Assessing the Number of Components Sample Clauses

Assessing the Number of Components. Hypothesis Test- ing The most common hypothesis test for assessing the number of components in a mix- ture model relies on the likelihood ratio test statistic [54]. Consider a hypothesis test of H0 : C = C0 versus HA : C = C1, where C1 > C0 and C0 represents the true-order of a C-component mixture model. Further, let ψˆ0 and ψˆ1 represent the maximum likelihood estimates of ψ under the null and alternative hypothesis, respectively. By definition, the likelihood ratio test statistic is TLR = 2 {logL(ψˆ1) − logL(ψˆ0)} . Unfortunately, for a mixture model, the likelihood ratio test statistic does not follow its traditional chi-square asymptotic distribution under the null hypothesis. When the null hypothesis holds, the parameter vector is on the boundary of the parameter space and in a non-identifiable subspace. As a result, the regularity conditions required for TLR to have a chi-square asymptotic distribution break down. Although some theoretical results exist for determining the distribution of the likelihood ratio test statistic under the null hypothesis for specific mixture models, McLachlan(1987)[52] proposed a more general bootstrapping approach for the likelihood ratio test. A less popular alternative for hypothesis testing proposed by Liang and Xxxxxxx(1999)[45] is the score test. In their paper, Liang and Xxxxxxx consider testing a two-component mixture model against a one-component mixture model. They let y1, . . . , yn be inde- pendent observations with fi∗ denoted as the probability density function for the ith observation of the form fi∗ (yi; α, θ) = αfi (yi; θ) + (1 − α) fi (yi; θ0) . Here, fi is the pdf, θ0 is known to investigators, α ∈ [0, 1], and the hypothesis of interest tests H0 : α = 0. Note that, under the null hypothesis, θ is meaningless and, thus, the standard asymptotic results required for likelihood ratio hypothesis testing are not applicable. Now, the score function for α evaluated at α = 0 is n n S(θ) = S (θ) = i . 0 ∑ ∑ { f (y ; θ) }