# Endogeneity and Partial effect interpretation of coefficients

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 $x_2$

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.

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.

• 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. Jun 1 '16 at 11:05
• @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 Jun 1 '16 at 11:08
• 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. Jun 1 '16 at 11:09
• @Anoldmaninthesea. Added a response. – Alecos Papadopoulos Jun 1 '16 at 11:42
• @Anoldmaninthesea. Offended? Now I am offended! :) – Alecos Papadopoulos Jun 1 '16 at 18:39