The Confidence Cost of Pairwise Distribution Learning

Pairwise conditional access to an unknown distribution returns a draw from the distribution restricted to any requested pair. At constant confidence, this weak oracle is as sample-efficient for labeled total-variation learning as ordinary sampling, up to constants. We show that this equivalence fails at high confidence. For a distribution on \(n\) labels, accuracy \(\varepsilon\), and failure probability \(\delta\), we prove the exact minimax query complexity \(\Theta(n\log(1/\delta)/\varepsilon^2)\). The corresponding ordinary-sampling complexity is \(\Theta((n+\log(1/\delta))/\varepsilon^2)\). Thus confidence amplification costs up to a factor \(n\) under pairwise access. The upper bound reuses one trajectory of a pairwise-winner Markov chain whose stationary law is the target. Although its worst-case relaxation time is linear in \(n\), we prove that its empirical distribution has the i.i.d.-scale stationary risk \(O(\sqrt{n/T})\). The proof compares its conductances with a minimum-mass graph. After sorting the target probabilities, this graph has an explicit Helmert eigensystem, and a capped-mass Green-function bound telescopes without any minimum-mass or dynamic-range assumption. A geometric median of independent trajectories gives the high-confidence result. The lower bound uses a heavy label and uniform light labels. Every adaptive pair query then carries only \(O(\varepsilon^2/n)\) KL information, which makes the multiplicative confidence cost unavoidable.