학술논문

Matching Observations to Distributions: Efficient Estimation via Sparsified Hungarian Algorithm
Document Type
Conference
Source
2019 57th Annual Allerton Conference on Communication, Control, and Computing (Allerton) Communication, Control, and Computing (Allerton), 2019 57th Annual Allerton Conference on. :368-375 Sep, 2019
Subject
Aerospace
Bioengineering
Communication, Networking and Broadcast Technologies
Computing and Processing
Power, Energy and Industry Applications
Robotics and Control Systems
Signal Processing and Analysis
Transportation
Maximum likelihood estimation
Bipartite graph
Standards
Runtime
Probability distribution
Linear programming
Language
Abstract
Suppose we are given observations, where each observation is drawn independently from one of k known distributions. The goal is to match each observation to the distribution from which it was drawn. We observe that the maximum likelihood estimator (MLE) for this problem can be computed using weighted bipartite matching, even when n, the number of observations per distribution, exceeds one. This is achieved by instantiating n duplicates of each distribution node. However, in the regime where the number of observations per distribution is much larger than the number of distributions, the Hungarian matching algorithm for computing the weighted bipartite matching requires $O(n^{3})$ time. We introduce a novel randomized matching algorithm that reduces the runtime to $\tilde{O}(n^{2})$ by sparsifying the original graph, returning the exact MLE with high probability. Next, we give statistical justification for using the MLE by bounding the excess risk of the MLE, where the loss is defined as the negative $\log$-likelihood. We test these bounds for the case of isotropic Gaussians with equal covariances and whose means are separated by a distance $\eta$, and find (1) that $\gg \log k$ separation suffices to drive the proportion of mismatches of the MLE to 0, and (2) that the expected fraction of mismatched observations goes to zero at rate $O((\log k)^{2}/\eta^{2})$.