I know that "-1" indicates to remove the intercept term. But my question is why intercept term CAN NOT be treated as a variable term as we place a column consited of 1 in the predictor matrix.
If I stick to make a comparison between a model with intercept and one without intercept on adjusted r2 term, now I think the strategy is always to use another definition of r-square or adjusted r-square, in which r-square=sum((y.hat)^2)/sum((y)^2). Am I in the right way? Thanks Li Junjie 2007/5/19, Paul Lynch <[EMAIL PROTECTED]>: > > In case you weren't aware, the meaning of the "-1" in y ~ x - 1 is to > remove the intercept term that would otherwise be implied. > --Paul > > On 5/17/07, Àî¿¡½Ü <[EMAIL PROTECTED]> wrote: > > Hi, everybody, > > > > 3 questions about R-square: > > ---------(1)----------- Does R2 always increase as variables are added? > > ---------(2)----------- Does R2 always greater than 1? > > ---------(3)----------- How is R2 in summary(lm(y~x-1))$r.squared > > calculated? It is different from (r.square=sum((y.hat-mean > > (y))^2)/sum((y-mean(y))^2)) > > > > I will illustrate these problems by the following codes: > > ---------(1)----------- R2 doesn't always increase as variables are > added > > > > > x=matrix(rnorm(20),ncol=2) > > > y=rnorm(10) > > > > > > lm=lm(y~1) > > > y.hat=rep(1*lm$coefficients,length(y)) > > > (r.square=sum((y.hat-mean(y))^2)/sum((y-mean(y))^2)) > > [1] 2.646815e-33 > > > > > > lm=lm(y~x-1) > > > y.hat=x%*%lm$coefficients > > > (r.square=sum((y.hat-mean(y))^2)/sum((y-mean(y))^2)) > > [1] 0.4443356 > > > > > > ################ This is the biggest model, but its R2 is not the > biggest, > > why? > > > lm=lm(y~x) > > > y.hat=cbind(rep(1,length(y)),x)%*%lm$coefficients > > > (r.square=sum((y.hat-mean(y))^2)/sum((y-mean(y))^2)) > > [1] 0.2704789 > > > > > > ---------(2)----------- R2 can greater than 1 > > > > > x=rnorm(10) > > > y=runif(10) > > > lm=lm(y~x-1) > > > y.hat=x*lm$coefficients > > > (r.square=sum((y.hat-mean(y))^2)/sum((y-mean(y))^2)) > > [1] 3.513865 > > > > > > ---------(3)----------- How is R2 in summary(lm(y~x-1))$r.squared > > calculated? It is different from (r.square=sum((y.hat-mean > > (y))^2)/sum((y-mean(y))^2)) > > > x=matrix(rnorm(20),ncol=2) > > > xx=cbind(rep(1,10),x) > > > y=x%*%c(1,2)+rnorm(10) > > > ### r2 calculated by lm(y~x) > > > lm=lm(y~x) > > > summary(lm)$r.squared > > [1] 0.9231062 > > > ### r2 calculated by lm(y~xx-1) > > > lm=lm(y~xx-1) > > > summary(lm)$r.squared > > [1] 0.9365253 > > > ### r2 calculated by me > > > y.hat=xx%*%lm$coefficients > > > (r.square=sum((y.hat-mean(y))^2)/sum((y-mean(y))^2)) > > [1] 0.9231062 > > > > > > Thanks a lot for any cue:) > > > > > > > > > > -- > > Junjie Li, [EMAIL PROTECTED] > > Undergranduate in DEP of Tsinghua University, > > > > [[alternative HTML version deleted]] > > > > ______________________________________________ > > R-help@stat.math.ethz.ch mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > > > > -- > Paul Lynch > Aquilent, Inc. > National Library of Medicine (Contractor) > -- Junjie Li, [EMAIL PROTECTED] Undergranduate in DEP of Tsinghua University, [[alternative HTML version deleted]]
______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
