# AIC versus Likelihood Ratio Test in Model Variable Selection

The software that I am currently using to build a model compares a "current run" model to a "reference model" and reports (where applicable) both a chi-squared p-value based on likelihood ratio tests and AIC values for each model. I know that one advantage of AIC over likelihood ratio tests is that AIC can be compared on non-nested models. However, I am not aware of any reason why AIC couldn't or shouldn't be compared on nested models. In my model, when comparing nested models for variable selection, I'm finding several cases where the likelihood ratio test and the AIC comparison are suggesting opposite conclusions.

Since both are based on likelihood calculations, I'm struggling to interpret these results. However the documentation of my software says (without explaining),

"If two models are nested (i.e. one is a sub-set of the other) then the more usual chi-squared test is the most appropriate to use. If the models are not nested, the AIC can be used..."

Can anyone elaborate on this and/or explain why AIC is not as helpful as likelihood ratio tests on nested models?

• The best answer I can think of is that the likelihood ratio test is an actual test, and tests for the statistical significance of the added variable being tested in the nested model. So when appropriate, one might favor LRT more. Feb 26, 2016 at 20:57
• Seems to be answered her: stats.stackexchange.com/questions/20441/… Feb 26, 2016 at 21:08
• user42719, have I answered your question, or is there something else you need to know? Mar 1, 2016 at 8:30
• No reason for doing model selection vs. specifying a full model and using it was given. And note that AIC is just using a different $\alpha$-level cutoff for stepwise model building, so using AIC does not solve the serious problems with stepwise analysis. May 8, 2016 at 18:16
• @Greenparker It doesn't answer the question because the link is about nested and non-nested. Here, we care only nested. Feb 16, 2017 at 2:36