Manuscript Psy03-A@1 · Psy03-A.tex · theorem · b30eb6e9

Theorem: Conditional ranking certificate

Matched-budget comparisons and statistical certificates › A conditional exact-target ranking · lines 1005–1034 · proof included or linked

Psy03-A@1#thm:ranking, Psy03-A@1#eq:ranking

Suppose selected candidates have estimates $Q_{0}, Q_{1}$ , with simultaneous error bounds

\abs Q_{m} - Q_{m c_{m}} \leq r_{m} .

Under Assumption ass:target and tolerance matching,

q_{1} (d_{*}) - q_{0} (d_{*}) \geq Q_{1} - Q_{0} - r_{0} - r_{1} - (L_{0} + L_{1}) τ - ϵ_{0} - ϵ_{1} .

A positive remainder certifies the oriented ranking.

Without an exact-target response calibration, the simpler condition

Q_{1} - Q_{0} > r_{0} + r_{1}

certifies the ranking of the two actual selected checkpoints, not a universal equal-dose curve.

Proof. Apply the two estimate bounds and the two calibration bounds, then subtract.

Displayed formulas in this passage

unlabeled · line 1009
$\abs Q_{m} - Q_{m c_{m}} \leq r_{m} .$
Psy03-A@1#eq:ranking · line 1013
$q_{1} (d_{*}) - q_{0} (d_{*}) \geq Q_{1} - Q_{0} - r_{0} - r_{1} - (L_{0} + L_{1}) τ - ϵ_{0} - ϵ_{1} .$
unlabeled · line 1024
$Q_{1} - Q_{0} > r_{0} + r_{1}$