

BIC vs. LRT and singularity 

Message/Author 

Gareth posted on Wednesday, August 26, 2009  2:30 am



I am using both BIC and LRT to help me decide the number of latent classes in a GMM, following the Nylund paper. Males #, BIC, LMRLRT 2 63157.780 0.2544 3 62329.038 0.2567 4 61835.977 0.6700 5 61649.647 0.0112 6 61488.656 0.6649 7 61460.095 0.0002 8 61476.898 0.2398 Females #, BIC, LMRLRT 2 60697.091 0.0000 3 60413.812 0.0000 4 59599.679 0.1862 5 59518.102 0.0188 6 59237.200 0.0000 7 (singularity) According to the BIC information, this suggests a sixclass solution for males. LRT seems to suggest either 5 or 6 classes, because the p value dips twice. However, in females I see no dip in the BIC values. LRT suggests a 5class solution, but I encounter problems at 7 classes (e.g. singularity) so I have not been able to check LRT yet here. I have two questions: 1. Are the low p values for 2 and 3 classes in females meaningful i.e. might they be suggesting 3 classes, or is there some other reason why they are very low? 2. If there are empty cells resulting from binary variables which cause singularity at 7 classes, is there a way to find out which binary variables are causing the problem? When I look for the parameters which Mplus has fixed, they do not appear in the parameter specification. I should also say that all of the solutions so far are meaningful and interpretable. 


For males, the LRMLRT suggests one class. For females it suggests three classes. BIC does not seem to be helpful here. You can use TECH10 to assess model fit by looking at the univeriate, bivariate, and response pattern residuals. 

Gareth posted on Thursday, August 27, 2009  3:25 am



I see, meaning that support is found for k classes when there is a low p value for k1 compared to high p value in the k+1 run. However, I do not understand why 1 class is sufficient for males. Wouldn't 5 or 7 be suggested according to this criterion? 


The pvalue for the two class solution is .25 indicating that a oneclass solution should be chosen. You look for the first instance of a pvalue greater than .05 and select the solution with one less class. 

csulliva posted on Monday, September 21, 2009  9:37 am



In the context of LC modeling, can the BIC value be used to distinguish between model specifications (in addition to different numbers of classes)? For example, in a situation where one is uncertain as to whether models using a censored distribution are more or less appropriate than those with a linear, continuous distribution, could their respective BIC values be used as one information point for assessing their differences? 


The BIC's for continuous and censored are not on the same scale and cannot be compared. Following are the technical reasons which I quote: "For BIC to be comparable the models have to have the same domain (i.e. plausible values) for the dependent variables. This is not the case for continuous v.s. censored. There is a second requirement for BIC to be comparable which is a bit more complicated  all the points with positive mass have to be the same, which is also not the same." 

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