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 sho
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 know n v
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
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 more
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 wit
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 b
ssage -
From: Jim Silverton
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 a>0 and
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 greatl
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
> unde
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 + b2*b2*
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