Continual Mean Estimation Under User-Level Privacy

Anand Jerry George;Lekshmi Ramesh;Aditya Vikram Singh;Himanshu Tyagi
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引用次数: 0

Abstract

We consider the problem of continually releasing an estimate of the population mean of a stream of samples that is user-level differentially private (DP). At each time instant, a user contributes a sample, and the users can arrive in arbitrary order. Until now these requirements of continual release and user-level privacy were considered in isolation. But, in practice, both these requirements come together as the users often contribute data repeatedly and multiple queries are made. We provide an algorithm that outputs a mean estimate at every time instant t such that the overall release is user-level $\varepsilon $ -DP and has the following error guarantee: Denoting by $m_{t}$ the maximum number of samples contributed by a user, as long as $\tilde {\Omega }(1/\varepsilon)$ users have $m_{t}/2$ samples each, the error at time t is $\tilde {O}(1/\sqrt {t}+\sqrt {m}_{t}/t\varepsilon)$ . This is a universal error guarantee which is valid for all arrival patterns of the users. Furthermore, it (almost) matches the existing lower bounds for the single-release setting at all time instants when users have contributed equal number of samples.
用户级隐私下的连续平均值估计
我们考虑的问题是,如何持续发布用户级差异保密(DP)样本流的总体均值估计值。在每个时间瞬间,用户都会贡献一个样本,而且用户可以以任意顺序到达。到目前为止,人们一直在孤立地考虑持续发布和用户级隐私的要求。但在实践中,由于用户经常重复提供数据并进行多次查询,这两个要求会同时出现。我们提供了一种算法,它能在每个时间瞬间 t 输出一个平均估计值,从而使整体发布达到用户级 $\varepsilon $ -DP,并具有以下误差保证:用 $m_{t}$ 表示用户贡献的最大样本数,只要 $\tilde {\Omega }(1/\varepsilon)$ 用户每人有 $m_{t}/2$ 样本,时间 t 的误差就是 $\tilde {O}(1/\sqrt {t}+\sqrt {m}_{t}/t\varepsilon)$ 。这是一个通用误差保证,对所有用户的到达模式都有效。此外,当用户贡献的样本数量相等时,它(几乎)与现有的单次发布设置在所有时间时刻的下限相匹配。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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CiteScore
8.20
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