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The code below generates series y, which by design is clearly non-stationary. The ADF test below was run with 12 lags to yield (what visually appear to be) uncorrelated residuals and it would have us conclude that y is stationary. What went wrong here?

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enter image description here

set.seed(100)
y<-rep(NA,100)
for (i in 1:100) {
 y[i]<-rnorm(1,mean=0,sd=i)
}
par(mfrow=c(1,3))
plot(y,type="l",main="y")
  
u<-urca::ur.df(y=y, type = "none",lags=12)
summary(u)
forecast::Acf(u@res,lag.max=70,type="correlation",main="ACF",xlab="")
forecast::Acf(u@res,lag.max=70,type="partial",main="PACF",xlab="")
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2 Answers 2

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The ADF test correctly concluded that the series does not have a unit root. The test does not say anything about stationarity beyond the mean of the series. Your series is nonstationary due to variance, not mean, so no wonder the ADF test did not react to that.

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The augmented Dickey-Fuller (ADF) test does not have an alternate hypothesis that the data "are stationary." Rather, the ADF tests for evidence that the coefficient $\beta$ in the below equation is not equal to 0 (equivalent to testing whether $\rho=1$ in the un-augmented Dickey Fuller test):

$$\Delta y_{t} = \alpha + \beta y_{t-1} + \delta t + \zeta_1 \Delta y_{t-1} + \dots + \zeta_{k}\Delta y_{t-k} + \varepsilon_{t}\text{where }\varepsilon \sim \mathcal{N}(0,\sigma^{2})$$

While it is true that a $\rho = 1$ (or $\rho = -1$) imply nonstationarity of a time series' data, but note that it is possible for $\beta=0$, and yet for $\sigma^{2} = f(t)$, so the ADF's usefulness in providing evidence that data are stationary is not robust to this issue.

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    $\begingroup$ This is an important point to clarify, Alexis. Book after book after rejecting the null hypothesis in ADF, conclude the series is stationary or trend-stationary when it really should say perhaps: mean-stationary or trend-mean-stationary. $\endgroup$ Aug 25, 2021 at 18:29
  • $\begingroup$ Regarding the phrasing of your last sentence: ADF is not robust to this issue. I think I understand what you mean, but on the other hand the ADF test showed itself to be robust to misspecification of variance; the test's result that primarily depends on the mean specification was as expected in spite of time-varying variance. $\endgroup$ Aug 26, 2021 at 14:01
  • $\begingroup$ @RichardHardy Beautiful! Thank you! See my edit. (And feel free to make further suggestions if that does not help enough or makes things worse. :) $\endgroup$
    – Alexis
    Aug 26, 2021 at 15:24

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