https://www.machinelearningplus.com/time-series/augmented-dickey-fuller-test/
In 2), why do we only care about $\alpha=1$ to get a unit root. I guess if $\alpha<1$ we have some mean reversion, but why dont we care about $\alpha>1$?
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1$\begingroup$ FYI $\endgroup$– SeymourCommented Feb 26, 2020 at 21:55
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It's not that "we don't care". Mathematically, it is not possible to test the null hypothesis $\alpha > 1$, i.e. the model is explosive, against the alternative, say, $\alpha \leq 1$.
One can characterize the distribution of, say, the Dickey-Fuller-type $t$-statistic for each $\alpha > 1$. The limit distribution is a Cauchy distribution. However, the normalizing factor of the statistic depends on $\alpha$, in contrast to the stationary ($|\alpha| < 1$) case.
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$\begingroup$ The null hypothesis of the ADF test is presence of unit root, $\alpha=1$. The distribution of the test statistic under the null does not change if we consider alternative $\alpha<1$ or $\alpha>1$. Why should we care about the distribution of the test statistic under the alternative (except for power analysis)? Is it probably difficult to define the $p$-value for the two-sided alternative? $\endgroup$ Commented Feb 27, 2020 at 6:31
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$\begingroup$ "except for power analysis"---power is the only thing that matters, for any test. Unit roots tests already have low power against one-sided stationary alternative. A two-sided test would have even lower power. (p-values are not an issue, they're easily obtainable.) $\endgroup$– MichaelCommented Feb 27, 2020 at 17:35
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$\begingroup$ It's also not set in stone that one must take $\alpha = 1$ as the null. Tests where $\alpha = 1$ is the alternative, like KPSS, exists because power issues with unit root tests. For the explosive case, such complementary test would not be possible. $\endgroup$– MichaelCommented Feb 27, 2020 at 17:37
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$\begingroup$ I understand the part of the question but why dont we care about $\alpha>1$ as considering that as the alternative, not the null. As far as I understand from your comment, technically this would not be a problem, just as well as $H_1\colon \alpha\neq 1$? $\endgroup$ Commented Feb 27, 2020 at 20:29
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$\begingroup$ And now I realize I misread your post initially. I was thinking of $H_0\colon\alpha=1$ and considering different alternatives, while you were focusing on $H_0\colon\alpha>1$. $\endgroup$ Commented Feb 27, 2020 at 20:46