Shouldn't the
AIC(full) = 275.93
match the output of of the AIC when the step() function runs with the full model which is -9.86 below
Start: AIC=-9.86
y ~ x + x2
Df Sum of Sq RSS AIC
- x2 1 0.03672 85.372 -11.8147
- x 1 1.03869 86.374 -10.6479
<none> 85.336 -9.8578
Step: AIC=-11.81
y ~ x
Df Sum of Sq RSS AIC
- x 1 1.004 86.376 -12.646
<none> 85.372 -11.815
Step: AIC=-12.65
y ~ 1
Call:
lm(formula = y ~ 1, data = data)
Coefficients:
(Intercept)
-0.03719
here is the full code:
set.seed(101)
y = rnorm(100)
x = rnorm(100)
x2 = rnorm(100)
data = data.frame(y = y, x = x, x2 = x2)
null = lm(y~1, data = data)
full = lm(y~x+x2, data = data)
#step(null, scope= list(lower = null, upper = full) , direction="backward", trace = TRUE)
step(full, direction="backward", trace = TRUE)
AIC(full)
You would need extractAIC instead of AIC
extractAIC(lm(y~x+x2, data = data), scale=0)
#OR
#extractAIC(full, scale=0)
If you refer the documentation ?AIC & ?extractAIC it clearly says that
The log-likelihood and hence the AIC/BIC is only defined up to an additive constant. Different constants have conventionally been used for different purposes and so extractAIC and AIC may give different values (and do for models of class "lm": see the help for extractAIC).
Hope this helps!