Number of components in a mixture

August 5, 2011

(This article was first published on Xi'an's Og » R, and kindly contributed to R-bloggers)

I got a paper (unavailable online) to referee about testing for the order (i.e. the number of components) of a normal mixture. Although this is an easily spelled problem, namely estimate k in

\sum_{i=1}^k p_{ik} \mathcal{N}(\mu_{ik},\sigma^2_{ik}),

I came to the conclusion that it is a kind of ill-posed problem. Without a clear definition of what a component is, i.e. without a well-articulated prior distribution, I remain unconvinced that k can be at all estimated. Indeed, how can we distinguish between a k component mixture and a (k+1) component mixture with an extremely small (in the sense of the component weight) additional component? Solutions ending up with a convenient chi-square test thus sound unrealistic to me… I am not implying the maths are wrong in any way, simply that the meaning of the test and the nature of the null hypothesis are unclear from a practical and methodological perspective. In the case of normal (but also Laplace) mixtures, the difficulty is compounded by the fact that the likelihood function is unbounded, thus wide open to over-fitting (at least in a non-Bayesian setting). Since Ghosh and Sen (1985), authors have come up with various penalisation functions, but I remain openly atheistic about the approach! (I do not know whether or not this is related with the summer season, but I have received an unusual number of papers to referee lately, e.g., handling three papers last Friday, one on Saturday, and yet another one on Monday morning. Interestingly, about half of them are from  non-statistical journals!)

Filed under: Books, R, Statistics, University life Tagged: Bayesian Core, chi-square test, mixture estimation, penalisation, refereeing, testing of hypotheses

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