On Dec 5, 7:08 am, djh <halitsk...@att.net> wrote: > Here is the table for the covariances AubC and AubqC for the > regressions Rub = c on (u,e,u*e) and Rubq = c on (e,u,u*e,u^2) > respectively. > a1 a3 b1 b47 c1 c2 "H-L > C S C S c S c S C S C S Het" > > N1 AubC H L H L H L H L H L H L 6 > AubqC H L H L H L H L H L H L 6 > > N2 AubC H L H L H L H L H L H L 6 > AubqC H L H L H L H L H L H L 6 > > N3 AubC L L L L H H H L H H H L 2 > AubqC L L H H L H L L H H H L 1 > > R1 AubC L H L L L L H H H L H H 1 > AubqC L H L L L L H H H L H H 1 > > R2 AubC L H L L H L L H H L H H 2 > AubqC L H H L L L L H H L H H 2 > > R3 AubC L H H L L H L L H H H L 2 > AubqC L H H L L H L H H H L H 1 > > Note that this time, the ?het? singularity is ?H-L Het-ness?, rather > than ?L-H Hetness?, as was the case for the average slopes Auq, Aubu, > Aubqu in the last table posted. > > Quite a remarkable result, at least in my naive and ignorant opinion.
Sometimes the easiest way to answer such questions is via simulation instead of analysis. Here is a Monte Carlo estimate, based on 10^6 trials, of the distribution of "Het" when all 12! orderings of the 12 sample values are equally likely:
