Abstract:Directed acyclic graphs are used to model the causal structure of a system. ``Causal discovery'' describes the problem of learning this structure from data. When data is an aggregate from multiple sources (populations or environments), global confounding obscures conditional independence properties that drive many causal discovery algorithms. For this reason, existing causal discovery algorithms are not suitable for the multiple-source setting. We demonstrate that, if the confounding is of bounded cardinality (i.e. the data comes from a limited number of sources), causal discovery can still be achieved. The feasibility of this problem is governed by a trade-off between the cardinality of the global confounder, the cardinalities of the observed variables, and the sparsity of the causal structure.
Abstract:We consider the problem of identifying, from its first $m$ noisy moments, a probability distribution on $[0,1]$ of support $k<\infty$. This is equivalent to the problem of learning a distribution on $m$ observable binary random variables $X_1,X_2,\dots,X_m$ that are iid conditional on a hidden random variable $U$ taking values in $\{1,2,\dots,k\}$. Our focus is on accomplishing this with $m=2k$, which is the minimum $m$ for which verifying that the source is a $k$-mixture is possible (even with exact statistics). This problem, so simply stated, is quite useful: e.g., by a known reduction, any algorithm for it lifts to an algorithm for learning pure topic models. In past work on this and also the more general mixture-of-products problem ($X_i$ independent conditional on $U$, but not necessarily iid), a barrier at $m^{O(k^2)}$ on the sample complexity and/or runtime of the algorithm was reached. We improve this substantially. We show it suffices to use a sample of size $\exp(k\log k)$ (with $m=2k$). It is known that the sample complexity of any solution to the identification problem must be $\exp(\Omega(k))$. Stated in terms of the moment problem, it suffices to know the moments to additive accuracy $\exp(-k\log k)$. Our run-time for the moment problem is only $O(k^{2+o(1)})$ arithmetic operations.