Oral Session: Competitive Distribution Estimation: Why is Good-Turing Good

Опубликовано 22 июня 2016, 19:28

k n (3+o(1))/n^{1/3} \tilde{\mathcal{O}}(\min(k/n,1/\sqrt n)) \ge\tilde\Omega(\min(k/n,1/ n^{2/3})) \ge\tilde\Omega(\min(k/n,1/\sqrt{n})) Estimating distributions over large alphabets is a fundamental machine-learning tenet. Yet no method is known to estimate all distributions well. For example, add-constant estimators are nearly min-max optimal but often perform poorly in practice, and practical estimators such as absolute discounting, Jelinek-Mercer, and Good-Turing are not known to be near optimal for essentially any distribution.We describe the first universally near-optimal probability estimators. For every discrete distribution, they are provably nearly the best in the following two competitive ways. First they estimate every distribution nearly as well as the best estimator designed with prior knowledge of the distribution up to a permutation. Second, they estimate every distribution nearly as well as the best estimator designed with prior knowledge of the exact distribution, but as all natural estimators, restricted to assign the same probability to all symbols appearing the same number of times.Specifically, for distributions over k symbols and n samples, we show that for both comparisons, a simple variant of Good-Turing estimator is always within KL divergence of (3+o(1))/n 1/3 from the best estimator, and that a more involved estimator is within O ~ (min(k/n,1/n √ )) . Conversely, we show that any estimator must have a KL divergence ≥Ω ~ (min(k/n,1/n 2/3 )) over the best estimator for the first comparison, and ≥Ω ~ (min(k/n,1/n √ )) for the second.