 Finite-calibration guards for reciprocal endpoint randomization

Finite-calibration guards for reciprocal endpoint randomization must specify both a valid endpoint decision and how calibration uncertainty enters its probability guarantee. An arbitrary positive floor avoids division by zero but does not establish nominal inclusion, as the example in near-zero probability estimates in reciprocal plug-in rules demonstrates.

One finite-sample construction uses an upper confidence bound $U(J)$ for the atom probability $p$, based on independent calibration count $J\sim \mathrm{Binomial}(M,p)$. Supposewhere $0<\gamma <1$ is a calibration failure budget. An exact one-sided binomial bound can be obtained by inversion: for $j<M$, choose $U(j)$ satisfying ${\mathrm{Pr}}_{U(j)}(J\le j)=\gamma$, and put $U(M)=1$. This is a confidence statement over calibration samples, not a posterior probability for a fixed $p$ after observing one count. The pinned reference documents the exact binomial proportion-interval method.^[ref]

For a fixed upper duration endpoint with actual cumulative probability $b$, setThe rule is defined and $0\le r\le 1$. On the event $p\le U(J)$, its conditional included probability is $b-ap/q(J)\ge b-a$. This calculation also requires independence of the calibration, future duration and auxiliary endpoint draw as stipulated by the parent model.

For an original zero-count example, take $M=20$ and $\gamma =0.01$. Since ${\mathrm{Pr}}_p(J=0)=(1-p{)}^{20}$, inversion gives $U(0)=1-0.01^{1/20}\approx 0.20567$. With $a=0.005$, $b=0.995$, and stipulated true $p=0.02$, the zero-count conditional included probability is approximately $0.99451$. The bound is much larger than the raw estimate zero, reflecting the uncertainty of a small calibration sample rather than inventing a measured atom fraction.

The calibration confidence level is not free. Since included probabilities are nonnegative even on the failure event, a general joint guarantee from this argument is onlyAt conditional target $0.99$ and calibration confidence $0.99$, this lower guarantee is $0.9801$, not $0.99$. To certify an unconditional target $t$ using this argument, the conditional target must be at least $t/(1-\gamma )$, provided it can fit below $b$, or the failure budget must be reduced. Any upper-endpoint model error also needs separate accounting.