Faster Rates for Private Adversarial Bandits

• URL: http://arxiv.org/abs/2505.21790v1
• Date: Tue, 27 May 2025 21:42:10 GMT
• Title: Faster Rates for Private Adversarial Bandits
• Authors: Hilal Asi, Vinod Raman, Kunal Talwar,
• Abstract summary: We design new differentially private algorithms for the problems of adversarial bandits and bandits with expert advice.<n>For adversarial bandits, we give a simple and efficient conversion of any non-private bandit algorithm to a private bandit algorithm.<n>For bandits with expert advice, we give the first differentially private algorithms, with expected regret $Oleft(fracsqrtNTsqrtepsilonright), Oleft(fracsqrtKTlog(N)log(KT)epsilonright)$, and $tildeOleft(
• Score: 31.74205580746007
• License: http://creativecommons.org/licenses/by/4.0/
• Abstract: We design new differentially private algorithms for the problems of adversarial bandits and bandits with expert advice. For adversarial bandits, we give a simple and efficient conversion of any non-private bandit algorithm to a private bandit algorithm. Instantiating our conversion with existing non-private bandit algorithms gives a regret upper bound of $O\left(\frac{\sqrt{KT}}{\sqrt{\epsilon}}\right)$, improving upon the existing upper bound $O\left(\frac{\sqrt{KT \log(KT)}}{\epsilon}\right)$ for all $\epsilon \leq 1$. In particular, our algorithms allow for sublinear expected regret even when $\epsilon \leq \frac{1}{\sqrt{T}}$, establishing the first known separation between central and local differential privacy for this problem. For bandits with expert advice, we give the first differentially private algorithms, with expected regret $O\left(\frac{\sqrt{NT}}{\sqrt{\epsilon}}\right), O\left(\frac{\sqrt{KT\log(N)}\log(KT)}{\epsilon}\right)$, and $\tilde{O}\left(\frac{N^{1/6}K^{1/2}T^{2/3}\log(NT)}{\epsilon ^{1/3}} + \frac{N^{1/2}\log(NT)}{\epsilon}\right)$, where $K$ and $N$ are the number of actions and experts respectively. These rates allow us to get sublinear regret for different combinations of small and large $K, N$ and $\epsilon.$

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