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?


1 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




Now this is simply a regression


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) { 

> 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 $\alpha_1=2$.

  • $\begingroup$ +1 - What is your definition of spurious though? If you include a time trend, e.g. $Y_t = a_0 + a_1X_t + a_2t + e_t$, $a_1$ should go to zero asymptotically. $\endgroup$
    – Andy W
    Oct 3, 2013 at 14:49
  • $\begingroup$ Spurious - showing relationship when it's not there. Like a regression between the two not cointegrated unit roots. In this case the relationship is clear. $\endgroup$
    – mpiktas
    Oct 3, 2013 at 16:59
  • $\begingroup$ They are related. They both have the time trend in common. If we detrend them then the relationship disappears. But for level variables the relationship is the common trend. You cannot argue that the variables which share common trend are not related. $\endgroup$
    – mpiktas
    Oct 8, 2013 at 14:15
  • $\begingroup$ +1 - Although I would argue it is possible to argue that the variables with common trend are not related. As schoolkids get older they get better at math (on average anyway) and they get taller, but in a very important sense being taller is not related to being better at math. $\endgroup$ Oct 18, 2019 at 13:01

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