In accordance with Venables and Ripley, SAS documentation and other
sources AIC with sigma^2 unknown is calculated as:
AIC = -2LL + 2* #parameters = n log(RSS/n) + 2p
For the fitness data:
(http://support.sas.com/ctx/samples/index.jsp?sid=927), SAS gets an AIC
of 64.534 with model oxygen = runtime. (SAS STAT User's Guide. Chapter
61. pp 3956, the REG Procedure). This value of AIC accords with p = 2.
When I run the same problem in R ver 2.5.1, I get
> rt.glm =glm(oxy ~ runtime, data=fitness)
> rt.glm
Call: glm(formula = oxy ~ runtime, data = fitness)
Coefficients:
(Intercept) runtime
82.422 -3.311
Degrees of Freedom: 30 Total (i.e. Null); 29 Residual
Null Deviance: 851.4
Residual Deviance: 218.5 AIC: 154.5
I get very close to what R gets if the constant term is included in
-2LL, (31*Log(2*pi)+n-1), divide RSS by n-1 and the number of parameters
is 3 (the predictor, the intercept and the error term)
> 31 * (log(2*pi)+log(sum(rt.glm$res^2)/30)) + 30 + 2 * 3
[1] 154.5248
> AIC(rt.glm)
[1] 154.5083
3 questions:
1) Why the discrepancy between SAS and R?
2) Why the slight difference between my calculation in R and R's AIC?
3) How should AIC be computed if row weights are used in the linear model?
Thanks!
-joe yarmus
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