Timeline for Can I compare a negative AIC with a positive AIC?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2016 at 22:01 | vote | accept | lusicat | ||
| Dec 21, 2016 at 21:58 | answer | added | usεr11852 | timeline score: 6 | |
| Dec 21, 2016 at 21:37 | history | edited | Richard Hardy |
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| Dec 21, 2016 at 19:17 | comment | added | lusicat | Some miracle happened...my AIC for all models are positive now!!!!!! | |
| Dec 21, 2016 at 17:11 | comment | added | lusicat | From R, md_best<-lm(I(O3^lambda)~temp+humidity+ibh,data=ozone). It appears this guy didn't do something like (mor^.18-1)/(.18*exp(mean(log(mor)))^(.18-1) in the linear regression. | |
| Dec 21, 2016 at 16:19 | comment | added | Ben Bolker | ? not sure what this last comment means ? BTW, negative vs positive is a red herring, e.g see emdbolker.wikidot.com/faq | |
| Dec 21, 2016 at 16:00 | comment | added | lusicat | Umm..this guys on the website has md_best<-lm(I(O3^lambda)~temp+humidity+ibh,data=ozone)... | |
| Dec 21, 2016 at 15:58 | comment | added | Jarle Tufto | Yes, that's right. | |
| Dec 21, 2016 at 15:54 | comment | added | lusicat | Ok. Thank you, so the website did not do all this transformation, so my regression becomes lm((mor^.18-1)/(.18*exp(mean(log(mor)))^(.18-1))~x)? | |
| Dec 21, 2016 at 15:52 | comment | added | Jarle Tufto |
Instead of mor^.18 you need to use (mor^.18-1)/(.18*exp(mean(log(mor)))^(.18-1)) as your response.
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| Dec 21, 2016 at 15:50 | comment | added | lusicat | So if I find my lambda = 0.1818182 ...my new linear regression becomes fit = lm(mor^0.1818182~x) under the Box-Cox transformation. | |
| Dec 21, 2016 at 15:34 | comment | added | Jarle Tufto | Subtracting one and dividing by $\lambda \text{GM}^{\lambda-1}$ is only a linear transformation so ommiting this (as in the example you link to) does not affect R-square values. To compare maximum log-likelihoods and AICs, however, you need to use the complete formula in wikipedia. | |
| Dec 21, 2016 at 15:27 | comment | added | lusicat | I followed the steps here: yimizhao.com/single-post/2015/04/10/… | |
| Dec 21, 2016 at 15:18 | comment | added | Jarle Tufto |
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.
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| Dec 21, 2016 at 15:09 | comment | added | lusicat | fit = lm(mor^0.1818182~x) My model ended up giving me a lambda of 0.1818182. | |
| Dec 21, 2016 at 15:06 | comment | added | Jarle Tufto | Did you include the geometric mean as described at en.wikipedia.org/wiki/Power_transform#Definition? Only then are the AIC-values compareable. | |
| Dec 21, 2016 at 15:01 | history | asked | lusicat | CC BY-SA 3.0 |