I have seen in several papers that the $p$-value for the Dickey-Fuller (DF) test is reported for the test including trend and constant, then without the trend, and in the end in the absence of both. Can anybody explain to me why one would report all the results?
I thought the researcher would only report the result with the correct specification of the deterministic regressors a la Elder & Kennedy (2001).
But the more important question is, why does Stata not give me a $p$-value for the DF test if trend and constant are suppressed, and how can I obtain it?
Best Answer
As for your first question, yes, the applied researcher should think about the correct specification of the deterministic part. It is definitely not appropriate (something you did not suggest, to be sure!) to try all three and pick the preferred result.
As for the simulation of the critical values in the case without constant, see this post (disclosure: of mine).
There, we see that the test statistic (with superscript $n$ for "no" constant) converges in distribution to $$ t^n\Rightarrow \frac{W(1)^2-1}{2 \left\{\int_0^1W(r)^2dr\right\}^{1/2}} $$ under the null. It is useful to write $$ t^n\Rightarrow \frac{W(1)^2-W(0)^2-1}{2 \left\{\int_0^1W(r)^2dr\right\}^{1/2}} $$ as $W(0)=0$ because we may then write all three cases in a unified format $$ t^i\Rightarrow \frac{W^i(1)^2-W^i(0)^2-1}{2 \left\{\int_0^1W^i(r)^2dr\right\}^{1/2}} $$ where $i\in\{n,\mu,\tau\}$ and $$W^\mu(r)\equiv W(r)-\int_0^1W(s)d s$$ as well as $$W^\tau(r)\equiv W(r)-(4-6r)\int_0^1W(s)d s-(12r-6)\int_0^1sW(s)d s$$ That is, depending on whether you include nothing, a constant, or also a time trend, the distribution of the test statistic changes, where the functionals naturally reflect the type of detrending performed. (There are also conditions on how the series is actually generated. For example, if the series has drift, not accouting for this with a trend in the test regression would invalidate the results for $t^n$ and $t^\mu$.)