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    $\begingroup$ There are sometimes substantive grounds to force a regression through the origin. That can make sense as a reason for insisting on $y = bx$. After all, the laws of physics met at introductory level often have this form. I would need a stronger story for insisting on $y = b_1 x_1 + b_2 x_2 + \cdots + b_p x_p$ and omitting a $b_0$. $\endgroup$
    – Nick Cox
    Commented Nov 25, 2023 at 12:18
  • $\begingroup$ Hi Nick, Thanks! Not sure if I really get you right. So if I regress y on x without a constant, the projection will still be orthogonal? And what about the other two facts? I assume that exogeneity does only matter for causal interpretation, but not in terms of the construction. And the third one I am still not sure. I think applying the law of iterated expectation yields always 0, if x is exogenous. So that would bring me back to the second statement. Long story short: Is û always uncorrelated with the explanatory variables? $\endgroup$
    – Marlon Brando
    Commented Dec 2, 2023 at 14:13
  • $\begingroup$ stats.stackexchange.com/questions/474102/… The first answer also tells that the fact is only true, when we have a constant in the model.... not sure what to think right now.... $\endgroup$
    – Marlon Brando
    Commented Dec 2, 2023 at 15:29
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    $\begingroup$ Since you understand the geometry, you don't need to ask about orthogonality: you know the residuals are orthogonal to all vectors in the space spanned by the regressor variables. (That's called the Normal Equations.) What might be confusing is that when that space contains no nonzero constant vectors, orthogonality does not imply uncorrelated. $\endgroup$
    – whuber ♦
    Commented Dec 5, 2023 at 14:21
  • $\begingroup$ So in other words, orthogonality holds always? But $xi' ûi = 0$ only holds if we have a constant in the model? Is that not a contradiction? $\endgroup$
    – Marlon Brando
    Commented Dec 9, 2023 at 16:13