I want to apply the unit root test to the original equation before writing it in the Dickey-Fuller form.

The original equation is: $$ Y_t=b Y_{t-1} + u_t. $$ I want to test for unit root directly by testing if $b=1$. I read in Enders's book "Applied Time Series Econometrics" that this is equivalent to the Dickey-Fuller test. However he didn't mention what critical value to compare to.

My $t$-test for $b=1$ hypothesis is $-1$. Can you please help me with the critical value?

Assume, the critical value is $-1.95$; what conclusion about unit root can I draw from my test?

  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – gung - Reinstate Monica Oct 3 '16 at 0:33
  • $\begingroup$ It is not from a textbook and it is not for a course I'm taking rn. I'm reviewing my econometrics notes before I tutor a student. $\endgroup$ – Abed111 Oct 3 '16 at 0:43
  • $\begingroup$ @Abed, was my answer helpful, or do you need some extra elaboration? (Just checking since there was no response from you.) $\endgroup$ – Richard Hardy Oct 11 '16 at 18:49
  • $\begingroup$ Testing that (standardized) coefficient against the Dickey-Fuller distribution is the Dickey-Fuller test (not "equivalent to"). You can find the critical values in a Dickey-Fuller table. Testing it against the $t$-distribution is wrong, because it does not have a $t$-distribution under the null of a unit root, as mentioned below. $\endgroup$ – Chris Haug Feb 24 '17 at 15:10

Regarding your second question:
If the critical value were indeed $-1.95$ (which I don't think it is as the estimator distribution is nonstandard under the null of a unit root), a test statistic of $-1$ would not allow rejecting the null hypothesis of $b=1$ at the confidence level corresponding to the critical value -- because the test statistic is less extreme (closer to zero) than the critical value: $-1.95<-1<0$.

| cite | improve this answer | |

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.