Abstract
Testing the number of components in finite normal mixture models is a long-standing challenge because of its nonregularity. This article studies likelihood-based testing of the number of components in normal mixture regression models with heteroscedastic components. We construct a likelihood-based test of the null hypothesis of m0 components against the alternative hypothesis of m0 + 1 components for any m0. The null asymptotic distribution of the proposed modified EM test statistic is the maximum of m0 random variables that can be easily simulated. The simulations show that the proposed test has very good finite sample size and power properties. Supplementary materials for this article are available online.
Highlights
Finite mixtures of normal distributions and regressions have been used in numerous empirical applications in diverse fields such as biological, physical, and social sciences, including economics and finance
The number of components is an important parameter in applications of finite mixture models
Despite its importance, testing for the number of components in normal mixture regression models has been a long-standing unsolved problem because the standard asymptotic analysis of the likelihood ratio test (LRT) statistic breaks down due to problems such as non-identifiable parameters and the true parameter being on the boundary of the parameter space
Summary
Finite mixtures of normal distributions and regressions have been used in numerous empirical applications in diverse fields such as biological, physical, and social sciences, including economics and finance (see, e.g., Kon, 1984; Tucker, 1992; Venkataraman, 1997; Quandt and Ramsey, 1978; Kon and Jen, 1978; Conway and Deb, 2005). This paper develops a likelihood-based testing procedure of the null hypothesis of m0 components against the alternative hypothesis of m0 + 1 components for a general m0 ≥ 1 in heteroscedastic normal mixture regression models. To this end, we introduce a new reparameterization that substantially simplifies the analysis. Let := denote “equals by definition.” For a k × 1 vector a and a function f (a), let ∇af (a) denote the k × 1 vector of the derivative (∂/∂a)f (a), and let ∇aa f (a) denote the k × k vector of the derivative (∂/∂a∂a )f (a)
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