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Error-budget allocation between prevalence and detection calibration assigns the allowed failure probability of a joint confidence bound between the specimen survey and its detection-calibration experiment. Under the model $r=qp$, an upper bound $r_U$ with failure probability at most $\alpha$ and a positive lower bound $q_L$ with failure probability at most $\beta$ give $p_U=\min \{1,r_U/q_L\}$. The union bound guarantees coverage at least $1-\alpha -\beta$. A total budget $\gamma$ therefore permits any prespecified positive allocation satisfying $\alpha +\beta =\gamma$; equal allocation is only one choice.

For independent specimens with zero detections among $n$ and all successful detections among $m$ known-positive calibration specimens, exact binomial distribution tail inversion givesThe tail construction is the one-sided method described by NIST.^[ref] For this anticipated outcome pattern, the uncapped bound becomesMaking $\alpha$ larger tightens the numerator but leaves less failure budget for calibration, which lowers $q_L$ and weakens the denominator. The two effects compete.

An original planning example sets $n=37$, $m=20$ and $\gamma =0.05$. Equal allocation gives $U(0.025)\approx 0.11411$. A fixed alternative allocation $\alpha =0.043$, $\beta =0.007$ gives approximately $0.10448$, still with at least 95% joint coverage. For a differentiable interior optimum of this specific objective,The example is minimized near $\alpha =0.04295$. It is not a universal optimal split for other counts or detection models.

Prespecified versus data-selected confidence allocations is crucial. Choosing the most favorable allocation after inspecting many outcomes does not inherit this fixed-allocation coverage proof. Design the allocation before observation, or use a construction whose validity includes the selection rule. Calibration must also address the same detection difficulty as the survey; redistributing failure probabilities does not repair an inappropriate calibration population.

Extremal calibration for nested confidence families obtains simultaneous protection for many candidate bounds by checking their most demanding marginal endpoints. It uses the ordering of a nested family rather than treating every candidate as an unrelated comparison. The simultaneous event then also protects a choice made after observing the data.

Consider the detection model $r=qp$, with upper confidence bounds $r_U(\alpha )$ decreasing in their failure budget $\alpha$ and lower bounds $q_L(\beta )$ increasing in $\beta$. For a prescribed set of candidate allocations $D$, let ${\alpha }_{\ast }$ and ${\beta }_{\ast }$ be the largest respective budgets, with valid bounds defined at those endpoints. Onevery candidate satisfiesThe union bound gives simultaneous coverage at least $1-{\alpha }_{\ast }-{\beta }_{\ast }$. Neither independence of the estimates nor a finite number of candidates is needed for this common-event proof.

For the line segment $\beta ={\gamma }_0-\alpha$, $a\le \alpha \le b$, the endpoints are ${\alpha }_{\ast }=b$ and ${\beta }_{\ast }={\gamma }_0-a$. The protected failure budget is therefore ${\gamma }_0+(b-a)$. The width of the candidate interval consumes part of the simultaneous budget; it cannot be ignored simply because every point on the segment has the same marginal budget sum.

An original exact-tail example has zero detections in $n=37$ and successful detection of all $m=20$ measurement calibration positives. Binomial distribution tail inversion gives $r_U(\alpha )=1-{\alpha }^{1/37}$ and $q_L(\beta )={\beta }^{1/20}$. NIST describes the one-sided tail construction and simultaneous error-budget principle.^[ref] For candidates $(0.01,0.03)$ and $(0.02,0.02)$, the bounds are about 0.13946 and 0.12201. Their extremal endpoint combination $(0.02,0.03)$ gives about 0.11956 and at least 95% coverage. It is a tighter bound with the larger combined budget, and its validity follows directly from the endpoint event.

The construction requires the stated nesting and valid endpoint bounds. A family whose order reverses or whose measurement calibration changes across candidates needs a new argument. If the lower endpoint bound for detection is zero, the prevalence upper bound is one, rather than an undefined division.

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.

Joint coverage in detection-calibrated prevalence bounds controls the chance that both the observed-event bound and its detection calibration are valid. A prevalence calculation that substitutes an estimated detection probability into a sampling bound does not automatically retain the original confidence level.

In an original model with no false detections, prevalence $p$ and detection probability $q>0$ give observed-event probability $r=qp$. Let $r_U$ be an upper confidence bound with coverage at least $1-\alpha$, and $q_L>0$ a lower confidence bound with coverage at least $1-\beta$. On the event that both bounds cover,The union bound gives $\mathrm{Pr}(r>r_U\text{ or }q<q_L)\le \alpha +\beta$. Hence $p_U=\min \{1,r_U/q_L\}$ has coverage at least $1-\alpha -\beta$. This argument does not require independence between the two estimates, but it does require valid marginal bounds for the stated model. If no positive lower bound on $q$ is established, the resulting prevalence bound may be uninformative.

Exact extreme-count examples follow by inverting binomial distribution probabilities. NIST describes exact interval construction from binomial tails and distinguishes one-sided limits.^[ref] Zero detections among $n$ specimens gives $r_U=1-{\alpha }^{1/n}$. Detecting all $m$ independently known positive calibration specimens gives $q_L={\beta }^{1/m}$. These are one-sided extreme-case formulas derived from $(1-r{)}^n$ and $q^m$.

For stipulated $n=37$, $m=20$ and $\alpha =\beta =0.025$, the combined coverage is at least 95% and $p_U\approx 0.114$. Choosing two separate 95% bounds instead guarantees only at least 90% joint coverage by this argument; if their failures are independent, the joint success probability is $0.95^2=0.9025$. Neither is a 95% joint guarantee.

Error-budget allocation between prevalence and detection calibration need not split the allowed failure probability equally. A prespecified allocation can reflect which experiment supplies the weakest information. Calibration specimens must also represent the detection difficulty of the target specimens; a formally valid interval for an easier calibration set does not establish the relevant $q$.

A one-sided Cantelli confidence bound with a bias budget combines a validated bound on the estimator's systematic displacement with a one-sided variance inequality. Suppose $|E_{\theta }\hat{\theta }-\theta |\le B$, where $B\ge 0$ is deterministic, and a nonnegative statistic satisfies $\hat{V}\ge V_{\theta }={\mathrm{Var}}_{\theta }(\hat{\theta })$ almost surely. For $0<\alpha <1$, defineThen ${\mathrm{Pr}}_{\theta }(\theta \ge L)\ge 1-\alpha$. This is a lower confidence bound, with an unbounded upper side.

For the proof, put $Y=\hat{\theta }-E_{\theta }\hat{\theta }$. If $\theta <L$, the bias bound impliesFor positive $V_{\theta }$, Cantelli's inequality bounds this event by $V_{\theta }/(V_{\theta }+a^2)=\alpha$ at the last displayed threshold $a$. For zero variance, $Y=0$ almost surely and the conclusion follows directly. A random $\hat{V}$ may depend on the centre: the samplewise comparison is what supports the event inclusion.

For an original numerical bound, let $\hat{\theta }=2.0$, $B=0.2$, $\hat{V}=0.04$, and $\alpha =0.1$. The square-root term is $0.6$, so $L=1.2$ is a 90 percent lower bound under these assumptions. Reversing the inequality gives the separately valid upper bound $U=2.8$. Each has 90 percent one-sided coverage; reporting both endpoints together only guarantees 80 percent by the union bound. To guarantee a joint 90 percent interval this way, allocate error probabilities summing to $0.1$ across its two sides.

This construction complements the two-sided result in confidence bounds from a samplewise variance bound. It needs a bias bound and an almost-sure variance bound, not merely a conservative variance estimator in the expected-value sense. A coverage interpretation must follow the actual tail argument, as emphasized by NIST's discussion of expanded uncertainty.^[ref]

Prespecified versus data-selected confidence allocations distinguishes a fixed procedure's coverage from coverage after choosing among procedures using the same observations. For detection-calibrated prevalence, a fixed pair $(\alpha ,\beta )$ with $\alpha +\beta =\gamma$ gives a bound with coverage at least $1-\gamma$ when the two marginal bounds are valid. Taking the smallest result from many such allocations requires another argument: the selected result fails if any candidate producing that small result fails.

An original counterexample makes the issue exact. For an unknown location parameter $\theta$, suppose the observed pair is $X=\theta +U$, $Y=\theta -U$, where $U$ has a uniform distribution on $[0,1]$. Two upper confidence bounds areEach separately covers $\theta$ with probability 0.95. The first fails when $U<0.05$, and the second when $U>0.95$. Choosing the smaller observed bound givesThe selection uses only the reported bounds, yet loses the individual 95% coverage. These artificial observations illustrate a selection effect; they are not a model of actual anatomical calibration data.

For $M$ prespecified candidate allocations, let each candidate's total failure budget be at most $\gamma /M$. The union bound then controls the failure of any candidate by at most $\gamma$, so even the selected minimum retains coverage at least $1-\gamma$. This is a direct Bonferroni correction; NIST presents the corresponding simultaneous-coverage principle.^[ref] For two candidates and a 5% overall budget, each candidate receives at most 2.5%, including both its prevalence and detection-calibration contributions.

A continuum of possible allocations is not covered by assigning each the original finite-candidate budget. Uniform coverage over confidence-allocation choices requires a jointly valid construction or a proof that includes the selection rule. A split chosen from independent pilot data can instead be treated as fixed conditional on that pilot, provided the required coverage is valid for every resulting split. The source of the allocation and the data used for the reported bound are therefore part of the statistical procedure.

Ruggero De Paulis is a surgeon whose work includes aortic root reconstruction and the design of a sinus-containing conduit. UniCamillus identifies cardiac surgery as his field and credits him with the Valsalva graft.^[ref] The original 2000 paper gives his full byline and explicitly attributes the modified conduit's design to RDP.^[ref]

The design addresses a geometric distinction between a straight tube and the bulging aortic sinuses. In the report, differently oriented corrugations in a short graft segment allow circumferential expansion near the root while the adjoining tube retains a different directional response. The resulting shape includes artificial sinus-like spaces and a narrowed sinotubular junction.^[ref] This is an example of using structural geometry to control a device's mechanical response, rather than specifying only one nominal tube diameter.

The initial report described nine operations and used visual inspection, echocardiography and angiography to assess the reconstructed geometry. It reported no mortality or morbidity in that series.^[ref] Those observations document an initial series; they do not by themselves establish a zero underlying event probability.

An original binomial distribution illustration quantifies that distinction. If nine independent procedures from a specified population have zero events, a one-sided 95% upper confidence bound under the stipulated model isThis is a mathematical illustration of sample-size limitations, not an estimate of an individual patient's risk or a comparison between treatments. Follow-up duration and the event definition also belong to the observed endpoint.

With Andrea Salica, De Paulis later coauthored the detailed 2019 account of aortic valve and root anatomy.^[ref] The papers connect explicit geometry, a fabricated conduit and measurements of the resulting reconstruction.

If $X\sim \mathrm{Bin}(n,p)$ and $X=0$ is observed, the maximum-likelihood estimate of $p$ is zero. A one-sided $1-\alpha$ upper confidence bound solves $P_{p_U}(X=0)=(1-p_U{)}^n=\alpha$, so $p_U=1-{\alpha }^{1/n}$. For large $n$, $p_U\simeq -\log (\alpha )/n$, giving approximately $3/n$ at 95% confidence. This describes uncertainty after no observed events, rather than proving that the underlying probability is zero. It assumes independent observations with a common event probability; clustered data require a different uncertainty calculation.