Comparison of models by AIC with or without log transformation on Y

You cannot compare the AIC or BIC when fitting to two different data sets i.e. 𝑌 and 𝑍. You only can compare two models based on AIC or BIC just when fitting to the same data set. Have a look at Model Selection and Multi-model Inference: A Practical Information-theoretic Approach (Burnham and Anderson, 2004). They mentioned my answer on page 81 (section 2.11.3 Transformations of the Response Variable):

Investigators should be sure that all hypotheses are modeled using the same response variable (e.g., if the whole set of models were based on log(y), no problem would be created; it is the mixing of response variables that is incorrect).

Akaike (1978, pg. 224) describes how the AIC can be adjusted in the presence of a transformed outcome variable to enable model comparison. He states: “the effect of transforming the variable is represented simply by the multiplication of the likelihood by the corresponding Jacobian to the AIC ... for the case of log{𝑦(𝑛)+1}, it is −2 ⋅∑log{𝑦(𝑛)+1}, where the summation extends over 𝑛=1,2,...,𝑁.”

Akaike, H. 1978. "On the likelihood of a time series model," Journal of the Royal Statistical Society, Series D (The Statistician), 27(3/4), pp. 217–235.

Let do a toy example:

seedrates <- data.frame(rate = c(50, 75, 100, 125, 150), 
                        grain = c(21.2, 19.9, 19.2, 18.4, 17.9))
quad.lm <- lm(grain~poly(rate,2), data=seedrates)
loglin.lm <- lm(log(grain)~log(rate), data=seedrates)
AIC(quad.lm, loglin.lm)

We need to add sum(2*log(seedrates$grain)) = 29.6 to the AIC for the loglinear model (or, subtract it from the AIC for the quadratic model).

AIC(quad.lm, loglin.lm) + matrix(ncol=2, c(0,0,0, sum(2*log(seedrates$grain))))
          df  AIC
quad.lm    4 -4.1
loglin.lm  3 -7.6

Take a look at https://stats.stackexchange.com/questions/48714/prerequisites-for-aic-model-comparison

From Ben Bolker answer in List:

We need -2 * sum(log( d( log(x/(1-x)), "x" )))

Being super-lazy and using sympy

from sympy import *

x = Symbol("x")

simplify(diff(log(x/(1-x)), x))

## -1/(x*(x-1)) = 1/(x*(1-x))

taking -2*log() of this we get

2*sum(log(x*(1-x)))

seedrates <- data.frame(rate = c(50, 75, 100, 125, 150),

                           grain = c(21.2, 19.9, 19.2, 18.4, 17.9)) |>

   transform(pgrain = grain/100, logit_grain = qlogis(grain/100))

m0 <- lm(pgrain~1, data=seedrates)

m1 <- lm(logit_grain ~ 1, data = seedrates)

AIC(m0)

with(seedrates, AIC(m1) + 2*sum(log(pgrain*(1-pgrain))))

These are actually slightly different because of Jensen's inequality (the predicted values mean(pgrain) and plogis(mean(logit_grain)) are not quite the same), but close enough that I think the computation is done correctly.