1
$\begingroup$

In the regression model

$Y$ = $\beta_0$ + $\beta_1X_1$ + $\beta_2X_2$ +.......+ $\beta_kX_k$ + $\epsilon$

where $\epsilon$ = $\delta_0X_2$ + $\lambda$

Will this also be the case of endogeneity since E($\epsilon|X$) is not independent of $X_2$?

$\endgroup$
3
$\begingroup$

If assume the true model is

$$Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \epsilon$$

and that $\epsilon = \delta_0 X_2 + \lambda$ it follows that

$$Y = \beta_0 + \beta_1 X_1 + (\beta_2+\delta_0) X_2 + \lambda$$

and estimating this model using the OLS estimator

$$\hat \theta_{OLS} = \left( \frac{1}{N} \sum_i \mathbf x_i \mathbf x_i^\top\right)^{-1} \left( \frac{1}{N} \sum_i \mathbf x_i y_i\right)$$

where $\mathbf x_i = (1,x_{i1},x_{i2})^\top$ will result in consistent estimator $\hat \theta_1$ as an estimator of $\beta_0$ and $\hat \theta_2$ is estimator of $\beta_1$ and $\hat \theta_3$ as an estimate of $(\beta_2 + \delta_0)$ if

$$\mathbf E[\lambda \lvert \mathbf x] = 0$$

however the fact that $\epsilon = \delta_0 X_2 + \lambda$ means that you cannot identiy $\beta_2$ nor $\delta_0$ but only the combined effect $\beta_2 + \delta_0$.

Simulation in R

N<-10000
x1 <- rnorm(N)
x2 <- rnorm(N)


beta_0 <- 2
beta_1 <- 1
beta_2 <- -1


delta_0 <- 2
lambda <- rnorm(N)
epsilon <- delta_0 * x2 + lambda

y <- beta_0 + beta_1*x1 + beta_2 * x2 + epsilon

model<-lm(y~x1 + x2)
summary(model)


beta_0
beta_1
delta_0 + beta_2
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.