7
$\begingroup$

I understand that if we have a simple model such as:

$$Y_t=\rho Y_{t-1}+\epsilon_t$$

where $\rho$ is less than one in absolute value then we have a stationary process. If $\rho$ equals one then we have a unit-root and we can use the augmented Dickey Fuller test to test against the presence of a unit-root and thus for stationarity.

What confuses me is what if $\rho$ is equal to, say, 1.5. The data will certainly not be stationary. If I simulate a $Y_t$ in R with 120 observations and $\rho$ equal to 1.5, we get process that looks exponential. If I then run a simple OLS regression of $Y_t$ on $Y_{t-1}$ I get a correct estimate of $\rho$ and a bad estimate of the intercept:

set.seed(15) 
n=120
error=rnorm(n, mean = 0, sd = 1);
b=1.5
y=vector(length=120)
for (i in 2:n){
y[1]=error[1]
y[i]=b*y[i-1]+error[i];
}
data=cbind(yt,lag(yt,-1))
data=data[-1,]
data=data[-120,]
results1=lm(data[,1]~data[,2])
summary(results1)
ts.plot(results1$residuals)
acf(results1$residuals)

The plot of the ACF of the residuals looks OK as well. It seems that OLS does a good job at capturing the true model (the $\rho$ coefficient, not the intercept). This is something that I thought cannot happen with nonstationary series.

Furthermore, and this is what puzzles me the most. I always thought that to test stationarity we would just use the augmented Dickey Fuller test. But this is just a test of the null of unit root. The null is that $\rho$ equals one and the alternative is that $\rho$ is less than one, so that it is stationary. If I run it on this artificial data it cannot reject the presence of unit root. Is this enough to say that the data is not stationary? I guess my confusion is, how can we test for stationarity when we don't know if the coefficient is bigger than 1?

I suspect my question is very silly but I cannot get my head around this for the moment. Any suggestions?

$\endgroup$
  • $\begingroup$ Tests are of the null hypothesis, and for the alternative hypothesis. $\endgroup$ – Alexis Jan 18 at 18:29
7
$\begingroup$

I think your understanding is quite correct. The issue is, as you noticed, that the DF test is a left-tailed test, testing $H_0:\rho=1$ against $H_1:|\rho|<1$, using a standard t-statistic

$$ t=\frac{\hat\rho-1}{s.e.(\hat\rho)} $$ and negative critical values ($c.v.$) from the Dickey-Fuller distribution (a distribution that is skewed to the left). For example, the 5%-quantile is -1.96 (which, btw, is only spuriously the same as the 5% c.v. of a normal test statistic - it is the 5% quantile, this being a one-sided test, not the 2.5%-quantile!), and one rejects if $t< c.v.$. Now, if you have an explosive process with $\rho>1$, and OLS correctly estimates this, there is of course no way the DF test statistic can be negative, as $t>0$, too. Hence, it won't reject against stationary alternatives, and it shouldn't.

Now, why do people typically proceed in this way and should they?

The reasoning is that explosive processes are thought to be unlikely to arise in economics (where the DF test is mainly used), which is why it is typically of interest to test against stationary alternatives.

That said, there is a recent and burgeoning literature on testing the unit root null against explosive alternatives, see e.g. Peter C. B. Phillips, Yangru Wu and Jun Yu, International Economic Review 2011: EXPLOSIVE BEHAVIOR IN THE 1990s NASDAQ: WHEN DID EXUBERANCE ESCALATE ASSET VALUES?. I guess the title of the paper already provides motivation for why this might be interesting. And indeed, these tests proceed by looking at the right tails of the DF distribution.

Finally (your first question actually), that OLS can consistently estimate an explosive AR(1) coefficient is shown in work like Anderson, T.W., 1959. On asymptotic distributions of estimates of parameters of stochastic difference equations. Annals of Mathematical Statistics 30, 676–687.

$\endgroup$
  • $\begingroup$ Explosive process if $\rho < -1$ also. $\endgroup$ – Alexis Jan 18 at 16:47
  • $\begingroup$ OK, but in that case, the reasoning that a left-tailed DF test does not have power does not seem so convincing anymore, I'd say. $\endgroup$ – Christoph Hanck Jan 18 at 17:13
  • $\begingroup$ Fair point about the test, but important to bear in mind that explosions can result from ever increasing oscillation (i.e. $\rho<-1$), as when 'correcting' processes are delayed in an oscillating system. $\endgroup$ – Alexis Jan 18 at 18:26
  • $\begingroup$ Relatedly, shouldn't the null be $H_{0}:|\rho| =1$? $-1$ and $1$ are both unit root, and I have never seen ADF or DF as teasts of the null that the unit root is specifically positive as you have written it (although it may well be applied in circumstances only considering positive values). $\endgroup$ – Alexis Jan 18 at 18:30
  • $\begingroup$ I, on the contrary, have only ever seen the DF test being applied to the null $\rho=1$, and that is the case under which its properties are derived. Of course, $\rho=-1$ also implies a unit root (although I am not sure when it might arise in practice), but that would call for another testing approach, and OP's question was about DF tests. See, e.g., top of p. 815 in jstor.org/stable/2171846?seq=3#metadata_info_tab_contents $\endgroup$ – Christoph Hanck Jan 19 at 5:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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