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SCIENCE · INVESTIGATE AND EXPLAIN

Update an occupancy claim after repeated nondetection

An occupied site may produce a nondetection because organisms are cryptic, inactive or outside the sampled portion. Occupancy concerns whether a site is used or contains the species under a defined period and scale; detection concerns the observation process conditional on that state. Treating every zero as absence confounds these two quantities.

Learning outcome: Compare one and four visits at p=0.2 and p=0.8. Separate conditional detection from marginal all-zero probability, and describe one real condition violating independent visits.

Starting recap

Separate target population, sampling frame, inclusion probability, detection and ecological state. Identify the quantity and assumptions before interpreting a corrected estimate.

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Try

At p=0.4 and three visits, an occupied site is missed with probability 0.6³=0.216. Marginal all-zero probability is 0.4+0.6×0.216=0.5296. Conditional occupancy is 0.1296/0.5296≈0.2447, not zero.

Open the experiment

Predict

Do three nondetections prove absence when detection is imperfect?

Experiment

Compare one and four visits at p=0.2 and p=0.8. Separate conditional detection from marginal all-zero probability, and describe one real condition violating independent visits.

Interpret repeated nondetection under explicit assumptions

Worked teaching model · synthetic values · Assigned prior occupancy 0.6; constant independent conditional detection p, no false positives, site occupancy closed across visits. P(occupied|allzero)=0.6(1−p)^k/[0.4+0.6(1−p)^k]. This is a stipulated probability exercise, not an occupancy model fitted to data.

Reset keeps your prediction, captures and practice history.

Original synthetic system; units differ by row
QuantityValueUnit
Detection at least once if occupied0.784probability
All-zero probability if occupied0.216probability
Marginal all-zero probability0.5296probability
Occupancy probability after all zero0.244713probability

Nondetection updates an occupancy claim but does not automatically prove absence.

Repeated visits require closure and a defensible conditional detection model. Heterogeneity, correlated visits and false positives need richer designs; this example does not estimate either p or occupancy from observations.

Predict, change one control, and save contrasting captures. Reconstruct the balance with units; then explain a process the ledger deliberately leaves out. Display precision is not measurement precision.

Nondetection updates an occupancy claim but does not automatically prove absence.

Repeated visits require closure and a defensible conditional detection model. Heterogeneity, correlated visits and false positives need richer designs; this example does not estimate either p or occupancy from observations.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define the ecological target

An occupied site may produce a nondetection because organisms are cryptic, inactive or outside the sampled portion. Occupancy concerns whether a site is used or contains the species under a defined period and scale; detection concerns the observation process conditional on that state. Treating every zero as absence confounds these two quantities.

Specify the observation model

For a closed occupied site with independent visits and constant detection probability p, the probability of no detections in k visits is (1−p)^k. The chance of at least one detection is its complement. More visits generally increase information, but repeated visits under nearly identical unfavorable conditions may be correlated rather than independent.

Work through the sampling logic

The exercise stipulates prior occupancy probability 0.6 and no false-positive detections. An all-zero record can arise from an unoccupied site, probability 0.4, or an occupied site missed on every visit, probability 0.6(1−p)^k. Normalize the latter by the sum to obtain occupancy conditional on all zeros. This is a supplied probability model, not an occupancy fit to real observations.

Challenge the inference

Actual occupancy studies use repeat detection histories across sites to estimate occupancy and detection under a specified model. Closure can fail when use changes during visits, and detection may vary by habitat, observer or weather. Misidentification creates false positives that require verification or richer models. A number from the simplified formula cannot solve these design problems.

Design a defensible survey

Balance more sites against more repeat visits according to the question and expected detection. Record effort, timing, method, observer and nondetection as an actual observation, not missing data. Do not disclose sensitive species locations or manipulate organisms for this virtual exercise. Absence claims should specify the survey conditions and residual uncertainty.

Worked original packet

At p=0.4 and three visits, an occupied site is missed with probability 0.6³=0.216. Marginal all-zero probability is 0.4+0.6×0.216=0.5296. Conditional occupancy is 0.1296/0.5296≈0.2447, not zero.

Investigation

Compare one and four visits at p=0.2 and p=0.8. Separate conditional detection from marginal all-zero probability, and describe one real condition violating independent visits.

Changed-case transfer

With prior occupancy 0.5, p=0.5 and two visits, compute probability of at least one detection if occupied and occupancy after all zeros.

Feedback

Conditional at-least-one detection is 0.75. All-zero occupied weight is 0.125; unoccupied weight is 0.5, giving posterior occupancy 0.125/0.625=0.2. The result depends on the specified prior and detection model.

Sources and evidence boundary

Original instructional prose, values and exercises. Context checked 30 October 2026: primary agency source. All observations are synthetic. No organism collection, trapping, tagging, release or unsafe-site access is requested; a real study requires qualified design, permission and supervision. Public feedback is practice, not independently verified mastery.

Apply

The event represented by (1−p)^k is:

Public practice checks a specific claim. It does not establish independent mastery.

With prior occupancy 0.5, p=0.5 and two visits, compute probability of at least one detection if occupied and occupancy after all zeros.

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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