[R] How to compare linear models with intercept and those withoutintercept using minimizing adjs R^2 strategy
Lucke, Joseph F
Joseph.F.Lucke at uth.tmc.edu
Mon May 21 16:53:35 CEST 2007
One issue is whether you want your estimators to be based on central
moments (covariances) or on non-central moments. Removing the intercept
changes the statistics from central to non-central moments. The
adjusted R2, by which I think you mean Fisher's adjusted R2, is based on
central moments (ratio of unbiased estimators of variances---central
moments). So if you remove the intercept, you must re-derive the
adjusted R2 for non-central moments --- you can't just plug in the
number of independent variables as zero.
-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of ???
Sent: Sunday, May 20, 2007 8:53 PM
To: r-help at stat.math.ethz.ch
Subject: [R] How to compare linear models with intercept and those
withoutintercept using minimizing adjs R^2 strategy
Dear R-list,
I apologize for my many emails but I think I know how to desctribe my
problem differently and more clearly.
My question is how to compare linear models with intercept and those
without intercept using maximizing adjusted R^2 strategy.
Now I do it like the following:
> library(leaps)
> n=20
> x=matrix(rnorm(n*3),ncol=3)
> b=c(1,2,0)
> intercept=1
> y=x%*%b+rnorm(n,0,1)+intercept
>
> var.selection=leaps(cbind(rep(1,n),x),y,int=F,method="adjr2")
> ##### Choose the model with maximum adjr2
> var.selection$which[var.selection$adjr2==max(var.selection$adjr2),]
1 2 3 4
TRUE TRUE TRUE FALSE
Actually, I use the definition of R-square in which the model is without
a intercept term.
Is what I am doing is correct?
Thanks for any suggestion or correction.
--
Junjie Li, klijunjie at gmail.com
Undergranduate in DEP of Tsinghua University,
[[alternative HTML version deleted]]
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