Nelsen 18 is not Schur-increasing in its parameter #
At threshold v=1/9, with q=θ/(u-1), S=e^q+e^{θ/(v-1)}, L=log S and p=e^q/S,
the conditional CDF on the positive region L<-θ equals
e^q q²/(S L²)=p(1+(-log p)/(-L))². For θ=4 this is at most
p(1-log p/4)² ≤ (2/5)(1+log(5/2)/4)² ≤ 5/8, because p<1-e^{-1/2}≤2/5.
For θ=2, at the point with e^q=e^{-9/4}/4 (so p=1/5) it is
(9/4+2log 2)²/(5(9/4-log(5/4))²)>5/8. The convex test z ↦ (z-5/8)₊ therefore
separates the two conditional CDFs, so C₂ ≤_∂S C₄ fails.
Instances For
Equations
- Verification.N18Schur.F θ K x = 1 + θ / Real.log (Verification.N18Schur.S θ K x)
Instances For
Equations
Instances For
On the positive region the CDF section is differentiable with derivative G.
On the zero region the CDF section is locally zero.
The bound for θ=4 #
h(p)=p(1-log p/4)² is monotone on (0,1).
The witness for θ=2 #
The witness point x* = 1 + 2/q* with q* = -9/4 - log 4.
Equations
- Verification.N18Schur.qStar = -(9 / 4) - Real.log 4
Instances For
Equations
Instances For
The separation #
The threshold v=1/9.
Equations
Instances For
For θ=4, the conditional CDF at v=1/9 is at most 5/8 almost everywhere.