Consider the problem of binary hypothesis testing. Given $Z$ coming from either $\mathbb P^{\otimes m}$ or $\mathbb Q^{\otimes m}$, to decide between the two with small probability of error it is sufficient, and in many cases necessary, to have $m\asymp1/\epsilon^2$, where $\epsilon$ measures the separation between $\mathbb P$ and $\mathbb Q$ in total variation ($\mathsf{TV}$). Achieving this, however, requires complete knowledge of the distributions and can be done, for example, using the Neyman-Pearson test. In this paper we consider a variation of the problem which we call likelihood-free hypothesis testing, where access to $\mathbb P$ and $\mathbb Q$ is given through $n$ i.i.d. observations from each. In the case when $\mathbb P$ and $\mathbb Q$ are assumed to belong to a non-parametric family, we demonstrate the existence of a fundamental trade-off between $n$ and $m$ given by $nm\asymp n_\sf{GoF}^2(\epsilon)$, where $n_\sf{GoF}(\epsilon)$ is the minimax sample complexity of testing between the hypotheses $H_0:\, \mathbb P=\mathbb Q$ vs $H_1:\, \mathsf{TV}(\mathbb P,\mathbb Q)\geq\epsilon$. We show this for three families of distributions, in addition to the family of all discrete distributions for which we obtain a more complicated trade-off exhibiting an additional phase-transition. Our results demonstrate the possibility of testing without fully estimating $\mathbb P$ and $\mathbb Q$, provided $m \gg 1/\epsilon^2$.

翻译：考虑二元假设检验问题。给定来自$\mathbb P^{\otimes m}$或$\mathbb Q^{\otimes m}$的观测$Z$，要在两假设之间以较小的错误概率进行判决，充分且在许多情况下必要条件是$m\asymp1/\epsilon^2$，其中$\epsilon$衡量$\mathbb P$与$\mathbb Q$在总变差距离（$\mathsf{TV}$）上的分离程度。然而，实现这一目标需要完全了解分布，例如可通过Neyman-Pearson检验达成。本文研究该问题的一个变体，称为不依赖似然函数的假设检验，其中通过来自$\mathbb P$和$\mathbb Q$的$n$个独立同分布观测来访问二者。当$\mathbb P$和$\mathbb Q$假定属于非参数族时，我们证明存在$n$与$m$之间的基本权衡关系，由$nm\asymp n_\sf{GoF}^2(\epsilon)$给出，其中$n_\sf{GoF}(\epsilon)$为检验假设$H_0:\,\mathbb P=\mathbb Q$与$H_1:\,\mathsf{TV}(\mathbb P,\mathbb Q)\geq\epsilon$的极小极大样本复杂度。我们针对三种分布族展示了这一结果，此外对于所有离散分布族，我们得到了一个更复杂的权衡关系，该关系呈现出额外的相变现象。我们的结果表明，在无需完全估计$\mathbb P$和$\mathbb Q$的情况下即可进行检验，前提是满足$m \gg 1/\epsilon^2$。