Starting with the 'residual maker' defined by M in: $e= y-\hat{y} = Y-X(X'X)^{-1}X'Y = [I-X(X'X)^{-1}X']Y =MY$

where e is the regression residual.

one common equality i see relating the regression residual to the error (denoted $\epsilon$) is:

$e = MY = M[X \beta+\epsilon] = M\epsilon$

my question is.. why isnt this term = 0 if we are always assuming X is exogenous when making these derivaions? Isn't $X'\epsilon =0$? Or is this saying that in the particular sample, $X'\epsilon \neq 0$?

  • $\begingroup$ Please see my answer below. If you have questions, I am happy to elaborate. If you are satisfied with the answer, please accept. $\endgroup$
    – dlnB
    Commented Apr 25, 2020 at 15:23

1 Answer 1


It's not that $X'\epsilon = 0$, but rather that $E[X'\epsilon] = 0,$ which is an assumption of OLS (strict exogeneity). It follows that $E[M \epsilon] = 0$, which tells us that $E[e]=0$. By construction, however, $X'e=0$:

$$X'e = X'Y - X'X(X'X)^{-1}X'Y = X'Y-X'Y = 0.$$


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