Corollary: Finite sample sufficiency for the ideal benchmark

Suppose the tested and deployed two-leg channels are ideal, calibration bias is zero, and all setting counts equal $N$. For a common tolerance $\epsilon >0$, the sufficient condition

ensures $U_{\nu }\le \epsilon$ on the simultaneous score-coverage event.

Proof. For the ideal channel, $J({\Phi }_{\nu ,k})=J({\mathcal{T}}_k)$. Score coverage gives ${\hat{\delta }}_{\nu ,k}\le R_{\nu ,k}/2$. Thus $U_{\nu ,k}\le n{R}_{\nu ,k}$. Here

The displayed count makes this upper bound at most $\epsilon$.