5
$\begingroup$

I have an equation of the form(all vectors): $y=X_1\beta_1+X_2\beta_2+u$.

I'm interested in knowing if the beta OLS estimators and respective residual for this equation are the same as for when we apply OLS to the following equations:

  1. $P_{X_1}y=P_{X_1}X_2\beta_2+v$
  2. $P_Xy=X_1\beta_1+X_2\beta_2+v$,

where the $P_Z$ are the usual definition of projection matrices, using $Z$.

So, I've tried using the FWL theorem, and I've got respectively:

  1. $\hat\beta_2 = (X_2' P_{X_1}X_2)^{-1}X_2'P_{X_1}y$, and $\hat v = (I- P_{X_1}X_2(X_2'P_{X_1}X_2)^{-1}X_2'P_{X_1})P_{X_1}y$. I was wondering if I miscalculated $\hat u$ since looking at equation 1, since both $y$ and $X_2\beta_2$ are projected in to the space spanned by columns of $X_1$, the residuals would be zero.
  2. $\hat\beta_2 = (X_2' M_{X_1}X_2)^{-1}X_2'M_{X_1}P_X y$, and $\hat v = (I- M_{X_1}X_2(X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1})P_{X}y$. However, I do not see how the estimate for $\beta_2$ is equal in both cases, since if you notice that applying OLS to equation 2, we get $\hat \beta=(X'X)^{-1}X'P_X y=(X'X)^{-1}X'y$.

Any help would be appreciated.

Edit1: well, I found out how to do the 2nd point. We have to notice that $M_{X_1}P_X=(I-P_{X_1})P_X=P_X-P_{X_1}=P_X'-P_{X_1}'=(M_{X_1}P_X)'=P_X'M_{X_1}'=P_X M_{X_1}$ and that $ X_2'P_X=(P_X X_2)=X_2'$. As to the 1st point I have no idea...

$\endgroup$
  • $\begingroup$ It seems to me that you are confusing the Projection matrix $P$ with the Annihilator/residual maker matrix $M = I-P$. $\endgroup$ – Alecos Papadopoulos Feb 26 '16 at 18:09
  • $\begingroup$ @AlecosPapadopoulos Thanks for the interest. Why do you say that? $\endgroup$ – An old man in the sea. Feb 27 '16 at 8:30
3
$\begingroup$

What we know from FWL theorem, is that the regression

$$M_1y = M_1X_2\beta_2 + M_1u \tag{1}$$

will give the same estimates for $\beta_2$ as the full regression

$$y = X_1\beta_1 +X_2\beta_2 + u \tag{2}$$

where

$$M_1 = I - P_1 = I - X_1(X_1'X_1)^{-1}X_1'$$

is the so-called annihilator or residual-maker matrix. The estimator from $(1)$ is

$$\hat \beta_2 = (X_2'M_1X_2)^{-1}X_2'M_1y \tag{3}$$

So it boils down to examine whether the estimator from the specification

$$P_1y = P_1X_2\beta_2 + w \tag{4}$$

which is

$$\tilde \beta_2 = (X_2'P_1X_2)^{-1}X_2'P_1y \tag{5} $$

will be the same as $\hat \beta_2$.

Well,

$$(2),(3) \implies \hat \beta_2 - \beta_2 = (X_2'M_1X_2)^{-1}M_1u \tag{6}$$

while

$$ (2), (5) \implies \tilde \beta_2 -\beta_2 = (X_2'P_1X_2)^{-1}X_2'X_1\beta_1+ (X_2'P_1X_2)^{-1}X_2'P_1u \tag{7}$$

Given that $(6)$ and $(7)$ involve arbitrary exogenous quantities ($u, \beta_1$) I don't see how they could be equal, except by zero-probability chance.

Even if in our sample $X_1$ and $X_2$ are orthogonal (which would eliminate the first term in $(7)$ but which, with observational data, is a joke to even mention), then the two would be unbiased under strict exogeneity -but this is as far as similarities appear to go here.

$\endgroup$
  • $\begingroup$ Alecos, thanks for your answer. I think you're just stating the FWL theorem. What I was trying to do was to use the FWL theorem and check if I could obtain some conclusions out of its usage... $\endgroup$ – An old man in the sea. Feb 27 '16 at 16:30
  • $\begingroup$ In the first equation, those results are not from FWL. They're just the usual OLS. I did use FWL on the second equation and on the initial eq. too, though. $\endgroup$ – An old man in the sea. Feb 27 '16 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.