  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.
