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Are all (non-explosive) time series either stationary around a deterministic trend or random walks?

If I run the ADF test and I can't reject the null of non-stationarity does it imply the series is a random walk?

In particular, if I run the ADF after detrending a series but it still gives me non-stationarity, does it imply that the series is a random walk or may I just have misspecified the trend? (the problem is that it is always possible to find a trend complicated enough that the series always looks stationary).

I know these are a lot of questions but they are all related (by my poor understanding of time series). No need to answer all of them.

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  • $\begingroup$ I describe a counterexample at stats.stackexchange.com/a/282875/919. As far as the "random walks" parts of your question go, maybe that depends on what you mean by this term. After all, you can view any series $(y_t)$ as given by an initial value $y_0$ and accumulating the successive differences $y_t - y_{t-1}$ for $t=1,2,3,\ldots.$ That's a very general form of random walk, but probably too general to be useful. What conditions do you impose on your random walks? $\endgroup$ – whuber May 6 at 21:35
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Are all (non-explosive) time series either stationary around a deterministic trend or random walks?

No. All non-explosive AR(1) processes with unit roots are random-walks, but this doesn't generalize to other processes or even AR processes with longer lags.

Any AR process where the characteristic equation has a unit root is non-stationary. For example: $y_{t} = 1.5 y_{t-1} - 0.5 y_{t-2} + \epsilon_t$ is non-stationary. Because $1 - 1.5z + 0.5z^2 = 0.5(z-1)(z-2)$ so z=1 is a root.

But only $y_{t} = y_{t-1} + \epsilon_t$ is a random walk.

In particular, if I run the ADF after detrending a series but it still gives me non-stationarity, does it imply that the series is a random walk or may I just have misspecified the trend?

You could have mis-specified the trend, but it may not be a random walk even if you have correctly specified it.

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