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Why is not the Dickey-Fuller test applied directly on :

$Y_t = \rho Y_{t-1} + u_t$

instead of :

$\Delta Y_t = (\rho-1) Y_{t-1} + u_t$.

Many papers apply the Dicker-Fuller on the first difference operator and later use Augmented Dickey Fuller also with difference operators but not explaining why not dealing directly with $Y_t$.

Christoph Hanck
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Bastiat
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1 Answers1

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You are right that this is pure convention.

The only practical reason (I am aware of) is that the unit root null hypothesis $H_0:\rho=1$ is evidently equivalent to $H_0:\rho-1=0$. Hence, the default t-statistic produced by standard software packages directly yields the Dickey-Fuller statistic.

Recall, however, that the critical values and p-values supplied by standard packages are typically taken from the standard normal or t-distribution, which are not the right ones for the Dickey-Fuller test.

Christoph Hanck
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    Thanks. I'm missing something however. When applied on an Autoregressive model of order 1,

    $\Delta S_t = \alpha + \beta t + (\phi -1) S_{t-1} + e_t$

    we have the following hypothesis :

    Null Hypothesis: A unit root is present in the autoregressive model (of order 1), the process is non stationary.

    Alternative Hypothesis: The model is stationary.

    Here what is stationary ? $\Delta S_t$ or $S_t$ or both ?

     – Bastiat Oct 16 '18 at 11:49
  • In $S_t$, as that is what you are testing the unit root null hypothesis on, and taking differences and tesing a zero coefficient in the transformed model is just an equivalent reformulation of the null. – Christoph Hanck Oct 16 '18 at 12:37