SVD provably recovers the uncorrupted nearest neighbor from noisy data when σ is O(1/k^{1/4}), with a matching lower bound showing the threshold is necessary.
Hence kX i=1 (uT i y)2 =||X|| 2 = 2σ2 · kX i=1 Xi√ 2σ2 2 = 2σ2Z whereZ∼χ 2 k
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SVD Provably Denoises Nearest Neighbor Data
SVD provably recovers the uncorrupted nearest neighbor from noisy data when σ is O(1/k^{1/4}), with a matching lower bound showing the threshold is necessary.