What Rejecting the ADF Unit-Root Null Says About Stationarity

Summary

The document asks how the Augmented Dickey-Fuller test's unit-root null relates to stationarity in an autoregressive process. The response clarifies that the test is framed around a unit root, with an alternative in which the relevant root is less than one. Rejecting the null is presented as evidence against a unit root and in favor of stationarity, rather than as proof that every root has been located individually.

The explanation is brief and its wording about polynomial roots is not fully precise. In an AR model, stationarity depends on the locations of all characteristic roots under the chosen polynomial convention, while the standard ADF test targets a unit root in the tested process and has a specified alternative. The document gives no derivation, test-statistic distribution, lag-selection guidance, or discussion of deterministic terms and finite-sample limitations. Readers should treat the answer as an introductory distinction between a unit-root null and stationarity, not a complete account of root conditions.

Key ideas

• The ADF null hypothesis tests for a unit root in the series.
• Rejecting the unit-root null supports the specified stationary alternative.
• Stationarity conditions for an autoregressive model involve characteristic-root locations and depend on polynomial convention.
• The response does not give the full test setup, derivation, or implementation guidance.
• The stated root interpretation is brief and requires care when generalized to all roots.

Tags

# Augmented Dickey-Fuller Questions


# Augmented Dickey-Fuller Questions












I've been searching in bibliography about this test applied to an AR(p) model. $$Q(L)(Y_{t})=c+\epsilon_{t}$$

Where L represent the Lag Operator and $Q=1-\phi_{1}x-.....-\phi_{p}x^{p}$ is the polynomial expression associated to the model.

I know that if $Q(r)=0$ implies $|r|>1$, then the process is stationary (at least in weak sense).

My question is: Why the Null Hypothesis of Augmented Dickey-Fuller test is stated as: "$r=1$ is a root of the polynomial"? Rejecting that hypothesis implies that every single root of Q lies outside the unit circle??

I'm new at this area so every recommendation or suggestion will be useful. Thanks.

## Answer by Alejandro Andrade (score 1)

https://quant.stackexchange.com/a/22782

EDITED

Your interpretation is wrong. If r>1 (not in absolute value) the series follows and explosive and therefore is not stationary. If you reject the unit root it means that the series does not have a unit root, because as it is stated in the comments the hypothesis is h0: r=1 H1: r<1. so rejecting means that all every single root of Q lies inside the unit circle.

Thanks Richard Hardy for the correction

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.