[Gretl-users] None-uniform distribution of the p-value from ADF test in gretl when there is an intercept or trend?

Thanks Allin. That is interesting indeed. In other words, with an uniform distribution of the p-value [when y(t) = y(-1)], there is only 5% changes that we do reject the null of random walk at 5% level of significance. However, for other parameters of the intercept or trend, the p-value distribution is not uniform. So their size is incorrect? For example, when I set both the intercept and trend parameters to 0.1, I get the attached table. It seems that there is only a 0.18% chance that I will reject the null (reading the 2nd column of the table). I’m confused. Fred The following frequency table illustrates my point: y = 0.1 +y(-1) + 0.1*t y = y(-1) p-val ct p-val c p-val nc p-val ct p-val c p-val nc p-val bin cum. cum. cum. cum. cum. cum. 0.000 0.04% 0.00% 0.00% 2.24% 2.27% 2.28% 0.025 0.08% 0.00% 0.00% 4.51% 4.52% 4.44% 0.050 0.18% 0.00% 0.00% 7.07% 6.80% 6.79% 0.075 0.28% 0.00% 0.00% 9.57% 9.26% 9.02% 0.100 0.35% 0.00% 0.00% 12.09% 11.70% 11.31% 0.125 0.41% 0.00% 0.00% 14.23% 14.17% 13.88% 0.150 0.47% 0.00% 0.00% 16.51% 16.65% 15.94% 0.175 0.51% 0.00% 0.00% 18.78% 19.24% 18.24% 0.200 0.65% 0.00% 0.00% 21.47% 21.59% 20.47% 0.225 0.79% 0.00% 0.00% 24.05% 24.12% 22.92% 0.250 0.93% 0.00% 0.00% 26.28% 26.66% 25.40% 0.275 1.15% 0.00% 0.00% 28.71% 29.02% 27.85% 0.300 1.31% 0.00% 0.00% 30.81% 31.51% 30.49% 0.325 1.53% 0.00% 0.00% 33.17% 33.77% 33.15% 0.350 1.71% 0.00% 0.00% 35.60% 36.16% 35.29% 0.375 1.93% 0.00% 0.00% 38.22% 38.81% 37.77% 0.400 2.19% 0.00% 0.01% 40.91% 41.46% 40.51% 0.425 2.40% 0.00% 0.01% 43.10% 43.93% 42.97% 0.450 2.67% 0.00% 0.02% 45.68% 46.36% 45.83% 0.475 2.92% 0.00% 0.02% 48.26% 48.99% 48.31% 0.500 3.22% 0.00% 0.03% 50.85% 51.40% 50.64% 0.525 3.60% 0.00% 0.06% 53.34% 53.76% 53.21% 0.550 3.87% 0.00% 0.12% 55.91% 56.12% 55.55% 0.575 4.22% 0.00% 0.16% 58.54% 58.48% 57.88% 0.600 4.73% 0.00% 0.32% 60.83% 60.69% 60.57% 0.625 5.12% 0.00% 0.49% 63.59% 63.01% 62.98% 0.650 5.63% 0.00% 0.81% 66.29% 65.69% 65.55% 0.675 6.13% 0.00% 1.39% 68.74% 68.24% 68.23% 0.700 6.71% 0.00% 2.30% 71.49% 70.59% 71.14% 0.725 7.35% 0.00% 3.71% 74.05% 72.90% 73.77% 0.750 8.06% 0.00% 6.04% 76.53% 75.43% 76.29% 0.775 9.06% 0.00% 9.63% 79.17% 77.91% 79.17% 0.800 10.03% 0.00% 15.20% 81.86% 80.50% 81.73% 0.825 11.09% 0.00% 23.84% 84.31% 83.13% 84.38% 0.850 12.67% 0.00% 36.05% 86.71% 85.75% 86.74% 0.875 14.80% 0.00% 51.83% 89.00% 88.25% 89.39% 0.900 17.99% 0.00% 71.48% 91.66% 91.06% 92.07% 0.925 22.44% 0.00% 89.52% 94.19% 93.94% 94.69% 0.950 31.22% 0.00% 99.00% 96.75% 96.80% 97.43% 1.000 100.00% 100.00% 100.00% 100.00% 100.00% 100.00% # simulation of unit root process # y(t)=b0+rho*y(t-1)+b1*time+error # there are b0, rho, and b1 parameters to set scalar N = 500 # series length scalar R = 10000 # repeats or resampling nulldata N --preserve scalar rho = 1 # set rho from 0.0 to 1.0 scalar b0 = 0.1 # set intercept b0 = any value scalar b1 = 0.1 # set trend b1 to any value string outfilename = sprintf("unitroot,N=%d,b0=%3.1f,b1=%3.1f,rho=%3.2f.gdt",N,b0,b1,rho) loop R --progressive --quiet series e = normal(0,10) # error vector series y = 0 y = b0 + rho*y(-1) + b1*index + e ols y const y(-1) index --quiet scalar b_c = $coeff(const) scalar b_rho = $coeff[2] scalar b_t = $coeff[3] adf -1 y --ct --quiet scalar ct_p = $pvalue scalar ct_t = $test adf -1 y --c --quiet scalar c_p = $pvalue scalar c_t = $test adf -1 y --nc --quiet scalar nc_p = $pvalue scalar nc_t = $test print b_c b_rho b_t ct_p ct_t c_p c_t nc_p nc_t store "@outfilename" b_c b_rho b_t ct_p ct_t c_p c_t nc_p nc_t endloop open "@outfilename" #freq b_c --plot=display #freq b_rho --plot=display #freq b_t --plot=display freq ct_p --plot=display --min=0 --binwidth=0.025 #freq ct_t --plot=display freq c_p --plot=display --min=0 --binwidth=0.025 #freq c_t --plot=display freq nc_p --plot=display --min=0 --binwidth=0.025 #freq nc_t --plot=display