Learning Sums of Independent Integer Random Variables

Abstract

Let S = X[subscript 1]+···+X[subscript n] be a sum of n independent integer random variables X[subscript i], where each X[subscript i] is supported on {0, 1, ..., k - 1} but otherwise may have an arbitrary distribution (in particular the Xi's need not be identically distributed). How many samples are required to learn the distribution S to high accuracy? In this paper we show that the answer is completely independent of n, and moreover we give a computationally efficient algorithm which achieves this low sample complexity. More precisely, our algorithm learns any such S to ε-accuracy (with respect to the total variation distance between distributions) using poly(k, 1/ε) samples, independent of n. Its running time is poly(k, 1/ε) in the standard word RAM model. Thus we give a broad generalization of the main result of [DDS12b] which gave a similar learning result for the special case k = 2 (when the distribution S is a Poisson Binomial Distribution). Prior to this work, no nontrivial results were known for learning these distributions even in the case k = 3. A key difficulty is that, in contrast to the case of k = 2, sums of independent {0, 1, 2}-valued random variables may behave very differently from (discretized) normal distributions, and in fact may be rather complicated - they are not log-concave, they can be Θ(n)-modal, there is no relationship between Kolmogorov distance and total variation distance for the class, etc. Nevertheless, the heart of our learning result is a new limit theorem which characterizes what the sum of an arbitrary number of arbitrary independent {0, 1, ... , k-1}-valued random variables may look like. Previous limit theorems in this setting made strong assumptions on the “shift invariance” of the random variables Xi in order to force a discretized normal limit. We believe that our new limit theorem, as the first result for truly arbitrary sums of independent {0, 1, ... - k-1}-valued random variables, is of independent interest.