interpreting random intercepts when no fixed intercept?present
Hello,
On Tue, Feb 12, 2013 at 04:10:59PM +0000, Ben Bolker wrote:
? Hmmm. I think in order to answer this question I'd have to ? figure out what model.matrix() is doing when we use ? [ordered factor]+0 in a formula. I thought I knew but now ? I don't think I do ... According to its code, it also uses the default contrasts, but for ordered factor it is contr.poly and not contr.treatment. ? > d <- data.frame(f=ordered(rep(1:5,10)),y=runif(50)) ? > options(digits=3) ? > coef(lm(y~f,data=d)) ? (Intercept) f.L f.Q f.C f^4 ? 0.525 -0.064 0.154 0.144 -0.116 ? > coef(lm(y~f+0,data=d)) ? f1 f2 f3 f4 f5 ? 0.589 0.651 0.360 0.428 0.599 ? > coef(lm(y~f,data=d,contrasts=list(f=contr.treatment))) ? (Intercept) f2 f3 f4 f5 ? 0.5885 0.0625 -0.2286 -0.1602 0.0101 With your example:
d <- data.frame(f=ordered(rep(1:5,10)),y=runif(50)) options(digits=3) coef(lm(y~f,data=d))
(Intercept) f.L f.Q f.C f^4
0.4936 -0.0775 -0.1209 0.0614 -0.0233
coef(lm(y~f+0,data=d))
f1 f2 f3 f4 f5 0.456 0.600 0.542 0.474 0.397
coef(lm(y~f,data=d,contrasts=list(f=contr.poly)))
(Intercept) f.L f.Q f.C f^4
0.4936 -0.0775 -0.1209 0.0614 -0.0233
coef(lm(y~f+0,data=d,contrasts=list(f=contr.poly)))
f1 f2 f3 f4 f5 0.456 0.600 0.542 0.474 0.397 I imagine the default choice of contr.poly is to have this separation between linear, quadratic... terms (orthogonal polynomials?), building contrasts assuming equally-spaced X values... ? (It would probably be better to use an example with a clear ? linear and quadratic term and nothing else, for clarity) ? I think the answer to this is going to have to involve more ? searching into how model.matrix() parameterizes these models. ? Basically, once you know how the fixed effects are parameterized, ? you can interpret what it means to add a zero-mean random-effects ? offset to it ... Generally, would it mean by forcing a null fixed intercept means that we assume that the population average is 0 for the intercept, but vary from a patient to another? Of course, assuming a null (mean) intercept strongly depends on the coding of the quantitative (or ordered) predictor, so can be easily misleading I guess... Hope this help,
Emmanuel CURIS
emmanuel.curis at parisdescartes.fr
Page WWW: http://emmanuel.curis.online.fr/index.html