The following table is derived in the obvious way from the rankings of "m" in CI plots of the regressions of two of our average slopes against all available singleton lengths (Average slope ?Aube? is for the e coefficient of the regression c on (u,e,u*e) and average slope Aubqe is for the e coefficient in the regression c on (e,u,u*e,u^2).
a1 a3 b1 b47 c1 c2 C S C S c S c S C S C S "Het"
N1 Aube H H L L H H H H L L L L 0 Aubqe H H L L H H H H L L L L 0
N2 Aube L L L H H H H H L L H L 2 Aubqe L H L H H L H H L L H L 4
N3 Aube L L H H L L H H H H L L 0 Aubqe L L H H L L H H H H L L 0
R1 Aube H L H H H L H L L L L H 4 Aubqe H L H H H H H L L L L H 3
R2 Aube L L L H L H L L H H H H 2 Aubqe L L L H H H L L H H L H 2
R3 Aube L H L H L L H H H H L L 2 Aubqe L H L H L L H H H H L L 2
For example, the N1 Aube row was derived from this set of values:
a) both a1 values are among the highest 6 b) both a3 values are among the lowest 6 c) both b1 values are among the highest 6 d) both b47 values are among the highest 6 e) both c1 values are among the lowest 6 f) both c2 values are among the lowest 6
So here?s the question:
What is the probability of ?het = 0?, as in the case of the N1 Aube and Aubqe rows and the N3 Aube and Aubqe rows. That is, what is the probability that every one of the six folds will have the same value (H or L) for both S and C?
The reason why this question is so important is that the above data paradigm is kind of the ?inverse? of the one we saw for average slope Auq, where ?het = 6 for N1 and 5 for N2. And in that sense, the scientifically interesting thing about Aube and Aubqe is that the solution to the mystery of their behavior lies in the Sherlock Holmes story whose plot turns on the fact that the dog DIDN?T bark:
