equality constraints in lmer/lme4
No, think about the algebra, if we start with: y = b0 + b1*x1 + b2*x2 + ... then the constraint that b1 = b2 gives us: y = b0 + b1*x1 + b1*x2 + ... then factoring: y = b0 + b1*(x1 + x2) + ... Or we can demonstrate with a basic simulation example (here the true common slope is 2):
x1 <- rnorm(100, 10, 3) x2 <- rnorm(100, 100, 5) y <- 2*x1 + 2*x2 + rnorm(100,0,1) lm(y ~ x1 + x2)
Call:
lm(formula = y ~ x1 + x2)
Coefficients:
(Intercept) x1 x2
-1.862 2.034 2.015
lm( y ~ I(x1+x2) )
Call:
lm(formula = y ~ I(x1 + x2))
Coefficients:
(Intercept) I(x1 + x2)
-2.282 2.021
lm( y ~ I( (x1+x2)/2 ) )
Call:
lm(formula = y ~ I((x1 + x2)/2))
Coefficients:
(Intercept) I((x1 + x2)/2)
-2.282 4.041
So you can see the averaging is not needed (or you could average, but
then you would need to multiply by 2 because we are going to make
predictions based on b1*x1+b1*x2, so the 2's would cancel).
This is just linear regression, but the concept should hold for lme as
well (if you want more convincing, simulate an lme style problem and
try it).
On Mon, Oct 21, 2013 at 1:46 PM, David Winsemius <dwinsemius at comcast.net> wrote:
On Oct 21, 2013, at 10:44 AM, Greg Snow wrote:
If X1 and X2 are both numeric variables then Outcome ~ I( X1 + X2 ) + (1|Subject)
Would that be: Outcome ~ I( (X1 + X2)/2 ) + (1|Subject) # ? -- David.
should give you what you need. If both are categorical, then you need to create a set of variables that represent the combination (make sure that you understand what that combination represents). On Sun, Oct 20, 2013 at 7:52 PM, Jean-Philippe Laurenceau <jlaurenceau at psych.udel.edu> wrote:
Dear R-sig-ME list--
When specifying the following lmer model, I get intercept fixed and random effects, a fixed effect for the X1 predictor, and a fixed effect for the X2 predictor.
fm <- lmer ( Outcome ~ X1 + X2 + ( 1 | Subject ), data = mydata)
My question: is there a way to ask lme4 to re-estimate this model but set an equality constraint on the effects of X1 and X2, such that their estimates would be equal to each other?
Thanks for your time, J-P
Jean-Philippe Laurenceau, Ph.D.
Department of Psychology
University of Delaware
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David Winsemius Alameda, CA, USA
Gregory (Greg) L. Snow Ph.D. 538280 at gmail.com