Convergence Rates for Empirical Bayes Two-Action Problems II. Continuous Case

Abstract

For a general discussion of empirical Bayes problems and motivation of the present paper see Section 1 of the previous paper [1]. In that paper we studied the convergence to Bayes optimality and its rate properties for empirical Bayes two-action problems in certain discrete exponential families. This paper continues that investigation for the continuous case. Under appropriate conditions, Theorems 3 and 4 yield convergence rates to Bayes risk of $O(n^{-\beta})$ for $0 < \beta < 1$, for the $(n + 1)\mathrm{st}$ stage risk of the continuous case empirical Bayes procedures of Section 2. These theorems provide, for the continuous case, convergence rate results for the empirical Bayes procedures of the general type considered by Robbins [5] and Samuel [6] for two different parameterizations of a model. The rate results given here in the continuous case involve upper bounds and are weaker than the discrete case results in [1] wherein exact rates are reported. Specifically, in Section 2 we present the two cases to be considered and define the appropriate empirical Bayes procedures for each. Section 3 gives some technical lemmas and Section 4 establishes the asymptotic optimality (the asymptotic Bayes property) of the procedures introduced. The main results on rates, Theorems 3 and 4, are given in Section 5. Section 6 examines in detail two specific examples--the negative exponential and the normal distributions--and gives corollaries to Theorems 3 and 4 which state convergence rates depending on moment properties of the unknown prior distribution of the parameters. Section 7 gives an example with $\beta$ arbitrarily close to 1 in the rate $O(n^{-\beta})$. The model we consider is the following. Let $f_\lambda(x)$ be a family of Lebesgue densities indexed by a parameter $\lambda$ in an interval of the real line. As in [1], we wish to test the hypothesis $H_1: \lambda \leqq c \mathrm{vs}. H_2: \lambda > c$ with the loss function being \begin{align*}L_1(\lambda) &= 0 \quad\text{if}\quad \lambda \leqq c \\ &= b(\lambda - c) \quad\text{if}\quad \lambda > c \\ L_2(\lambda) &= b(c - \lambda) \quad\text{if}\quad \lambda \leqq c \\ &= 0 \quad\text{if}\quad \lambda > c\end{align*} where $L_i(\lambda)$ indicates the loss when action $i$ (deciding in favor of $H_i$) is taken, $i = 1, 2$ and $b$ is a positive constant. Let $\delta(x) = \mathrm{Pr}\{\text{accepting} H_1 \mid X = x\}$ be a randomized decision rule for the above two-action problem. If $G = G(\lambda)$ is a prior distribution on $\lambda$, then the risk of the (randomized) decision procedure $\delta$ under prior distribution $G$ is given as in [1] by, \begin{align*}\tag{1}r(\delta, G) &= \int\int \{L_1(\lambda)f_\lambda(x)\delta(x) + L_2(\lambda)f_\lambda(x)(1 - \delta(x))\} dx dG(\lambda) \\ &= b \int \alpha(x)\delta(x) dx + C_G\end{align*} where $C_G = \int L_2(\lambda) dG(\lambda)$ and \begin{equation*}\tag{2}\alpha(x) = \int \lambda f_\lambda(x) dG(\lambda) - cf(x)\end{equation*} with \begin{equation*}\tag{3}f(x) = \int f_\lambda(x) dG(\lambda).\end{equation*} From (1) it is clear that a Bayes rule (the minimizer of (1) given $G$) is \begin{align*}\tag{4}\delta_G(x) &= 1\quad\text{if} \alpha(x) \leqq 0 \\ &= 0\quad\text{if} \alpha(x) > 0.\end{align*} Hence, the minimal attainable risk knowing $G$ (the Bayes risk) is \begin{equation*}\tag{5}r^\ast(G) = \inf_\delta r(\delta, G) = r(\delta_G, G).\end{equation*}