An animal was identified in one survey and its identifier is absent in the next. The record does not automatically establish that the animal died. Survival, movement, loss of the identifier and the observation process can all affect whether the same individual appears again.
The gold standard of zoology in this problem is to distinguish the animal’s biological state from the evidence available to identify it. This guide uses original probability models to show how tag retention can be confused with survival and what additional observations might separate them.
Begin with The Gold Standard of Zoology. The earlier encounter-probability guide concerns unequal chances of observation. Here the individual may be present and observable while its original identity has become unavailable.
All populations, probabilities and tables below are invented teaching examples. They are for supplied-data reasoning, not instructions to capture, handle or mark wildlife. Real animal research requires appropriate expertise, welfare safeguards and applicable authorisation.
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Distinguish survival, availability, tag retention and detection through original probability cases, dependent-loss tables and conditional bounds.
Chapter contents
Understand the evidence · Chapters 1–4
Work the model · Chapters 5–8
Test assumptions · Chapters 9–12
Practise and repair · Chapters 13–16
Report and continue · Chapters 17–20
CHAPTER 1 OF 20 · Understand the evidence
1. Define the event that the record actually contains
A record of “identifier detected” combines several conditions. The animal must be alive at the relevant time, available within the observation system, retain an identifiable mark or working tag, and be detected and correctly identified.
The absence of that record is the complement of the combined event. It can occur when any required condition fails. That is why “not recorded again” is a broader observation than “dead”.
Start by naming the observation unit. Is it a tag code received by an antenna, a visual mark seen by an observer, an individual photographed with recognisable features, or a recovered identifier? Different records have different failure mechanisms and may support different inferences.
A useful classroom exercise gives four fictional histories with the same final zero: an animal leaves the surveyed area, a visible tag is lost, a receiver fails and an animal dies. Ask which histories the available record can distinguish. If all four produce the same observation, a biological conclusion needs additional evidence.
CHAPTER 2 OF 20 · Understand the evidence
2. Build the probability chain with conditional meanings
Let S be the probability of surviving an interval. Let A be the probability of being available at the later survey conditional on survival. Let R be the probability of retaining a usable identifier conditional on survival and availability. Let p be detection and correct identification probability conditional on the previous conditions.
With these sequential definitions, the probability of an identifiable re-encounter is q = S × A × R × p. This is a chain-rule decomposition. It does not require an additional claim that the four events are independent.
If instead someone supplies marginal probabilities for tag retention and survival from separate populations, multiplying them may require assumptions about dependence and applicability. The conditional meanings cannot be omitted merely because the symbols look familiar.
The model also stipulates no false positive identification and a consistent observation interval. If one animal’s code can be assigned to another, the observation process needs more structure. Write the assumptions before treating the product as an explanation of the data.
CHAPTER 3 OF 20 · Understand the evidence
3. Work an original single-identifier example
Suppose S = 0.90, A = 0.80, R = 0.95 and p = 0.75 under the conditional definitions above. Then q = 0.90 × 0.80 × 0.95 × 0.75 = 0.513.
For a cohort of 1,000 originally identified animals following this stipulated common model, the expected number of identifiable re-encounters is 513. Expected counts are not promises about a realised sample. Random variation can produce a different observed total.
Calling 513/1,000 the survival rate would confuse the combined observation probability with S. The model’s survival probability is 0.90, while q is lower because availability, retention and detection also enter.
The arithmetic makes the distinction visible without claiming that the supplied probabilities are known in real fieldwork. In practice, estimating those components is part of the scientific task. A calculation using assumed values is a conditional prediction, not an empirical estimate of every parameter.
CHAPTER 4 OF 20 · Understand the evidence
4. Show why one re-encounter fraction is not enough
Even if q were known exactly, the product alone does not generally identify all its factors. Many combinations of S, A, R and p can produce the same value. A larger cohort can reduce sampling variability without resolving that structural ambiguity.
For a simpler construction, set A = p = 1 and suppose q = 0.72. One possible explanation is S = 0.80 and R = 0.90. Another is S = 0.90 and R = 0.80. The same identifiable re-encounter probability supports different survival probabilities when retention is unknown.
This is not a reason to abandon the study. It identifies what additional evidence would be useful: a retention study, an independent identification method, repeated encounter histories or a suitable model connecting several types of observation.
Ask learners to produce two valid parameter pairs themselves. Doing so demonstrates non-identifiability more clearly than a general warning about uncertainty. Then ask which proposed new observation would distinguish their two explanations.
In a simple setting with permanent departure from the observation area, let F be the probability of remaining within the relevant area conditional on survival. The combined probability S × F describes being alive and still part of the observable population under that model.
This explains why apparent survival and true survival need not coincide. An individual that survives but permanently leaves the system may have no later local record, just as a dead individual does. The study design determines whether those mechanisms can be separated.
Tag retention adds a further distinction. A locally present survivor may no longer carry the mark used to connect its later observation with its original history. The loss concerns identity evidence, even if the animal itself is observed as an unidentified individual.
Do not attach the full product S × A × R × p indiscriminately to every published “apparent survival” parameter. Different statistical models define parameters and handle detection differently. Our product is an explicit one-interval observation model that helps explain which mechanisms can enter a missing record.
CHAPTER 6 OF 20 · Work the model
6. Distinguish mark loss, tag failure and identity error
A visible mark can disappear, an electronic tag can stop functioning, and a working identifier can be recorded incorrectly. These mechanisms have different evidence trails. Combining them in a single “loss” category may be acceptable for a particular question, but the report should say so.
Imagine a supplied dataset in which a tag is physically present but its code is not received. Calling that event “tag loss” would misdescribe the known evidence. Conversely, a tag detected after becoming detached from its animal may not establish that the animal is present.
The definition of a usable identifier must therefore match the biological question. A tag signal, an attached tag and a correctly identified live animal are related observations, not interchangeable ones.
For learning, use a record with separate columns for biological state, physical attachment, identifier function and observation result. Some columns may be unknown. Leaving them unknown preserves the scientific question better than filling them with a preferred explanation.
CHAPTER 7 OF 20 · Work the model
7. Calculate the benefit of two independent identifiers
Suppose an available surviving animal begins with two identifiers. Let each have retention probability r = 0.90 over the interval, and stipulate independent retention under the relevant conditions. Identity is preserved if at least one remains usable.
The probability that both are lost is (1 − r)² = 0.01. Therefore the probability of retaining at least one is 1 − 0.01 = 0.99. This exceeds the single-identifier retention probability of 0.90 under the stated model.
The full set of conditional outcomes is: both retained with probability 0.81, only the first retained with probability 0.09, only the second retained with probability 0.09, and neither retained with probability 0.01. These probabilities sum to one.
Two identifiers do not automatically establish independence. They may share a failure mechanism or be affected by the same animal behaviour. The calculation shows the benefit under a declared assumption and prepares the next question: what happens if losses are dependent?
Keep the marginal retention probability of each identifier at 0.90. Now stipulate the joint probabilities in the dependent model below. Each identifier still appears equally reliable when considered by itself.
| Identifier outcome | Independent-loss model | Dependent-loss model |
|---|---|---|
| Both retained | 0.81 | 0.88 |
| First only | 0.09 | 0.02 |
| Second only | 0.09 | 0.02 |
| Neither retained | 0.01 | 0.08 |
In the dependent model, the first identifier’s retention probability is 0.88 + 0.02 = 0.90, and the second’s is also 0.90. Yet the chance of preserving identity through at least one identifier is 0.92, rather than 0.99.
The difference comes from joint behaviour. Knowing each marginal reliability does not determine the probability of simultaneous failure. A design that uses two identifiers must consider whether their loss mechanisms provide useful independence.
This is an original counterexample, not a claim about any species or tag system. Its purpose is to show exactly which assumption makes the independent calculation optimistic in this constructed case.
If animals are recognised only through the two identifiers, individuals that lose both cannot automatically be placed in the “neither retained” category. Their original identities are missing. They may disappear from the identifiable dataset entirely.
Consequently, counting only recognised animals can give a distorted view of loss. The observed table is conditioned on remaining recognisable unless another method identifies the animals whose two marks have disappeared.
For the independent example, the probabilities among animals retaining at least one identifier would be 0.81/0.99 for both, 0.09/0.99 for first only and 0.09/0.99 for second only. These conditional proportions omit the unrecognised one percent.
This observation explains why a seemingly complete table needs a denominator check. Ask, “Could an animal with neither identifier enter this dataset as its original identity?” If the answer is no, its absence cannot be used as direct evidence that neither identifier was ever lost.
Suppose a supplied teaching dataset includes a separate reference identity that remains available for the relevant subset. It can connect later observations with original individuals even when one or both study identifiers are gone.
In a constructed group of 100 known living, available individuals, the reference identifies 88 with both study identifiers, two with only the first, two with only the second and eight with neither. This reveals the dependent-loss table directly for that stipulated group.
The reference method also needs a definition and validation. A photograph may be uncertain, a separate mark may fail, and reference coverage may differ across animals. Calling something “permanent” in a model does not establish that it is infallible in the field.
The USGS research summary on dependent mark loss describes integrating mark-loss and survival models and using a permanently marked subset to address dependence. Our table is an original simplified illustration of why such auxiliary identity evidence matters.
CHAPTER 11 OF 20 · Test assumptions
11. Quantify a survival correction conditionally
Return to the example q = 0.513. If A = 0.80, R = 0.95 and p = 0.75 are independently justified for the exact population and interval, then S = q/(A × R × p) = 0.90.
If retention is incorrectly set to one while the other values remain fixed, the inferred S becomes 0.513/(0.80 × 0.75) = 0.855. In this constructed model, ignoring retention lowers the recovered survival value from 0.90 to 0.855.
The direction follows the assumptions: some failures to re-encounter are attributed to survival when retention was actually below one. Other model errors can behave differently. Do not turn this particular demonstration into a universal statement about the size or direction of every field-study bias.
The division also requires a non-zero denominator. If a proposed combination produces S greater than one, the values are incompatible with this probability model. Do not cap the result at one and declare the inference repaired. Investigate the estimates, definitions and model assumptions.
CHAPTER 12 OF 20 · Test assumptions
12. Bound survival when retention is only bounded
Suppose q = 0.54, A = 0.80 and p = 0.75 are stipulated, while R is known only to lie between 0.90 and 1.00. The combined A × p is 0.60, so S = 0.90/R.
At R = 1.00, S = 0.90. At R = 0.90, S = 1.00. Under the declared model and bounds, survival lies between 0.90 and 1.00. The endpoints reverse because S decreases as R increases when q, A and p stay fixed.
These are conditional deterministic bounds. They do not include sampling error in q or uncertainty in A and p. They are not a 95% confidence interval. Adding statistical language without an actual probability model would overstate what the exercise establishes.
This is a useful middle ground between an unjustified point estimate and giving up entirely. Partial information can support a qualified range. The range must retain the assumptions that made it possible and should not be presented as more precise than the evidence.
CHAPTER 13 OF 20 · Practise and repair
13. Use repeated histories without treating every zero as death
A history such as 1, 0, 1 records identification, non-identification and later identification across three occasions. Under reliable identity and no resurrection, the final identification shows that the middle zero did not establish death before the final occasion.
The history could reflect temporary unavailability, non-detection or another observation failure. It can help inform a suitable model of encounter probabilities. It does not automatically tell us what happened to individuals with history 1, 0, 0, whose later state remains unresolved.
Repeated histories contain more information than a single final fraction, but only through their connection with a model. Temporary emigration, mark recovery, state changes and observation effort may need explicit treatment depending on the study.
The USGS multistate tag-loss research summary explains incorporating identifier states and additional encounter information. The learning lesson is to preserve the full history rather than compressing every individual into “returned” or “did not return” too early.
A useful ledger retains the original identifier, encounter occasion, observation method, recorded mark state, location or availability information, effort and confidence in the identification. Add corrections with a traceable reason rather than silently replacing earlier entries.
Separate “not surveyed”, “surveyed but not detected”, “detected without a usable identity” and “identified with a missing study tag”. These categories describe different evidence. Collapsing them into a single zero can discard information relevant to survival and retention.
For a classroom dataset, include an explicit legend for every code. A blank cell should not be interpreted as a confirmed absence unless the data contract says so. Ask another learner to reconstruct one individual’s history using only the ledger and legend.
This is also a lesson in respectful scientific reporting. The purpose is to explain what happened to the evidence, not to force every uncertainty into a dramatic biological conclusion. A careful unknown can be more useful than a confident but unsupported label.
CHAPTER 15 OF 20 · Practise and repair
15. Diagnose and repair common reasoning errors
Error one: “Seventy percent were identified again, so thirty percent died.” Repair by listing the conditions needed for identification. Ask which of those conditions the study measures independently.
Error two: “Each tag retains ninety percent of the time, so two tags always preserve identity ninety-nine percent of the time.” Repair by writing the independence assumption and constructing the dependent table. Marginal retention alone does not determine joint loss.
Error three: “No observed animal had lost both tags, so double loss never occurred.” Repair by asking whether such an animal could be recognised and included in the dataset. The missing category may be unobservable under the identification design.
Error four: “A corrected survival estimate above one should be rounded down to one.” Repair by treating it as a compatibility warning. Check definitions and supplied probabilities rather than hiding the contradiction.
For each error, record the first unsupported inference and a changed example that exposes it. Repeating the same multiplication is unlikely to repair a misunderstanding about what the denominator or event means.
Task A: stipulate S = 0.80, A = 0.90, R = 0.95 and p = 0.50 under the sequential definitions. Find q. Answer: 0.342. In a model cohort of 500, the expected identifiable re-encounter count is 171.
Task B: A = p = 1 and q = 0.81. Supply two different S/R pairs. Answers include S = 0.90 with R = 0.90, and S = 0.81 with R = 1.00. The single product does not distinguish them.
Task C: two independent identifiers each have retention 0.80. Find the probability at least one remains. Answer: 1 − 0.20² = 0.96. State independence and the relevant conditional population.
Task D: both retained has probability 0.75, first only 0.05, second only 0.05 and neither 0.15. Find each marginal retention and identity retention. Answer: each marginal is 0.80; at least one is retained with probability 0.85. Compare with Task C.
Task E: a history is 1, 0, 1. Did the middle zero prove death? Answer: no. Reliable later identification is incompatible with that interpretation under the stated biological assumptions.
CHAPTER 17 OF 20 · Report and continue
17. Build a learning progression around evidence
Beginner learners can start with event trees and small whole-number cohorts. Ask how an animal can fail to appear in a later identifiable record. The goal is to distinguish a biological state from an observation.
At the next level, introduce conditional probabilities and the product model. Require every factor to be defined in words. Then use two different parameter pairs to show why one product can leave several explanations possible.
For extension, compare independent and dependent identifier loss, examine the unobserved zero-identifier group and derive conditional bounds. These tasks connect probability with scientific study design rather than turning the topic into arithmetic alone.
A parent can ask, “What do we know about this animal, and what do we know only about its record?” A teacher can ask, “Which extra observation would separate these explanations?” The questions invite careful reasoning without requiring a learner to conduct animal handling or field tagging.
CHAPTER 18 OF 20 · Report and continue
18. Preserve subgroup structure when averaging probabilities
A population can contain groups with different survival and identifier retention. Multiplying the population’s average survival by its average retention does not generally reproduce the average identifiable re-encounter probability when those quantities are associated across groups.
For an original construction, let half the animals have S = 0.90 and R = 0.90, while the other half have S = 0.60 and R = 0.60. Set availability and detection to one in both groups. Their identifiable re-encounter probabilities are 0.81 and 0.36, giving a population average q = 0.585.
The average S is 0.75 and the average within-group R is also 0.75. Their product is 0.5625, which differs from 0.585. The shortcut has discarded the association between the group-specific quantities.
The conditional chain remains valid when its factors are defined for the actual pooled population. Among survivors in this construction, the correctly pooled retention is 0.585/0.75 = 0.78, because the higher-survival group contributes more survivors. Multiplying 0.75 by 0.78 recovers q = 0.585.
This example is valuable because it distinguishes a mathematical identity from a careless choice of averages. The chain rule did not fail. The proposed pooled retention of 0.75 did not have the conditional meaning required by the chain.
Ask learners to calculate the expected numbers in each subgroup of a model cohort of 200. The first group contributes 81 identifiable survivors and the second 36, giving 117 in total. The survivor counts are 90 and 60, so retention among all 150 survivors is 117/150 = 0.78. Whole-number expectations make the conditioning visible.
CHAPTER 19 OF 20 · Report and continue
19. Check whether auxiliary evidence represents the target group
An independent identity method can reveal loss in a reference subset, but that subset may differ from the main study group. Animals observed in a closely monitored area may have different movement, retention or detection patterns from those elsewhere.
Suppose a fictional reference subset has retention 0.98 while the broader target group has retention 0.90. If the target has S = 0.80 and availability and detection are one, its identifiable re-encounter probability is q = 0.72. Dividing by the subset’s 0.98 yields about 0.7347, rather than the target’s stipulated survival 0.80.
The reference calculation is arithmetically correct but applied to the wrong conditional population. The lesson is to examine why the subset’s retention evidence should transfer. Similar equipment alone does not establish that the relevant conditions and animals are comparable.
Useful questions include how the subset was chosen, whether missing reference identities were recorded, whether observation effort differed and whether retention changed over the interval. The report should identify limitations rather than presenting a convenient auxiliary estimate as universally applicable.
This does not make auxiliary evidence unhelpful. It makes its scope part of the inference. A carefully characterised subset may support a stratified model or a qualified sensitivity analysis, even when a single pooled correction would be unjustified.
For teaching, change only the subset retention while holding the target construction fixed. Ask the learner to predict the direction of the inferred-survival error before calculating. Using an overly high retention value attributes more of the target’s missing records to survival in this simplified model, lowering the recovered S. The explanation should retain those model conditions.
CHAPTER 20 OF 20 · Report and continue
20. Answer the questions and report the next step
Does a missing tag establish a dead animal? No. It describes identifier evidence. Survival needs an appropriate biological and observation model.
Do two identifiers solve every loss problem? No. Joint loss, dependence and the ability to recognise animals with neither identifier all matter. Additional identification evidence can improve what is estimable.
Does a larger sample automatically separate survival and retention? No. It can improve precision for an observed fraction while leaving several parameter combinations consistent with it.
What should a report say? State the interval, cohort, observation event, conditional factor definitions, assumptions and unresolved mechanisms. Label supplied teaching probabilities clearly. Report bounds or competing explanations when the evidence cannot support a unique value.
Continue through Zoology, encounter probabilities and the Gold Standard series. The next scientific question is which observation would distinguish the animal’s fate from a change in our ability to recognise it.