I have linear time series regression model where the dependent variable Y is bounded between 0 and 1.

Using classical unit root tests (dickey-fuller and kpss), results would make you conclude that Y is non-stationary (cannot reject $H_0$ in dickey-fuller and kpss rejects).

However, at the same time the mean of Y cannot be infinite since it is bounded; and I assume that for the same reason neither the variance can (right?).

Recalling the definition of weak stationarity, I am probably missing something related to the autocovariances of Y as well - they should depend only on the lag, but I have no formal test for this.

Is then Y really non-stationary because of unit-root, even if it is bounded?

  • 1
    $\begingroup$ You probably want to ask about a particular form of nonstationarity -- presence of a unit root. If so, you could articulate that and make the question more concrete. $\endgroup$ Commented Oct 26, 2016 at 16:11
  • $\begingroup$ I see no reason why a variable should not drift from (say) 0 to 1 or from 1 to 0 over a lengthy period. Human lives offer many examples! $\endgroup$
    – Nick Cox
    Commented Oct 26, 2016 at 16:51
  • 3
    $\begingroup$ Take any time series dataset whatsoever. Rescale its values (via an affine transformation) to lie between 0 and 1. Being a mere change of measurement scale, this will leave almost all meaningful properties unaffected, including the results of unit root tests. Therefore boundedness is wholly unrelated to (evidence concerning) stationarity. $\endgroup$
    – whuber
    Commented Oct 26, 2016 at 16:52
  • $\begingroup$ Very useful comment, @whuber. Many thanks. I was wondering also if there is some theoretical result demonstrating that stationarity properties are preserved, for example, under any monotonic transformation/class of functions. Any guess/reference? $\endgroup$
    – Giuseppe
    Commented Nov 2, 2016 at 8:47

3 Answers 3


2nd-order stationarity requires constant mean, not just finite mean, constant variance, not just finite variance, and covariances depending only on the lag-length and not on the time index.

A stochastic process comprised of elements that have bounded support, will have a finite mean but not necessarily constant, a finite variance but not necessarily constant.

The existence of a unit root alone allows for a constant mean, but it gives an ever increasing variance and covariance.

So while a bounded variable may be non-stationary, postulating a priori that it has a bounded support does not allow for the existence of a unit root.

But, as @whuber commented, every observed data set is necessarily "bounded" (it has an observed minimum and an observed maximum), and by re-scaling it can be made to range in $[0,1]$.

So performing a unit-root test on a data set that is assumed to come from a variable with bounded support is invalid.


Unit root tests are very sensitive. Many bounded variables can be deemed nonstationary. If you look at the underlying calculations of the Dickey Fuller or ADF test, the core calculation is: Change in Y = Coefficient*Y Lag 1. And, the Coefficient has to be very statistically significant and negative. In plain English (besides any unit root language obfuscation) this is really about mean revertion. Y has to zig zag very rapidly up and down around the mean to be adequately mean reverting to pass the Dickey Fuller/ADF test. If it means revert, but not quite fast enough it will be deemed to have a unit root according to those tests.

In summary, your variable even though it is bounded does have a unit root. The question is whether this unit root is truly problematic or not? The first thing to look at is whether your variable is fully detrended or not. I suspect it is not. If you transform your variable to a change in that variable from one period to another, the unit root may disappear (or not). If it does not disappear, given that your variable is fully detrended now, your unit root may not pause much of a problem.


Non-stationarity does not imply infinite mean or variance. The infinite mean is obviously not stationary, but not the other way. A process with changing mean can be seen as non-stationary, at least in unconditional sense.

Autoregressive processes are stationary in long run, but in short horizon they look like unit root processes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.