We consider the federated frequency estimation problem, where each user holds a private item $X_i$ from a size-$d$ domain and a server aims to estimate the empirical frequency (i.e., histogram) of $n$ items with $n \ll d$. Without any security and privacy considerations, each user can communicate its item to the server by using $\log d$ bits. A naive application of secure aggregation protocols would, however, require $d\log n$ bits per user. Can we reduce the communication needed for secure aggregation, and does security come with a fundamental cost in communication? In this paper, we develop an information-theoretic model for secure aggregation that allows us to characterize the fundamental cost of security and privacy in terms of communication. We show that with security (and without privacy) $\Omega\left( n \log d \right)$ bits per user are necessary and sufficient to allow the server to compute the frequency distribution. This is significantly smaller than the $d\log n$ bits per user needed by the naive scheme, but significantly higher than the $\log d$ bits per user needed without security. To achieve differential privacy, we construct a linear scheme based on a noisy sketch which locally perturbs the data and does not require a trusted server (a.k.a. distributed differential privacy). We analyze this scheme under $\ell_2$ and $\ell_\infty$ loss. By using our information-theoretic framework, we show that the scheme achieves the optimal accuracy-privacy trade-off with optimal communication cost, while matching the performance in the centralized case where data is stored in the central server.
翻译:我们考虑的频率估算问题, 即每个用户持有一个以美元大小为美元域的私人项目 $X_ 美元, 而一个服务器的目的是用美元 美元 美元 。 我们考虑的是, 每个用户都可以使用 $\ log d 位元向服务器传送它的项目。 然而, 安全聚合协议的简单应用需要每个用户$d\ log n 位元。 我们能否减少安全聚合所需的通信, 安全是否带来通信的基本成本? 在本文中, 我们开发了一个安全集合的经验频率( 直方图), 从而让我们能够描述安全和隐私的基本成本 。 我们显示, 在安全( 没有隐私) $mega\ left (n log d\ right) 的情况下, 每个用户都需要使用 $\ log n commission 协议, 并且让服务器能够计算频率分布。 这大大小于 $\ 美元 compax 每个用户所需的 美元 。 但是, 我们开发了一个 $\ rental commal commal commal commissueal commission roup roup roup roup roup roup 。