I'm having trouble understanding the concept of comparing models of evolution with LRT. I have following models: M0 (neutrality), M1A (negative selection) and M2a (positive selection) with following results from Maximum Likelihood:

M0: w=0.81; lnL=-5702.55 M1a: wo=0.50, po=0.65; w1=1.0, p1=0.35; lnL=-5650.21 M2a: wo=0.46, po=0.64; w1=1.0, p1=0.25; w2=2.38, p2=0.11; lnL=-5632.10.

I want to do the Likelihood Ratio Test to compare the models and make assumptions about the selective pressures on the analyzed sequences. So, first I compare M0 and M1a and calculate LRT=2(lnL1-lnL0)=104.68.

Now I want to take a look at a Chi table, but what am I using there? Am I supposed to used the po value for the M1a model? And if that value exceeds the value of the p-value in the table then the simpler model is chosen?


The answer to this question depends strongly on how you are formulating the different models. Are you for example computing this from dN/dS from a sequence alignment? Or based on explicit phylogenetic models? Those can have quite different formulations. The model names you are using out sound like they might come from PAML, but it's not clear. There are some exercises for this specific case elsewhere.

I am assuming that what you are really asking is how many degrees of freedom to use when e.g. computing a p value.

For this question, you have to think about the LRT a little. What it is really asking is "do I explain the data better by adding parameters to a model (while accounting for the fact that likelihood will by definition increase with more parameters)?"

So what you need to first establish is what the parameters of your models are. Assuming that the models are "nested", you can just use the difference in the number of parameters as the degrees of freedom. "nested" in this context means that the base model (e.g. H0, probably M0 for you) doesn't have parameters that are missing for the alternative model (which is assumed to have more parameters).

More flexible than the LRT are things like the AIC, which gets rid of Chi square and just computes a value from likelihood and model parameter number. This works well in practice and is nice because it's more flexible and a little less convoluted model comparison technique than the LRT. You can additionally use the AIC to compare non-nested models, which doesn't really work for the LRT.

I hope that this helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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