m.linear <- lm(y ~ x, data = data)
m.exponential <- lm(log(y) ~ x, data = data)
m.asymptotic <- drm(y ~ x, data = data, fct = DRC.asymReg())
m.logistic <- drm(y ~ x, data = data, fct = L.3())
m.gompertz <- drm(y ~ x, data = data, fct = G.3())

I want to use AIC value to determine the best fitted model, however, the AIC(m.exponential) seems not correct.

  • 2
    $\begingroup$ With one response and one predictor, why lean on AIC? Some combination of plotting data and fitted response and scientific knowledge should get you where you want to be. $\endgroup$
    – Nick Cox
    Feb 26 at 10:06
  • $\begingroup$ The problem now is that the fit of a rising curve results in a logistic model. It seems to be the first half of the fit and the logistic model, so it's impossible to tell if it's accurate or not. $\endgroup$
    – user408308
    Feb 27 at 3:28
  • $\begingroup$ I have seen that problem quite often. $\endgroup$
    – Nick Cox
    Feb 27 at 7:59

1 Answer 1


You can't compare AIC between a model fitted to the original data and another model fitted to logarithmized data.

In addition, Be extremely careful about comparing AICs between models fitted using different tools, like lm() versus drm(). Different implementations have different definitions of "the" AIC.

I recommend you compare predictive accuracy on holdout data, or on out-of-bag data in a bootstrap. (Remember the bias correction if you are estimating conditional means using logged data.)

  • 1
    $\begingroup$ What usually differs is the definition of the log-likelihood of which AIC is a function. $\endgroup$
    – Roland
    Feb 27 at 7:11

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