2
$\begingroup$

I have continuous data with a point mass at zero, so my plan is to use a two-part model where I first model whether an observation is zero or non-zero in a logistic regression and then model the positive values by a Gamma GLM.

How do I compute the AIC for this model?

I'm using R and the glm statement, both GLMs give me an AIC, what is the overall AIC to compare this to e.g. a simple Normal Distribution glm?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

according to http://www.stata-journal.com/sjpdf.html?articlenum=st0040 I can just add up the log likelihoods of both parts.

Improve this answer
$\endgroup$
6
  • $\begingroup$ I don't see anything about AIC in the article that you linked $\endgroup$
    – jeramy townsley
    Commented Nov 13, 2017 at 23:14
  • $\begingroup$ @jeramytownsley it says the sum of the loglikelihoods of the hurdle is the sum of both individuals models (p. 182), and the AIC is based on the loglikelihood. $\endgroup$
    – spore234
    Commented Nov 14, 2017 at 13:18
  • $\begingroup$ It still doesn't seem to answer the question that was asked, in which case it should have been a comment not an answer. You have made a number of assumptions. 1) What is the AIC formula? There are a number that come up on a search-which is right for this situation? All require k--what is k here? 2) I don't see the two separate LogL in the paper you cite--where do you get these? 3) Your citation refers to Stata documentation--the question is about R. What R program would provide this information? $\endgroup$
    – jeramy townsley
    Commented Nov 14, 2017 at 14:04
  • $\begingroup$ @jeramytownsley I would be more than happy to accept a better answer if you come up with one. $\endgroup$
    – spore234
    Commented Nov 14, 2017 at 14:09
  • $\begingroup$ I don't think it's possible to use AIC or LogL to do this kind of comparison--between one-part vs two-part. I think you would have to use other approaches, like plotting predicted vs observed, etc. As you note, you can sum the LogL of both parts to get the model LogL which means it is a fundamentally different thing than single-part GLM. Have you looked at zcpglm in CPLM? It won't give you the AIC, but it will give you a single LogL for a two-part model with a semicontinuous DV, so I think you can just use -2LogL+2k for AIC. K includes both intercepts, and all covariates from both models. $\endgroup$
    – jeramy townsley
    Commented Nov 14, 2017 at 14:58

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.