Lemma: A concrete calibration conversion

Suppose the actual common pre-analyzer state $\sigma$ obeys $d_{\mathrm{tr}}(\sigma ,J_{{\mathcal{E}}_0})\le \delta$. Let the complete score observable $B_j$ satisfy

Then a valid bias allowance is

Proof. Replacing $B_j$ by $A_j$ changes the fidelity combination by at most ${\sum }_j{\mu }_j/4$. The ideal combination on $\sigma$ is $\mathrm{Tr}(P_2\sigma )$. Since $P_2$ is an effect, its difference from $\mathrm{Tr}(P_2{J}_{{\mathcal{E}}_0})$ is at most $\delta$.