$$ \begin{aligned} y_t &= a + bx_t + u_t, \\ u_t &= \phi u_{t-1} + e_t \end{aligned} $$

where $ e_t$ follows a White Noise process. Let Breusch-Godfrey LM test statistic be strictly greater than the relevant $\chi^2$ critical value.

I had a similar question in my problem set where it asks me to determine the stationarity of $y_t$. In the solution it concluded that since there exist autocorrelation in the given model, it implies that $\phi < 1$.

I am suspicious of this argument and want to know if there is a hidden assumption that was made before. Any idea will be appreciated.

  • $\begingroup$ Clearly, if the parameter equals $1$, your error term $u_t$ is the infinite sum over all past $e_t$ (i.e., $u_t = \sum_{i=0}^{\infty} e_i$), and then the variance of the error term $u_t$ explodes. Thus, $u_t$ is not stationary and then, neither is $y_t$. It's a strange choice to use the BGLM Test for this, because it assumes normality of $e_t$. You could have achieved a more general (but asymptotic) result using the ADF test. $\endgroup$
    – Jeremias K
    Commented Apr 17, 2016 at 18:46
  • $\begingroup$ Oh, for the record, the conclusion that the process is stationary because there exists autocorrelation is incorrect. In the nonstationary model with $|\phi| < 1$, you would still have (severe!) autocorrelation. $\endgroup$
    – Jeremias K
    Commented Apr 17, 2016 at 18:48
  • $\begingroup$ i think the reason for not using ADF is that it thinks unnecessary to check it since it does already have AC problem .it does not trust standard errors of explanatory variables as it uses them in ADF. $\endgroup$
    – Esat Koç
    Commented Apr 17, 2016 at 18:55
  • $\begingroup$ @JeremiasK, are you sure BGLM assumes normality of errors? Would you have a reference for that? $\endgroup$ Commented Apr 18, 2016 at 18:16
  • $\begingroup$ Yeah, you can look it up in 'Econometric theory' of Davidson or 'Econometric Analysis' by Greene $\endgroup$
    – Jeremias K
    Commented Apr 19, 2016 at 19:14

1 Answer 1


Under weak stationarity, it must be that the covariance function $C(Y_t,Y_{t+\tau})$ depends only on $\tau$, the mean function $E(Y_t)$ is constant for all $t$, and these moments exist (are finite!).

Assume exogeneity of $U_t$ with respect to $X_\tau$ for all $\tau$. Then you can write

$C(Y_t,Y_{t+\tau}) = b^2C(X_t,X_{t+\tau}) + C(U_t,U_{t+\tau})$. Assume $X_t$ is a stationary process, you get

$C(Y_t,Y_{t+\tau}) = k_X(\tau) + C(U_t,U_{t+\tau})$. Substitution gives

$C(Y_t,Y_{t+\tau}) = k_X(\tau) + C(U_t,\phi^{\tau}U_t + \sum^{\tau-1}_{l=0}\phi^{l-1}e_{t+\tau-l}) = k_X(\tau) + C(U_t,\phi^{\tau}U_t)$, since $e_t$ is white noise. The last term is crucial for the weak stationarity of $Y_t$. You have $C(U_t,\phi^{\tau}U_t) = \phi^{\tau}C(U_t,U_t) = \phi^{\tau}V(U_t)$.

How can you obtain $V(U_t)$? Transform $U_t = \phi L U_{t} +e_t$ into $U_t(1-\phi L) = e_t$. Hence, $V(U_t) = V(\frac{e_t}{1-\phi L}) = V(\sum_{l=0}^{\infty}\phi^{l}e_{t-l})$, where the second equality holds only if $|\phi|\leq 1$.

Now, from the white noise assumption, you can simplify this last expression into

$\sum_{l=0}^{\infty} \phi^{2l} V(e_{t}) = \frac{1}{1-\phi^2}\sigma_e$. For $\phi=1$, you obtain that $V(U_t)$ is not finite, hence $C(Y_t,Y_{t+\tau})$ is not finite, and the process $Y_t$ is not stationary.

This shows that autocorrelated process may not be stationary.

However, it is also true that uncorrelated processes may not be stationary. Consider the simple model: $Y_t = t + e_t$. $E(Y_t) = t$, hence depends on $t$ which violates the stationarity assumption.


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.