<pre><code>IF y = (x1)(β1), we can work out (β1) directly.
I believe FWL is thinking of that way, so it changes to use residuals;
Residuals ε of x1 on x2 and ε of y on x2 are both linear independent to x2.
</code></pre>
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2024-01-19T11:11:40.817Z
2024-01-30T10:31:28.467Z
2024-01-30T10:31:28.467Z
Mou
<p>What we know from FWL theorem, is that the regression</p>

<p>$$M_1y = M_1X_2\beta_2 + M_1u \tag{1}$$</p>

<p>will give the same estimates for $\beta_2$ as the full regression</p>

<p>$$y = X_1\beta_1 +X_2\beta_2 + u \tag{2}$$ </p>

<p>where </p>

<p>$$M_1 = I - P_1 = I - X_1(X_1'X_1)^{-1}X_1'$$ </p>

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

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

<p>So it boils down to examine whether the estimator from the specification </p>

<p>$$P_1y = P_1X_2\beta_2 + w \tag{4}$$</p>

<p>which is</p>

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

<p>will be the same as $\hat \beta_2$.</p>

<p>Well,</p>

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

<p>while </p>

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

<p>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.</p>

<p>Even if in our <em>sample</em> $X_1$ 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.</p>

CC BY-SA 3.0
2016-02-27T16:30:23.313Z
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...
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CC BY-SA 3.0
2016-02-27T20:09:57.993Z
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.
An old man in the sea.
CC BY-SA 3.0
2016-02-27T14:37:42.823Z
2016-02-28T16:18:04.943Z
2016-02-28T16:18:04.943Z
Alecos Papadopoulos
<p>I have an equation of the form(all vectors): $y=X_1\beta_1+X_2\beta_2+u$.</p>

<p>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:</p>

<ol>
<li>$P_{X_1}y=P_{X_1}X_2\beta_2+v$</li>
<li>$P_Xy=X_1\beta_1+X_2\beta_2+v$, </li>
</ol>

<p>where the $P_Z$ are the usual definition of projection matrices, using $Z$.</p>

<p>So, I've tried using the FWL theorem, and I've got respectively:</p>

<ol>
<li>$\hat\beta_2 = (X_2' P_{X_1}X_2)^{-1}X_2'P_{X_1}y$, 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. </li>
<li>$\hat\beta_2 = (X_2' M_{X_1}X_2)^{-1}X_2'M_{X_1}P_X y$, 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 $\hat \beta=(X'X)^{-1}X'P_X y=(X'X)^{-1}X'y$.</li>
</ol>

<p>Any help would be appreciated.</p>

<p>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...</p>

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2016-02-26T18:09:41.327Z
It seems to me that you are confusing the Projection matrix $P$ with the Annihilator/residual maker matrix $M = I-P$.
Alecos Papadopoulos
CC BY-SA 3.0
2016-02-27T08:30:04.513Z
@AlecosPapadopoulos Thanks for the interest. Why do you say that?
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<econometrics>
The Frisch-Waugh-Lovell Theorem: an exercise
An old man in the sea.
# The Frisch-Waugh-Lovell Theorem: an exercise

(Asked by: username_0 on 2016-02-26)

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...

**Tags:** econometrics

> **Comments:**
> **username_1**: It seems to me that you are confusing the Projection matrix $P$ with the Annihilator/residual maker matrix $M = I-P$.  
> **username_0**: @username_1 Thanks for the interest. Why do you say that?  

---

## 2 answers

### Answer by username_1 on 2016-02-27. Score: 8 | Accepted answer

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.

> **Comments:**
> **username_0**: username_1, 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...  
> **username_0**: 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.  

---

### Answer by username_2 on 2024-01-19. Score: 0

```
IF y = (x1)(β1), we can work out (β1) directly.
I believe FWL is thinking of that way, so it changes to use residuals;
Residuals ε of x1 on x2 and ε of y on x2 are both linear independent to x2.

```

[![enter image description here](https://i.sstatic.net/eHBGs.png)](https://i.sstatic.net/eHBGs.png)


---

