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Is it valid to say that the Dickey-Fuller test, tests for a random walk?

Since the AR(1) process $Y_{t} = \rho Y_{t-1} + e_{t}$ with $\rho = 1$ is the same as the random walk. (Next value is maximum correlated with the previous since $\rho = 1$ + the unpredicted term.

And as wikipedia says, a unit root is present if $\rho = 1$. The model is non-stationary, which we know random walks are.

The reason for asking this is that it feels like an easy way of explaining it and intuitive.

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  • $\begingroup$ the short answer is Yes. Most people use augmented DF (ADF) test though. Also, look at the different settings such as trend stationarity $\endgroup$ – Aksakal Jul 15 '15 at 16:56
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It would perhaps be important to point also that the DF and ADF test for a Random Walk/Unit Root, by setting the existence of the unit root as the null hypothesis.

This was one of the criticisms towards these tests, since they "break from tradition": if we "suspect" the existence of a unit root ($\rho =1$ instead of, say, $\rho =0.99$), then the established approach would be to set as the null hypothesis the "trend-stationarity" hypothesis, and then attempt to reject it at the conventional significance levels (the latter reflecting an attempt to "preserve" the null, by keeping the Type I error low, i.e. by keeping low the probability of false rejection of the null).

The most well known test of this second approach (where the null hypothesis is the absence of a unit root), is the Kwiatkowski, Phillips, Schmidt and Shin (1992) one (KPSS).

The current consensus appears to be that a researcher should use both approaches, and see whether they agree in their conclusions. Namely, if the (A)DF test cannot reject its null, while the KPSS rejects its null, the data provide evidence in two different ways that the series has a unit root/ is a Random Walk.

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  • $\begingroup$ Alecos made an excellent comment. Another problem is that the ADF unit root test ( using the AR(p) is different depending on whether one includes an intercept and or a trend term. Knowing what to use is tricky and the result can change depending on what you include. This is part of the reason why Perron and I think NG developed their test. $\endgroup$ – mlofton Jul 15 '15 at 17:47

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