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Prerequisites: Understanding partitioned least squares and its solution.
Setup
Given the linear model:
y=Xb+e
We can partition the columns of X into X1 and X2 to get:
y=X1b1+X2b2+e
Partitioned Least Squares Result
Recall that b2=(X~2′X~2)−1X~2′y~ and that each variable with a tilde is the result of being partialled out by M1=I−X1(X1′X1)−1X1′.
Applying result to inclusion of a column of 1’s
Consider the case where X1=ι, then each column of X~2 is the form:
x~2=(M1)x2=(I−ι(ι′ι)−1ι′)x2=M0x=x¨
We also have M0y so y~=y¨ so all variables in the regression are effectively converted to deviations from their means in the regression. (See here for more on M0)
Which brings us to the result including a coefficient has on the solution to regression:
b2=(X¨2′X¨2)−1X¨2′y¨
Deriving the coefficient on ι
Recall that b1=(X1′X1)−1X1′(y−X2b2).
Also, notice that b1 is just a a scalar b1 (because X1 is just one column).
We have:
b1=(ι′ι)−1ι′(y−X2b2)=(ι′ι)−1(ι′y−ι′X2b2)=(1/n)(nyˉ−nxˉ′b2)=yˉ−xˉ′b2