Sampling limits in anatomical variant frequencies

Sampling limits in anatomical variant frequencies distinguish how often a structure was observed in a specimen set from how common it is in a defined population. The denominator must identify the independent anatomical unit: people, hearts, limbs or images are different units. Preserved material and the detection procedure also determine what can be observed.

A primary example dissected 37 formalin-fixed adult human hearts, finding the small cardiac vein in 54% of specimens and an approximately 1 mm venous connection in 19%.^[ref] Those percentages describe that study's material; they do not establish a universal frequency across populations or measurement methods.

Under an original independent, representative-sampling model with perfect detection, a binomial distribution gives $X\sim \mathrm{Bin}(n,p)$. If no examples of a specified variant are seen, $\mathrm{Pr}(X=0)=(1-p{)}^n$. Inverting the one-sided test at level $\alpha$ givesFor a hypothetical zero observations among 37 specimens and $\alpha =0.05$, the upper limit is approximately 0.0778, not zero. The familiar approximation $-\log (\alpha )/n$ is about 0.0810 here. These are illustrations, not unreported results from the anatomical paper.

Detection-limited anatomical prevalence bounds add another assumption explicitly. If a real variant is detected with probability $q$ and false detections are absent, the observed-event probability is $qp$. Zero observed cases therefore gives $p_{\mathrm{upper}}=\min \{1,(1-{\alpha }^{1/n})/q\}$ when $q$ is known. At stipulated $q=0.5$, the example limit becomes approximately 0.156. Unknown detection probability cannot be repaired by reporting a more precise count.

Repeated photographs of the same heart do not create independent new hearts. Counting every image as an independent specimen would artificially narrow a confidence interval. Selection of unusually accessible, preserved or clinically referred specimens can also produce bias that a binomial distribution does not quantify. A useful anatomical report states selection, preservation, observation criteria and the denominator alongside its frequency; uncertainty calculations then make their sampling and detection assumptions visible.

Detection-limited anatomical prevalence bounds account for the fact that a structure can exist without being recognized by the observation method. Zero observed variants constrain the probability of observing a variant; translating that into a population frequency requires a detection model. In anatomical research, specimen preservation, resolution and the definition of a positive observation belong to that model.

A primary coronary-venous study dissected 37 formalin-fixed hearts and reported frequencies of visible venous structures.^[ref] Its specimen count is a concrete setting for explaining sampling, but the detection probabilities below are hypothetical, not estimates from that paper.

Let $p$ be population prevalence and $q$ the probability of detecting a present variant, with no false detections. For independent specimens, the observed count has a binomial distribution with event probability $r=qp$. Zero detections among $n$ gives the one-sided upper boundAt $n=37$, $\alpha =0.05$ and known $q=0.5$, this gives $p_U\approx 0.156$. The confidence statement depends on $q$ being known, not merely on a convenient assumed value.

Joint coverage in detection-calibrated prevalence bounds matters when $q$ is estimated. Suppose a calibration gives a lower bound $q_L>0$ with coverage at least $1-\beta$, while $r_U$ has coverage at least $1-\alpha$. Whenever both bounds cover, $p\le r_U/q_L$; the union bound guarantees combined coverage at least $1-\alpha -\beta$, without requiring independence of the two bounds. Allocating $\alpha =\beta =0.025$ and stipulating a calibrated lower bound $q_L=0.5$ gives $p_U\approx 0.190$ for the same zero count.

False detections change the relation to $r=f+(q-f)p$, where $f$ is the false-detection probability for absent structures. If $q=f$, the observed frequency carries no information about prevalence. Simply dividing by $q$ is then invalid. Detection controls should therefore distinguish missed structures from false recognitions.

These bounds do not repair unrepresentative specimen selection or treating repeated images as independent bodies. Report the anatomical unit, detection validation and target population alongside a confidence interval; more decimal places cannot substitute for those assumptions.

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 $q$.

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.

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.

Uniform coverage over confidence-allocation choices means that one probability statement protects every allocation in a specified family simultaneously. For upper bounds $B_{\lambda }(X)$ on a parameter $\theta$, the required statement isIt implies coverage for the minimum selected after observing the data. Separate statements ${\mathrm{Pr}}_{\theta }\{\theta \le B_{\lambda }(X)\}\ge 1-\gamma$ for each fixed $\lambda$ do not imply the simultaneous statement. The difference is the order of the probability and the requirement to cover every choice.

For a finite set of $M$ candidates, assigning each a failure budget $\gamma /M$ and applying the union bound provides uniform protection. This is the simultaneous-coverage principle behind the Bonferroni correction.^[ref] A continuum requires a family-level argument rather than multiplying one finite-candidate budget by an undefined number of choices.

An original construction exploits nested one-sided bounds in the detection model $r=qp$. Suppose $r_U(\alpha )$ decreases as its failure budget $\alpha$ increases, and $q_L(\beta )$ increases as $\beta$ increases. Choose allocations $\alpha \in [a,b]$, $\beta ={\gamma }_0-\alpha$, with $0<a\le b<{\gamma }_0$. On the common eventevery candidate ratio $r_U(\alpha )/q_L({\gamma }_0-\alpha )$ bounds $p$. The union bound controls this common event's failure probability byThis extremal calibration for nested confidence families protects the entire interval of allocations using just its most demanding marginal endpoints.

For example, take ${\gamma }_0=0.04$, $a=0.01$ and $b=0.02$. Each fixed allocation has at least 96% coverage, while the common-event argument guarantees at least 95% coverage for every allocation simultaneously. Selecting the smallest bound within that prespecified interval is therefore protected at the latter level, without assuming independence of the marginal bounds.

The proof depends on the nesting, valid marginal bounds and a specified family. Enlarging the family after inspecting the observations needs corresponding protection. Where a lower detection bound can be zero, the prevalence upper bound must revert to the uninformative value one rather than divide by zero.

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.