3
$\begingroup$

Suppose I have two I(1) time series X and Y, and I want to know whether X and Y are "related" (for some definition of "related").

The standard cointegration approach defines relationship as cointegration, and says that X and Y are cointegrated if some linear combination of X and Y is stationary. To test whether X and Y are cointegrated, you perform a regression on X and Y, and test for stationarity of the residual errors.

It seems to me like another approach might be to difference the I(1) time series X and Y, to get new I(0) time series X' and Y', and to use a standard linear regression relationship test on X' and Y' (i.e., perform a regression to get Y' = aX' + b, and use a t-test to see whether a is significantly non-zero). You could then define X and Y to be related if X' and Y' pass this test.

Is this second approach valid, or do you get spurious relationships? What's the difference between this approach and the cointegration approach, or what are the advantages of the cointegration definition?

$\endgroup$
4
$\begingroup$

If time series are cointegrated they admit VECM representation according to Granger Representation theorem. This is scantily explained in this wikipedia page. So if we have I(1) process:

$$X_t=\mu+\Phi D_t+\Pi_1 X_{t-1}+...+\Pi_p X_{t-p}+\varepsilon_t$$

it admits VECM representation

$$\Delta X_t=\mu+\Phi D_t+\Pi X_{t-p}+ \Gamma_1 \Delta X_{t-1}+...+\Gamma_p \Delta X_{t-p+1}+\varepsilon_t$$

What this means is that if you difference time series and do a linear regression as per your second approach you are not including the cointegration term $\Pi X_{t-p}$. So your regression suffers from omitted variable problem, which in turn makes the test you are trying to use not viable.

$\endgroup$
2
  • $\begingroup$ The link for the example is dead. I would be curious to see what it was all about. $\endgroup$ Oct 30 '14 at 20:45
  • $\begingroup$ Yes the link is dead, and I forgot what I found there. I am removing the note. $\endgroup$
    – mpiktas
    Oct 31 '14 at 12:40

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.