```Date: Tue, 4 Dec 2001 09:28:12 -0500 Reply-To: "Todd, Mike" Sender: "SAS(r) Discussion" From: "Todd, Mike" 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 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?? ------------------------------ ><><><><><><><><><><><><><><><><><><><>< 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 ```

Back to: Top of message | Previous page | Main SAS-L page