regression coefficients

dear all, How can I compare regression coefficients across three (or 
more) groups?

Thank you very much

	  I don't know of a simply function to do what you want, but I can give 
you part of the standard log(likelihood ratio) theory:

	  Suppose b[i]|s ~ N.r(b, s^2*W[i]), i = 1, ..., k.  Then the 
log(likelihood) is a sum of k terms of the following form:

	  l[i] = (-0.5)*(r*log(2*pi*s^2)+log|W[i]|
	      +(s^-2)*t(b[i]-b)%*%solve(W[i]%*%(b[i]-b)

By differentiating with respect to b and setting to 0, we get the 
maximum likelihood estimate for b as follows:

	  b.hat = solve(sum(solve(W[i]))%*%sum(solve(W[i])%*%b[i])

In words:  b.hat = weighted average with weights inversely proportional 
to the variances.  Then log(likelihood ratio) is as follows:

	  log.LR = sum((s^-2)*t(b[i]-b.hat)%*%solve(W[i])%*%(b[i]-b.hat))

This problem should be in most good books on multivariate analysis.  I 
would guess that log.LR probably has an F distribution with numerator 
degrees of freedom = r*(k-1) and with denominator degrees of freedom = 
degrees of freedom in the estimate of s.  However, I don't remember for 
sure.  It's vaguely possible that this is an "unsolved" problem.  In the 
latter case, you should have all the pieces here to generate a Monte 
Carlo.

lamack lamack wrote:

dear all, How can I compare regression coefficients across three (or 
more) groups?

Why is s assumed known and common to the k groups?  I doubt if that is 
what was meant (although it was too imprecise to be at all sure).

If `common' is a viable assumption, you can just fit a model with by-group
regressions vs one with a common regression (which seems to be what you
are testing) and use anova().

If not, the case k=2 encompasses the Welch t-test so exact distribution
theory is not going to be possible, but by fitting a common model and
three separate models and summing the -2log-lik for the latter you can
easily get the LT test and refer it to its `standard' (asymptotic)
Chi-squared distribution.

On Tue, 20 May 2003, Spencer Graves wrote:

  I don't know of a simply function to do what you want, but I can give 
you part of the standard log(likelihood ratio) theory:

  Suppose b[i]|s ~ N.r(b, s^2*W[i]), i = 1, ..., k.  Then the 
log(likelihood) is a sum of k terms of the following form:

  l[i] = (-0.5)*(r*log(2*pi*s^2)+log|W[i]|
      +(s^-2)*t(b[i]-b)%*%solve(W[i]%*%(b[i]-b)

By differentiating with respect to b and setting to 0, we get the 
maximum likelihood estimate for b as follows:

  b.hat = solve(sum(solve(W[i]))%*%sum(solve(W[i])%*%b[i])

In words:  b.hat = weighted average with weights inversely proportional 
to the variances.  Then log(likelihood ratio) is as follows:

  log.LR = sum((s^-2)*t(b[i]-b.hat)%*%solve(W[i])%*%(b[i]-b.hat))

This problem should be in most good books on multivariate analysis.  I 
would guess that log.LR probably has an F distribution with numerator 
degrees of freedom = r*(k-1) and with denominator degrees of freedom = 
degrees of freedom in the estimate of s.  However, I don't remember for 
sure.  It's vaguely possible that this is an "unsolved" problem.  In the 
latter case, you should have all the pieces here to generate a Monte 
Carlo.

You have assumed s is known, in which case it is a Chi-squared 
distribution.  If s is unknown, you need to maximize over it to get an LR 
test (separately under the null and the alternative).

lamack lamack wrote:

dear all, How can I compare regression coefficients across three (or 
more) groups?