Re: Simple Question on R-Squared
- From: Karl Ove Hufthammer <Karl.Hufthammer@xxxxxxxxxxx>
- Date: Fri, 09 Feb 2007 10:56:37 +0100
TonyTudor@xxxxxxxxx:
Say, R-Squared=89%, this is interpreted as : 89% of the variation in Y
is explained by the Regression.
My question is: what "variation" in Y are we talking about here ? Is
it the variance(Yi) that is reduced or ........?
Y has a certain variance. The question is: How much of this variance is due
to the linear association with X, and how much is not (the residual
variance). You want much of the variance to be due to the linear
association (otherwise X wouldn't be of must use in predicting Y).
You assume a linear model Y = alpha + beta × X + eps, where X and eps are
independent. R² is simpliy an estimate of Var(E(Y|X))/Var(Y) =
Var(a + beta × X)
--------------------------- (*)
Var(alpha + beta × X + eps)
which you can write beta² × Var(X) / Var(Y). (This is also equal to the
squared correlation.)
In the regression, you estimate beta with b, and Var(X) and Var(Y) by Sxx
and Syy, respectively, and get the estimate R² = b² × Sxx / Syy.
When you have more than one regressor (explantory variable) in a linear
model, R² is still an estimate of Var(E(Y|X))/Var(Y); (*) will just have
more terms. I just showed the simple case here.
Also note that the interepretation of R² as the square of a correlation is
*only* meaningful when both X and Y are random. If X is fixed/chosen, you
can get whatever value of R² you want just by chosing your x values
carefully. (If their empirical variance is large, you get R² values close
to 1, if its small, you get values close to 0.)
--
Karl Ove Hufthammer
.
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- Simple Question on R-Squared
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