When I have a two mixed models (lme function) with different df then ANOVA summary shows the p-value of likelihood ratio test as following :

         Model df   AIC      BIC      logLik   Test  L.Ratio  p-value
fsModel3     1 14 -1429.098 -1367.059 728.5489                        
fsModel1     2 15 -1428.438 -1361.968 729.2192 1 vs 2 1.340498  0.2469

However, when I compared two or more models with the same df ( same random effect ) only different fixed effects ( same number of fixed effects ). My ANOVA table look like this :

         Model df    AIC       BIC     logLik
fsModel2     1 14 -1428.469 -1366.431 728.2346
fsModel3     2 14 -1429.098 -1367.059 728.5489
fsModel4     3 14 -1428.886 -1366.847 728.4429

My intuition tell me that then I am selecting the model with the lowest AIC/BIC. Am I right ? Is there a reason why the L.Ratio can't be calculated on models with same df ?

  • 1
    $\begingroup$ You can calculate likelihood ratio on models with same df (non-nested models), and the intuition behind likelihood ratios should still be valid, but the usual mathematics is not: You cannot any more expect a chisquare distribution, not even asymptotically, as the ratio can vary on both sides of one! $\endgroup$ Commented Jan 8, 2019 at 23:07
  • $\begingroup$ Thank you for your comment. Then is there a way how to compare two (non-nested models) and obtain meaningful p-value ( different test ) ? As you probably know, in research papers everybody want to see a p-value ( don't ask me why :) ). For me the AIC/BIC is sufficient, however I am affraid that for my collegues not. $\endgroup$
    – user210804
    Commented Jan 9, 2019 at 8:36

1 Answer 1


Likelihood ratio tests are only appropriate on nested models.

If your models have the same df, they must not be nested: indeed, you state that they have different fixed effects.

You can still compare AIC/BIC, but your models are all sufficiently similar in AIC and BIC that under normal conventions there isn't much meaningful difference between them.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.