Hello,
Your problem is y=bX+epsilon
It can be transformed into: epsilon^2=(y-bX)^2
Standard (unconstrained) regressions are about minimizing the variance of
epsilon, ie (y-bX)^2.
In your case, you need to minimize again the quantity (y-bX)^2 with your
constraints on b=(b1,...,b5). Solve.QP should just do that for you.
HTH,
Samuel
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On Behalf Of Jackie Chen
Sent: 20 January 2011 16:31
To: R-help at r-project.org
Subject: [R] Constrained Regression
Hi everyone,
I'm trying to perform a linear regression y = b1x1 + b2x2 + b3x3 + b4x4 +
b5x5 while constraining the coefficients such that -3 <= bi <= 3, and the
sum of bi =1. I've searched R-help and have found solutions for
constrained regression using quadratic programming (solve.QP) where the
coefficients are between 0 and 1 and sum to 1, but unfortunately do not
understand it well enough to adapt to my problem. Is there a way to do
this using the lm function or do I absolutely need to use solve.QP? And
if I need to use solve.QP, how would I modify the Boston data example to
my problem?
Thanks so much.
Jackie
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