Date: Tue, 4 Dec 2001 09:28:12 -0500
Reply-To: "Todd, Mike" <todd@PSYCHIATRY.UCHC.EDU>
Sender: "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From: "Todd, Mike" <todd@PSYCHIATRY.UCHC.EDU>
Subject: Re: regression model
Content-Type: text/plain; charset="iso-8859-1"
Angel:
I'm not sure what your exact need is. If you have a simple case where i and
j differ by a consistent "distance", then you can use the LAGn function to
create an x_j vector from the original x_i scores. A brief example where i
- j = 2:
/*CREATING ORIGINAL DATA SET*/
data a;
input id y x_i;
cards;
01 4 5
02 3 6
03 6 3
04 5 2
05 2 6
;
/*CREATING DATA SET CONTAINING X_I, X_J, */
/*AND XIXJ = X_I * X_J WHERE I - J = 2 */
data b;
set a;
x_j = lag2(x_i);
xixj = x_i * x_j;
run;
This yields a data set that looks like this:
ID Y X_I X_J XIXJ
01 4 5 . .
02 3 6 . .
03 6 7 5 35
04 5 8 6 48
05 2 9 7 63
You can then run a regression model for y_i = b_0 + b_1(x_i * x_j)
like so:
proc reg data=b;
model y = x_i x_j xixj;
run;
Though, of course, the model here would be based on only the last 3 cases,
AND the multiplicative (interaction) term would be highly correlated with
its component linear terms (x_i, x_j). Using mean-centered x_i and x_j
scores in computing xixj will help with the collinearity issue--send another
note if you'd like help with centering.
Hope this helps.
-Mike
------------------------------
Date: Tue, 4 Dec 2001 09:40:07 +0800
From: Angel Sung <angel_sok@HOTMAIL.COM>
Subject: regression model
I would like to fit the following model
y_ij = b_0 + b_1(x_i * x_j ) for different i, j
How can SAS do that??
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Michael Todd, Ph.D.
University of Connecticut Health Center
Alcohol Research Center-MC2103
Department of Psychiatry
263 Farmington Avenue
Farmington CT 06030-2103
Voice: 860.679.5468
Fax: 860.679.5464
E-mail: todd@psychiatry.uchc.edu
