3
$\begingroup$

I am reading this PDF:

https://warwick.ac.uk/fac/soc/economics/staff/gboero/personal/hand2_cointeg.pdf

where on pages 4 and 5 it says that if the residuals are stationary, the OLS regression is superconsistent even if the Y and X variables are non-stationary.

My question then is why is it necessary to estimate the Error Correction Model (Second step)? Isn't the stationarity of residuals a sufficient condition?

$\endgroup$
2
$\begingroup$

A rejection of the null of the Engle-Granger test indeed establishes cointegration (up to a type I error, of course).

But estimating the error correction model tells you interesting things about the economic question of interest. E.g., which variable adjusts to deviations from the equilibrium relationship in the previous period so as to restore convergenve towards the long-run cointegration vector.

Why negative $\alpha$? Suppose the error-correction equation is $\Delta y_t=\alpha(y_{t-1}-\beta x_{t-1})+e_t$. When $y_{t-1}-\beta x_{t-1}>0$, it means that the lagged value $y_{t-1}$ is "too large" relative to the cointegrating relationship, so that if $y_t$ is to contribute to restoring the equilibrium relationship, its change in $t$ must be negative, whence $\alpha<0$ is required.

$\endgroup$
6
  • $\begingroup$ Thank you Christoph. Also, on page 5 it says that the adjustment coefficient α must be negative. What happens if it is not negative? $\endgroup$
    – adrCoder
    Jan 24 '20 at 9:19
  • 1
    $\begingroup$ I made an edit which hopefully helps. $\endgroup$ Jan 24 '20 at 9:31
  • $\begingroup$ Is it possible to have stationary residuals but α>0? What happens in this case? Does it mean there is no cointegration i.e. the OLS is not superconsistent? Or the fact that the residuals are stationary imply that α will be >0? $\endgroup$
    – adrCoder
    Jan 24 '20 at 9:42
  • 1
    $\begingroup$ I would recomment to have a look for Granger representation theorem, which tells us about the relationship between cointegration and existence of an error-correction relationship. $\endgroup$ Jan 24 '20 at 9:43
  • 2
    $\begingroup$ @adrCoder, here are some answers regarding signs of the adjustment coefficient: 1, 2, 3. $\endgroup$ Jan 24 '20 at 11:46

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.