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Re: st: R: How much of variation in dep var is explained by various sets of variables?


From   "G. Dai" <dgecon@gmail.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: R: How much of variation in dep var is explained by various sets of variables?
Date   Sun, 4 Apr 2010 19:00:45 -0700

no you can't. one deep problem with any such kind of decomposition is
the order dependence, that is
the percentage of variation depends on the order of adding variables.

On Sun, Apr 4, 2010 at 8:33 AM, Carlo Lazzaro
<carlo.lazzaro@tiscalinet.it> wrote:
> Dear Adrian,
> As far as I know, in case of linearity (as OLS should imply) this issue can
> be addressed via ANCOVA (please, see Briggs A, Sculpher M, Claxton K.
> Decision Modelling for Health Economic Evaluation. Oxford: Oxpord University
> press, 2006: 130-132).
>
> Kind Regards,
> Carlo
>
> -----Messaggio originale-----
> Da: owner-statalist@hsphsun2.harvard.edu
> [mailto:owner-statalist@hsphsun2.harvard.edu] Per conto di kokootchke
> Inviato: domenica 4 aprile 2010 2.35
> A: statalist
> Oggetto: st: How much of variation in dep var is explained by various sets
> of variables?
>
> Dear all,
>
> I would like to know if it's possible to determine how much of the variation
> in the dependent variable is explained by different sets of variables. For
> instance, suppose I have:
>
> (1) y = a*x1 + b*x2 + c*x3 + d*z1 + d*z2 + d*z3 + d*z4
> (2) y = e*x1 + f*x2 + g*x3
> (3) y = h*z1 + i*z2 + j*z3 + k*z4
>
> If I run these regressions by OLS, I obtain, say, R-sq = 0.30, 0.20, 0.15,
> respectively. Is it possible to determine what percentage of the variation
> in y in (1) is explained by the x's and what percentage is explained by the
> z's?
>
> I read some of the threads on this issue and I found some notes on partial
> correlation and the -pcorr- command. I read Richard Williams's notes and it
> seems like you can determine the proportion of the variation in y focusing
> on one variable at a time... but I don't know if it's possible to do it by
> sets of variables.
>
> Thank you very much for your help.
>
> Best,
> Adrian
>
>
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