Date: Dec 5, 2012 11:52 PM
Author: Ray Koopman
Subject: Re: Holy Moly, were you right about covariances for Rub and Rubq !!!!

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:

Het Prob
0 .021822
1 0
2 .389284
3 0
4 .519354
5 0
6 .069540

Here is an estimate of the distribution of "L-H Het" and "H-L Het",
again based on 10^6 trials:

Het Prob
0 .152731
1 .331035
2 .308409
3 .151769
4 .048634
5 .006345
6 .001077