如何在stepAIC中计算AIC

Question

Here is a very simple lm model from ?lm

ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group <- gl(2,10,20, labels=c("Ctl","Trt"))
weight <- c(ctl, trt)
lm.D9 <- lm(weight ~ group)

If I use stepAIC to lm.D9, on the very first line it says AIC = -12.58

require(MASS)
stepAIC(lm.D9)

If I use AIC directly on lm.D9, it gives a different value 46.17648

AIC(lm.D9)

My question is why the 2 AIC values are different. Thanks!

Answer 1

AIC is only defined up to an arbitrary constant. As long as the same value of the constant is used when comparing AICs for different models, it doesn't matter. If you look at ?extractAIC and ?AIC , you'll find the formulas used by both methods.

Basically, either use extractAIC or AIC , but not both at the same time.

Answer 2

This was annoying me, so I decided to work it out from first principles.

Re-fit the model:

d <- data.frame(weight=
                c(ctl=c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14),
                  trt=c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)),
                group=gl(2,10,20, labels=c("Ctl","Trt")))
lm.D9 <- lm(weight ~ group, d)

Values returned by standard accessors:

(AIC1 <- AIC(lm.D9))
> 46.17468
(LL1 <- logLik(lm.D9))
> -20.08824 (df=3)

Reconstruct from first principles:

n <- nrow(d)
ss0 <- summary(lm.D9)$sigma
ss <- ss0*(n-1)/n
(LL2 <- sum(dnorm(d$weight,fitted(lm.D9),
                 sd=ss,log=TRUE)))
> -20.08828

This is a tiny bit off, haven't found the glitch.

Number of parameters:

npar <- length(coef(lm.D9))+1 


(AIC2 <- -2*LL2+2*npar)
> 46.1756

Still off by more than numeric fuzz, but only about one part in a million.

Now let's see what stepAIC is doing:

MASS::stepAIC(lm.D9)  ## start: AIC = -12.58
extractAIC(lm.D9)     ## same value (see MASS::stepAIC for details)
stats:::extractAIC.lm ## examine the code


RSS1 <- deviance(lm.D9)   ## UNSCALED sum of squares
RSS2 <- sum((d$weight-fitted(lm.D9))^2)  ## ditto, from first principles
AIC3 <- n*log(RSS1/n)+2*2 ## formula used within extractAIC

You can work out the formula used above from sigma-hat=RSS/n -- or see Venables and Ripley MASS for the derivation.

Add missing terms: uncounted variance parameter, plus normalization constant

(AIC3 + 2 - 2*(-n/2*(log(2*pi)+1)))

This is exactly the same (to 1e-14) as AIC1 above

Answer 3

Thank you @benbolker for detailed answer. You mentioned:

This is a tiny bit off, haven't found the glitch.

I looked into it and found that if you modify this line:

ss <- ss0*(n-1)/n

to this:

ss <- sqrt( (ss0)^2 * (n - length(coef(lm.D9))) / n )

then the result would be exactly the same.