R help - Oct 2010 - Constrained Regression

Hello everyone,
I have 3 variables Y, X1 and X2. Each variables lies between 0 and 1. I want
to do a constrained regression such that a>0 and (1-a) >0

for the model:

Y = a*X1 + (1-a)*X2

I tried the help on the constrained regression in R but I concede that it
was not helpful.

Any help is greatly appreciated
-- 
Thanks,
Jim.

	[[alternative HTML version deleted]]

What is the stochastic mechanism that generates the data? In other words, what
distribution is Y, conditioned on X1 and X2, supposed to be?

You can also get `a' if you do not wanrt to specify the probability
mechanism by just doing a least squares fitting, but then making inferences can
be a bit tricky.

Ravi.

____________________________________________________________________

Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University

Ph. (410) 502-2619
email: rvaradhan at jhmi.edu


----- Original Message -----
From: Jim Silverton <jim.silverton at gmail.com>
Date: Sunday, October 31, 2010 2:45 am
Subject: Re: [R] Constrained Regression
To: r-help at r-project.org

> Hello everyone,
>  I have 3 variables Y, X1 and X2. Each variables lies between 0 and 1. 
> I want
>  to do a constrained regression such that a>0 and (1-a) >0
>  
>  for the model:
>  
>  Y = a*X1 + (1-a)*X2
>  
>  I tried the help on the constrained regression in R but I concede 
> that it
>  was not helpful.
>  
>  Any help is greatly appreciated
>  -- 
>  Thanks,
>  Jim.
>  
>  	[[alternative HTML version deleted]]
>  
>  ______________________________________________
>  R-help at r-project.org mailing list
>  
>  PLEASE do read the posting guide 
>  and provide commented, minimal, self-contained, reproducible code.

On Oct 31, 2010, at 2:44 AM, Jim Silverton wrote:
> Hello everyone,
> I have 3 variables Y, X1 and X2. Each variables lies between 0 and  
> 1. I want
> to do a constrained regression such that a>0 and (1-a) >0
>
> for the model:
>
> Y = a*X1 + (1-a)*X2

It would not accomplish the constraint that a > 0 but you could  
accomplish the other constraint within an lm fit:

X3 <- X1-X2
lm(Y ~ X3 + offset(X2) )

Since beta1 is for the model Y ~ 1 + beta1(X1- X2) + 1*X2)
                              Y ~ intercept + beta1*X1 + (1 -beta1)*X2

... so beta1 is a.

In the case beta < 0 then I suppose a would be assigned 0. This might  
be accomplished within an iterative calculation framework by a large  
penalization for negative values. In a reply (1) to a question by  
Carlos Alzola in 2008 on rhalp, Berwin Turlach offered a solution to a  
similar problem ( sum(coef) == 1 AND coef non-negative). Modifying his  
code to incorporate the above strategy (and choosing two variables for  
which parameter values might be inside the constraint boundaries) we  
get:

library(MASS)   ## to access the Boston data
   designmat <- model.matrix(medv~I(age-lstat) +offset(lstat),  
data=Boston)
   Dmat <-crossprod(designmat, designmat); dvec <- crossprod(designmat,
   Boston$medv)
   Amat <- cbind(1, diag(NROW(Dmat)))
   bvec <- c(1,rep(0,NROW(Dmat)))
   meq <- 1
library(quadprog)
   res <- solve.QP(Dmat, dvec, Amat, bvec, meq)

 > zapsmall(res$solution)
[1] 0.686547 0.313453

Turlach specifically advised against any interpretation of this  
particular result which was only contructed to demonstrate the  
mathematical mechanics.
>
> I tried the help on the constrained regression in R but I concede  
> that it
> was not helpful.
I must not have that package installed because I got nothing that  
appeared to be useful with ??"constrained regression" .


David Winsemius, MD
West Hartford, CT

1) http://finzi.psych.upenn.edu/Rhelp10/2008-March/155990.html