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Is it ok to use LRT (likelyhood ratio test) to compare two mixed-models that have the same number of degrees of freedom? They differ just by adding spatial restriction to one of them that is not changing amount of dfs.

Ex 

Model 1- df 1206, logL = -296.78
Model 2 - df 1206, logL = -289.8

I used to do comparison like this

pchisq(-2 * -(-296.78+289.8), 1, lower.tail = F)

But here there is no difference in dfs so I do not know if writing 1 is right.

Mateusz1981
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  • AFAIK you can't use the LRT for non-nested models. I do not know what exactly you mean by a spatial restriction, but if the dfs do not change, the LRT is out of the question. Moreover, which model will be your null-hypothesis model? Most of the times with the LRT, this is the one with least amount of parameters/dfs. So if you do use df=1 and find a significant improvement of model fit, which model is rejected (and remember that this should be clear before seeing the logL)? – IWS Jan 13 '17 at 09:55
  • the models are nested so there are blocks, plots and rows and columns. by spatial restriction I ment I used spatial autocorrelation as such correlation is present in the experiment: model with h0 is the model without considering autocorrelation – Mateusz1981 Jan 13 '17 at 10:07
  • And you are sure this does not use any dfs? If you assumed some sort of correlation structure this often costs dfs to estimate the correlations. I would recommend you to extend your question with additional information on the exact models (structure of the mixed model; correlaiton structure assumed, etc.) – IWS Jan 13 '17 at 10:23
  • I am never sure :) but the random structure is the same. The only difference is that in the second model I set up the autocorrelation error structure. I have not observed changes in the dfs running both. – Mateusz1981 Jan 13 '17 at 10:31

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