probability of the complementary event to that of (17) may be bounded above by

$$ \mathbb{P}\left[\inf_{X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon)} \hat{\alpha}_{n}^{\varepsilon}(\kappa) \le \underline{\alpha}(\kappa)\right] + \mathbb{P}[\hat{\alpha}(\kappa) \ge \bar{\alpha}(\kappa)] $$

For the first term, notice that

$$ \mathbb{P}\left[\inf_{X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon)} \hat{\alpha}_{n}^{\varepsilon}(\kappa) \le \underline{\alpha}(\kappa)\right] \le \mathbb{P}\left[\inf_{X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon)} \sum_{i=1}^{n} w(\underline{\alpha}(\kappa) | X_i^{\varepsilon} - \kappa|) \le n - \eta\right], $$

where we used Lemma B.1 and the fact that $w$ is non-increasing to obtain the last inequality. We are then able to replace $X_i^{\varepsilon}$ with $X_i$ by noting that

$$ \sum_{i=1}^{n} w(\underline{\alpha}(\kappa) | X_i^{\varepsilon} - \kappa|) - \sum_{i=1}^{n} w(\underline{\alpha}(\kappa) | X_i - \kappa|) = \sum_{i: X_i \neq X_i^{\varepsilon}} [w(\underline{\alpha}(\kappa) | X_i^{\varepsilon} - \kappa|) - w(\underline{\alpha}(\kappa) | X_i - \kappa|)] \geq -\varepsilon n. $$

The same reasoning applies to bound $\mathbb{P}[\hat{\alpha}(\kappa) \ge \bar{\alpha}(\kappa)]$, except that we do not have to deal with the contamination. Thus, we may control the probability of $\hat{\alpha} \notin I_\alpha(\kappa)$ by controlling the concentration of sums of independent random variables through Bernstein's inequality. See Appendix B for details.

The remaining terms $\sup_{\alpha \in I_\alpha(\kappa)} |b(\alpha, \kappa)|$, $\|F_\kappa\|_{L_2(P)}$ and $\sup_{\alpha \in I_\alpha(\kappa)} m(\alpha) = m(\underline{\alpha}(\kappa))$ are now bounded by employing the appropriate definition of the set $I_\alpha(\kappa)$ presented in Lemma 3.4.

**Lemma 3.5 (Bounding bias and variance).** Under Assumptions 1 to 3, let $I_\alpha(\kappa)$ and $\eta$ be as in Lemma 3.4. Set $c_w := \sup_{t \ge 0} tw(t)$. It holds that

$$ \sup_{\alpha \in I_\alpha(\kappa)} |b(\alpha, \kappa)| \le \inf_{q \in (1, p]} \nu_q \left( \frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1 + 2\xi)\varepsilon \right)^{1-\frac{1}{q}} + |\kappa - \mu| \left( \frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1 + 2\xi)\varepsilon \right), \\ m(\underline{\alpha}(\kappa)) \le \inf_{q \in (1, p]} c_w (\nu_q + |\kappa - \mu|) \left( \frac{c}{n} \ln \frac{4}{\delta} + \frac{\xi}{7}\varepsilon \right)^{-\frac{1}{q}}, \\ \|F_\kappa\|_{L_2(P)}^2 \le \inf_{1 < q \le 2 \wedge p} 2\nu_q^q (m(\underline{\alpha}(\kappa))^{2-q} + |\kappa - \mu|^{2-q}) + 2|\kappa - \mu|^2 \left( \frac{\bar{c} \ln \frac{4}{\delta}}{n} + (1 + 2\xi)\varepsilon \right). $$
The inequalities for sup<sub>(α,κ)∈Λ</sub> |b(α, κ)| and ‖F‖<sub>L<sub>2</sub>(P)</sub> follow from the implicit definitions of α(κ), ᾱ(κ). Since m(α(κ)) depends directly on α(κ) through the identity m(α(κ)) = c<sub>w</sub>/α(κ) (see (10)), we have to explicitly lower-bound α(κ). Assumption 3 enables us to use the implicit definition of α(κ) through w in order to obtain the desired explicit bound. See Appendix B for details.

The bounds presented thus far refer to the estimator ̂μ unaffected by adversarial contamination. The following lemma shows that the error introduced by contamination given by | ̂μ<sup>ε</sup> − ̂μ | is well-controlled.

**Lemma 3.6 (Bounding the error introduced by contamination).** Let any κ ∈ ℝ, η ∈ (0, n), ε ∈ [0, 1/2). Consider any sample X<sub>1:n</sub> and X<sub>1:n</sub><sup>ε</sup> ∈ A(X<sub>1:n</sub>, ε). Then, it holds that

$$
|\hat{\mu}^{\epsilon}(\kappa) - \hat{\mu}(\kappa)| \le \left( \epsilon + \frac{2n - \sum_{i=1}^{n} w(\hat{\alpha}^{\epsilon}(\kappa) |X_i^{\epsilon} - \kappa|) - \sum_{i=1}^{n} w(\hat{\alpha}(\kappa) |X_i - \kappa|)}{n} \right) (m(\hat{\alpha}(\kappa)) + m(\hat{\alpha}^{\epsilon}(\kappa))).
$$

Proof. We will omit the dependence on κ throughout the proof to simplify notation. The proof follows directly by the definitions of ̂μ, ̂μ<sup>ε</sup>, ̂α, and ̂α<sup>ε</sup>:

$$
\begin{align*}
|\hat{\mu}^{\epsilon} - \hat{\mu}| &= \left| \frac{1}{n} \sum_{i=1}^{n} (X_i - \kappa) w(\hat{\alpha} | X_i - \kappa|) - (X_i^{\epsilon} - \kappa) w(\hat{\alpha}^{\epsilon} | X_i^{\epsilon} - \kappa|) \right| \\
&\le \epsilon(m(\hat{\alpha}) + m(\hat{\alpha}^{\epsilon})) + \frac{1}{n} \sum_{i=1}^{n} |X_i - \kappa| |w(\hat{\alpha} | X_i - \kappa|) - w(\hat{\alpha}^{\epsilon} | X_i - \kappa|)|.
\end{align*}
$$

In the last inequality, we used the definition sup<sub>x∈ℝ</sub>|xw(αx)| = m(α) to bound the summands corresponding to the contaminated points. For the remaining term, notice that, for any a, b ∈ [0, 1], it holds that |a - b| ≤ a(1 - b) + b(1 - a). Therefore ∑<sub>i=1</sub><sup>n</sup>|X<sub>i</sub> - κ||w(α̂|X<sub>i</sub> - κ|) - w(α̂<sup>ε</sup>|X<sub>i</sub> - κ|)| is upper-bounded by

$$
\sum_{i=1}^{n} |X_i - \kappa| w(\hat{\alpha} | X_i - \kappa|) (1 - w(\hat{\alpha}^{\epsilon} | X_i^{\epsilon} - \kappa|)) + \sum_{i=1}^{n} |X_i - \kappa| w(\hat{\alpha}^{\epsilon} | X_i - \kappa|) (1 - w(\hat{\alpha} | X_i - \kappa|)).
$$

Bounding |X_i - κ|w(α̂|X_i - κ|) ≤ m(α̂) and |X_i - κ|w(α̂<sup>ε</sup>|X_i - κ|) ≤ m(α̂<sup>ε</sup>) for all i and combining the resulting bounds, we obtain the desired result. □

Under Assumption 1, Lemma B.1 yields, for all κ ∈ ℝ and X<sub>1:n</sub><sup>ε</sup> ∈ A(X<sub>1:n</sub>, ε) with probability 1,

$$
n - \eta - 1 - \lfloor n\varepsilon \rfloor \le \sum_{i=1}^{n} w(\hat{\alpha}(\kappa; X_{1:n}^{\varepsilon}) |X_i^{\varepsilon} - \kappa|),
$$

and so

$$
|\hat{\mu}^{\epsilon}(\kappa) - \hat{\mu}(\kappa)| \le \left( \epsilon + \frac{2\eta + 2 + 2n\epsilon}{n} \right) (m(\hat{\alpha}(\kappa)) + m(\hat{\alpha}^{\epsilon}(\kappa))).
$$

In addition, when ᾱ(κ) < inf<sub>X<sub>1:n</sub><sup>ε</sup>∈A(X<sub>1:n</sub>,ε)</sub> α̂<sup>ε</sup>(κ) ≤ α̂(κ), it holds that the error | ̂μ<sup>ε</sup> − ̂μ | is bounded by (6ε + 4η+4/n) m(α̂(κ)). Then, for η as in Lemma 3.4, we may use the bounds on m(α̂(κ)) from Lemma 3.5 to control the error introduced by contamination, concluding that, with high probability,

$$
\sup_{X_{1:n}^{\epsilon} \in \mathcal{A}(X_{1:n}, \epsilon)} |\hat{\mu}^{\epsilon} - \hat{\mu}| \le \left( (10 + 4\xi)\epsilon + \frac{8}{n} \ln \frac{4}{\delta} \right) c_w (\nu_p + |\kappa - \mu|) \left( \frac{c}{n} \ln \frac{4}{\delta} + \frac{\xi}{7} \epsilon \right)^{-\frac{1}{p}}.
$$

Theorem 3.1 then follows by combining the previous results. Details are presented in Appendix B.

4 Properties of shrinkage estimators

In this section, we formally state and prove the desirable properties of the shrinkage estimators ̂μ( ̂κ) presented in Section 1.1.3.

**Proposition 4.1 (Affine-equivariance).** Assume ̂κ depends on the sample Y<sub>1:m</sub> and is affine-equivariant, that is, ̂κ(λY<sub>1:m</sub> + t) = λ̂κ(Y<sub>1:m</sub>) + t, for every λ ̸= 0, t ∈ ℝ, where λX<sub>1:n</sub> + t = (λX<sub>1</sub> + t, . . . , λX<sub>n</sub> + t). Denote by ̂μ(X<sub>1:n</sub>, Y<sub>1:m</sub>) = ̂μ( ̂κ(Y<sub>1:m</sub>); X<sub>1:n</sub>) the shrinkage estimator from (8) applied to the base estimate ̂κ(Y<sub>1:m</sub>) and the sample X<sub>1:n</sub>. Then, ̂μ(λ(X<sub>1:n</sub>, Y<sub>1:m</sub>) + t) = λ̂μ(X<sub>1:n</sub>, Y<sub>1:m</sub>) + t.
*Proof.* Note that

$$
\begin{align*}
\hat{\mu}(\lambda(X_{1:n}, Y_{1:m}) + t) &= \hat{\kappa}(\lambda Y_{1:m} + t) \\
&\quad + \frac{1}{n} \sum_{i=1}^{n} (\lambda X_i + t - \hat{\kappa}(\lambda Y_{1:m} + t)) w(\hat{\alpha}(\lambda(X_{1:n}, Y_{1:m}) + t) | \lambda X_i + t - \hat{\kappa}(\lambda Y_{1:m} + t)|) \\
&= \lambda \hat{\kappa}(Y_{1:m}) + t + \frac{\lambda}{n} \sum_{i=1}^{n} (X_i - \hat{\kappa}(Y_{1:m})) w(\hat{\alpha}(\lambda(X_{1:n}, Y_{1:m}) + t) | \lambda | X_i - \hat{\kappa}(Y_{1:m})|) \\
&= \lambda \left( \hat{\kappa}(Y_{1:m}) + \frac{1}{n} \sum_{i=1}^{n} (X_i - \hat{\kappa}(Y_{1:m})) w(\hat{\alpha}(\lambda(X_{1:n}, Y_{1:m}) + t) | \lambda | X_i - \hat{\kappa}(Y_{1:m})|) \right) + t.
\end{align*}
$$

It is then enough to check that $|\lambda|\hat{\alpha}(\lambda(X_{1:n}, Y_{1:m}) + t) = \hat{\alpha}(X_{1:n}, Y_{1:m})$. Recall that $\hat{\alpha}(X_{1:n}, Y_{1:m})$ is such that

$$
\begin{align*}
\hat{\alpha}(X_{1:n}, Y_{1:m}) &= \inf \left\{ \alpha > 0 : \sum_{i=1}^{n} w(\alpha |X_i - \hat{\kappa}(Y_{1:m})|) \le n - \eta \right\} \\
&= \inf \left\{ \alpha > 0 : \sum_{i=1}^{n} w \left( \frac{\alpha}{|\lambda|} |\lambda X_i + t - \hat{\kappa}(\lambda Y_{1:m} + t)| \right) \le n - \eta \right\} \\
&= |\lambda| \inf \left\{ \alpha > 0 : \sum_{i=1}^{n} w (\alpha |\lambda X_i + t - \hat{\kappa}(\lambda Y_{1:m} + t)|) \le n - \eta \right\} = |\lambda| \hat{\alpha}(\lambda(X_{1:n}, Y_{1:m}) + t). \quad \Box
\end{align*}
$$

**Proposition 4.2 (Distributional limit of shrinkage estimators).** Let $\eta$ be fixed and take $\hat{\mu}_n = \hat{\mu}(\hat{\kappa}_n; X_{1:n})$ as in (8). Under Assumption 1, further assume that $\nu_{p+\gamma} < \infty$ for some $p > 1, \gamma > 0$, that $(\hat{\kappa}_n - \mu)n^{-\frac{1}{p}} \to 0$ in probability and that $\hat{\kappa}_n$ is independent from $X_{1:n}$. Then, $n^{1-\frac{1}{p}}(\hat{\mu}_n - \mu) \xrightarrow{d} D$ for some distribution D if, and only if, $n^{1-\frac{1}{p}}(\bar{X} - \mu) \xrightarrow{d} D$.

*Proof.* Consider $n^{1-1/p}(\hat{\mu}_n - \bar{X})$. We show that it converges in probability to zero:

$$
\begin{align*}
|n^{1-\frac{1}{p}}(\hat{\mu}_n - \bar{X})| &= \left| n^{-\frac{1}{p}} \left( n - \sum_{i=1}^{n} w(\hat{\alpha}_n |X_i - \hat{\kappa}_n|) \right) (\hat{\kappa}_n - \mu) + n^{-\frac{1}{p}} \sum_{i=1}^{n} (X_i - \mu) (w(\hat{\alpha}_n |X_i - \hat{\kappa}_n|) - 1) \right| \\
&\le (\eta+1) n^{-\frac{1}{p}} |\hat{\kappa}_n - \mu| + (\eta+1) n^{-\frac{1}{p}} \max_{1 \le i \le n} |X_i - \mu|,
\end{align*}
$$

where $\hat{\alpha}_n = \hat{\alpha}(\hat{\kappa}_n; X_{1:n})$ and we used Lemma B.1 with $\varepsilon = 0$ to attain the last inequality (which holds almost surely). To finish the proof, it is enough to show that $n^{-1/p} \max_{1 \le i \le n} |X_i - \mu|$ converges in probability to zero. To that end, notice that

$$
\begin{align*}
\mathbb{P} \left[ n^{-\frac{1}{p}} \max_{1 \le i \le n} |X_i - \mu| > t \right] &= 1 - \left( 1 - \mathbb{P} \left[ |X_1 - \mu| > t n^{\frac{1}{p}} \right] \right)^n \\
&= 1 - \left( 1 - \mathbb{P} \left[ |X_1 - \mu|^{p+\gamma} > t^{p+\gamma} n^{\frac{p+\gamma}{p}} \right] \right)^n \\
&\le 1 - \left( 1 - \frac{\nu_{p+\gamma}^{p+\gamma}}{t^{p+\gamma} n^{1+\frac{\gamma}{p}}} \right)^n \xrightarrow{n \to \infty} 0. \quad \Box
\end{align*}
$$

**Proposition 4.3** ($\hat{\mu}$ preserves breakdown point). Under Assumption 3, let $\eta \ge n\varepsilon^*(\hat{\kappa}, Y_{1:m})$. Then $\varepsilon^*(\hat{\mu}(\hat{\kappa}), (X_{1:n}, Y_{1:m})) \ge \varepsilon^*(\hat{\kappa}, Y_{1:m})$.

*Proof.* We need to show that, when $\varepsilon < \varepsilon^*(\hat{\kappa}, Y_{1:m})$,

$$
\sup_{\substack{X_{1:n} \in \mathcal{A}(X_{1:n}, \varepsilon) \\ Y_{1:m} \in \mathcal{A}(Y_{1:m}, \varepsilon)}} |\hat{\mu}(\hat{\kappa}(Y_{1:m}^{\varepsilon}); X_{1:n}^{\varepsilon}) - \hat{\mu}(\hat{\kappa}(Y_{1:m})); X_{1:n})| < \infty.
$$

Denote $\hat{\kappa}^{\varepsilon} = \hat{\kappa}(Y_{1:m})$ and $\hat{\kappa} = \hat{\kappa}(Y_{1:n}^{\varepsilon})$. We use that $|\hat{\mu}(\hat{\kappa}^{\varepsilon}; X_{1:n}^{\varepsilon}) - \hat{\mu}(\hat{\kappa}; X_{1:n})| \le |\hat{\mu}(\hat{\kappa}^{\varepsilon}; X_{1:n}^{\varepsilon}) - \hat{\mu}(\hat{\kappa}^{\varepsilon}; X_{1:n})| + |\hat{\mu}(\hat{\kappa}^{\varepsilon}; X_{1:n}) - \hat{\mu}(\hat{\kappa}; X_{1:n})|$. For the first part, we use Lemma 3.6 and the fact that $w$ is non-negative to attain the inequality

$$
|\hat{\mu}(\hat{\kappa}^{\varepsilon}; X_{1:n}^{\varepsilon}) - \hat{\mu}(\hat{\kappa}^{\varepsilon}; X_{1:n})| \le (\varepsilon + 2) (m(\hat{\alpha}(\hat{\kappa}^{\varepsilon}; X_{1:n})) + m(\hat{\alpha}(\hat{\kappa}^{\varepsilon}; X_{1:n})))
$$
We then use Assumption 3 to conclude that, for any $\kappa \in \mathbb{R}$,

$$ 
\begin{aligned}
n - \eta &\ge \sum_{i=1}^{n} w(\hat{\alpha}(\kappa; X_{1:n}^{\varepsilon}) | X_{i}^{\varepsilon} - \kappa|) \ge \sum_{i=1}^{n} w(\hat{\alpha}(\kappa; X_{1:n}^{\varepsilon}) | X_{i} - \kappa|) - n\varepsilon \\
&\ge n(1 - \varepsilon) - \hat{\alpha}(\kappa; X_{1:n}^{\varepsilon})^p \sum_{i=1}^{n} |X_{i} - \kappa|^p.
\end{aligned}
$$

which implies the bound

$$ (\hat{\alpha}(\hat{\kappa}^{\varepsilon}; X_{1:n}^{\varepsilon}))^{-p} \le (\eta - n\varepsilon)^{-1} \sum_{i=1}^{n} |X_i - \hat{\kappa}^{\varepsilon}|^p \le 2^{p-1}(\eta - n\varepsilon)^{-1} \left( \sum_{i=1}^{n} |X_i - \hat{\kappa}|^p + n|\hat{\kappa} - \hat{\kappa}^{\varepsilon}|^p \right). $$

It is then immediate that

$$ \sup_{\substack{X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon) \\ Y_{1:m}^{\varepsilon} \in \mathcal{A}(Y_{1:m}, \varepsilon)}} (\hat{\alpha}(\hat{\kappa}^{\varepsilon}; X_{1:n}^{\varepsilon}))^{-1} < \infty, $$

because $\hat{\kappa}$ has breakdown point greater than $\varepsilon$ and $\eta \ge n\varepsilon^*(\hat{\kappa}, Y_{1:m}) > n\varepsilon$ by assumption. It follows that

$$ \sup_{\substack{X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon) \\ Y_{1:m}^{\varepsilon} \in \mathcal{A}(Y_{1:m}, \varepsilon)}} m(\hat{\alpha}^{\varepsilon}(\hat{\kappa}^{\varepsilon})) < \infty, $$

since $m(\alpha) = c_w \alpha^{-1}$. The bound for $m(\hat{\alpha}(\hat{\kappa}^{\varepsilon}; X_{1:n}))$ is obtained similarly.

We now focus on the term $|\hat{\mu}(\hat{\kappa}^{\varepsilon}; X_{1:n}) - \hat{\mu}(\hat{\kappa}; X_{1:n})|$. Its supremum is easily bounded:

$$ \sup_{\substack{X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon) \\ Y_{1:m}^{\varepsilon} \in \mathcal{A}(Y_{1:m}, \varepsilon)}} |\hat{\mu}(\hat{\kappa}^{\varepsilon}; X_{1:n}) - \hat{\mu}(\hat{\kappa}; X_{1:n})| \le \sup_{\substack{X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon) \\ Y_{1:m}^{\varepsilon} \in \mathcal{A}(Y_{1:m}, \varepsilon)}} |\hat{\kappa} - \hat{\kappa}^{\varepsilon}| + 2n \max_{1 \le i \le n} |X_i| < \infty. $$

The proof is complete by collecting the bounds. $\square$

<figure><img src="image_3.png" /></figure>

# 5 Examples of estimators

In this section, we show that suitable choices of $w: [0, \infty) \to [0, 1]$ in (8) lead to known robust estimators or variants thereof. In particular, the theoretical results in Section 3 can be directly applied to establish their concentration bounds. Assumptions 2 and 3 are shown to hold in each example below.

**Example 5 (The asymmetric trimmed mean).** The choice of $w(t) = 1_{t<1}$ yields

$$ \hat{\mu} = \frac{|\eta|}{n} \hat{\kappa} + \frac{1}{n} \sum_{i=1}^{n-|\eta|} X_{(i)}, $$

where $(\cdot)$ is a permutation such that $|X_{(1)} - \hat{\kappa}| \le |X_{(2)} - \hat{\kappa}| \le \dots \le |X_{(n)} - \hat{\kappa}|$. This estimator is similar to the classical trimmed mean, but instead of discarding the $k$ largest and the $k$ smallest points in the sample before averaging, it discards the ones further from the base estimate $\hat{\kappa}$. Assumption 2 follows from $\sup_{t \ge 0} tw(t) = 1 < \infty$. Regardless of which $p$-th moment is available, Assumption 3 is satisfied. Thus, the estimator achieves optimal concentration in an appropriate regime of $\delta$ depending on the concentration of $\hat{\kappa}$ (see Theorem 3.1). The classical trimmed mean [Oliveira et al., 2025] attains sub-Gaussian concentration rates and optimal bounds for the infinite variance scenario. In these cases it has been established that $\delta_{\min}$ is of order $e^{-c'n}$ for some $c' > 0$, which is optimal. The estimator proposed in (9) in this case takes the form of

$$ \hat{\mu}_W = \frac{1}{n - |\eta|} \sum_{i=1}^{n - |\eta|} X_{(i)}, $$

effectively excluding $\hat{\kappa}$ from its explicit formula and using it only to guide the reweighting scheme.

**Example 6 (Winsorized mean).** If $w(t) = 1 \wedge t^{-1}$, then

$$ \hat{\mu} = \hat{\kappa} + \frac{1}{n} \left( M(n_+ - n_-) + \sum_{|X_i - \hat{\kappa}| \le M} X_i - \hat{\kappa} \right) \quad \text{and} \quad \hat{\alpha}^{-1} = M, \qquad (18) $$
where $n_+ = |\{i: X_i - \hat{\kappa} > M\}|$, $n_- = |\{i: X_i - \hat{\kappa} < -M\}|$ and $M$ is the (unique) solution of

$$n - \eta = |\{i: |X_i - \hat{\kappa}| \le M\}| + \sum_{|X_i - \hat{\kappa}| > M} \frac{M}{|X_i - \hat{\kappa}|} \quad (19)$$

The resulting estimator is the same regardless of $p$. It corresponds to the Winsorized mean (Huber [1964]), but with the threshold $M$ chosen via (19). Assumption 3 follows directly for all $p \in (1, 2]$ and, since $\sup_{t \ge 0} tw(t) = 1 < \infty$, Assumption 2 holds. This function attains the slowest possible decay in the sense that $\lim_{t \to \infty} tw(t) \in (0, \infty)$ and any decay of a slower order would violate Assumption 2.

**Example 7 (Lee and Valiant [2022]).** Taking $w(t) = (1 - t^p)_+$ yields the estimator $\hat{\mu}$ such that $(\hat{\alpha}, \hat{\mu})$ solves

$$\left\{ \begin{array}{l} \sum_{i=1}^{n} \min(\alpha^p |X_i - \hat{\kappa}|^p, 1) = \eta, \\ \mu = \hat{\kappa} + \frac{1}{n} \sum_{i=1}^{n} (X_i - \hat{\kappa})(1 - \min(\alpha^p |X_i - \hat{\kappa}|^p, 1)). \end{array} \right.$$

Lee and Valiant [2022] introduced this estimator for the case $p = 2$ and showed that if $\hat{\kappa}$ is sub-Gaussian, then $\hat{\mu}$ is sub-Gaussian with asymptotically optimal constants as $(\frac{1}{n} \ln \frac{1}{\delta}, \delta) \to (0, 0)$. In our framework, this estimator satisfies Assumption 3 with equality and Assumption 2 follows for every $p \in (1, 2]$ since $\sup_{t \ge 0} tw(t) = \sup_{t \ge 0} t(1 - t^p)_+ = p(p+1)^{-(p+1)/p} < \infty$. Therefore, we generalize the original estimator from Lee and Valiant [2022] to $p \in (1, 2]$.

Beyond well-studied estimators, our framework also facilitates the search for new ones with possibly different desirable properties. For example, we may look for shrinkage estimators that do not shrink large sample points as aggressively as the trimmed mean but more aggressively than the Winsorized mean. One way to do this is by tuning the $w$ function, as it determines the shrinkage associated with each sample point.

**Example 8 (Fast decay).** A possible choice of estimator that shrinks large sample points polynomially would entail choosing $w(t) = (1 + t^p)^{-1}$. It satisfies Assumptions 2 and 3 for all $p \in (1, 2]$ since $\sup_{t \ge 0} t(1 + t^p)^{-1} \le 1$ and $(1 + x)^{-1} \ge (1 - x)_+$ for all $x \in \mathbb{R}$. Notice that this choice of $w$ interpolates between Example 6 and Example 7 in the sense that $(1 - x^p) \le \frac{1}{1+x^p} \le 1 \wedge x^{-1}$ for all $x \ge 0$. In fact, we observe empirically that this interpolation also takes place in terms of performance, as is detailed in the appendix.

# 6 Computational experiments

We now empirically study the impact of shrinking. We show that improvements are not only in terms of order (see Theorem 3.1), but also hold for small values of $n$. A GitHub repository with code to reproduce all figures and tables is made available at https://github.com/lucasresenderc/shrink.

## 6.1 Improved concentration

Our first experiment is related to Theorem 1.2 and aims to study whether the proposed shrinkage estimators $\hat{\mu}(\hat{\kappa})$ indeed improve upon the performance of their base estimators $\hat{\kappa}$ in practice. To that end, we consider four base estimators: the empirical mean $\bar{X}$, the sample median $M$, the trimmed mean TM with trimming level $k = \lceil \ln \frac{1}{\delta} \rceil$ [Oliveira et al., 2025], and the median-of-means MoM with $K = \lceil \ln \frac{1}{\delta} \rceil$ buckets [Lerasle and Oliveira, 2011]. For this experiment, we compute these base estimators on a sample of size $m = 25$, whereas the shrinkage estimators $\hat{\mu}$ are computed on an independent sample of size $n = 475$, using the shrinkage functions given by $w(t) = e^{-t^p}$, $w(t) = (1 + t^p)^{-1}$, $w(t) = 1 \wedge t^{-1}$, $w(t) = \mathbf{1}_{t<1}$, and $w(t) = (1 - t^p)_+$, where $p = 2$ since we only consider distributions with finite variance (a standard normal distribution N, a skewed normal SN, a Student's $t$-distribution with 2.01 degrees of freedom T, and a skewed $t$-distribution with 2.01 degrees of freedom ST). We fix $\delta = 0.05$ and $\eta = \ln \frac{1}{\delta}$.

To evaluate the performance of each estimator, we compute the $1 - \delta$-quantile of the errors computed over 10000 independent trials for the four different distributions. We then calculate the relative difference of this quantile with respect to that of the base estimate over the full sample of size $N = 500$. The results are presented in Table 1.

We show that, in most cases, shrinking does not hurt the performance of the trimmed mean, and consistently improves the median-of-means. For both the empirical mean and the sample median, $\hat{\mu}(\hat{\kappa})$ dramatically improves their performance except when the base estimator is already excellent (e.g., the empirical mean under light-tails and the sample median under symmetrical, heavy-tailed distributions, see Figure 1). These results corroborate Theorem 1.2 and suggest that shrinking is a generally beneficial procedure. Finally, note that estimators with similar features (such as those with $w(t) = (1 - t^p)_+$, $(1 + t^p)^{-1}$, and $e^{-t^p}$, which decay quickly for small $t$) also display similar performances. These shrinkage functions provide the best results under skewed distributions, while those with plateaus for small $t$ (such as $w(t) = 1 \wedge t^{-1}$ and $w(t) = \mathbf{1}_{t<1}$) perform better under symmetric distributions.
<figure><img src="image_4.png" /><figcaption>Figure 1: Empirical $1 - \delta$ confidence interval errors plotted against sample sizes, for different distributions (skew/symmetry is represented by the columns, while tail weight is represented by the rows), and mean estimators. The shrinkage estimator based on the sample median has excellent performance across all distributions and sample sizes considered: it always outperforms the median-of-means and is competitive against the trimmed mean.</figcaption></figure>

In order to compare between the different base estimators, we also display in Figure 1 the error for each base estimator and, for the sake of a clearer graph, only the shrunk version of the sample median with $w(t) = (1 + t^p)^{-1}$, since the performance of the shrinkage estimators is similar, as Table 1 suggests.

<table><caption>Table 1: Relative difference (in %) of the empirical $1 - \delta$ confidence interval error of the shrunk estimators (for each $w$ (line) and $\hat{\kappa}$ (column)) with respect to the base estimator $\hat{\kappa}$ under different distributions. The entries corresponding to the two best performing shrinkage functions compared to the base estimator for each fixed base estimator and distribution are highlighted in bold.</caption><thead><tr><th rowspan="2"></th><th colspan="4">X&#772;</th><th colspan="4">M</th><th colspan="4">TM</th><th colspan="4">MoM</th></tr><tr><th>N</th><th>SN</th><th>T</th><th>ST</th><th>N</th><th>SN</th><th>T</th><th>ST</th><th>N</th><th>SN</th><th>T</th><th>ST</th><th>N</th><th>SN</th><th>T</th><th>ST</th></tr></thead><tbody><tr><td>w(t) = e<sup>-t<sup>p</sup></sup></td><td>1</td><td><b>2</b></td><td>-33</td><td>-7</td><td>-19</td><td><b>-69</b></td><td>55</td><td><b>-65</b></td><td>1</td><td><b>-1</b></td><td>4</td><td>-7</td><td><b>-12</b></td><td><b>-11</b></td><td>-31</td><td>-5</td></tr><tr><td>w(t) = (1 + t<sup>p</sup>)<sup>-1</sup></td><td>1</td><td><b>2</b></td><td>-32</td><td>-9</td><td>-19</td><td><b>-69</b></td><td>59</td><td><b>-66</b></td><td>1</td><td><b>-1</b></td><td>6</td><td>-9</td><td><b>-12</b></td><td><b>-11</b></td><td>-30</td><td>-7</td></tr><tr><td>w(t) = 1 &amp; t<sup>-1</sup></td><td><b>0</b></td><td>9</td><td><b>-38</b></td><td>2</td><td><b>-19</b></td><td>-67</td><td><b>43</b></td><td>-62</td><td><b>1</b></td><td>5</td><td>-4</td><td>-0</td><td>-12</td><td>-5</td><td><b>-37</b></td><td>4</td></tr><tr><td>w(t) = t<sub>t&lt;1</sub></td><td>2</td><td>11</td><td><b>-37</b></td><td>8</td><td>-17</td><td>-66</td><td><b>47</b></td><td>-60</td><td>2</td><td>7</td><td>-2</td><td>6</td><td>-10</td><td>-3</td><td><b>-35</b></td><td>10</td></tr><tr><td>w(t) = (1 - t<sup>p</sup>)<sub>+</sub></td><td><b>1</b></td><td>2</td><td>-34</td><td>-5</td><td><b>-19</b></td><td>-69</td><td>53</td><td>-64</td><td><b>1</b></td><td>-1</td><td>2</td><td>-5</td><td>-12</td><td>-11</td><td>-32</td><td>-3</td></tr></tbody></table>

## 6.2 Splitting the sample

We now aim to study the effect of splitting the sample on the performance of the shrinkage estimators. Splitting the sample is a natural way to fulfill Assumption 4. Nevertheless, it is not clear how much of the available data should be dedicated to the base estimator, or even if computing the base estimator and the shrinkage estimator on the full sample (ignoring Assumption 4) is the better approach. To evaluate the performance of shrinkage estimators with varying sample splitting strategies, we compute the errors of these estimators as described in Section 6.1 for varying split ratios given by $\lfloor m/N \rfloor$, along with fully utilizing the sample for both computations. The resulting performances are displayed in Table 2, for $\delta = 0.05$, $\eta = \ln \frac{1}{\delta}$, and $N = 500$.

These results suggest that, as long as we want to keep a splitting stratety, we should dedicate as few sample points as possible to the base estimator. A closer look at small values for $m$ (and, thus, of split ratios) suggests that there often is
an optimal value of $m$ greater than 1. See Figures 2 and 3. For Figure 2, the full sample size is kept at $N = 500$, while $N$ varies from 100 to 1000 in Figure 3. To obtain Figure 3, we computed the error and the value of $m$ that minimizes the error (denoted by $m^*$) for 50 independent trials and plotted its mean and error band determined by the 2.5% and 97.5% quantiles. The resulting graph shows that $m^*$ remains approximately constant even as $N$ increases. This corroborates Theorem 1.1, which states that $\hat{\mu}(\hat{\kappa})$ attains optimal concentration rates as long as $\hat{\kappa}$ is not too far from the population mean.

<table><caption>Table 2: Relative error of $\hat{\mu}(\hat{\kappa})$ with respect to $\hat{\kappa}$ (found in the first row of each block) for different distributions and split ratios (columns). NA means that both $\hat{\kappa}$ and $\hat{\mu}(\hat{\kappa})$ were computed on the same full sample. The performance seems to decrease as the split ratio increases. Fully utilizing the sample for both computations yields the best results, although splitting the data with 5% of the data dedicated to the base estimator leads to similar performances.</caption><thead><tr><th></th><th></th><th colspan="3">N</th><th colspan="3">SN</th><th colspan="3">T</th><th colspan="3">ST</th></tr><tr><th></th><th></th><th>NA</th><th>0.05</th><th>0.5</th><th>0.95</th><th>NA</th><th>0.05</th><th>0.5</th><th>0.95</th><th>NA</th><th>0.05</th><th>0.5</th><th>0.95</th></tr></thead><tbody><tr><td>X</td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td></tr><tr><td>w(t) = e<sup>-t<sup>p</sup></sup></td><td></td><td>-0</td><td>1</td><td>34</td><td>221</td><td>1</td><td>3</td><td>37</td><td>253</td><td>-34</td><td>-32</td><td>-17</td><td>49</td></tr><tr><td>w(t) = (1 + t<sup>p</sup>)<sup>-1</sup></td><td></td><td>-0</td><td>1</td><td>34</td><td>226</td><td>1</td><td>3</td><td>37</td><td>247</td><td>-32</td><td>-31</td><td>-15</td><td>55</td></tr><tr><td>w(t) = 1 &amp; t<sup>-1</sup></td><td></td><td>0</td><td>1</td><td>30</td><td>205</td><td>8</td><td>12</td><td>55</td><td>250</td><td>-39</td><td>-38</td><td>-24</td><td>42</td></tr><tr><td>w(t) = 1<sub>t&lt;1</sub></td><td></td><td>2</td><td>2</td><td>33</td><td>224</td><td>11</td><td>15</td><td>61</td><td>361</td><td>-38</td><td>-37</td><td>-22</td><td>52</td></tr><tr><td>w(t) = (1 - t<sup>p</sup>)<sub>+</sub></td><td></td><td>-0</td><td>1</td><td>34</td><td>217</td><td>1</td><td>3</td><td>38</td><td>266</td><td>-35</td><td>-33</td><td>-19</td><td>49</td></tr><tr><td>M</td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td></tr><tr><td>w(t) = e<sup>-t<sup>p</sup></sup></td><td></td><td>-21</td><td>-20</td><td>6</td><td>155</td><td>-70</td><td>-69</td><td>-59</td><td>22</td><td>52</td><td>57</td><td>92</td><td>245</td></tr><tr><td>w(t) = (1 + t<sup>p</sup>)<sup>-1</sup></td><td></td><td>-21</td><td>-20</td><td>6</td><td>159</td><td>-70</td><td>-69</td><td>-59</td><td>18</td><td>56</td><td>60</td><td>97</td><td>256</td></tr><tr><td>w(t) = 1 &amp; t<sup>-1</sup></td><td></td><td>-21</td><td>-20</td><td>4</td><td>143</td><td>-68</td><td>-67</td><td>-54</td><td>27</td><td>41</td><td>45</td><td>78</td><td>228</td></tr><tr><td>w(t) = 1<sub>t&lt;1</sub></td><td></td><td>-19</td><td>-19</td><td>5</td><td>158</td><td>-67</td><td>-66</td><td>-52</td><td>59</td><td>43</td><td>46</td><td>83</td><td>244</td></tr><tr><td>w(t) = (1 - t<sup>p</sup>)<sub>+</sub></td><td></td><td>-21</td><td>-20</td><td>6</td><td>152</td><td>-70</td><td>-69</td><td>-59</td><td>28</td><td>51</td><td>54</td><td>90</td><td>239</td></tr><tr><td>TM</td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td></tr><tr><td>w(t) = e<sup>-t<sup>p</sup></sup></td><td></td><td>-0</td><td>1</td><td>34</td><td>221</td><td>-4</td><td>-1</td><td>31</td><td>240</td><td>3</td><td>5</td><td>29</td><td>132</td></tr><tr><td>w(t) = (1 + t<sup>p</sup>)<sup>-1</sup></td><td></td><td>-0</td><td>2</td><td>34</td><td>226</td><td>-4</td><td>-1</td><td>31</td><td>233</td><td>6</td><td>8</td><td>33</td><td>140</td></tr><tr><td>w(t) = 1 &amp; t<sup>-1</sup></td><td></td><td>0</td><td>1</td><td>30</td><td>205</td><td>3</td><td>7</td><td>48</td><td>238</td><td>-4</td><td>-3</td><td>19</td><td>119</td></tr><tr><td>w(t) = 1<sub>t&lt;1</sub></td><td></td><td>2</td><td>2</td><td>33</td><td>224</td><td>6</td><td>10</td><td>54</td><td>344</td><td>-3</td><td>-2</td><td>23</td><td>132</td></tr><tr><td>w(t) = (1 - t<sup>p</sup>)<sub>+</sub></td><td></td><td>-0</td><td>1</td><td>34</td><td>217</td><td>-4</td><td>-1</td><td>32</td><td>253</td><td>2</td><td>4</td><td>28</td><td>129</td></tr><tr><td>MoM</td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td><td>0</td><td></td><td></td></tr><tr><td>w(t) = e<sup>-t<sup>p</sup></sup></td><td></td><td>-14</td><td>-12</td><td>16</td><td>178</td><td>-14</td><td>-12</td><td>17</td><td>201</td><td>-31</td><td>-30</td><td>-14</td><td>56</td></tr><tr><td>w(t) = (1 + t<sup>p</sup>)<sup>-1</sup></td><td></td><td>-14</td><td>-12</td><td>16</td><td>181</td><td>-14</td><td>-12</td><td>17</td><td>196</td><td>-29</td><td>-29</td><td>-12</td><td>61</td></tr><tr><td>w(t) = 1 &amp; t<sup>-1</sup></td><td></td><td>-13</td><td>-13</td><td>13</td><td>164</td><td>-8</td><td>-5</td><td>33</td><td>199</td><td>-37</td><td>-36</td><td>-20</td><td>48</td></tr><tr><td>w(t) = 1<sub>t&lt;1</sub></td><td></td><td>-11</td><td>-11</td><td>15</td><td>182</td><td>-5</td><td>-2</td><td>38</td><td>295</td><td>-36</td><td>-35</td><td>-18</td><td>57</td></tr><tr><td>w(t) = (1 - t<sup>p</sup>)<sub>+</sub></td><td></td><td>-14</td><td>-12</td><td>16</td><td>175</td><td>-13</td><td>-12</td><td>17</td><td>215</td><td>-32</td><td>-31</td><td>-15</td><td>55</td></tr></tbody></table>

## 6.3 Robustness under contamination

To evaluate robustness under contamination, we sample our data as in Sections 6.1 and 6.2 and randomly substitute $[\epsilon m]$ of the sample points destined for the computation of $\hat{\kappa}$ and $[\epsilon n]$ of the sample points destined for the computation of $\hat{\mu}(\hat{\kappa})$ by $10^6$. We display the performance of the base estimators under this contamination as well as that of $\hat{\mu}$ in Table 3. The results confirm in practice that our shrinkage estimators $\hat{\mu}$ indeed retain robustness under contamination as long as the base estimator is robust. Is is noteworthy that the worst-performing shrinkage estimators under heavy contamination are given by those based on the shrinkage function $w(t) = 1 \wedge t^{-1}$, which has the slowest asymptotic decay contemplated by our theory.

## 6.4 Shrinkage functions that violate Assumptions 2 and 3

This subsection is dedicated to evaluating the performance of shrinkage estimators that employ shrinkage functions $w$ that violate either Assumption 2 or Assumption 3, or both. Specifically, we consider $w(t) = \frac{1}{\ln(e+t^2)}$, which violates only Assumption 2, $w(t) = 1 - \sqrt{1 - (1 - t)^2}$, which violates only Assumption 3, as well as $w(t) = \frac{1}{\ln(e+t)}$ and $w(t) = \frac{1}{1+\sqrt{t}}$, which violate both assumptions.
<figure><img src="image_5.png" /><figcaption>Figure 2: Error of shrinkage estimators plotted against split ratio for different base estimators (rows), distributions (columns) and shrinkage functions (curves). The solid blue line represents the performance of the base estimator computed on the full sample. The best performing split ratios are consistently below 10% ($m = 50$).</figcaption></figure>

<figure><img src="image_6.png" /><figcaption>Figure 3: Best performing sample size $m^*$ for base estimators $\bar{X}$ and $M$ (rows), plotted against $N$, for different distributions (columns) and shrinkage functions (curves), for fixed $\delta = 0.05$. The value $m^*$ does not increase as $N$ increases, in accordance with Theorem 1.1.</figcaption></figure>

The computational experiments are similar to the one whose results are presented in Table 1, where we add these violating estimators to the previously considered ones. The results, presented in Table 4, indicate that, while these violating shrinkage estimators still often improve upon the base estimators, they tend to perform worse than the ones that satisfy both assumptions. In particular, the estimators that have derivative $-\infty$ at 0, given by $w(t) = (1 + \sqrt{t})^{-1}$ and $w(t) = 1 - \sqrt{1 - (1-t)^2}_{+}$, often perform the worst suggesting that, at least in the considered scenarios, Assumption 3, which controls the decay close to the origin, is more relevant than Assumption 2, which controls the asymptotic decay.
<table><caption>Table 3: Base 10 logarithm of the errors of the mean estimators (rows) for different distributions and contamination levels (columns). The two best performing estimator for each distribution and contamination level are highlighted in bold. All estimators except for those based on the sample mean are reasonably robust to contamination, with the shrinkage estimators $\hat{\mu}$ often attaining the best performance under heavy contamination.</caption><thead><tr><th></th><th colspan="4">N</th><th colspan="4">SN</th><th colspan="4">T</th><th colspan="4">ST</th></tr><tr><th></th><th>0.0</th><th>0.05</th><th>0.1</th><th>0.2</th><th>0.0</th><th>0.05</th><th>0.1</th><th>0.2</th><th>0.0</th><th>0.05</th><th>0.1</th><th>0.2</th><th>0.0</th><th>0.05</th><th>0.1</th><th>0.2</th></tr></thead><tbody><tr><td>X</td><td><b>-1.1</b></td><td>4.7</td><td>5.0</td><td>5.3</td><td><b>-1.0</b></td><td>4.7</td><td>5.0</td><td>5.3</td><td>-0.5</td><td>4.7</td><td>5.0</td><td>5.3</td><td>-0.3</td><td>4.7</td><td>5.0</td><td>5.3</td></tr><tr><td>w(t) = 1<sub>t<1</sub></td><td>-1.0</td><td>3.7</td><td>4.1</td><td>4.6</td><td>-0.9</td><td>3.7</td><td>4.1</td><td>4.6</td><td>-0.7</td><td>3.7</td><td>4.1</td><td>4.6</td><td>-0.3</td><td>3.7</td><td>4.1</td><td>4.6</td></tr><tr><td>w(t) = 1 &amp; t<sup>-1</sup></td><td>-1.1</td><td>3.9</td><td>4.4</td><td>4.9</td><td>-0.9</td><td>3.9</td><td>4.4</td><td>4.9</td><td>-0.7</td><td>3.9</td><td>4.4</td><td>4.9</td><td>-0.3</td><td>3.9</td><td>4.4</td><td>4.9</td></tr><tr><td>w(t) = (1 - t<sup>p</sup>)<sub>+</sub></td><td>-1.1</td><td>3.8</td><td>4.4</td><td>4.9</td><td><b>-1.0</b></td><td>3.8</td><td>4.4</td><td>4.9</td><td>-0.7</td><td>3.8</td><td>4.4</td><td>4.9</td><td>-0.3</td><td>3.8</td><td>4.4</td><td>4.9</td></tr><tr><td>w(t) = e<sup>-t<sup>p</sup></sup></td><td>-1.1</td><td>4.1</td><td>4.5</td><td>5.0</td><td>-1.0</td><td>4.1</td><td>4.5</td><td>5.0</td><td>-0.7</td><td>4.1</td><td>4.5</td><td>5.0</td><td>-0.3</td><td>4.1</td><td>4.5</td><td>5.0</td></tr><tr><td>w(t) = (1 + t<sup>p</sup>)<sup>-1</sup></td><td>-1.1</td><td>4.3</td><td>4.6</td><td>5.1</td><td>-1.0</td><td>4.3</td><td>4.6</td><td>5.1</td><td>-0.7</td><td>4.3</td><td>4.6</td><td>5.1</td><td><b>-0.3</b></td><td>4.3</td><td>4.6</td><td>5.1</td></tr><tr><td>M</td><td>-1.0</td><td>-0.8</td><td>-0.6</td><td>-0.4</td><td>-0.4</td><td>-0.5</td><td>-0.7</td><td>-0.5</td><td><b>-0.9</b></td><td><b>-0.7</b></td><td>-0.6</td><td>-0.3</td><td>0.1</td><td>0.1</td><td>0.0</td><td>-0.1</td></tr><tr><td>w(t) = 1<sub>t<1</sub></td><td>-1.0</td><td>-1.0</td><td>-0.9</td><td>-0.7</td><td>-0.9</td><td>-0.9</td><td>-0.9</td><td><b>-0.7</b></td><td>-0.7</td><td><b>-0.8</b></td><td><b>-0.7</b></td><td>-0.6</td><td>-0.3</td><td>-0.2</td><td>-0.2</td><td>-0.2</td></tr><tr><td>w(t) = 1 &amp; t<sup>-1</sup></td><td>-1.1</td><td>-0.7</td><td>-0.5</td><td>-0.2</td><td>-0.9</td><td>-0.7</td><td>-0.4</td><td>-0.2</td><td>-0.7</td><td>-0.2</td><td>0.1</td><td>0.3</td><td>-0.3</td><td>-0.0</td><td>0.3</td><td>0.5</td></tr><tr><td>w(t) = (1 - t<sup>p</sup>)<sub>+</sub></td><td>-1.1</td><td>-1.0</td><td>-1.0</td><td>-0.7</td><td>-1.0</td><td>-0.9</td><td>-0.9</td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td><b>-0.7</b></td><td>-0.6</td><td>-0.3</td><td>-0.3</td><td>-0.3</td><td>-0.2</td></tr><tr><td>w(t) = e<sup>-t<sup>p</sup></sup></td><td>-1.1</td><td>-1.0</td><td>-1.0</td><td>-0.7</td><td>-1.0</td><td>-0.9</td><td>-0.9</td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td>-0.6</td><td>-0.3</td><td>-0.3</td><td>-0.3</td><td>-0.2</td></tr><tr><td>w(t) = (1 + t<sup>p</sup>)<sup>-1</sup></td><td>-1.1</td><td>-1.0</td><td>-1.0</td><td>-0.7</td><td>-1.0</td><td>-0.9</td><td>-0.9</td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td>-0.6</td><td>-0.3</td><td>-0.3</td><td>-0.3</td><td>-0.2</td></tr><tr><td>TM</td><td><b>-1.1</b></td><td>-0.7</td><td>-0.5</td><td>-0.3</td><td>-1.0</td><td>-0.8</td><td>-0.6</td><td>-0.3</td><td>-0.7</td><td>-0.4</td><td>-0.3</td><td>-0.1</td><td>-0.3</td><td><b>-0.4</b></td><td><b>-0.4</b></td><td>-0.1</td></tr><tr><td>w(t) = 1<sub>t<1</sub></td><td>-1.0</td><td>-1.0</td><td>-0.9</td><td>-0.7</td><td>-0.9</td><td>-0.9</td><td><b>-0.9</b></td><td><b>-0.8</b></td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td>-0.6</td><td>-0.3</td><td>-0.3</td><td>-0.2</td><td>-0.2</td></tr><tr><td>w(t) = 1 &amp; t<sup>-1</sup></td><td>-1.1</td><td>-0.7</td><td>-0.5</td><td>-0.2</td><td>-0.9</td><td>-0.7</td><td>-0.4</td><td>-0.2</td><td>-0.7</td><td>-0.2</td><td>0.1</td><td>0.3</td><td>-0.3</td><td>-0.0</td><td>0.3</td><td>0.5</td></tr><tr><td>w(t) = (1 - t<sup>p</sup>)<sub>+</sub></td><td>-1.1</td><td><b>-1.0</b></td><td>-1.0</td><td>-0.7</td><td>-1.0</td><td>-1.0</td><td>-0.9</td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td>-0.6</td><td>-0.3</td><td>-0.3</td><td>-0.3</td><td>-0.3</td></tr><tr><td>w(t) = e<sup>-t<sup>p</sup></sup></td><td>-1.1</td><td><b>-1.0</b></td><td><b>-1.0</b></td><td><b>-0.7</b></td><td>-1.0</td><td><b>-1.0</b></td><td>-0.9</td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td>-0.6</td><td>-0.3</td><td>-0.3</td><td>-0.3</td><td><b>-0.3</b></td></tr><tr><td>w(t) = (1 + t<sup>p</sup>)<sup>-1</sup></td><td>-1.1</td><td>-1.0</td><td><b>-1.0</b></td><td><b>-0.7</b></td><td>-1.0</td><td><b>-1.0</b></td><td><b>-0.9</b></td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td>-0.7</td><td>-0.6</td><td>-0.3</td><td><b>-0.3</b></td><td>-0.3</td><td><b>-0.3</b></td></tr><tr><td>MoM</td><td>-1.0</td><td>-0.3</td><td>-0.4</td><td>-0.2</td><td>-0.9</td><td>-0.2</td><td>-0.3</td><td>-0.1</td><td>-0.6</td><td>-0.1</td><td>-0.1</td><td>0.0</td><td>-0.3</td><td>0.1</td><td>0.0</td><td>0.2</td></tr><tr><td>w(t) = 1<sub>t<1</sub></td><td>-1.0</td><td>-0.9</td><td>-0.9</td><td>-0.7</td><td>-0.9</td><td>-0.9</td><td>-0.9</td><td>-0.4</td><td>-0.7</td><td>-0.6</td><td>-0.6</td><td>-0.5</td><td>-0.3</td><td>-0.3</td><td>-0.3</td><td>-0.2</td></tr><tr><td>w(t) = 1 &amp; t<sup>-1</sup></td><td>-1.1</td><td>-0.7</td><td>-0.5</td><td>-0.1</td><td>-0.9</td><td>-0.6</td><td>-0.4</td><td>-0.2</td><td><b>-0.7</b></td><td>-0.2</td><td>0.1</td><td>0.3</td><td>-0.3</td><td>0.0</td><td>0.4</td><td>0.5</td></tr><tr><td>w(t) = (1 - t<sup>p</sup>)<sub>+</sub></td><td>-1.1</td><td>-1.0</td><td>-0.8</td><td>-0.6</td><td>-1.0</td><td>-0.9</td><td>-0.9</td><td>-0.6</td><td>-0.7</td><td>-0.6</td><td>-0.5</td><td><b>-0.6</b></td><td>-0.3</td><td>-0.3</td><td>-0.3</td><td>-0.2</td></tr><tr><td>w(t) = e<sup>-t<sup>p</sup></sup></td><td>-1.1</td><td>-1.0</td><td>-0.8</td><td>-0.5</td><td>-1.0</td><td>-0.9</td><td>-0.7</td><td>-0.6</td><td>-0.7</td><td>-0.6</td><td>-0.5</td><td>-0.5</td><td>-0.3</td><td>-0.3</td><td>-0.2</td><td>-0.2</td></tr><tr><td>w(t) = (1 + t<sup>p</sup>)<sup>-1</sup></td><td>-1.1</td><td>-1.0</td><td>-0.9</td><td>-0.6</td><td>-1.0</td><td>-0.9</td><td>-0.9</td><td>-0.5</td><td>-0.7</td><td>-0.6</td><td>-0.5</td><td><b>-0.6</b></td><td><b>-0.3</b></td><td>-0.3</td><td><b>-0.3</b></td><td>-0.2</td></tr></tbody></table>

<table><caption>Table 4: Relative error of the shrinkage estimator for a given shrinkage function (rows) with respect to the corresponding base estimator and distribution (columns). The shrinkage functions that have derivative $-\infty$ at 0 (thus violating Assumption 3) display the worst performances across different base estimators and distributions.</caption><thead><tr><th></th><th colspan="4">X&#773;</th><th colspan="4">M</th><th colspan="4">TM</th><th colspan="4">MoM</th></tr><tr><th></th><th>N</th><th>SN</th><th>T</th><th>ST</th><th>N</th><th>SN</th><th>T</th><th>ST</th><th>N</th><th>SN</th><th>T</th><th>ST</th><th>N</th><th>SN</th><th>T</th><th>ST</th></tr></thead><tbody><tr><td>w(t) = 1 &amp; t<sup>-1</sup></td><td>1</td><td><b>12</b></td><td>-38</td><td><b>1</b></td><td>-20</td><td><b>-67</b></td><td>45</td><td><b>-62</b></td><td>1</td><td><b>7</b></td><td>-3</td><td><b>0</b></td><td>-13</td><td><b>-5</b></td><td>-36</td><td><b>3</b></td></tr><tr><td>w(t) = (1 + t<sup>p</sup>)<sup>-1</sup></td><td>1</td><td><b>3</b></td><td>-31</td><td>-10</td><td>-20</td><td><b>-69</b></td><td>60</td><td>-65</td><td>2</td><td><b>-1</b></td><td>8</td><td>-9</td><td>-12</td><td><b>-12</b></td><td>-29</td><td>-8</td></tr><tr><td>w(t) = (ln(e + t))<sup>-1</sup></td><td>2</td><td>2</td><td>-11</td><td>-11</td><td>-19</td><td>-70</td><td>107</td><td>-67</td><td>2</td><td>-3</td><td>40</td><td>-13</td><td>-12</td><td>-12</td><td>-7</td><td>-10</td></tr><tr><td>w(t) = (ln(e + t<sup>2</sup>))<sup>-1</sup></td><td>1</td><td>3</td><td>-28</td><td>-11</td><td>-20</td><td>-69</td><td>66</td><td>-66</td><td>2</td><td>-1</td><td>12</td><td>-11</td><td>-12</td><td>-12</td><td>-25</td><td>-10</td></tr><tr><td>w(t) = 1 - &radic;(1 - (1 - t)<sup>2</sup>)<sub>+</sub></td><td><b>2</b></td><td>2</td><td>-2</td><td>-3</td><td><b>-19</b></td><td>-70</td><td><b>129</b></td><td>-64</td><td><b>2</b></td><td>-3</td><td><b>54</b></td><td>-5</td><td><b>-12</b></td><td>-12</td><td><b>2</b></td><td>-2</td></tr><tr><td>w(t) = (1 + &radic;(t))<sup>-1</sup></td><td><b>2</b></td><td>2</td><td>-2</td><td>-3</td><td><b>-19</b></td><td>-70</td><td><b>129</b></td><td>-64</td><td><b>2</b></td><td>-3</td><td><b>54</b></td><td>-5</td><td><b>-12</b></td><td>-12</td><td><b>2</b></td><td>-1</td></tr></tbody></table>

# 7 Discussion

This paper formalizes the intuition that downweighting sample points far from a location estimate enhances robustness in mean estimation through improvement of concentration rates, in particular achieving sub-Gaussian behavior in certain $(n, \delta)$ regimes. Unlike most works in robust statistics that focus on a single estimator, our results apply to a broad class of mean estimators. Robust mean estimation underpins various statistical tasks — including uniform mean estimation [Minsker and Ndaoud, 2021], estimation under arbitrary norms [Lugosi and Mendelson, 2019b], regression [Audibert and Catoni, 2011], and covariance estimation [Oliveira and Rico, 2022] — yet the computational tractability of existing methods remains a significant challenge. This limitation has motivated research into computationally efficient alternatives [Cherapanamjeri et al., 2019, Diakonikolas et al., 2020, Hopkins, 2020]. We believe that the flexibility in choosing $w$ could lead to robust procedures with improved computational properties. Furthermore, our experiments reveal that different choices of $w$ excel in distinct regimes of sample size and confidence level $1 - \delta$,
exhibiting varying sensitivities to the base estimator. Thus, a refinement of our results beyond order bounds could guide the selection of w for specific applications.

A Useful Lemmata and Definitions

**Lemma A.1 (Bernstein's inequality).** Let $Y_1, \dots, Y_n$ be i.i.d. random variables with mean zero and bounded by 1. Then,

$$ \mathbb{P}\left(\frac{1}{n}\sum_{i=1}^{n} Y_i \geq \sigma\sqrt{\frac{2}{n}\ln\frac{1}{\delta}} + \frac{1}{3n}\ln\frac{1}{\delta}\right) \geq 1 - \delta. $$

**Definition A.2 (VC-subgraph dimension).** Given a class of functions $\mathcal{F}$ mapping from $\mathcal{X}$ to $\mathbb{R}$, the VC-subgraph dimension of $\mathcal{F}$ is the largest integer $d$ such that there exist $d$ pairs in $\mathcal{X} \times \mathbb{R}$ $(x_1, y_1), \dots, (x_d, y_d)$ for which the set of subgraphs $\{(x, y) \in \mathcal{X} \times \mathbb{R} : f(x) > y\} : f \in \mathcal{F}\}$ can realize all $2^d$ possible labelings. That is, for every subset $S \subseteq \{1, \dots, d\}$, there exists a function $f \in \mathcal{F}$ such that $f(x_i) > y_i$ if and only if $i \in S$.

**Lemma A.3 (Concentration of suprema of empirical processes under VC assumptions).** Let $P$ be a distribution over a set $\mathcal{X}$ and $\mathcal{G}$ be a class of integrable functions $g: \mathcal{X} \to \mathbb{R}$. Assume that $\sup_{g \in \mathcal{G}} \|g - Pg\|_{\infty} \le L$ for some $L > 0$. Let $G(x) = \sup_{g \in \mathcal{G}} |g(x)|$ be the envelope of $\mathcal{G}$ and assume $\mathcal{G}$ is a VC-subgraph class with dimension $d$. Assume also that exists a countable $\mathcal{H} \subset \mathcal{G}$ such that every $g \in \mathcal{G}$ is the $P$-pointwise a.s. and $L_1(P)$ limits of functions $(h_k)_{k=1}^{\infty} \subset \mathcal{H}$. For every $\delta \in (0, 1)$, it holds with probability at least $1 - \delta$, that

$$ \begin{aligned} \sup_{g \in \mathcal{G}} \left| (\hat{P}_n - P) g \right| &\le K \|G\|_{L_2(P)} \sqrt{\frac{d}{n}} + \sqrt{\frac{2}{n} \ln \frac{1}{\delta} \left( \sigma^2 + 2LK \|G\|_{L_2(P)} \sqrt{\frac{d}{n}} \right)} + \frac{L}{3n} \ln \frac{1}{\delta} \\ &\le 2K \|G\|_{L_2(P)} \sqrt{\frac{d}{n}} + \sigma \sqrt{\frac{2}{n} \ln \frac{1}{\delta}} + \frac{4L}{3n} \ln \frac{1}{\delta} \end{aligned} $$

where $K > 0$ is an absolute constant and $\sigma^2 = \sup_{g \in \mathcal{G}} P(g - Pg)^2$.

*Proof.* Since $\mathcal{G}$ is pointwise $P$-a.s. and $L_1(P)$ limit of functions in $\mathcal{H}$ we have

$$ \sup_{g \in \mathcal{G}} \left| (\hat{P}_n - P) g \right| = \sup_{h \in \mathcal{H}} \left| (\hat{P}_n - P) h \right| \quad \text{a.s.} $$

On the countable family $\mathcal{H}$, apply Bousquet's version of Talagrand's inequality [Bousquet, 2002, Theorem 2.3] and use the entropy bound on the expectation of the empirical process (Lemma A.5). $\square$

Theorem A.2 is closely related to the following definition of covering number. This relation yields the proof of Lemma A.5.

**Definition A.4 (Covering number).** Let $(\mathcal{X}, d)$ be a metric space. Given $\varepsilon > 0$, the $\varepsilon$-covering number of $\mathcal{X}$ with respect to the metric $d$, denoted by $\mathcal{N}(\mathcal{X}, d, \varepsilon)$, is the smallest number of elements $x_1, \dots, x_n \in \mathcal{X}$ such that for every $x \in \mathcal{X}$, there exists some $x_i$ satisfying $d(x, x_i) \le \varepsilon$. In other words, $\mathcal{N}(\mathcal{F}, d, \varepsilon)$ is the minimal number of points required to form an $\varepsilon$-net that covers $\mathcal{X}$ with respect to the given metric.

**Lemma A.5 (Integral entropy bound).** Let $X, X_1, \dots, X_n$ be i.i.d., $\mathcal{X}$-valued random variables with distribution $P$ and $\mathcal{F}$ a family of real functions defined in $\mathcal{X}$. Assume $\mathcal{F}$ is VC-subgraph with dimension $d$ and an envelope function $F$ (that is, $|f(x)| \le F(x)$ for all $f \in \mathcal{F}$ and $x \in \mathcal{X}$).

$$ \mathbb{E} \sup_{f \in \mathcal{F}} \left| \frac{1}{n} \sum_{i=1}^{n} f(X_i) - \mathbb{E} f(X) \right| \leq K \|F\|_{L_2(P)} \sqrt{\frac{d}{n}} $$

where $K > 0$ is an absolute constant.

*Proof.* A symmetrization argument followed by an application of Dudley's theorem (see the proof of Theorem 8.3.23 in Vershynin [2018]) implies

$$ \begin{aligned} \mathbb{E} \sup_{f \in \mathcal{F}} \left| \frac{1}{n} \sum_{i=1}^{n} f(X_i) - \mathbb{E} f(X) \right| &\le \frac{C}{\sqrt{n}} \mathbb{E} \int_{0}^{2\|F\|_{L_2(\mathbb{P}_n)}} \sqrt{\ln \mathcal{N}(\mathcal{F}, \|\cdot\|_{L_2(\mathbb{P}_n)}, \varepsilon)} d\varepsilon \\ &\le \frac{C}{\sqrt{n}} \|F\|_{L_2(P)} \int_{0}^{2} \sup_{Q} \sqrt{\ln \mathcal{N}(\mathcal{F}, \|\cdot\|_{L_2(Q)}, \varepsilon \|F\|_{L_2(Q)})} d\varepsilon, \end{aligned} \quad (20) $$
where the supremum is taken over distributions $Q$. We proceed by using that $F$ is an envelope of $\mathcal{F}$ to bound (proof in Lemma A.6)

$$ \|f - g\|_{L_2(Q)} \le 2 \|F\|_{L_2(Q)} \|\mathbf{1}_{\mathrm{sg}(f)} - \mathbf{1}_{\mathrm{sg}(g)}\|_{L_2(Q')} $$

It then follows that

$$ \mathcal{N}(\mathcal{F}, \|\cdot\|_{L_2(Q)}, \varepsilon \|F\|_{L_2(Q)}) \le \mathcal{N}(\{\mathbf{1}_{\mathrm{sg}(f)} : f \in \mathcal{F}\}, \|\cdot\|_{L_2(Q')}, \varepsilon/2) \le \left(\frac{4}{\varepsilon}\right)^{C'd}, $$

where the last inequality follows by [Vershynin, 2018, Theorem 8.3.18] and $\mathrm{sg}(f) := \{(x, t) \in \mathcal{X} \times \mathbb{R} : f(x) > t\}$. Plugging this inequality into (20) concludes the proof. $\square$

**Lemma A.6.** Let $\mathcal{F}$ be a family of functions from $\mathcal{X}$ to $\mathbb{R}$ and $F$ be its envelope. Let $Q$ be any distribution over $\mathcal{X}$. Then, for any $f, g \in \mathcal{F}$,

$$ \|f - g\|_{L_2(Q)} \le 2 \|F\|_{L_2(Q)} \|\mathbf{1}_{\mathrm{sg}(f)} - \mathbf{1}_{\mathrm{sg}(g)}\|_{L_2(Q')}, $$

where $Q'$ is a distribution over $\mathcal{X} \times \mathbb{R}$.

*Proof.* We follow the proof presented in Zhou [2020].

$$ \begin{aligned} \int |f - g|^2 dQ &\le \int 2F(x) |f(x) - g(x)| dQ(x) = \int 2F(x) \left( \int |\mathbf{1}_{\mathrm{sg}(f)}(x, t) - \mathbf{1}_{\mathrm{sg}(g)}(x, t)| dt \right) dQ(x) \\ &= 2 \iint_{|t| \le F(x)} F(x) |\mathbf{1}_{\mathrm{sg}(f)}(x, t) - \mathbf{1}_{\mathrm{sg}(g)}(x, t)| dt dQ(x) \\ &= 2 \iint_{|t| \le F(x)} F(x) dQ(x) dt \\ &\qquad \times \iint |\mathbf{1}_{\mathrm{sg}(f)}(x, t) - \mathbf{1}_{\mathrm{sg}(g)}(x, t)|^2 \frac{F(x) \mathbf{1}_{|t| \le F(x)}}{\iint_{|t| \le F(x)} F(x) dQ(x) dt} dt dQ(x) \\ &= 4 \|F\|_{L_2(Q)}^2 \|\mathbf{1}_{\mathrm{sg}(f)} - \mathbf{1}_{\mathrm{sg}(g)}\|_{L_2(Q')}^2, \end{aligned} $$

where $Q'$ is absolutely continuous with respect to $Q \times \text{Lebesgue}$ and its density is given by

$$ \frac{F(x) \mathbf{1}_{|t| \le F(x)}}{\iint_{|t| \le F(x)} F(x) dQ(x) dt} \square $$

## B Technical proofs

**Lemma B.1.** Assume $X_{1:n} \sim P^{\otimes n}$ with $P$ atom-free. Fix $\kappa \in \mathbb{R}$. Then the following inequality holds almost surely for all $X_{1:n} \in \mathcal{A}(X_{1:n}, \varepsilon)$:

$$ n - \eta - 1 - \lfloor n\varepsilon \rfloor \le \sum_{i=1}^{n} w(\hat{\alpha}(\kappa; X_{1:n}^{\varepsilon}) |X_i^{\varepsilon} - \kappa|) \le n - \eta. $$

*Proof.* For the purpose of this proof, denote $\hat{\alpha} = \hat{\alpha}(\kappa; X_{1:n}^{\varepsilon})$ Define

$$ S_{X_{1:n}^{\varepsilon}}(\alpha) := \sum_{i=1}^{n} w(\alpha | X_i^{\varepsilon} - \kappa|). $$

The upper bound follows directly by the definition of $\hat{\alpha}$ and right-continuity of $w$:

$$ S_{X_{1:n}^{\varepsilon}}(\hat{\alpha}) = \lim_{\alpha \downarrow \hat{\alpha}} S_{X_{1:n}^{\varepsilon}}(\alpha) \le n - \eta $$

We proceed by proving the lower bound. Assume that $S_{X_{1:n}^{\varepsilon}}(\hat{\alpha}) < n - \eta - 1 - \lfloor n\varepsilon \rfloor$, with positive probability. By definition of $\hat{\alpha}$, $\lim_{\alpha \uparrow \hat{\alpha}} S_{X_{1:n}^{\varepsilon}}(\alpha) \ge n - \eta$. Thus,

$$ \lim_{\alpha \uparrow \hat{\alpha}(\kappa)} S_{X_{1:n}^{\varepsilon}}(\alpha) - S_{X_{1:n}^{\varepsilon}}(\hat{\alpha}) > 1 + \lfloor n\varepsilon \rfloor $$
and $\hat{\alpha}$ is a discontinuity of $\alpha \mapsto w(\alpha |X_i^\varepsilon - \kappa|)$ for at least $2 + \lfloor n\varepsilon \rfloor$ different indices. Since $X_{1:n}^\varepsilon \neq X_{1:n}$ holds for at most $\lfloor n\varepsilon \rfloor$ indices, then there are at least two different indices $i, j$ such that $\hat{\alpha}$ is a discontinuity of $\alpha \mapsto w(\alpha |X_l - \kappa|)$ for $l \in \{i, j\}$. Let $\mathcal{D}$ be the set of discontinuities of $w$, which we know to be countable since $w$ is non-increasing. We will now show that, because $P$ has no point mass, the event

$$D_{k,l} := \{(\hat{\alpha} |X_k - \kappa|, \hat{\alpha} |X_l - \kappa|) \in \mathcal{D} \times \mathcal{D}\}$$

has probability zero, where $k \neq l$. Denote $\mathcal{D}/\mathcal{D} := \{\frac{t_1}{t_2} : (t_1, t_2) \in \mathcal{D} \times \mathcal{D}\}$, which is well defined since $0 \notin \mathcal{D}$ ($w$ is defined on $[0, \infty)$ and right-continuous) and is countable. Notice that

$$D_{k,l} \subseteq \left\{ \frac{|X_k - \kappa|}{|X_l - \kappa|} \in \mathcal{D}/\mathcal{D} \text{ and } |X_l - \kappa| > 0 \right\}.$$

The proof is complete once we observe that $\frac{|X_k - \kappa|}{|X_l - \kappa|}$ has no point mass because $X_k, X_l$ are independent, $|X_s - \kappa|$ has no point mass for $s \in \{k, l\}$, and is almost surely positive. We finish the proof with an union bound argument on the pairs $(k, l)$, showing that $\mathbb{P}(\cup_{k \neq l} D_{k,l}) = 0$, which implies that $\hat{\alpha}$ is a discontinuity of $\alpha \mapsto w(\alpha |X_i - \kappa|)$ for at most one index almost surely. $\square$

**Lemma B.2.** For every $\kappa \in I_{\hat{\kappa}}$, the family

$$\mathcal{F}_{\kappa} = \{x \mapsto \kappa + (x - \kappa)w(\alpha |x - \kappa|) - \mu : \alpha > 0\}$$

has VC-subgraph dimension 1.

*Proof.* Without loss of generality, we may show that the family

$$\mathcal{F}' = \{x \mapsto xw(\alpha |x|) : \alpha > 0\}$$

has VC-subgraph dimension 1. Assume by contradiction that $((x_1, t_1), (x_2, t_2))$ is shattered by

$$\{(x, t) \in \mathbb{R}^2 : g(x) > t\} : g \in \mathcal{F}'\}.$$

Notice that $x \in \{x_1, x_2\}$ is not equal to zero, since that would imply that, for every $g \in \mathcal{F}', g(x) = 0$. This leads to two cases

* $x_1x_2 < 0$: In this case, assume that $x_1 < 0$ and $x_2 > 0$. Since $((x_1, t_1), (x_2, t_2))$ is shattered, there exist $\alpha_1, \alpha_2 > 0$ such that

$$\begin{cases} x_1w(\alpha_1|x_1|) > t_1 & \text{and } x_1w(\alpha_2|x_1|) \le t_1 \\ x_2w(\alpha_1|x_2|) > t_2 & \text{and } x_2w(\alpha_2|x_2|) \le t_2 \end{cases} \Rightarrow \alpha_1 > \alpha_2, \\ \Rightarrow \alpha_1 < \alpha_2$$

which is a contradiction. The implications above are a result of $w$ being non-increasing and $|x_1|, |x_2| > 0$.

* $x_1x_2 > 0$: Analogously to the first case, there exist $\alpha_1, \alpha_2 > 0$ such that

$$\begin{cases} x_1w(\alpha_1|x_1|) > t_1 & \text{and } x_1w(\alpha_2|x_1|) \le t_1 \\ x_2w(\alpha_1|x_2|) \le t_2 & \text{and } x_2w(\alpha_2|x_2|) > t_2 \end{cases}.$$

This implies a contradiction both in the case $x_1, x_2 > 0$ and $x_1, x_2 < 0$. $\square$

*Proof of Theorem 3.3.* By hypothesis (15), we have that with probability at least $1 - \frac{3\delta}{4}$,

$$\begin{align} |\hat{\mu}(\kappa) - \mu| &\le \sup_{\alpha \in I_{\alpha}(\kappa)} |b(\alpha, \kappa)| + \sup_{\alpha \in I_{\alpha}(\kappa)} |s(\alpha, \kappa) - \mu - b(\alpha, \kappa)| \nonumber \\ &= \sup_{\alpha \in I_{\alpha}(\kappa)} |b(\alpha, \kappa)| + \sup_{f \in \mathcal{F}_{\kappa}} \left| (\hat{P}_n - P) f \right|. \tag{21} \end{align}$$

We can use Lemma A.3 with $\mathcal{G} = \mathcal{F}_{\kappa}$ to bound the second term of (21). We proceed by checking the assumptions of Lemma A.3. First, since $\alpha \mapsto m(\alpha)$ is non-increasing,

$$\begin{align*} \sup_{f \in \mathcal{F}_{\kappa}} \|f - Pf\|_{\infty} &= \sup_{\alpha \in I_{\alpha}(\kappa)} \sup_{x \in \mathbb{R}} |(x - \kappa)w(\alpha |x - \kappa|) - \mathbb{E}(X - \kappa)w(\alpha |X - \kappa|)| \\ &\le \sup_{\alpha \in I_{\alpha}(\kappa)} 2m(\alpha) = 2m(\underline{\alpha}(\kappa)). \end{align*}$$
Define $\mathcal{H} := \{x \mapsto \kappa + (x - \kappa)w(\alpha|x - \kappa)| - \mu : \alpha \in I_\alpha(\kappa) \cap \mathbb{Q}\} \subseteq \mathcal{F}_\kappa$. $\mathcal{H}$ is a countable subset of $\mathcal{F}_\kappa$ and, since $w$ is right-continuous and non-increasing, every $h \in \mathcal{H}$ has a countable set of discontinuities. Combining this with Assumption 1 implies that each function $f \in \mathcal{F}_\kappa$ is the P-a.s. pointwise limit of functions $(h_i)_{i=1}^\infty \in \mathcal{H}$. Since each $f \in \mathcal{F}_\kappa$ is bounded by $m(\underline{\alpha}(\kappa)) + |\kappa - \mu|$, the dominated convergence theorem ensures that $f$ is also the $L_1(P)$ limit of functions $(h_i)_{i=1}^\infty \in \mathcal{H}$. Thus, a straightforward application of Lemma A.3 gives

$$ \mathbb{P} \left[ \sup_{\alpha \in I_\alpha(\kappa)} |s(\alpha, \kappa) - \mu - b(\alpha, \kappa)| > (2K + \sqrt{2}) \|F_\kappa\|_{L_2(P)} \sqrt{\frac{1}{n} \ln \frac{4}{\delta}} + \frac{8m(\underline{\alpha}(\kappa))}{3n} \ln \frac{4}{\delta} \right] \le \frac{\delta}{4}, $$

where the VC-subgraph dimension of $\mathcal{F}_\kappa$ is bounded by Theorem B.2. The result then follows by integrating over $\hat{\kappa}$. $\square$

*Proof of Lemma 3.4.* Let $c, \bar{c} > 0$ such that

$$ \begin{aligned} 1 - c &\ge \sqrt{2c} + \frac{2}{3}, \\ \bar{c} - 2 &\ge \sqrt{2\bar{c}} + 2\xi^{-1} + \frac{13}{3}, \end{aligned} $$

To see that there is such a solution, it is enough to solve for $c, \bar{c}$. Since $\frac{\bar{c} \ln \frac{4}{\delta}}{n} + (1 + 2\xi)\varepsilon < 1$, it holds that $1 - \frac{c \ln \frac{4}{\delta}}{\bar{c}} - \frac{\xi}{2} \varepsilon, 1 - \frac{\bar{c} \ln \frac{4}{\delta}}{n} - (1 + 2\xi)\varepsilon \in (0, 1)$. Then, by Assumption 1, $\underline{\alpha}(\kappa), \bar{\alpha}(\kappa)$ are well defined, because $\alpha \mapsto \mathbb{E}[w(\alpha | \hat{X} - \kappa)]$ is continuous (as $P$ has no point mass) and, by Assumption 2, $\lim_{\alpha \to +\infty} \mathbb{E}[w(\alpha | X_i - \kappa)] = 0$.

Recall the definition given by (7). Using that $w$ is right-continuous, takes values on $[0, 1]$ and that $P$ is atom-free, we can bound (Lemma B.1))

$$ n - \eta - 1 \le \sum_{i=1}^{n} w(\hat{\alpha}(\kappa) | X_i - \kappa|) \le \sup_{X_{1:n} \in \mathcal{A}(X_{1:n}, \varepsilon)} \sum_{i=1}^{n} w(\hat{\alpha}^{\varepsilon}(\kappa) | X_i^{\varepsilon} - \kappa|) \le n - \eta, \text{ a.s.} $$

Since $\alpha \mapsto \frac{1}{n} \sum_{i=1}^n w(\alpha | X_i - \kappa|)$ is a non-increasing function on $\alpha$, in order to show the inequalities $\underline{\alpha}(\kappa) < \inf_{X_{1:n} \in \mathcal{A}(X_{1:n}, \varepsilon)} \hat{\alpha}^{\varepsilon}(\kappa) \le \bar{\alpha}(\kappa) < \bar{\alpha}(\kappa)$, it suffices to obtain

$$ \frac{1}{n} \sum_{i=1}^{n} w(\bar{\alpha}(\kappa) | X_i - \kappa|) < 1 - \frac{\eta+1}{n} < 1 - \frac{\eta}{n} < \inf_{X_{1:n} \in \mathcal{A}(X_{1:n}, \varepsilon)} \frac{1}{n} \sum_{i=1}^{n} w(\underline{\alpha}(\kappa) | X_i^{\varepsilon} - \kappa|) $$

with high probability. We want to bound

$$ \begin{aligned} & \mathbb{P} \left[ \frac{1}{n} \sum_{i=1}^{n} w(\bar{\alpha}(\kappa) | X_i - \kappa|) - \mathbb{E}w(\bar{\alpha}(\kappa) | X_1 - \kappa|) \ge 1 - \frac{\eta+1}{n} - \mathbb{E}w(\bar{\alpha}(\kappa) | X_1 - \kappa|) \right] \\ & = \mathbb{P} \left[ \frac{1}{n} \sum_{i=1}^{n} w(\bar{\alpha}(\kappa) | X_i - \kappa|) - \mathbb{E}w(\bar{\alpha}(\kappa) | X_1 - \kappa|) \ge \frac{\bar{c}-1}{n} \ln \frac{4}{\delta} - \frac{1}{n} + \xi\varepsilon \right]. \end{aligned} $$

We apply Theorem A.1, which implies that this probability is lesser than $\frac{\delta}{4}$ if

$$ \frac{\bar{c}-1}{n} \ln \frac{4}{\delta} - \frac{1}{n} + \xi\varepsilon \ge \sigma \sqrt{\frac{2}{n} \ln \frac{4}{\delta}} + \frac{1}{3n} \ln \frac{4}{\delta}, $$

where $\sigma^2 := \mathbb{E}|w(\bar{\alpha}(\kappa)|X - \kappa|) - \mathbb{E}w(\bar{\alpha}(\kappa)|X - \kappa|)|^2$. We bound

$$ \begin{aligned} \sigma^2 &= \mathbb{E}[w(\bar{\alpha}(\kappa)|X - \kappa|) - \mathbb{E}w(\bar{\alpha}(\kappa)|X - \kappa|)]^2 = \mathbb{E}w(\bar{\alpha}(\kappa)|X - \kappa|^2) - (\mathbb{E}w(\bar{\alpha}(\kappa)|X - \kappa|))^2 \\ &\le 1 - \mathbb{E}[w(\bar{\alpha}(\kappa)|X - \kappa|)] = \frac{\bar{c} \ln \frac{4}{\delta}}{n} + (1 + 2\xi)\varepsilon. \end{aligned} $$

Thus, it suffices to prove that

$$ \frac{\bar{c}-2}{n} \ln \frac{4}{\delta} + \xi\varepsilon \ge \frac{\sqrt{2\bar{c}}}{n} \ln \frac{4}{\delta} + \sqrt{\frac{2(1+2\xi)\varepsilon}{n} \ln \frac{4}{\delta}} + \frac{1}{3n} \ln \frac{4}{\delta}. $$
Let us consider two different scenarios. The first is $\sqrt{\frac{2(1+2\xi)\varepsilon}{n}} \ln \frac{4}{\delta} \le \xi\varepsilon$. In this case, the bound is satisfied whenever

$$ \bar{c} - 2 \ge \sqrt{2\bar{c}} + \frac{1}{3}. $$

The second scenario concerns the inequality $\sqrt{\frac{2(1+2\xi)\varepsilon}{n}} \ln \frac{4}{\delta} > \xi\sqrt{\varepsilon}$. Then, the desired inequality follows if

$$ \bar{c} - 2 \ge \sqrt{2\bar{c}} + 2\xi^{-1} + \frac{13}{3}, $$

which is the stronger condition and holds by the definition of $\bar{c}$. Arguing similarly for the lower bounds on $\tilde{\alpha}$, we would like to bound

$$ \begin{aligned} \mathbb{P}\left[\frac{1}{n}\sum_{i=1}^{n} w(\underline{\alpha}(\kappa)|X_i^{\varepsilon} - \kappa|) \le 1 - \frac{\eta}{n}\right] &\le \mathbb{P}\left[\frac{1}{n}\sum_{i=1}^{n} w(\underline{\alpha}(\kappa)|X_i - \kappa|) \le 1 - \frac{\eta}{n} + \varepsilon\right] \\ &= \mathbb{P}\left[\frac{1}{n}\sum_{i=1}^{n} \left[w(\underline{\alpha}(\kappa)|X_1 - \kappa|) - w(\underline{\alpha}(\kappa)|X_i - \kappa|)\right] \ge \left[w(\underline{\alpha}(\kappa)|X_1 - \kappa|)\right] - 1 + \frac{\eta}{n} - \varepsilon\right] \end{aligned} $$

As before, this is bounded by $\frac{\delta}{4}$ if

$$ \frac{1-c}{n} \ln \frac{4}{\delta} + \frac{6\xi}{7}\varepsilon \ge \sigma \sqrt{\frac{2}{n} \ln \frac{4}{\delta}} + \frac{1}{3n} \ln \frac{4}{\delta}. $$

Here, $\sigma \le \sqrt{\frac{c \ln \frac{4}{\delta}}{n}} + \sqrt{\frac{\xi}{7}}\varepsilon$. In a similar manner as before, we consider two cases. The first is given by $\sqrt{\frac{2\xi\varepsilon}{7n}} \ln \frac{4}{\delta} \le \frac{6\xi}{7}\varepsilon$, while the second is given by $\sqrt{\frac{2\xi}{7n}} \ln \frac{4}{\delta} > \frac{6\xi}{7}\sqrt{\varepsilon}$. In both cases, it suffices to prove

$$ 1 - c \ge \sqrt{2c} + \frac{2}{3}, $$

which again is true by the definition of the constants. $\square$

<figure><img src="image_7.png" /></figure>

**Proof of Lemma 3.5.** We start decomposing the bias as

$$ \begin{aligned} \sup_{\alpha \in I_{\alpha}(\kappa)} |b(\alpha, \kappa)| &= \sup_{\alpha \in I_{\alpha}(\kappa)} |\mathbb{E}[(X - \kappa)w(\alpha|X - \kappa|)] + \kappa - \mu| \\ &\le \sup_{\alpha \in I_{\alpha}(\kappa)} |\mathbb{E}[(X - \mu)w(\alpha|X - \kappa|)]| \end{aligned} \quad (22) $$

$$ + \sup_{\alpha \in I_{\alpha}(\kappa)} |(\kappa - \mu)[1 - \mathbb{E}w(\alpha|X - \kappa|)]|, \quad (23) $$

where (22) estimates the bias introduced by shrinkage and (23) estimates the bias introduced by $\kappa$ weighted by how much $\kappa$ is taken into account in the final estimator. Now we bound each of these terms separately.

To bound (22) we start observing that $\mathbb{E}[(X - \mu)w(\alpha|X - \kappa|)] = -\mathbb{E}[(X - \mu)(1 - w(\alpha|X - \kappa|))]$ and use Hölder's inequality with $q = \frac{p}{p-1} > 1$ to get

$$ \sup_{\alpha \in I_{\alpha}(\kappa)} |\mathbb{E}[(X - \mu)w(\alpha|X - \kappa|)]| \le \nu_p \sup_{\alpha \in I_{\alpha}(\kappa)} \|1 - w(\alpha|X - \kappa|)\|_{L_q(P)}, $$

To bound the remaining norm we use the definition of $\bar{\alpha}(\kappa)$ and that $w$ takes values on $[0, 1]$:

$$ \sup_{\alpha \in I_{\alpha}(\kappa)} \mathbb{E}|1 - w(\alpha|X - \kappa|)|^q \le \sup_{\alpha \in I_{\alpha}(\kappa)} \mathbb{E}[1 - w(\alpha|X - \kappa|)] = \frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1 + 2\xi)\varepsilon $$

Thus,

$$ \sup_{\alpha \in I_{\alpha}(\kappa)} |\mathbb{E}[(X - \mu)w(\alpha|X - \kappa|)]| \le \nu_p \left( \frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1 + 2\xi)\varepsilon \right)^{1 - \frac{1}{p}}. $$

The bound on (23) is now straightforward:

$$ \sup_{\alpha \in I_{\alpha}(\kappa)} |(\kappa - \mu)[1 - \mathbb{E}w(\alpha|X - \kappa|)]| \le |\kappa - \mu| \left( \frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1 + 2\xi)\varepsilon \right). $$
Next, we bound $m(\underline{\alpha}(\kappa))$. To that end, we lower bound $\underline{\alpha}(\kappa)$ since

$$m(\alpha) = \sup_{x \in \mathbb{R}} |xw(\alpha|x)| = \alpha^{-1} \sup_{t \ge 0} tw(t) = c_w \alpha^{-1}. \quad (24)$$

Recall that, by definition, $\mathbb{E}[w(\underline{\alpha}(\kappa)|X-\kappa|)] = 1 - \frac{\epsilon}{n} \ln \frac{4}{\delta} - \frac{\xi}{7}\epsilon$. Then the property $w(t) \ge (1-t^p)_+ \ge 1-t^p$ implies $1 - \frac{\epsilon}{n} \ln \frac{4}{\delta} - \frac{\xi}{7}\epsilon \ge 1 - \underline{\alpha}(\kappa)^p \mathbb{E}[|X-\kappa|^p]$ and so

$$\underline{\alpha}(\kappa) \ge \left( \frac{\epsilon}{n} \ln \frac{4}{\delta} + \frac{\xi}{7} \epsilon \right)^{\frac{1}{p}} \frac{1}{\mathbb{E}[|X-\kappa|^p]^{\frac{1}{p}}} \ge \left( \frac{\epsilon}{n} \ln \frac{4}{\delta} + \frac{\xi}{7} \epsilon \right)^{\frac{1}{p}} \frac{1}{\nu_p + |\kappa - \mu|}.$$

By (24),

$$m(\underline{\alpha}(\kappa)) \le c_w (\nu_p + |\kappa - \mu|) \left( \frac{\epsilon}{n} \ln \frac{4}{\delta} + \frac{\xi}{7} \epsilon \right)^{-\frac{1}{p}}. \quad (25)$$

We may bound $\|F_\kappa\|_{L_2(P)}$ in the following way:

$$\begin{aligned} \|F_\kappa\|_{L_2(P)}^2 &= \mathbb{E} \sup_{\alpha \in I_\alpha(\kappa)} |(X - \kappa)w(\alpha |X - \kappa|) + \kappa - \mu|^2 \\ &= \mathbb{E} \sup_{\alpha \in I_\alpha(\kappa)} |(X - \mu)w(\alpha |X - \kappa|) + (\kappa - \mu)(1 - w(\alpha |X - \kappa|))|^2 \\ &\le 2\mathbb{E} \sup_{\alpha \in I_\alpha(\kappa)} |(X - \mu)w(\alpha |X - \kappa|)|^2 + 2\mathbb{E} \sup_{\alpha \in I_\alpha(\kappa)} |(\kappa - \mu)(1 - w(\alpha |X - \kappa|))|^2 \\ &\le \underbrace{2\mathbb{E} |(X - \mu)w(\underline{\alpha}(\kappa)|X - \kappa|)|^2}_{(I)} + \underbrace{2\mathbb{E} |(\kappa - \mu)(1 - w(\underline{\alpha}(\kappa)|X - \kappa|))|^2}_{(II)}. \end{aligned}$$

In the case $p \in (1, 2)$, we bound (I) using the centered $p$-th moment of $X$ and the fact that $xw(\alpha|x|$ is uniformly bounded over $x \in \mathbb{R}$:

$$2\mathbb{E}|X - \mu|^p |(X - \mu)w(\underline{\alpha}(\kappa)|X - \kappa|)|^{2-p} \le 2\nu_p^p (|\kappa - \mu|^{2-p} + m(\underline{\alpha}(\kappa))^{2-p}).$$

When $p \ge 2$, we bound it directly by $2\nu_2^2$. (II) is then bounded by means of the implicit definition of $\bar{\alpha}(\kappa)$:

$$2\mathbb{E} |(\kappa - \mu)(1 - w(\bar{\alpha}(\kappa)|X - \kappa|))|^2 \le 2|\kappa - \mu|^2 \mathbb{E}|1 - w(\bar{\alpha}(\kappa)|X - \kappa|)| = 2|\kappa - \mu|^2 \left( \frac{\bar{c} \ln \frac{4}{\delta}}{n} + (1 + 2\xi)\epsilon \right). \quad \square$$

*Proof of Theorem 3.1.* For this proof, denote $R := R_{\bar{\kappa}}(\frac{\delta}{4})$. Then, we may take $I_{\bar{\kappa}} = [\mu - R, \mu + R]$. We proceed by applying Theorem 3.3 and Lemma 3.5 with $I_\alpha(\kappa)$ as constructed in Lemma 3.4. First notice that Lemma 3.5 bounds $\sup_{\kappa \in I_{\bar{\kappa}}} \sup_{\alpha \in I_\alpha(\kappa)} |b(\alpha, \kappa)|$ by

$$\inf_{1<q \le p} \nu_q \left( \frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1+2\xi)\epsilon \right)^{1-\frac{1}{q}} + R \left( \frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1+2\xi)\epsilon \right) \le \inf_{1<q \le p} (\nu_q + R) \left( \frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1+2\xi)\epsilon \right)^{1-\frac{1}{q}},$$

where we used that $\frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1+2\xi)\epsilon < 1$ and that the infimum is over $q > 1$ for the second inequality. Next, we apply Theorem 3.3, the bound on $m(\underline{\alpha}(\kappa))$ from Lemma 3.5 and the fact that $\bar{c} > 1$ to conclude, with probability at least $1 - \delta$, that $|\hat{\mu} - \mu|$ is at most

$$\inf_{q \in (1, p]} \left( \frac{8\bar{c}^{-\frac{1}{q}}c_w}{3} + 1 \right) (\nu_q + R) \left( \frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1+2\xi)\epsilon \right)^{1-\frac{1}{q}} + \sup_{\kappa \in I_{\bar{\kappa}}} \left( 2K + \sqrt{2} \right) \|F_\kappa\|_{L_2(P)} \sqrt{\frac{1}{n} \ln \frac{4}{\delta}}. \quad (26)$$

Now we bound

$$\sup_{\kappa \in I_{\bar{\kappa}}} \|F_\kappa\|_{L_2(P)} \le \inf_{1<q \le 2 \wedge p} \sqrt{2\nu_q^{\frac{1}{2}}} \left( \sup_{\kappa \in I_{\bar{\kappa}}} m(\underline{\alpha}(\kappa))^{1-\frac{1}{q}} + R^{1-\frac{1}{q}} \right) + \sqrt{2R} \sqrt{\frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1+2\xi)\epsilon}. \quad (27)$$

We want to bound $\sup_{\kappa \in I_{\bar{\kappa}}} \|F_\kappa\|_{L_2(P)} \sqrt{\frac{1}{n} \ln \frac{4}{\delta}}$. The product of $\sqrt{\frac{1}{n} \ln \frac{4}{\delta}}$ with the second summand in (27) is bounded by

$$\inf_{1<q \le p} \sqrt{2R} \left( \frac{\bar{c}}{n} \ln \frac{4}{\delta} + (1+2\xi)\epsilon \right)^{1-\frac{1}{q}}.$$
For the first summand in (27), we use (25) to get

$$
\begin{align*}
& \inf_{1 < q \le 2 \wedge p} \nu_q^{\frac{q}{2}} \left( \sup_{\kappa \in I_{\hat{\kappa}}} m(\underline{\alpha}(\kappa))^{1-\frac{q}{2}} + R^{1-\frac{q}{2}} \right) \sqrt{\frac{2}{n} \ln \frac{4}{\delta}} \\
& \le \inf_{1 < q \le 2 \wedge p} \nu_q^{\frac{q}{2}} \left( c_w^{1-\frac{q}{2}} (\nu_q + R)^{1-\frac{q}{2}} \left( \frac{c}{n} \ln \frac{4}{\delta} + \frac{\xi}{7} \varepsilon \right)^{\frac{1}{2}-\frac{1}{q}} + R^{1-\frac{q}{2}} \right) \sqrt{\frac{2}{n} \ln \frac{4}{\delta}} \\
& \le \inf_{1 < q \le 2 \wedge p} \sqrt{\frac{2}{c}} ((c_w \lor 1) + 1) (\nu_q + \nu_q^{\frac{q}{2}} R^{1-\frac{q}{2}}) \left( \frac{c}{n} \ln \frac{4}{\delta} \right)^{1-\frac{1}{q}},
\end{align*}
$$

where we used $\frac{c}{n} \ln \frac{4}{\delta} + \frac{\xi}{7} \varepsilon \le 1$ and that $\frac{1}{2} - \frac{1}{q} \le 0$ in the last inequality. We conclude that, with probability at least $1 - \delta$, for all $X_{1:n}^{\varepsilon}$, $|\hat{\mu} - \mu|$ is at most

$$
C'_{w,\xi} \left\{ \inf_{1<q\le 2\wedge p} (\nu_q + \nu_q^{\frac{q}{2}} R^{1-\frac{q}{2}}) \left(\frac{1}{n}\ln\frac{4}{\delta}\right)^{1-\frac{1}{q}} + \inf_{1<q\le p} (\nu_q + R) \left(\frac{1}{n}\ln\frac{4}{\delta} + \varepsilon\right)^{1-\frac{1}{q}} \right\}
$$

for some $C'_{w,\xi} > 0$ depending only on $w$ and $\xi$.

The conclusion (11) is then proved by using the bound on the error

$$
\sup_{X_{1:n}^{\varepsilon} \in \mathcal{A}(X_{1:n}, \varepsilon)} |\hat{\mu}^{\varepsilon} - \hat{\mu}| \le \inf_{1 < q \le p} \left( 6\varepsilon + \frac{8}{n} \ln \frac{4}{\delta} \right) c_w(\nu_q + R) \left( \frac{c}{n} \ln \frac{4}{\delta} + \frac{\xi}{7} \varepsilon \right)^{-\frac{1}{q}},
$$

given by Lemma 3.6.

□

*Proof of Proposition 3.2.* The proof follows by a simple application of Bernstein’s inequality. Let $F^{-}(x) = \mathbb{P}[X_1 < x]$ and $Q_t^- = \sup\{x \in \mathbb{R}: F^{-}(x) \le t\}$. It holds that $F^{-}(Q_t^-) = t$ for $t \in (0, 1)$.

$$
\mathbb{P}[\hat{\kappa} \ge Q_t^-] \le \mathbb{P}\left[\frac{1}{n} \sum_{i=1}^{n} \mathbf{1}_{X_i \ge Q_t^-} \ge 1 - \gamma\right] = \mathbb{P}\left[\frac{1}{n} \sum_{i=1}^{n} \mathbf{1}_{X_i \ge Q_t^-} - (1-t) \ge -\gamma + t\right]
$$

Take $t$ to be $\gamma + \sqrt{\frac{1}{n} \ln \frac{2}{\delta}}$. Then,

$$
\mathbb{P}[\hat{\kappa} \ge Q_t^-] \le \mathbb{P}\left[\frac{1}{n} \sum_{i=1}^{n} \mathbf{1}_{X_i \ge Q_t^-} - (1-t) \ge \sqrt{\frac{1}{n} \ln \frac{2}{\delta}}\right] \le \frac{\delta}{2},
$$

as long as $\delta \ge 2e^{-7/16n}$. By Markov's inequality, we can bound $Q_t^-$ as follows:

$$
1 - t = \mathbb{P}[X_1 \ge Q_t^-] \le \frac{\nu_p^p}{|Q_t^- - \mu|^p}, \text{ if } Q_t^- \ge \mu
$$

$$
t = \mathbb{P}[X_1 < Q_t^-] \le \frac{\nu_p^p}{|Q_t^- - \mu|^p}, \text{ if } Q_t^- < \mu
$$

implying that $Q_t^- \le \mu + \nu_p \left[ \left( \gamma + \sqrt{\frac{1}{n} \ln \frac{2}{\delta}} \right) \wedge \left( 1 - \gamma - \sqrt{\frac{1}{n} \ln \frac{2}{\delta}} \right) \right]^{-1/p}$. In particular, if $\delta$ is greater than $2 \exp \left\{ -\frac{(1-\gamma)^2 n}{4} \right\}$, then $\hat{\kappa} - \mu \le \nu_p \left( \frac{2}{\gamma \wedge (1-\gamma)} \right)^{\frac{1}{p}}$. We provide a lower-bound similarly: let $F(x) = \mathbb{P}[X_1 \le x]$, $Q_t = \inf\{x \in \mathbb{R}: F(x) \ge t\}$. Then,

$$
\mathbb{P}[\hat{\kappa} \le Q_t] \le \mathbb{P}\left[\frac{1}{n} \sum_{i=1}^{n} \mathbf{1}_{X_i \le Q_t} \ge \frac{n-1}{n}\gamma\right] = \mathbb{P}\left[\frac{1}{n} \sum_{i=1}^{n} \mathbf{1}_{X_i \le Q_t} - t \ge \frac{n-1}{n}\gamma - t\right].
$$

Taking $t = \gamma - 2\sqrt{\frac{1}{n}\ln\frac{2}{\delta}}$ leads to $\mathbb{P}[\hat{\kappa} \le Q_t] \le \frac{\delta}{2}$ as before. Using Markov's inequality again yields the bound
$Q_t \ge \mu - \nu_p \left[ \left(\gamma - 2\sqrt{\frac{1}{n}\ln\frac{2}{\delta}}\right) \wedge \left(1 - \gamma + 2\sqrt{\frac{1}{n}\ln\frac{2}{\delta}}\right) \right]^{-1/p}$. If $\delta \ge \exp\left\{-\frac{\gamma^2 n}{16}\right\}$, $\hat{\kappa} - \mu \ge \nu_p \left(\frac{2}{\gamma \wedge (1-\gamma)}\right)^{\frac{1}{p}}$ holds
with high probability. Combining the lower and upper bounds, when $\delta \ge 2e^{-\frac{(1-\gamma)\wedge\gamma)^2 n}{16}$, with probability at least $1 - \delta$,

$$
|\hat{\kappa} - \mu| \le \nu_p \left( \frac{2}{\gamma \wedge (1-\gamma)} \right)^{\frac{1}{p}} \le C_{\gamma} \nu_p
$$

□
Finally, we show that the normalized shrinkage estimator $\hat{\mu}_W$ defined in (9) behaves as the non-normalized one:

**Proposition B.3.** Fix $\kappa \in \mathbb{R}$. Then $|\hat{\mu}(\kappa; X_{1:n}) - \hat{\mu}_W(\kappa; X_{1:n})| \le m(\hat{\alpha}(\kappa; X_{1:n}))^{\frac{n-S}{n}}$, where $S$ is defined by
$$S := \sum_{i=1}^n w(\hat{\alpha}(\kappa)|X_i - \kappa|).$$

*Proof.* By the definitions of $\hat{\mu}$ and $\hat{\mu}_W$, we have

$$|\hat{\mu}_W(\kappa; X_{1:n}) - \hat{\mu}(\kappa; X_{1:n})| = \left| \frac{1}{S} \sum_{i=1}^{n} X_i w(\hat{\alpha}(\kappa; X_{1:n})|X_i - \kappa|) - \kappa - \frac{1}{n} \sum_{i=1}^{n} (X_i - \kappa) w(\hat{\alpha}(\kappa; X_{1:n})|X_i - \kappa|) \right| \\ \le \left(\frac{1}{S} - \frac{1}{n}\right) \sum_{i=1}^{n} |X_i - \kappa| w(\hat{\alpha}(\kappa; X_{1:n})|X_i - \kappa|) \le m(\hat{\alpha}(\kappa; X_{1:n}))^{\frac{n-S}{n}}. \quad \square$$

It immediately follows that the concentration guarantee provided for $\hat{\mu}^{\epsilon}$ also holds for $\hat{\mu}_W^{\epsilon} = \hat{\mu}_W(\hat{\kappa}; X_{1:n}^{\epsilon})$ under the adversarial contamination scenario (with different constants), since $S \ge n - \eta - 1$ almost surely (Lemma B.1), $\inf_{X_{1:n}^{\epsilon} \in \mathcal{A}(X_{1:n}, \epsilon)} \hat{\alpha}^{\epsilon}(\kappa) \ge \underline{\alpha}(\kappa)$ with high probability (Lemma 3.4), and $m(\underline{\alpha}(\kappa)) \le c_w(\nu_p + |\kappa - \mu|) (\frac{\epsilon}{n} \ln \frac{4}{\delta} + \frac{\xi}{7} \epsilon)^{-\frac{1}{p}}$ (Lemma 3.5).
**Acknowledgments.** The authors thank Roberto I. Oliveira, Guillaume Lecué, and Arnak Dalalyan for their comments on previous versions of the manuscript.

**Funding.** A.C. was supported by a "FAPERJ Nota 10" grant (SEI-260003/004731/2025) from Fundação de Amparo à Pesquisa do Estado do Rio de Janeiro (FAPERJ). L.R. was partially supported by CNPq. P.O. was supported by grant SEI-260003/001545/2022 from FAPERJ.

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