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This question already has an answer here:

I understand the usual procedure to code categorical variables is to convert n categories into n-1 coded variables. For example, the categorical variable colour with levels red/green/blue could be coded as

         V1  V2 
red   -> 1   0
blue  -> 0   1
green -> 0   0

which in a regression setting means that the effect of green on the response is factored into the intercept.

I know that if we created an additional binary variable V3 such that green is coded

         V1  V2  V3 
red   -> 1   0   0
blue  -> 0   1   0
green -> 0   0   1

then we should fit a regression model with no intercept.

What happens if I take the latter coding (i.e. 3 variables V1, V2, V3 for 3 levels of colour) and fit a regression model with an intercept? I can't figure out why we shouldn't do this.

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marked as duplicate by Glen_b Oct 28 '15 at 4:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Because the three dummies add to the column of 1's for the intercept, making those four effects perfectly multicollinear. It's like trying to balance a sheet of plywood on a picket fence - there's not enough "information" in the line of points to keep it steady - the part along the fence is well-determined, but either side it flips up and down. To avoid this indeterminacy, you either need to eliminate a dummy or the intercept term. [This will be a duplicate. Hold on and I'll have a look.] $\endgroup$ – Glen_b Oct 28 '15 at 3:52
  • $\begingroup$ thanks, I found lots of posts about how to code dummy variables, but none explaining what happens if you add in an extra one. $\endgroup$ – Alex Oct 28 '15 at 3:55
  • $\begingroup$ Does this one (the reference to R doesn't alter the explanation) get at what you want? Also see some discussion of multicollinearity here. If you need something different from those, please clarify $\endgroup$ – Glen_b Oct 28 '15 at 4:02
  • $\begingroup$ Thanks, I think stats.stackexchange.com/questions/70699/… answers my question, I will just have to work out what it is saying. $\endgroup$ – Alex Oct 28 '15 at 4:09
  • $\begingroup$ I'll close this but if there's an outstanding issue that's not resolved at that post, modify your question here (with a link to that one if it helps) and flag to ask for it to be re-opened. $\endgroup$ – Glen_b Oct 28 '15 at 4:22