VIF values when no constant is present in a regression model
- From: "dfc@xxxxxxxxxxxxxxxx" <dfc@xxxxxxxxxxxxxxxx>
- Date: 25 May 2007 07:00:19 -0700
I've wriiten some code to perform a regression calcuation on specified
data and also calculate VIF values by taking the diagnonal values from
the inverse of the correlation matrix of independent variables. It
seemed to be working well until I ran my code for a model with no
intercept and compared the results with another program (SPSS). It
turns out the VIF values for each independent variable are different
when no intercept is included in the model.
Now I have two problems:
1) I don't understand why VIFs differ when no intercept is included in
the model.
The VIF is an indication of the strength of linear relationship it's
associated independent variable has with the other predictors. It
makes sense to me that this can be calculated using the correlation
matrix, where every predictor is included in the matrix. Why should
the inclusion of an intercept have any affect on this measure of
linear relationship?
Is the intercept also a predictor?
2) How does the VIF calculation change when there is no intercept in
the model?
(I can't include a constant in the correlation matrix because it will
have no valid correlation value)
I'm completely stumped so any help would be appreciated. Regards,
Darren.
.
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