Authors:
(1) Diptarka Chakraborty, National University of Singapore, Singapore
(2) Sourav Chakraborty, Indian Statistical Institute, Kolkata;
(3) Gunjan Kumar, National University of Singapore, Singapore;
(4) Kuldeep S. Meel, University of Toronto, Toronto. Table of Links Abstract and 1 Introduction 2 Notations and Preliminaries 3 Related Work 4 An Efficient One-Round Adaptive Algorithm and 4.1 High-Level Overview 4.2 Algorithm Description 4.3 Technical Analysis 5 Conclusion 6 Acknowledgements and References A Missing Proofs B An O(log log n)-query fully adaptive algorithm 4 An Efficient One-Round Adaptive Algorithm Our main contribution is a one-round adaptive tester for the equivalence testing problem in the COND model that makes at most O˜(log n) queries. In the remaining part of this paper, we will prove the above theorem. We start with a high-level idea behind our algorithm, and then we provide a formal description of the algorithm with a detailed analysis. 4.1 High-Level Overview The first attempt to design an equivalence tester is to pick a sample, say i, from P and compare the probability mass P(i) and Q(i) in P and Q respectively. If P and Q are far (in total variation distance), then with enough probability, the probability mass of i in both distributions is significantly different. However, the issue is that estimating the probability mass of the element i can be very expensive. To bypass this issue, the idea is to sample a subset S of the domain, hoping the following: • the probability mass of S in P is comparable to that of the mass of S in Q, and • the probability mass of i (in P) is similar to the probability mass of S. Assuming that all the above two statements hold, one can use conditional sampling (conditioned on S ∪i) to compare P(i)/P(S ∪ i) and Q(i)/Q(S ∪ i). Since, we assumed P(S) is similar to Q(S) so with high enough probability P(i)/P(S ∪ i) and Q(i)/Q(S ∪ i) will be different enough. And since we assumed that P(i) is comparable to P(S), one can estimate P(i)/P(S ∪ i) and can upper bound Q(i)/Q(S ∪ i) using only a few samples, which should be sufficient to distinguish P from Q if the the two distributions are far. But the issue is how to take care of the two assumptions. • (Case 2) There is large “tail probability,” because of which concentration is not possible. (The notion of tail probability is formalized in Section 4.3). This paper is available on arxiv under CC BY-NC-SA 4.0 DEED license. Authors: (1) Diptarka Chakraborty, National University of Singapore, Singapore (2) Sourav Chakraborty, Indian Statistical Institute, Kolkata; (3) Gunjan Kumar, National University of Singapore, Singapore; (4) Kuldeep S. Meel, University of Toronto, Toronto. Authors: Authors: (1) Diptarka Chakraborty, National University of Singapore, Singapore (2) Sourav Chakraborty, Indian Statistical Institute, Kolkata; (3) Gunjan Kumar, National University of Singapore, Singapore; (4) Kuldeep S. Meel, University of Toronto, Toronto. Table of Links Abstract and 1 Introduction Abstract and 1 Introduction 2 Notations and Preliminaries 2 Notations and Preliminaries 3 Related Work 3 Related Work 4 An Efficient One-Round Adaptive Algorithm and 4.1 High-Level Overview 4 An Efficient One-Round Adaptive Algorithm and 4.1 High-Level Overview 4.2 Algorithm Description 4.2 Algorithm Description 4.3 Technical Analysis 4.3 Technical Analysis 5 Conclusion 5 Conclusion 6 Acknowledgements and References 6 Acknowledgements and References A Missing Proofs A Missing Proofs B An O(log log n)-query fully adaptive algorithm B An O(log log n)-query fully adaptive algorithm 4 An Efficient One-Round Adaptive Algorithm Our main contribution is a one-round adaptive tester for the equivalence testing problem in the COND model that makes at most O˜(log n) queries. In the remaining part of this paper, we will prove the above theorem. We start with a high-level idea behind our algorithm, and then we provide a formal description of the algorithm with a detailed analysis. 4.1 High-Level Overview The first attempt to design an equivalence tester is to pick a sample, say i, from P and compare the probability mass P(i) and Q(i) in P and Q respectively. If P and Q are far (in total variation distance), then with enough probability, the probability mass of i in both distributions is significantly different. However, the issue is that estimating the probability mass of the element i can be very expensive. To bypass this issue, the idea is to sample a subset S of the domain, hoping the following: • the probability mass of S in P is comparable to that of the mass of S in Q, and • the probability mass of i (in P) is similar to the probability mass of S. Assuming that all the above two statements hold, one can use conditional sampling (conditioned on S ∪i) to compare P(i)/P(S ∪ i) and Q(i)/Q(S ∪ i). Since, we assumed P(S) is similar to Q(S) so with high enough probability P(i)/P(S ∪ i) and Q(i)/Q(S ∪ i) will be different enough. And since we assumed that P(i) is comparable to P(S), one can estimate P(i)/P(S ∪ i) and can upper bound Q(i)/Q(S ∪ i) using only a few samples, which should be sufficient to distinguish P from Q if the the two distributions are far. But the issue is how to take care of the two assumptions. • (Case 2) There is large “tail probability,” because of which concentration is not possible. (The notion of tail probability is formalized in Section 4.3). This paper is available on arxiv under CC BY-NC-SA 4.0 DEED license. This paper is available on arxiv under CC BY-NC-SA 4.0 DEED license. available on arxiv

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EquivTester Samples In Equivalence Testing Algorithm

SAMP Model Vs COND Model In Equivalence Testing

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An Efficient One-Round Adaptive Algorithm In Equivalence Testing

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