Date:         Tue, 6 Dec 2005 19:04:36 -0800
Reply-To:     shiling99@YAHOO.COM
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         shiling99@YAHOO.COM
Organization: http://groups.google.com
Subject:      Re: R Square
Comments: To: sas-l@uga.edu
In-Reply-To:  <200512062038.jB6KE12R020379@mailgw.cc.uga.edu>
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Suppose y= a+ b1*x + err.

err ~ normal(0, sigma)

Here is the proof.

R**2=SSreg/SStotal=sum over obs (Yhat-Ybar)**2/sum over obs (Y-Ybar)**2

then   E(SSreg)=sigma**2 + b1**2*Sxx     and   E(SSres)=(n-2)*sigma**2


Sxx=sum over obs (X-Xbar)**2

         E(SStotal)=E(SSreg) + E(SSres) = )=(n-1)*sigma**2 + b1**2*Sxx


So  R**2=(sigma**2 + b1**2*Sxx ) /[(n-1)*sigma**2 + b1**2*Sxx ]

Let w=b1**2*Sxx /n*sigma**2, then R**2 in terms w will be

     R**2=[n**(-1) + w] /[1-n**(-1)+w]

It is easy to show that the derivertive of R**2 w.r.t w is gt 0.

If you deem b1**2*Sxx /n*sigma**2 as a signal to noise ratio, higher
the ratio, better result/model.

Higher range of X is the same to say bigger Sxx.

Here is a simulation pgm.

data t1;
  do i = 1 to 100;
     x=rannor(123);
     y=2+2*x+rannor(123);
         output;
   end;
run;

proc reg data=t1;
  model y=x;
  where abs(x)<0.5;
run;

proc reg data=t1;
  model y=x;
  where abs(x)>0.5;
run;

HTH.