Why Stationarity Tests Reject Long Series but Accept Their Segments
Summary
The answer explains how a very long time series can fail a stationarity test even when shorter sections pass. It uses two stationary sequences with different means as an illustration: joining them creates a series whose mean changes at the transition, so the combined process is nonstationary.
The explanation suggests that segmentation can conceal structural changes. Each segment may look stationary on its own, while the full sample combines regimes with different properties. The example focuses on a shift in the mean and does not diagnose the particular data in the question or establish that every KPSS rejection reflects a regime change. Test power, sample behavior, and other forms of nonstationarity may also matter when interpreting results.
Key ideas
- A series formed by joining stationary segments with different means is nonstationary as a whole.
- Testing short segments separately can miss changes between segments.
- A full-sample stationarity rejection can reflect a shift in the mean across regimes.
- The illustrative explanation does not determine the cause of rejection in a particular dataset.
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# Testing for stationarity in large sample sizes
# Testing for stationarity in large sample sizes
I keep struggling with testing 9 samples if they are stationary. Each of these samples is a real valued time series with 714.000 values. If I use the KPSS test with the each compleete sample set, the hypothesis is rejected. But, if I split each sample in 20 parts of equal size, then test all these sample parts with KPSS in all most all cases KPSS accepts the sample parts to be stationary.
But, I cannot find any good explanation for this behaviour. Can you give me any explanation for it and maybe a reference?
## Answer by Ryogi (score 3)
https://quant.stackexchange.com/a/7218
Here is a possible explanation. Consider $X_t \sim N(0,1)$ and $Y_t \sim N(1,1)$. Then $(X_t)_0^n$ and $(Y_t)_0^n$ are realizations from stationary time series and I would expect the null hypothesis of stationarity not to be rejected (compatibly with the size of your test). Instead, the sample $(Z_t)_1^{2n} = (X_1, \dots, X_n, Y_1, \dots, Y_n)$ is drawn from a non-stationary process (the mean is not constant) and a test with enough power will in general reject the null of stationarity.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.