<p>I am not sure where the interpretation issue lies. If we look at a relation</p>

<p>$$y = b_0 + b_1x_1 + u \tag{1}$$</p>

<p>and we declare that we would like to estimate $b_0$ and $b_1$ but that $x_1$ is "endogenous", we are essentially revealing that we are interested in the partial effect of $x_1$ on $y$ in the relation</p>

<p>$$y = b_0 + b_1x_1 + \delta'\mathbf x{^*} + e \tag{2}$$</p>

<p>where $\mathbf x{^*}$ contains <em>all</em> the (known or unknown to us) variables that are correlated with $x_1$ (and have something to do with $y$ too), and therefore, $e$ is the <strong>conditional expectation function error</strong> (not just "any" error term), and as such, <em>by construction</em> uncorrelated with all the variables. Alas, $\mathbf x{^*}$ is unobservable or unavailable to us. Valid IV estimation will nevertheless allow us to estimate consistently $b_1$... which is the partial effect of $x_1$ on $y$ <em>in the above relation $(2)$</em>.</p>

<p>Even if we cannot account for endogeneity, the problem will be that our estimations will converge to someplace unknown. So while we will still be attempting to estimate the partial effect of $x_1$ in relation $(2)$, our numerical estimates will be unacceptable and suspect as regards their value (due to the deficiency of the estimation method) -and not because the desired (but unattainable) research target loses its fundamental interpretation. Our estimates will be "wrongly estimated partial effects". It is not the same as saying "they are not partial effects" (epistemologically speaking).</p>

<p><strong>ADDENDUM</strong> 
Responding on OP's comments and its enhancement of the question, nothing is stopping us from writing $\partial E(y \mid x, x^*)/\partial x$. At the same time our sample is realized, so is $x^*$ realized, irrespective of whether we know its realized value or not.
Now note that the coefficients/partial effects are assumed constant in this linear setting, <em>irrespective</em> of the variability in the regressors: the partial effect will be the same for all individuals in the sample, irrespective of their actual value for the observable regressor</p>

<p>$$\partial E(y_i \mid x_{1i}, x^*_{1i})/\partial x_{1i} = \partial E(y_j \mid x_{1j}, x^*_{1j})/\partial x_{1j} = \beta_1$$</p>

<p>per assumptions, where $i,j$ are two different observations/individuals in the sample.</p>

CC BY-SA 3.0
2016-06-01T11:05:50.857Z
Alecos, I've edited my question, and added some more info. Intuitively, I'm not sure I agree with your 1st paragraph. The partial effect of x1 is the effect we get on the response/dependent variable by changing x1 only, while keeping the rest constant. With endogeneity, if I change x1, keeping everything else constant(in your example we only know x1), we also change $x^*$...
An old man in the sea.
CC BY-SA 3.0
2016-06-01T11:08:23.020Z
@Anoldmaninthesea. Why you can perform the mental experiment "keep the other observable regressors constant" (which are also correlated with $x_1$), but you cannot perform the mental experiment "keep the unobservables constant"? With _experimental_ data you could certainly "command" the other observable regressors to be constant. With _observational_ data, you cannot.
Alecos Papadopoulos
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2016-06-01T11:09:23.513Z
Because we can only condition on observables. That's my available info set, right? (economically speaking. Mathematically, we could condition on those too)
An old man in the sea.
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2016-06-01T11:42:24.693Z
@Anoldmaninthesea. Added a response.
Alecos Papadopoulos
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2016-06-01T11:58:12.477Z
Alecos, but couldn't we write $x^*$ as a function of $x_1$? So that when we change $x_1$, we also change $x^*$ due to endogeneity... In this case, when doing the derivation with respect to $x_i$, we should also account for an extra term with $\frac{\partial E(y_j|x_{1j},x^*_{1j})}{\partial x^*_{1j}}\frac{\partial x^*_{1j})}{\partial x_{1j}}$
An old man in the sea.
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2016-06-01T12:00:22.913Z
Or should I think of partial effect as only a partial derivative?
An old man in the sea.
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2016-06-01T12:02:26.563Z
@Anoldmaninthesea.    We could also write the other observable regressors as a function of $x_1$, don't you think (since usually we expect them to be correlated)?
Alecos Papadopoulos
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2016-06-01T15:51:19.753Z
You're right, but then if we can also condition on unobservables, how does one explain the apparent paradox I allude to in the addendum to my question?
An old man in the sea.
CC BY-SA 3.0
2016-06-01T15:56:00.970Z
@Anoldmaninthesea. Look at my answer again. If we condition on unobservables, we don't have $u$ anymore, but the conditional expectation function error $e$, which is by construction uncorrelated with all the variables, observables or not.
Alecos Papadopoulos
CC BY-SA 3.0
2016-06-01T16:04:55.933Z
I've accepted your answer, Alecos. Thanks. Now, however, I'm left wondering what to think of page 73, of wooldridge's book (2nd edition)
An old man in the sea.
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2016-06-01T16:06:14.583Z
Please give chapter-subchapter, I do not have the 2nd edition I think.
Alecos Papadopoulos
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2016-06-01T16:08:07.183Z
It's Chapter 4- Single equation linear model and OLS estimation, subchapter 4.3.3 Models with interactions in unobservables: Random coefficients.
An old man in the sea.
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2016-06-01T16:16:24.410Z
@Anoldmaninthesea. It is not contradictory. It discusses an _extra_ difficulty, where we assume that there exist a _distinct_ effect on the dependent variable by the product of an observable and an unobservable ("interaction term"). This is a special case, not the general case -an in this special case, what Wooldridge writes, holds.
Alecos Papadopoulos
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2016-06-01T16:39:48.977Z
Does not the special case have to be conformable/consistent with the general case? unless it's an exception... But an exception is not a special case.
An old man in the sea.
CC BY-SA 3.0
2016-06-01T16:55:37.980Z
@Anoldmaninthesea. You seem to forget that English is not my native language. In my language, we can say "special case" and mean "a "special" case that leads to an exception" also! (exception is indeed the case here, in the sense that we can only go for "average partial effects")
Alecos Papadopoulos
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2016-06-01T18:19:23.220Z
Alecos, I hope you weren't offended. It was not my intention. thanks for your help.
An old man in the sea.
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2016-06-01T18:39:53.927Z
@Anoldmaninthesea. Offended? _Now_ I am offended! :)
Alecos Papadopoulos
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2016-06-01T02:45:54.273Z
2016-06-01T11:42:11.117Z
2016-06-01T11:42:11.117Z
Alecos Papadopoulos
<p>Is it possible to preserve the partial effect interpretation of the coefficients/parameters, when in the presence of endogeneity? I don't see how it's possible with the 2SLS... Wooldridge, in his graduate book speaks of a special case, in a random coefficient model, but where we can only talk of the Average Partial Effect (APE) interpretation.</p>

<p>Imagine we have the model: $y=\beta_0+\beta_1x_1+\beta_2x_2+u$. Where does the partial effect interpretation come from? Assuming $E(u|x_1,x_2)=0$, $E(y|x_1,x_2)=\beta_0+\beta_1x_1+\beta_2x_2$. Keeping everything known constant except for $x_2$(partial effect of $x_2$), i.e. $E(y|x_1,x_2=1)-E(y|x_1,x_2=0)=\beta_2$. Now if $x_2$ is endogenous, then $Cov(x_2,u)\neq 0 \Rightarrow E(u|x_1,x_2)\neq 0$</p>

<p>Then $E(y|x_1,x_2=1)-E(y|x_1,x_2=0)=\beta_2+E(u|x_1,x_2=1)-E(u|x_1,x_2=0)$.</p>

<p>Granted that IV estimate, when consistent, gives us $\beta_2$, but that's not the partial effect of $x_2$</p>

<p><strong>Addendum</strong>:</p>

<p>Alecos,
So let's be specific about the interaction between $x_2$ and $u$. Assume that $u=\beta_{\nu} x_2\ \nu+v$, where $\nu$ is unobserved, and $v$ is an error term such that $Cov(x_2,v)= 0$ </p>

<p>Then assuming that we can condition on the unobserved variable, we have 
$\frac{\partial E(y|x_1,x_2,q)}{\partial x_2}= \beta_2+\beta_{\nu}q$.
There's no way to estimate this effect, or is there?</p>

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

<econometrics><applied-econometrics>
Endogeneity and Partial effect interpretation of coefficients
An old man in the sea.
# Endogeneity and Partial effect interpretation of coefficients

(Asked by: username_0 on 2016-05-31)

Is it possible to preserve the partial effect interpretation of the coefficients/parameters, when in the presence of endogeneity? I don't see how it's possible with the 2SLS... Wooldridge, in his graduate book speaks of a special case, in a random coefficient model, but where we can only talk of the Average Partial Effect (APE) interpretation.

Imagine we have the model: $y=\beta\_0+\beta\_1x\_1+\beta\_2x\_2+u$. Where does the partial effect interpretation come from? Assuming $E(u|x\_1,x\_2)=0$, $E(y|x\_1,x\_2)=\beta\_0+\beta\_1x\_1+\beta\_2x\_2$. Keeping everything known constant except for $x\_2$(partial effect of $x\_2$), i.e. $E(y|x\_1,x\_2=1)-E(y|x\_1,x\_2=0)=\beta\_2$. Now if $x\_2$ is endogenous, then $Cov(x\_2,u)\neq 0 \Rightarrow E(u|x\_1,x\_2)\neq 0$

Then $E(y|x\_1,x\_2=1)-E(y|x\_1,x\_2=0)=\beta\_2+E(u|x\_1,x\_2=1)-E(u|x\_1,x\_2=0)$.

Granted that IV estimate, when consistent, gives us $\beta\_2$, but that's not the partial effect of $x\_2$

**Addendum**:

Alecos,
So let's be specific about the interaction between $x\_2$ and $u$. Assume that $u=\beta\_{\nu} x\_2\ \nu+v$, where $\nu$ is unobserved, and $v$ is an error term such that $Cov(x\_2,v)= 0$

Then assuming that we can condition on the unobserved variable, we have
$\frac{\partial E(y|x\_1,x\_2,q)}{\partial x\_2}= \beta\_2+\beta\_{\nu}q$.
There's no way to estimate this effect, or is there?

Any help would be appreciated.

**Tags:** econometrics, applied-econometrics


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## 1 answers

### Answer by username_1 on 2016-06-01. Score: 2 | Accepted answer

I am not sure where the interpretation issue lies. If we look at a relation

$$y = b\_0 + b\_1x\_1 + u \tag{1}$$

and we declare that we would like to estimate $b\_0$ and $b\_1$ but that $x\_1$ is "endogenous", we are essentially revealing that we are interested in the partial effect of $x\_1$ on $y$ in the relation

$$y = b\_0 + b\_1x\_1 + \delta'\mathbf x{^\*} + e \tag{2}$$

where $\mathbf x{^\*}$ contains *all* the (known or unknown to us) variables that are correlated with $x\_1$ (and have something to do with $y$ too), and therefore, $e$ is the **conditional expectation function error** (not just "any" error term), and as such, *by construction* uncorrelated with all the variables. Alas, $\mathbf x{^\*}$ is unobservable or unavailable to us. Valid IV estimation will nevertheless allow us to estimate consistently $b\_1$... which is the partial effect of $x\_1$ on $y$ *in the above relation $(2)$*.

Even if we cannot account for endogeneity, the problem will be that our estimations will converge to someplace unknown. So while we will still be attempting to estimate the partial effect of $x\_1$ in relation $(2)$, our numerical estimates will be unacceptable and suspect as regards their value (due to the deficiency of the estimation method) -and not because the desired (but unattainable) research target loses its fundamental interpretation. Our estimates will be "wrongly estimated partial effects". It is not the same as saying "they are not partial effects" (epistemologically speaking).

**ADDENDUM**
Responding on OP's comments and its enhancement of the question, nothing is stopping us from writing $\partial E(y \mid x, x^\*)/\partial x$. At the same time our sample is realized, so is $x^\*$ realized, irrespective of whether we know its realized value or not.
Now note that the coefficients/partial effects are assumed constant in this linear setting, *irrespective* of the variability in the regressors: the partial effect will be the same for all individuals in the sample, irrespective of their actual value for the observable regressor

$$\partial E(y\_i \mid x\_{1i}, x^\*\_{1i})/\partial x\_{1i} = \partial E(y\_j \mid x\_{1j}, x^\*\_{1j})/\partial x\_{1j} = \beta\_1$$

per assumptions, where $i,j$ are two different observations/individuals in the sample.

> **Comments:**
> **username_0**: Alecos, I've edited my question, and added some more info. Intuitively, I'm not sure I agree with your 1st paragraph. The partial effect of x1 is the effect we get on the response/dependent variable by changing x1 only, while keeping the rest constant. With endogeneity, if I change x1, keeping everything else constant(in your example we only know x1), we also change $x^*$...  
> **username_1**: @Anoldmaninthesea. Why you can perform the mental experiment "keep the other observable regressors constant" (which are also correlated with $x_1$), but you cannot perform the mental experiment "keep the unobservables constant"? With _experimental_ data you could certainly "command" the other observable regressors to be constant. With _observational_ data, you cannot.  
> **username_0**: Because we can only condition on observables. That's my available info set, right? (economically speaking. Mathematically, we could condition on those too)  
> **username_1**: @Anoldmaninthesea. Added a response.  
> **username_0**: Alecos, but couldn't we write $x^*$ as a function of $x_1$? So that when we change $x_1$, we also change $x^*$ due to endogeneity... In this case, when doing the derivation with respect to $x_i$, we should also account for an extra term with $\frac{\partial E(y_j|x_{1j},x^*_{1j})}{\partial x^*_{1j}}\frac{\partial x^*_{1j})}{\partial x_{1j}}$  
> **username_0**: Or should I think of partial effect as only a partial derivative?  
> **username_1**: @Anoldmaninthesea.    We could also write the other observable regressors as a function of $x_1$, don't you think (since usually we expect them to be correlated)?  
> **username_0**: You're right, but then if we can also condition on unobservables, how does one explain the apparent paradox I allude to in the addendum to my question?  
> **username_1**: @Anoldmaninthesea. Look at my answer again. If we condition on unobservables, we don't have $u$ anymore, but the conditional expectation function error $e$, which is by construction uncorrelated with all the variables, observables or not.  
> **username_0**: I've accepted your answer, Alecos. Thanks. Now, however, I'm left wondering what to think of page 73, of wooldridge's book (2nd edition)  
> **username_1**: Please give chapter-subchapter, I do not have the 2nd edition I think.  
> **username_0**: It's Chapter 4- Single equation linear model and OLS estimation, subchapter 4.3.3 Models with interactions in unobservables: Random coefficients.  
> **username_1**: @Anoldmaninthesea. It is not contradictory. It discusses an _extra_ difficulty, where we assume that there exist a _distinct_ effect on the dependent variable by the product of an observable and an unobservable ("interaction term"). This is a special case, not the general case -an in this special case, what Wooldridge writes, holds.  
> **username_0**: Does not the special case have to be conformable/consistent with the general case? unless it's an exception... But an exception is not a special case.  
> **username_1**: @Anoldmaninthesea. You seem to forget that English is not my native language. In my language, we can say "special case" and mean "a "special" case that leads to an exception" also! (exception is indeed the case here, in the sense that we can only go for "average partial effects")  
> **username_0**: Alecos, I hope you weren't offended. It was not my intention. thanks for your help.  
> **username_1**: @Anoldmaninthesea. Offended? _Now_ I am offended! :)  

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