Sum-of-squares meets square loss: Fast rates for agnostic tensor completion

30 May 2019  ·  Dylan J. Foster, Andrej Risteski ·

We study tensor completion in the agnostic setting. In the classical tensor completion problem, we receive $n$ entries of an unknown rank-$r$ tensor and wish to exactly complete the remaining entries. In agnostic tensor completion, we make no assumption on the rank of the unknown tensor, but attempt to predict unknown entries as well as the best rank-$r$ tensor. For agnostic learning of third-order tensors with the square loss, we give the first polynomial time algorithm that obtains a "fast" (i.e., $O(1/n)$-type) rate improving over the rate obtained by reduction to matrix completion. Our prediction error rate to compete with the best $d\times{}d\times{}d$ tensor of rank-$r$ is $\tilde{O}(r^{2}d^{3/2}/n)$. We also obtain an exact oracle inequality that trades off estimation and approximation error. Our algorithm is based on the degree-six sum-of-squares relaxation of the tensor nuclear norm. The key feature of our analysis is to show that a certain characterization for the subgradient of the tensor nuclear norm can be encoded in the sum-of-squares proof system. This unlocks the standard toolbox for localization of empirical processes under the square loss, and allows us to establish restricted eigenvalue-type guarantees for various tensor regression models, with tensor completion as a special case. The new analysis of the relaxation complements Barak and Moitra (2016), who gave slow rates for agnostic tensor completion, and Potechin and Steurer (2017), who gave exact recovery guarantees for the noiseless setting. Our techniques are user-friendly, and we anticipate that they will find use elsewhere.

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