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