= 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 $J\sim\operatorname{Binomial}(M,p)$. Suppose
$$
\Pr_p\{p\le U(J)\}\ge1-\gamma,
$$
where $0<\gamma<1$ is a calibration failure budget. An exact one-sided binomial bound can be obtained by inversion: for $j<M$, choose $U(j)$ satisfying $\Pr_{U(j)}(J\le j)=\gamma$, 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.<SciPy 1.16.2 documentation>[BinomTestResult.proportion_ci, confidence_level, exact method and returned bounds]{ref}
For a fixed upper duration endpoint with actual cumulative <probability> $b$, set
$$
q(J)=\max\{a,U(J)\},\qquad r(J)=1-\frac a{q(J)},\qquad0<a<b<1.
$$
The rule is defined and $0\le r\le1$. 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 $\Pr_p(J=0)=(1-p)^{20}$, inversion gives $U(0)=1-0.01^{1/20}\approx0.20567$. 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 only
$$
E C(J)\ge(1-\gamma)(b-a).
$$
At 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.
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