Calculation AIC without loglikelihood-function

Question

I want to calculate the AIC without calculating the loglikelihood-function (which seems complicated). If the residuals are normally distributed, this can be done, according to wikipedia, as follows:

$AIC_{\sigma} = n log(\sigma^2_Z) + 2k $,

where $\sigma^2_Z$ is the variance of the residuals, $n$ the number of samples and $k$ the number of parameters.

In order to test this I wrote a simple R-script:

library(MASS)
data(cats)
l<-lm(Hwt ~ Bwt,data=cats)

l.resi <- resid(l)
qqnorm(l.resi)
qqline(l.resi)
shapiro.test(l.resi)

n<-nrow(cats)
k<-2 # one cefficient + intercept
aic_s<- n*log(mean(l.resi^2) ) + 2*k 

print(paste("aic: ",AIC(l))) 
print(paste("aic_s: ",aic_self))

The script returns:

Shapiro-Wilk normality test

data:  l.resi
W = 0.9845, p-value = 0.1046

[1] "aic:  520.121593704369"
[1] "aic_s:  109.467296141423"

So, the residuals seem to be normally distributed. However, the AICs differ. Why is that?

Answer 1

That formula will only equal the true AIC up to a constant. So if $AIC_1$ is the actual AIC and $AIC_s$ is the self-computed AIC you have, then $AIC_1 = AIC_s+c$ for some constant $c$.

The formula you have is useful for model comparisons when you can't compute the log likelihood, since when you are doing model comparisons, you only need to know if the AIC for one model is higher than the AIC for another model (so the actual AIC isn't needed).