Re: [R] Test for equality of coefficients in multivariate multipleregression

[John Fox]

Dear Berwin,

Simply stacking the problems and treating the resulting observations as
independent will give you the correct coefficients, but incorrect
coefficient variances and artificially zero covariances.

The approach that I suggested originally -- testing a linear hypothesis
using the coefficient estimates and covariances from the multivariate linear
model -- seems simple enough. For example, to test that all three
coefficients are the same across the two equations,

 b - as.vector(coef(x.mlm))
 
 V - vcov(x.mlm)
 
 L - c(1, 0, 0,-1, 0, 0,
0, 1, 0, 0,-1, 0,
0, 0, 1, 0, 0,-1)
 L - matrix(L, nrow=3, byrow=TRUE)
 
 t(L %*% b)  %*% (L %*% V %*% t(L)) %*% (L %*% b)

The test statistic is chi-square with 3 df under the null hypothesis. (Note:
not checked carefully.)

(BTW, it's a bit unclear to me how much of this exchange was on r-help, but
I'm copying to r-help since at least one of Ulrich's messages referring to
alternative approaches appeared there. I hope that's OK.)

Regards,
 John


John Fox
Department of Sociology
McMaster University
Hamilton, Ontario
Canada L8S 4M4
905-525-9140x23604
http://socserv.mcmaster.ca/jfox 
 

 -Original Message-
 From: Berwin A Turlach 
 [mailto:[EMAIL PROTECTED] On Behalf Of Berwin 
 A Turlach
 Sent: Tuesday, July 18, 2006 9:28 PM
 To: Andrew Robinson
 Cc: Ulrich Keller; John Fox
 Subject: Re: [R] Test for equality of coefficients in 
 multivariate multipleregression
 
 G'day all,
 
  AR == Andrew Robinson [EMAIL PROTECTED] writes:
 
 AR I suggest that you try to rewrite the model system into a
 AR single mixed-effects model, [...] Why a mixed-effect 
 model, wouldn't a fixed effect be o.k. too?
 
 Something like:
 
  DF-data.frame(x1=rep(c(0,1),each=50),x2=rep(c(0,1),50))
  tmp-rnorm(100)
  DF$y1-tmp+DF$x1*.5+DF$x2*.3+rnorm(100,0,.5)
  DF$y2-tmp+DF$x1*.5+DF$x2*.7+rnorm(100,0,.5)
  x.mlm-lm(cbind(y1,y2)~x1+x2,data=DF)
  coef(x.mlm)
  y1  y2
 (Intercept) -0.08885266 -0.05749196
 x1   0.33749086  0.60395258
 x2   0.72017894  1.11932077
 
 
  DF2 - with(DF, data.frame(y=c(y1,y2)))
  DF2$x11 - with(DF, c(x1, rep(0,100)))
  DF2$x21 - with(DF, c(x2, rep(0,100)))
  DF2$x12 - with(DF, c(rep(0,100), x1))
  DF2$x22 - with(DF, c(rep(0,100), x2))
  DF2$x1 - with(DF, c(x1, x1))
  DF2$wh - rep(c(0,1), each=100)
  fm1 - lm(y~wh + x11 + x21 + x12 + x22, DF2)
  fm1
 
 Call:
 lm(formula = y ~ wh + x11 + x21 + x12 + x22, data = DF2)
 
 Coefficients:
 (Intercept)   wh  x11  x21  
 x12  x22  
-0.08885  0.03136  0.33749  0.72018  
 0.60395  1.11932  
 
  fm2 - lm(y~wh + x1 + x21 + x22, DF2)
  anova(fm2,fm1)
 Analysis of Variance Table
 
 Model 1: y ~ wh + x1 + x21 + x22
 Model 2: y ~ wh + x11 + x21 + x12 + x22
   Res.Df RSS  Df Sum of Sq  F Pr(F)
 1195 246.919
 2194 246.031   1 0.888 0.6998 0.4039
 
 
 Cheers,
 
 Berwin
 
 == Full address 
 Berwin A Turlach  Tel.: +61 (8) 6488 3338 
 (secr)   
 School of Mathematics and Statistics+61 (8) 6488 3383 
 (self)  
 The University of Western Australia   FAX : +61 (8) 6488 1028
 35 Stirling Highway   
 Crawley WA 6009e-mail: [EMAIL PROTECTED]
 Australiahttp://www.maths.uwa.edu.au/~berwin


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Re: [R] Test for equality of coefficients in multivariate multipleregression

[Berwin A Turlach]

G'day John,

 JF == John Fox [EMAIL PROTECTED] writes:

JF Simply stacking the problems and treating the resulting
JF observations as independent will give you the correct
JF coefficients, but incorrect coefficient variances
Yes, after Andrew's (off-list) answer I realised this too.  If I am
not mistaken, all variances/covariances should be off by a factor of
1/2 or something like that.

JF and artificially zero covariances.
Well, I must admit that I misread Ulrich's code for most of the day.
I hadn't realised that the variable `tmp' introduces a correlation
between `y1' and `y2' in his code:

 DF-data.frame(x1=rep(c(0,1),each=50),x2=rep(c(0,1),50))
 tmp-rnorm(100)
 DF$y1-tmp+DF$x1*.5+DF$x2*.3+rnorm(100,0,.5)
 DF$y2-tmp+DF$x1*.5+DF$x2*.7+rnorm(100,0,.5)

for some reason, my brain kept parsing this as generate *one* random
intercept for y1 and *one* random intercept for y2, not that each
individual observation has a random intercept.  Under the model that
my brain kept parsing, one would have zero covariances. :)

Now I understand why Andrew suggested the use of mixed models and
would go down that way too.  But I believe your approach is valid too.

JF (BTW, it's a bit unclear to me how much of this exchange was
JF on r-help,
Easy, all those that have r-help either in the TO: or CC: field.
Those were Ulrich's original message and the answer by you and Andrew,
I kept all my mails so far off-list.

JF but I'm copying to r-help since at least one of Ulrich's
JF messages referring to alternative approaches appeared there.
Yes, I noticed that and answered off-list.  In that message, if I read
it correctly, he had confused Andrew and me.

JF I hope that's OK.)
Sure, why not? :)

Cheers,

Berwin

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