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
should just do that for you.
HTH,
Samuel
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jackie Chen
Sent: 20 January 2011 16:31
To: R-help@r-project.org
Subject: [R] Constrained Regression
Hi everyone,
I'm trying to perform
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 a0 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
-
From: Jim Silverton jim.silver...@gmail.com
Date: Sunday, October 31, 2010 2:45 am
Subject: Re: [R] Constrained Regression
To: r-help@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 a0
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 a0 and (1-a) 0
for the model:
Y = a*X1 + (1-a)*X2
It would not accomplish the constraint that a 0 but
Have you tried the 'sos' package?
install.packages('sos') # if not already installed
library(sos)
cr - ???'constrained regression' # found 149 matches
summary(cr) # in 69 packages
cr # opens a table in a browser listing all 169 matches with links to
the help pages
However, I agree
On Oct 31, 2010, at 12:54 PM, Spencer Graves wrote:
Have you tried the 'sos' package?
I have, and I am taking this opportunity to load it with my .Rprofile
to make it more accessible. It works very well. Very clean display. I
also have constructed a variant of RSiteSearch that I find
I thought I would 'add' some meat to the problem I sent. This is all I know
(1) f = a*X1 + (1-a)*X2
(2) I know n values of f and X1 which happens to be probabilities
(3) I know nothing about X2 except that it also lies in (0,1)
(4) X1 is the probability under the null (fisher's exact test) and X2
in line
On 10/31/2010 6:26 PM, Jim Silverton wrote:
I thought I would 'add' some meat to the problem I sent. This is all I know
(1) f = a*X1 + (1-a)*X2
How do you know f = a*X1 + (1-a)*X2?
Why does this relationship make more sense than, e.g., log(f/(1-f)) =
a*X1 + (1-a)*X2?
(2) I
I would like to acknowledge the answers I received from Tom Filloon, Mike
Cheung and Berwyn Turlach.
Berwyn's response was exactly what I needed. Use solve.QP from the quadprog
package in R. S-Plus has the equivalent function solveQP in the NuOpt
module.
Berwyn's response is below
G'day Carlos,
Dear Carlos,
One approach is to use structural equation modeling (SEM). Some SEM
packages, such as LISREL, Mplus and Mx, allow inequality and nonlinear
constraints. Phantom variables (Rindskopf, 1984) may be used to impose
inequality constraints. Your model is basically:
y = b0 + b1*b1*x1 +
G'day Carlos,
On Mon, Mar 3, 2008 at 11:52 AM
Carlos Alzola [EMAIL PROTECTED] wrote:
I am trying to get information on how to fit a linear regression
with constrained parameters. Specifically, I have 8 predictors ,
their coeffiecients should all be non-negative and add up to 1. I
Dear list members,
I am trying to get information on how to fit a linear regression with
constrained parameters. Specifically, I have 8 predictors , their
coeffiecients should all be non-negative and add up to 1. I understand it is
a quadratic programming problem but I have no experience in the
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