I concur with previous posts.  Any likelihood ratio test has df =1 (or more) not df=0

Perhaps you are looking at output from saturated model?  This yields df=0 at last step.

Tryig to answer,

DF in general equals N-1, and in Chi Square it is the number of groups - 1. the lowerst number of gourps is 2 {ohterwise if only one group, then it is not a variable and in this case only the DF =0}. Therefore, No, the DF cant be 0.

The difference of -2LOG (likelihood) between two models is assumed Chi-squared distribution.

The sum of independent squared standard normal , i.e Zi 2 , is chi-squared distribution, the df= number of Zi, hence at lease one should be used. No zero DF Chi-squared random variable.

I concur with previous posts.  Any likelihood ratio test has df =1 (or more) not df=0

Perhaps you are looking at output from saturated model?  This yields df=0 at last step.

Agreed. Two other questions that might help: Are the models nested? Also, are you using restricted ML or full ML?

Thank you for all the helpful answers. The models were nested, differing by an interaction term. and I used full ML. Ah, oversaturated model -- perhaps that was the problem. I've chucked the analysis anyhow. Thanks again

Gina,

For what it's worth going forward, was the interaction term included with a path constrained to zero in the reduced model? If it wasn't included at all, then they were actually not nested.

Pat

Good point Theo, we don't need to burn a degree of freedom if we have parameter value from theory.  In biology the classic case is goodness of  fit test to Mendelian ratio (3:1) in dihybrid cross with a recessive gene (Mendel's  purple and white flowered pea plants).  Returning to the quest posted, I would avoid comparing two linear mixed models  where parameter estimates from the two models together exhaust the number of  cases. 

It is not clear what is your null hypothesis is. Perhaps, you can clarify your question.

Based on the previous comments and your question, I am guessing you are comparing two mixed effect with different covariate structures. If it is indeed the case, the LR test can be tricky since the parameter(s) you are testing is on the boundary and the usual large-sample theory does not apply. 

For example, if you are comparing a fixed-effect (no random effect) vs. a random-intercept model with the same fixed effects, then the LR statistic is approximately distributed as a 50-50 mixture of chi-square 0 and 1 degree of freedom.

If two or more models have a same number of parameters, then simply the model with a higher Log(L) is considered as the best model.

@ Pouya Zamani How large is large??

for me if the difference is less than 10 for the model with the same number of covairte we can take one of it.

How large is large?

It depends on whether the two models are nested or not. Nested means two models that differ in only one term. For example null model (1 parameter, the mean) versus full model (2 parameters, the mean and the slope of a regression). From the inital posting it is hard to tell.

"I am comparing two linear mixed effects models where one model had a max gradient of 0.006 during fitting triggering warnings of conversion (in case that is related). "

If the two models being compared are nested the support (lnLR) is all we need, no penalty for extra parameters, overfitting, etc. For example LRT G = 3.84 on one degree of freedom yields LR = exp(G/2) = 6.8 times more likely with the additional parameter than without. I chose 3.84 as it corresponds to the conventional 5% p-value.

In the absence of more information from Gina Joue, who initiated the thread, I don't think any of us can be of much further help.