We study the fundamental problem of estimating the mean of a $d$-dimensional distribution with covariance $\Sigma \preccurlyeq \sigma^2 I_d$ given $n$ samples. When $d = 1$, \cite{catoni} showed an estimator with error $(1+o(1)) \cdot \sigma \sqrt{\frac{2 \log \frac{1}{\delta}}{n}}$, with probability $1 - \delta$, matching the Gaussian error rate. For $d>1$, a natural estimator outputs the center of the minimum enclosing ball of one-dimensional confidence intervals to achieve a $1-\delta$ confidence radius of $\sqrt{\frac{2 d}{d+1}} \cdot \sigma \left(\sqrt{\frac{d}{n}} + \sqrt{\frac{2 \log \frac{1}{\delta}}{n}}\right)$, incurring a $\sqrt{\frac{2d}{d+1}}$-factor loss over the Gaussian rate. When the $\sqrt{\frac{d}{n}}$ term dominates by a $\sqrt{\log \frac{1}{\delta}}$ factor, \cite{lee2022optimal-highdim} showed an improved estimator matching the Gaussian rate. This raises a natural question: Is the $\sqrt{\frac{2 d}{d+1}}$ loss \emph{necessary} when the $\sqrt{\frac{2 \log \frac{1}{\delta}}{n}}$ term dominates? We show that the answer is \emph{no} -- we construct an estimator that improves over the above naive estimator by a constant factor. We also consider robust estimation, where an adversary is allowed to corrupt an $\epsilon$-fraction of samples arbitrarily: in this case, we show that the above strategy of combining one-dimensional estimates and incurring the $\sqrt{\frac{2d}{d+1}}$-factor \emph{is} optimal in the infinite-sample limit.

翻译：我们研究了在给定$n$个样本时估计$d$维分布均值的基本问题，该分布满足协方差$\Sigma \preccurlyeq \sigma^2 I_d$。当$d=1$时，\cite{catoni}提出了一种估计器，其误差为$(1+o(1)) \cdot \sigma \sqrt{\frac{2 \log \frac{1}{\delta}}{n}}$，置信概率为$1-\delta$，与高斯误差率匹配。对于$d>1$，一种自然的估计器通过输出一维置信区间的最小外接球中心，实现了$1-\delta$置信半径$\sqrt{\frac{2 d}{d+1}} \cdot \sigma \left(\sqrt{\frac{d}{n}} + \sqrt{\frac{2 \log \frac{1}{\delta}}{n}}\right)$，相比高斯速率产生了$\sqrt{\frac{2d}{d+1}}$倍的损失。当$\sqrt{\frac{d}{n}}$项主导$\sqrt{\log \frac{1}{\delta}}$因子时，\cite{lee2022optimal-highdim}提出了一种改进的估计器，匹配了高斯速率。这自然引发了一个问题：当$\sqrt{\frac{2 \log \frac{1}{\delta}}{n}}$项主导时，$\sqrt{\frac{2 d}{d+1}}$倍的损失是否\emph{必要}？我们证明答案是否定的——我们构造了一个估计器，在常数因子下改进了上述朴素估计器。我们还考虑了稳健估计，其中攻击者可以任意破坏$\epsilon$比例的样本：在此情况下，我们表明上述结合一维估计并引入$\sqrt{\frac{2d}{d+1}}$倍损失的策略在无限样本极限下\emph{是}最优的。