Re: Computation of AIC and AIC with weights
- From: joseph.yarmus@xxxxxxxxxx
- Date: Thu, 27 Sep 2007 06:20:04 -0700
On Sep 26, 11:22 am, joseph.yar...@xxxxxxxxxx wrote:
This message was also posted to R-help.
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 n log(RSS/n) + 2p and 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
Digging into the R-code behind AIC for linear models, I see:
AIC = nobs * (log(dev/nobs * 2 * pi) + 1) + 2 - sum(log(wt)) + 2 * p
dev = sum(wt * (y - mean(y))^2
For the unweighted case, this translates directly to -2LL with the
number of parameters including both intercept and error term (as
represented by the constant + 2) and the unknown
sigma-squared = sum((y - mean(y)^2)/ nobs (rather than nobs-1).
However, with weights, I am at a loss to understand the expression,
because, given
-2LL = nobs * (log(2 * pi * sigma^2) + sum(wt * (y - mean(y))^2/
sigma^2
if sigma^2 = sum(wt * (y - mean(y))/sum(wt)
then
-2LL = nobs * (log(2 * pi *dev/nobs) + log(nobs) - log(sum(wt)) +
sum(wt)
so if wt = 1 all is fine because
-2LL = nobs * (log(2 * pi * dev/nobs) + 1)
What am I missing? Thanks!
-joe yarmus
.
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