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I am trying to analyse a time series. I want to get only quantitative results (so, I'm excluding things like "looking at this plot we can note..." or "as you can see in the chart ...").

In my job, I analyse stationarity and persistence. First, I run ADF test and get "stationary" or "non-stationary" as results. Then, I need to work on persistence. To do so, I use ACF.

My question is: suppose I got "non-stationary" time series. Is it right to run ACF on it (without differencing)? I would like to comment upon stationarity and persistency without having to differenciate (so, just run tests on the original data and getting "asnwers" like "strong positive persistence", "weak negative persistence", ...).

Thanks to who will even just read my question

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  • $\begingroup$ How are you defining persistence? Degree of autocorrelation? $\endgroup$
    – Dayne
    Commented Dec 22, 2021 at 4:41

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Assuming by persistence you mean degree of autocorrelation.

ACF is simply a function that can be derived for any stochastic process, whether stationary or not. But the estimation of ACF for non-stationary, which is what you are doing, is the problem.

Generally, ACF is a function of both time, $t$ and lag, $h$:

$$\gamma(t,t+h) \equiv Cov[X_t,X_{t+h}]$$

For stationary processes, this is value is same for all $t$ and therefore, ACF is estimated only as a function of lag, $h$. If your process is however non-stationary then you cannot estimate the ACF from a given time series (a sample of stochastic process) as you will only have one value for each time point.

On the question of differencing, understand that it works for unit root processes. Non-stationarity can be due to other reasons as well (such as time trend, seasonality, etc.) which can even invalidate your ADF results. Assuming you are using ADF correctly and has identified a unit root (not just non-stationarity) then taking difference, testing for unit root again and then estimating ACF would be the right way to go.

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    $\begingroup$ Thank you very much for the clear explanation! $\endgroup$
    – user96624
    Commented Dec 22, 2021 at 20:02

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