TR16-194 Authors: Mohammad Bavarian, Badih Ghazi, Elad Haramaty, Pritish Kamath, Ronald Rivest, Madhu Sudan

Publication: 4th December 2016 01:50

Downloads: 711

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In the "correlated sampling" problem, two players, say Alice and Bob, are given two distributions, say $P$ and $Q$ respectively, over the same universe and access to shared randomness. The two players are required to output two elements, without any interaction, sampled according to their respective distributions, while trying to minimize the probability that their outputs disagree. A well-known protocol due to Holenstein, with close variants (for similar problems) due to Broder, and to Kleinberg and Tardos, solves this task with disagreement probability at most $2 \delta/(1+\delta)$, where $\delta$ is the total variation distance between $P$ and $Q$. This protocol has been used in several different contexts including sketching algorithms, approximation algorithms based on rounding linear programming relaxations, the study of parallel repetition and cryptography.

In this note, we give a surprisingly simple proof that this protocol is in fact tight. Specifically, for every $\delta \in (0,1)$, we show that any correlated sampling scheme should have disagreement probability at least $2\delta/(1+\delta)$. This partially answers a recent question of Rivest.

Our proof is based on studying a new problem we call "constrained agreement". Here, Alice is given a subset $A \subseteq [n]$ and is required to output an element $i \in A$, Bob is given a subset $B \subseteq [n]$ and is required to output an element $j \in B$, and the goal is to minimize the probability that $i \neq j$. We prove tight bounds on this question, which turn out to imply tight bounds for correlated sampling. Though we settle basic questions about the two problems, our formulation also leads to several questions that remain open.