3
$\begingroup$

The original model gives me an AIC of 75.43 and the Box-Cox transformed model gives me -189.2013. Does this me the Box-Cox transformed model is a much better model here?

Improve this question
$\endgroup$
13
  • 1
    $\begingroup$ Did you include the geometric mean as described at en.wikipedia.org/wiki/Power_transform#Definition? Only then are the AIC-values compareable. $\endgroup$
    – Jarle Tufto
    Commented Dec 21, 2016 at 15:06
  • 3
    $\begingroup$ You need to use the complete formula in wikipedia, otherwise you're comparing the likelihood of a model specifying the distribution of mor with a model specifying the distribution of mor^0.1818. These are not compareable. Using the complete formula, you get comparaable likelihoods and you'll probably see a much smaller difference in log likelihoods and AICs. $\endgroup$
    – Jarle Tufto
    Commented Dec 21, 2016 at 15:18
  • 2
    $\begingroup$ Instead of mor^.18 you need to use (mor^.18-1)/(.18*exp(mean(log(mor)))^(.18-1)) as your response. $\endgroup$
    – Jarle Tufto
    Commented Dec 21, 2016 at 15:52
  • 2
    $\begingroup$ ? not sure what this last comment means ? BTW, negative vs positive is a red herring, e.g see emdbolker.wikidot.com/faq $\endgroup$
    – Ben Bolker
    Commented Dec 21, 2016 at 16:19
  • 1
    $\begingroup$ Some miracle happened...my AIC for all models are positive now!!!!!! $\endgroup$
    – lusicat
    Commented Dec 21, 2016 at 19:17

1 Answer 1

Reset to default
6
$\begingroup$

Generally speaking: Yes, you can compare negative and positive values of AIC, it can happen. There is no reason why AIC should be exclusively non-negative; this has been covered already by some posts here and here.

Having said that, you should not compare AIC values from models with different response values. In that case the (log-) likelihoods calculated correspond to variables from different scales so their comparison can be totally misleading; I discuss this a bit with a bit more detail here.

Improve this answer
$\endgroup$

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.