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Is there a difference between right-tailed ADF and standard ADF tests in Eviews?

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  • $\begingroup$ To my knowledge, the difference is in the null/alt hypotheses. With the ADF test, we test the null of 'unit root' vs an alternative of 'no unit root'. With the right-tailed ADF, the null is 'unit root' versus an alternative of 'mildly explosive process' (i.e a process integrated of higher order than 1). In other words, you should be using a test like this if you expect your data to be characterised by a fairly explosive process. This is just on an intuitive level, but you should check all the details with Caspi (2017) jstatsoft.org/article/download/v081c01/1161 $\endgroup$
    – EB3112
    Mar 8 at 20:09

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Yes there is a difference. The right tailed test in Eviews is designed based on: Caspi, 2013.

If you're looking for just the operational difference between the two tests: The standard ADF test is also one sided with $H_0: \phi = 0$ against $H_1: \phi<0$ (so in practical terms the null hypothesis is $H_0 \geq 0$). Rejection of null would mean no unit root and the series is stationary.

For the right tailed test the alternative is $H_1: \phi >0$. So the rejection of null means that the process has no unit root but is non-stationary (explosive type).

I do want to mention that I do not know theoretical difference as in how would one go about designing such a test because the null distribution is same but the critical values for different alternatives need to different. But how, I don't know.

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  • $\begingroup$ may still have unit root class of non-stationarity: How so if the null of a unit root is being rejected? $\endgroup$ Mar 10 at 10:55
  • $\begingroup$ @RichardHardy: Yeah I am not fully sure ofy statement myself. By unit root class I mean that explosive series such as say $y_t = 1.2 y_{t-1} + \epsilon_t$ is also unit root type non stationarity because it will become stationary after differencing. In this example by first difference. $\endgroup$
    – Dayne
    Mar 10 at 14:23
  • $\begingroup$ I was actually planning of posting a question on CV because I am unable to see how the critical values be decided from df table. After all the null distribution is same. In usual tests, we take one side of the tail...I unable to directly see this here. $\endgroup$
    – Dayne
    Mar 10 at 14:29
  • $\begingroup$ I think this is a poor use of terminology, because the root of the characteristic polynomial is not 1. I suggest rephrasing. Moreover, I would think twice before differencing anything that does not have a unit. Instead of achieving stationarity, the result will be overdifferencing with all of its negative consequences. $\endgroup$ Mar 10 at 15:20
  • $\begingroup$ Point noted. Deleting the last part. $\endgroup$
    – Dayne
    Mar 11 at 8:47

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