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I am writing my master's thesis in quantile multilevel regression. My professor all of a sudden decided to change the subject of my thesis into something that could be called "quantile multilevel multivariate regression". He asked me to proof that MANOVA is a special case of mixed-models (saying that "someone has published something about it, I guess"). The problem is that I haven't found anything yet. We've never taken multivariate courses so I don't even know where to look at!

Any kind of help will be appreciated: a link to a paper/book where this proof is showed, or if you want to write it on your own, i'll be more than happy to see it!

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This is a really fun question! :D I don't know about the quantile part, I am sure that if you think about it a bit you'll come to that conclusion yourself regarding MANOVA and MLMVE.

To give you some pointers:

  1. Treat ANOVA as synonymous to regression (ANOVA is regression after all).
  2. Think of a MANOVA as a fully nested non-crossed two-level hierarchical mixed model, but then ignore the "lower" level groupings. What are you left with?

I haven't seen an exact proof myself, but yeah I am sure you can pull this off to some level. Sorry, I don't have the exact answer but try it, it is really thought provoking!

For a good targeted introduction to the subject try the following research report : Cross-classified and Multiple Membership Structures in Multilevel Models: An Introduction and Review (2006). I found the Behaviormetrika paper by Hwang and Takane An extended multivariate randon-effects growth curve model (2005) and the Metrika paper by Nummi and Mottonen On the analysis of multivariate growth curves (2000) also quite helpful. (I liked the Metrika et al. paper slightly more, but in case you don't have access to it, the Behaviormetrika is also quite good and is based in the same methodology)

Try to understand what Multilevel (or Hierarchical $^1$) Mixed Effects model does. MANOVA just comes out as a natural subcase of it by restricting some terms in the random effects covariance structure.

$^1$ those terms are not synonymous in general but for your MANOVA question I think they are; hierarchical models are multilevel models but not vice versa (multilevel models can be crossed).

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  • $\begingroup$ Thanks for the answer! Well, as I stated before, we've never taken any Multivariate course so I have to study it on my own (i guess it'll take a couple of days to fully understand how Manova works). It's really hard to prove that A is equal to B when you don't know what A is (Manova is A, in my situation). However, i do well know what mixed models are so at least 50% of the work is done! I wonder if there is some published work about it, I've only got two months left and I've yet to write the first page of my thesis! Thanks again for the answer $\endgroup$ – Davide Apr 22 '13 at 13:55
  • $\begingroup$ I haven't taken such a course either if that makes you feel any better. Check the reference I gave; you they are descent first step. As I mentioned: treating ANOVA as regression, will greatly simplify your life for your task. $\endgroup$ – usεr11852 Apr 22 '13 at 13:59
  • $\begingroup$ I've tried to study Manova a little bit on my own, and I've looked at your links. I get the goal of Manova and such, but I am not really sure about how to translate it into mixed-models. (I don't know how to put formula so it'll be very messy, sorry)First of all, given that ANOVA tests wheter Bi is different from 0 in the model Yij = M + Bi + Eij, i'd like to find the same equation for Manova. Since it is multivariate, the only thing that I have in mind is to use a notation like this: Yijk = M + Bi + Ck + Eijk, where i identifies the group, j the individual, k the Outcome variable. $\endgroup$ – Davide Apr 23 '13 at 12:14
  • $\begingroup$ A mixed model can be expressed in several ways, one of this is Y = Xb + Zc + E I remember there is also the possibility to specifiy a model in the form of Yijk= ... but i have my notes at home (8000 Km from here), so not exactly easy to get them. What do you think, is the manova equation right? I am not really sure to be honest. $\endgroup$ – Davide Apr 23 '13 at 12:16
  • $\begingroup$ Yes it is, but I wouldn't be bothered with the subscript-fest. Just put all your $Y$ in single vector (instead of a matrix), use appropriate Kronecker products to define X and Z matrices and the thing is there. $\endgroup$ – usεr11852 Apr 23 '13 at 14:05

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