[R-sig-ME] equality constraints in lmer/lme4

[Jean-Philippe Laurenceau]

Dear Ewart, David D., David W., and Greg--Thanks a lot for your helpful responses. I now see that the solution is crystal clear, especially with Greg's algebra showing how constraining the coefficients of two numeric predictors in a lm or lme to be the same as entering the sum of the two predictors. And, since it appears that these two models are nested, I can also test whether this equality constraint is tenable by examining a difference in the model deviances. 

One more follow-up if I may...I presume this approach would also hold for a binary or count outcome? J-P

Jean-Philippe Laurenceau, Ph.D.
Department of Psychology
University of Delaware

-----Original Message-----
From: Greg Snow [mailto:538280 at gmail.com] 
Sent: Monday, October 21, 2013 4:59 PM
To: David Winsemius
Cc: Jean-Philippe Laurenceau; r-sig-mixed-models at r-project.org
Subject: Re: [R-sig-ME] 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
>>>
>>>        [[alternative HTML version deleted]]
>>>
>>> _______________________________________________
>>> R-sig-mixed-models at r-project.org mailing list 
>>> https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
>>
>>
>>
>> --
>> Gregory (Greg) L. Snow Ph.D.
>> 538280 at gmail.com
>>
>> _______________________________________________
>> R-sig-mixed-models at r-project.org mailing list 
>> https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
>
> David Winsemius
> Alameda, CA, USA
>



--
Gregory (Greg) L. Snow Ph.D.
538280 at gmail.com