Re: [Gretl-devel] ADF p-value glitch?
Am 05.11.18 um 15:51 schrieb Allin Cottrell:
On Mon, 5 Nov 2018, Sven Schreiber wrote:
>
> open wgmacro.gdt
> adf 0 log(income) --ct # gives pos. test stat
>
> eval urcpval($test, $nobs, 1, 3)
> eval urcpval(abs($test), $nobs, 1, 3) # same
OK, this last line was foolish, I was still in my mind in the world of
negative ADF test stats, whereas here the point is exactly that it is
positive.
True, the test statistic is not in the rejection region. Nonetheless,
the p-value of 0.9971 gives the probability of obtaining an ADF tau
value less than $test under H0, doesn't it?
Well, is a p-value a probability? Suppose in general I run a one-sided
test with H0: b=0 and and H1: b<0. The decision rule is that whenever
the p-value < alpha (which is my own chosen significance level) then
reject H0 in favor of H1. Equivalently reject when the test statistic is
beyond my critical value, and beyond here with an underlying t-stat
would mean "to the left of".
The critical value is <0, and If I get bhat > 0, my t-stat can never be
to the left of the critical value. So one certainly shouldn't reject in
favor of H1. (Maybe in favor of some other hypo, but that is not on the
table here.) So how do I ensure this non-rejection with the p-value
output, i.e. prevent that p < alpha for any alpha \in (0,1)? Either set
p >= 1 or set it to NA.
So no, I don't think it is a probability in general, only in the
standard/easy cases can it be given such an interpretation.
Coming back to this DF case, if for \hat{(rho-1)} > 0 (or rhohat > 1)
one gets already a p-value of <1, what happens if by chance you land
exactly on the H0 value rho = 1? By monotonicity it would also be <1,
but surely if you have an estimate that equals your H0 value you want to
have a p-value of 1!
Well, the answer is given by urcpval(0, 0, 1, 3), which yields
0.99628822. This can't be right!
cheers,
sven