Spurious correlation

Question

I've read that if two time series, $Y_t$ and $X_t$, are trend stationary, then regressing $Y_t$ on $X_t$ results in a spurious regression because of an omitted time trend variable. Let $Y_t = \delta_0 + \delta_1t + u_t$ and $X_t = \gamma_0 + \gamma_1t + v_t$. I want to show that $Y_t$ is a linear function of $X_t$, a deterministic time trend and an error term. Can somebody please provide a mathematical proof of this?

Answer 1

The regression would not be spurious. If $Y_t=\delta_0+\delta_1 t+u_t$ and $X_t=\gamma_0+\gamma_1t+v_t$ then

$$t=\frac{1}{\gamma_1}X_t-\frac{\gamma_0}{\gamma_1}-\frac{1}{\gamma_1}v_t$$

and

$$Y_t=\delta_0-\frac{\delta_1\gamma_0}{\gamma_1}+\frac{\delta_1}{\gamma_1}X_t+u_t-\frac{\delta_1}{\gamma_1}v_t$$

Now this is simply a regression

$$Y_t=\alpha_0+\alpha_1X_t+\varepsilon_t$$

and it is possible to show that OLS estimates $\hat\alpha_0$ and $\hat\alpha_1$ are consistent and assymptoticaly normal with means $\delta_0-\frac{\delta_1\gamma_0}{\gamma_1}$ and $\frac{\delta_1}{\gamma_1}$ respectively, albeit with non-standard normalizing constants. The mathematical details can be found in this answer.

The consistency can be illustrated by the following code:

gend <- function(n) { 
    data.frame(x=1+2*1:n+rnorm(n),y=3+4*1:n+rnorm(n))
}

> set.seed(13)
> coef(lm(y~x,data=gend(10)))
(Intercept)           x 
  -1.291464    2.067586 
> coef(lm(y~x,data=gend(100)))
(Intercept)           x 
   1.396720    1.997408 
> coef(lm(y~x,data=gend(1000)))
(Intercept)           x 
  0.9864317   1.9999570 
> coef(lm(y~x,data=gend(10000)))
(Intercept)           x 
  0.9595726   2.0000065 

Here I generated two trend stationary variables with $\gamma_0=1$, $\gamma_1=2$, $\delta_0=3$ and $\delta_1=4$. As we see regression estimates approach the true values $\alpha_0=1$ and $\alpha_1=2$.