1
$\begingroup$

I can't conceive a regression which holds these three conditions. Assuming the properties of homoskedasticty and autocorrelation, the conditional mean of the error term would always be different from zero, like in the picture below.

enter image description here

Is there a case where these conditions hold simultaneously?

$\endgroup$

1 Answer 1

2
$\begingroup$

A picture of the type you draw would indeed indicate that there is serial correlation, but the "real" reason for this seems to be lack of exogeneity in that a linear regression is used to fit a data set that appears to be generated by some polynomial model, maybe of order 5 or so.

Suppose, as a simpler example that is however enough to make the point, that $y_t$ is actually generated as (the "true model") $$ y_t=\beta_1x_t+\beta_2x_t^2+u_t, $$ with $u_t$ independent of $x_t$, such that the regressors are exogenous. If we however attempt to model $y_t$ with a simpler function like $y_t=bx_t+v_t$, we face an issue of omitted variable bias. [Note that we leave out an intercept in both the true and the simpler linear model to keep the algebra short.]

To see this, add and subtract $bx_t$ to obtain $$ y_t=bx_t+(\beta_1-b)x_t+\beta_2x_t^2+u_t. $$ Hence, the residual $v_t$ in the linear model is given by $$ v_t=(\beta_1-b)x_t+\beta_2x_t^2+u_t $$ Then, $$E(v_t|x_t)=(\beta_1-b)x_t+\beta_2x_t^2\neq0,$$ so that the error is not independent of the regressor in the simple model.

We can say a little more given more specific assumptions. Suppose $x_t\sim N(\mu,1)$ and $b$ is the linear projection coefficient of the projection of $y_t$ on $x_t$ (this is what an OLS estimate would be consistent for, see here). In this case, we would obtain (see here) $$ b=\frac{E(x_ty_t)}{E(x_t^2)}=\frac{E(x_t(\beta_1x_t+\beta_2x_t^2+u_t))}{\mu^2+1}=\beta_1+\beta_2\frac{\mu^3+3\mu}{\mu^2+1} $$ Then, $$E(v_t|x_t)=-\beta_2\frac{\mu^3+3\mu}{\mu^2+1}x_t+\beta_2x_t^2$$ and one could play around with regions for $x_t$ where $E(v_t|x_t)>0$ and $E(v_t|x_t)<0$. E.g., for $\beta_2<0$ (and when $(\mu^3+3\mu)/(\mu^2+1)>0$), $E(v_t|x_t)>0$ when $x_t\in[0,(\mu^3+3\mu)/(\mu^2+1)]$.

Here is a numerical example.

n <- 10000
mu <- 1
x <- rnorm(n, mu)
beta_1 <- 4
beta_2 <- -1
y <- beta_1*x+beta_2*x^2 + rnorm(n,sd=.1)

plot(x,y, cex=.1, col="grey")
abline(lm(y~x-1))
abline(v=0, lty=2)
abline(v=(mu^3+3*mu)/(mu^2+1), lty=2)

enter image description here

Consider instead an example with "actual" serial correlation, like in $y_t=\beta t+u_t$, $u_t=\rho u_{t-1}+v_t$. Here, there is no relationship between the regressor and the error:

n <- 1000
u <- arima.sim(list(ar=0.9), n=n)
beta <- 0.01
x <- 1:n
y <- beta*x + u
plot(x, y, type="l", lwd=2, col="lightblue")
abline(a=0, b=beta)

enter image description here

For comparison, a data-generating process for your plot might be something like

n <- 60
x <- -n:n
y <- 0.1 + 0.02*x - .01*x^2 + .00003*x^3 + .0000001*x^4 + .0000001*x^5 + rnorm(1)
plot(x,y, type="l", lwd=2, col="lightblue")
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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