Consider the following Linear Regression Model.$$y_{it}=x_{it}\beta+\upsilon_{it}$$ where x is a scalar and $$cov(x_{it},\upsilon_{it})=0$$ It is know, however, that $cov(.)$ is a measure of linear association, in that two variables can be non linearly related with each other, but still have 0 covariance. My question is has the question of possible non-linear endogeneity been handled? What if the relationship between the error term and our included regressors is not linear, but still exists?

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    $\begingroup$ Bottom line is you are not fitting an optimal model, and not taking advantage of all the information provided by the dataset. Linear Regression, on top of exogeneity, also assumes that true relation is linear. At the end it boils down to whether you are happy with the results or not. If not, you are not limited to linear models. You can fit whatever you deem necessary. $\endgroup$ – Cagdas Ozgenc Jan 15 '16 at 10:21

I agree that the distinctions here are not easy. I will try to shed some light with an example. It will drop the $t$ index you use as I believe it is immaterial to your problem.

Suppose our model is $$ y_i=x_i\beta+u_i, $$ where $x\sim N(0,1)$ and $u_i=x_i^2-1$. Then, $$ cov(x,u)=E(x^3-x)-E(x)E(x^2-1)=0-0-0\cdot0=0, $$ due to lack of skewness ($E(x^3)=0$) of the normal distribution and properties of the chi-square random variable with one d.f. $x^2$, viz. $E(x^2)=1$.

So, the assumption is satisfied, although $x$ and $u$ are evidently not independent: $E(u|x)=E(x^2|x)-1=x^2-1$

Standard consistency proofs of OLS then tell us that OLS will be consistent for $\beta$.

Here is a little simulation to confirm:

n <- 1000
reps <- 2000

beta <- 2
estimates <- matrix(NA,reps)

for (i in 1:reps){
  x <- rnorm(n)
  u <- x^2-1
  y <- beta*x + u
  estimates[i] <- summary(lm(y~x))$coefficients[2]
 Min.   :1.651  
 1st Qu.:1.934  
 Median :2.001  
 Mean   :2.000  
 3rd Qu.:2.067  
 Max.   :2.380  

What it will not do without the stronger assumption $E(u_i|x_i)=0$ is estimate the partial effect of $x$ on $y$, which would be $$ \frac{\partial E(y|x)}{\partial x}=\beta+2x $$ Here is a plot of $y$ against $x$ for one realization of the simulation:

enter image description here

What you are then estimating consistently is the linear projection coefficient of a projection of $y$ on a constant and $x$. It is given by $$ \frac{cov(y,x)}{var(x)}=\frac{E((2x+x^2-1)x)-E(x)E(y)}{1}=E(2x^2)=2 $$

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    $\begingroup$ As a follow up question: if there exists non linear dependence between the error term and the regressors, we do not have a problem of endogeneity (in that the linear coefficient is consistently estimated). $\endgroup$ – ChinG Feb 26 '16 at 15:52
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    $\begingroup$ one thing this does show is the value of residual plots. if an analyst looked at that plot, they would immediately suspect some kind of non linear term for x was required $\endgroup$ – probabilityislogic Jan 23 at 10:46

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