Non Linear Endogeneity

Question

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?

Answer 1

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:

library(MASS)
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]
}
summary(estimates)
       V1       
 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:

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 $$ \frac{cov(y,x)}{var(x)}=\frac{E((2x+x^2-1)x)-E(x)E(y)}{1}=E(2x^2)=2 $$