Proposition: Different source weights with identical complete records

There is an admissible finite-kick calibration set containing two members with different source weights and identical complete laws. For that pair, every randomized binary test based on $n$ ideal independent records has errors satisfying ${\alpha }_0+{\alpha }_1=1$.

Proof. Take $\hat{n}=(0,1,0)$, $\gamma =1/4$, and three possible transition values

They are realized by $\varphi =\arcsin \sqrt{1/5}$ and ${\xi }_k=\arcsin \sqrt{t_k}-\varphi$. For an explicit admissible angle bound, choose

The other offset has smaller absolute value. For instance, the relevant elementary angle inequalities give

Hence $\varphi -\kappa >0$ and $\varphi +\kappa <0.72<\pi /4$. A compact set containing these fixed angles and the weight simplex satisfies (eq:calibration-set).

For the first member choose

Then $u^{(0)}=1/5,v^{(0)}=1/25$. For the second choose

Direct rational arithmetic gives $u^{(1)}=1/4,v^{(1)}=9/100$. Both members have

Their common complete law is

Let a randomized test declare the second member with probability $\psi (\mathbf{r})\in [0,1]$. The record-list laws are identical, so