The Gold Standard of Zero-Inflated Models | What Two Exposure Zero Fractions Can Bound About a Permanent Zero State

Let Λ be a nonnegative finite rate attached to an observational unit. Conditional on Λ and positive exposure t, the count has a Poisson distribution with mean Λt.

Assume the population distribution of Λ stays the same across the compared exposure conditions. The zero probability is q(t)=E[exp(−Λt)]. A point mass at Λ=0 contributes zeros at every exposure within this model.

These assumptions matter. If rates change during observation, populations differ between durations or detection changes with exposure, the same formula may not describe the records. Two reported zero fractions cannot justify the model that was needed to interpret them. State the observation rule, exposure units and comparability conditions before calculating a latent-state bound. Here we analyse a specified mathematical model so its information limits can be made exact and auditable.

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Let π=P(Λ=0). It is the fraction of the model population with rate exactly zero. All other rates may be arbitrarily small but strictly positive.

A very low positive rate can generate many zeros over a finite exposure. It is nevertheless different from a zero rate when the model is extended to arbitrarily long observation. Our question is whether two exact population zero probabilities determine π or only restrict it.

The word exact separates this identification exercise from sampling uncertainty. In real data, zero fractions estimate probabilities and need uncertainty analysis. We will add that caveat explicitly later. First, assume q(t) and q(2t) are known without error. If π remains ambiguous even under that idealisation, collecting more units at only those same two conditions cannot by itself remove the structural ambiguity of the reduced summaries.

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Fix t>0 and define X=exp(−Λt). Since Λ is nonnegative and finite, X lies in(0,1]. Rate zero corresponds exactly to X=1.

Write m=q(t)=E(X) and r=q(2t)=E(X²). The second equality follows because exp(−2Λt)=X². The desired π is now the probability mass at the endpoint X=1.

This transformation turns the exposure question into a bounded-moment problem. Two means, E(X) and E(X²), need not identify the entire distribution of X or its endpoint mass. They can still imply useful inequalities.

The transformation is one-to-one for positive finite t: Λ=−log(X)/t. Every constructed support point strictly between zero and one therefore corresponds to a finite positive rate. We will keep those support conditions visible rather than use an impossible infinite-rate component as though it were an ordinary Poisson population.

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Because X²≤X, we need r≤m. Because Var(X)=r−m² is nonnegative, we need r≥m². Also 0<m≤1 under the finite-rate assumptions.

For the nondegenerate interior case, use 0<m<1 and m²<r<m. These strict inequalities allow variation in X while excluding endpoint-only degeneracies that require separate treatment.

The constraints are model checks for exact population moments. Empirical estimates can violate them because of sampling error or because the model and observation conditions do not fit. A finite-sample violation is not automatically proof of a mechanism failure without an appropriate uncertainty analysis.

Before interpreting any fitted π, check that the supplied zero probabilities at least describe a compatible moment problem. An optimiser returning a coefficient does not eliminate an incompatible exposure definition, a changed cohort or a mistaken denominator in the zero fractions.

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Suppose m=1/2 and r=3/10. These satisfy m²=1/4<3/10<1/2, so they lie in the interior region.

The variance of X is 3/10−1/4=1/20. This is variation in the transformed rate, not the variance of the recorded count. Confusing those two random variables would attach the wrong interpretation to the number.

Our task is to identify possible distributions of X with those moments and compare their endpoint mass. We will construct one with π=0 and another with π=1/6. Both will have exactly the same q(t) and q(2t).

The numbers are deliberately simple so every moment can be checked independently. They do not describe observed percentages from a survey or a fitted zero-inflated model. Their purpose is to establish what two supplied summaries can and cannot imply even in an error-free mathematical setting.

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Let X take the two values (1/2)−sqrt(1/20) and(1/2)+sqrt(1/20), each with probability one-half. Numerically these are about 0.2763932023 and 0.7236067977.

Both values lie strictly between zero and one, so both corresponding rates −log(X)/t are finite and positive. Hence π=0.

Their mean is 1/2 by symmetry. Their second moment is(1/2)²+1/20=3/10. Thus this entirely positive-rate population produces the stipulated zero probabilities at t and 2t.

A high fraction of zeros at finite exposures is therefore compatible with no permanently zero-rate units in the model. This construction does not claim that real data came from two groups, or that a two-rate fit is scientifically preferable. It is an explicit counterexample to uniqueness of the endpoint mass from the two zero summaries.

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Now give X value one with probability 1/6 and value 2/5 with probability 5/6. The first group has rate zero; the second has positive rate −log(2/5)/t.

Its first moment is 1/6+(5/6)(2/5)=1/2. Its second moment is 1/6+(5/6)(4/25)=3/10. Therefore it has the same two zero probabilities as the previous positive-rate model, while π=1/6.

This is a ZIP-style rate distribution: a zero-rate component and one positive rate. It is not an estimate obtained by fitting the two positive-rate construction. Both models were constructed to satisfy the same two moment constraints.

The comparison establishes ambiguity of a particular reduced observation: zero probabilities at two exposures. It does not say that their full count probabilities at either exposure are identical. Positive-count shapes can carry additional information, as the existing mean-and-zero comparison guide explains.

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Separate the event X=1, whose probability is π, from the remaining population. The nonendpoint contributions satisfy E[X·1{X<1}]=m−π and E[X²·1{X<1}]=r−π.

Apply Cauchy–Schwarz to X·1{X<1} and 1{X<1}. It gives(m−π)²≤(r−π)(1−π).

Expanding and cancelling π² yields m²−r+π(1−2m+r)≤0. For m<1, the denominator 1−2m+r=E[(1−X)²] is positive, so π≤(r−m²)/(1−2m+r).

This bound uses both exposure moments. It is a model-based upper limit, not a fitted probability or a causal classification of observed zeros. The proof identifies every premise needed: a common bounded X, the endpoint interpretation and exact compatible moments. If the exposure model changes, the bound cannot simply be carried over unchanged.

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With m=1/2 and r=3/10, the numerator is 1/20 and the denominator is 3/10. Their ratio is 1/6.

Thus every rate distribution satisfying the declared two moments has π≤1/6. The ZIP-style construction attains that value, while the positive-rate construction attains zero. The two exact probabilities therefore permit at least this full range of endpoint interpretations rather than select one unique state fraction.

The upper bound is stronger than the simple inequality π≤q(t)=1/2. It does not turn into evidence that the upper limit is the actual value. A model chosen from the upper-bound construction is making an additional distributional choice.

When reporting such a result, use language such as “at most one-sixth under the stipulated fixed-rate mixture model.” Saying “one-sixth are permanently inactive” would replace a bound with an unsupported identification claim.

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For interior moments, define c=(m−r)/(1−m) and u=(r−m²)/(1−2m+r). Place mass u at X=1 and mass 1−u at X=c.

The moment inequalities imply 0<c<1 and 0<u<1. Direct substitution verifies E(X)=m and E(X²)=r. This construction reaches the upper bound, so no smaller universal upper bound can follow from those two moments alone under the stated class.

Equality in the Cauchy–Schwarz step also explains the construction: away from X=1, X must be constant for equality. The extremal distribution uses exactly one positive-rate value plus the endpoint.

Sharp means attainable within the permitted mathematical model class. It does not mean scientifically preferred or statistically most likely. Those judgments would require additional assumptions, observations or a specified fitting criterion. The extremal construction is valuable because it shows precisely what information is missing from the two summaries.

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The original symmetric construction proves π=0 for our chosen moments. More generally, interior moments can be represented using two support points strictly inside(0,1), so they need not force endpoint mass.

Let v=r−m², positive and smaller than m(1−m). Choose a positive a smaller than m−v/(1−m). Define b=m+v/(m−a). Then a<m<b<1.

Assign the weights that give mean m: probability (b−m)/(b−a) at a and(m−a)/(b−a) at b. Their variance is(m−a)(b−m)=v, so the second moment is r.

Both rates are finite and positive, giving π=0. This construction establishes a zero lower bound for the interior moment class without relying on symmetry. It also illustrates why allowing arbitrarily small positive rates matters: a bound on the positive-rate support would restrict which constructions remain admissible.

Any mixture of the two original constructed populations retains the same first and second moments, because both populations already have those moments. Giving weight w to the endpoint construction produces endpoint mass w/6. Thus every value from zero through one-sixth is attainable in this example, not merely its two endpoints.

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Under a supplied model with endpoint mass π, the conditional probability that an observed zero at exposure t came from that endpoint is π/m. In the original example, the upper bound gives π/m≤(1/6)/(1/2)=1/3.

The positive-rate construction gives zero for that conditional probability. The upper-bound construction gives one-third. Both share the same two zero probabilities.

These are conditional model quantities, not independent measurements of eligibility. The recorded zero does not reveal which latent rate generated it. A posterior class number can be useful within an assumed model, but its numerical precision does not make the mechanism empirically unique.

If the scientific question concerns a documented inactive or ineligible state, identify an observation that actually measures that state. Reusing the same zero count to define and then validate the class adds no independent evidence beyond the original model assumptions.

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At exposure 3t, q(3t)=E(X³). The two original constructions now disagree.

The symmetric positive-rate model gives(1/2)³+3(1/2)(1/20)=1/5. The endpoint construction gives 1/6+(5/6)(2/5)³=11/50=0.22.

Their agreement at t and 2t therefore did not force agreement at 3t. An additional exposure can distinguish these two stipulated proposals in principle, with a sampling design and uncertainty analysis appropriate to the task.

It does not follow that any three exposure probabilities identify every possible rate distribution. The new observation adds another moment constraint, and its information depends on where it lies and which model class remains allowed. The precise claim here is that the two constructed laws make different third-exposure predictions, not that a third measurement universally solves latent-rate inference.

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Apply Cauchy–Schwarz to X^(3/2) and X^(1/2). It gives[E(X²)]²≤E(X³)E(X), hence q(3t)≥r²/m for m>0.

For the original moments, the lower bound is(3/10)²/(1/2)=9/50=0.18. Both constructed third-exposure values, 0.20 and 0.22, satisfy it.

This inequality constrains a possible additional exposure probability without identifying the endpoint mass. Equality can involve support at zero in X for some moment combinations, which would correspond to an infinite rate rather than an ordinary finite Poisson component. Under a finite-rate model, a boundary value may be an infimum approached by very large rates instead of an attainable population.

Distinguish a necessary bound from its attainability conditions. Our main endpoint upper bound was attained with strictly positive nonendpoint X; this auxiliary lower moment bound has a different equality structure.

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Apply Cauchy–Schwarz to sqrt(1−X) and X sqrt(1−X). Their squared cross expectation gives(m−r)²≤(1−m)[r−q(3t)].

Therefore q(3t)≤r−(m−r)²/(1−m) when m<1. For the original moments, the upper bound is 0.3−0.04/0.5=0.22.

The endpoint construction attains that upper value. The positive-rate construction gives 0.20, inside the feasible range. This supplies an independent check on the third-exposure calculations and clarifies how moment constraints accumulate.

A real estimated probability outside these exact bounds may reflect sampling variation, changed populations, incompatible exposure definitions or model failure. It should not be repaired by silently clipping it to a boundary and then retaining the original interpretation. The numerical value, observation uncertainty and structural assumptions must be assessed together.

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Equality in the weighted Cauchy–Schwarz argument requires X to be constant on the population where X<1, apart from sets of zero probability. With interior m and r, that forces the two-support endpoint construction within the declared model class.

Thus an exact third moment at that special upper boundary supplies more information than an arbitrary additional interior moment. It can select the extremal structure under the assumptions used in the proof.

This is an ideal population-moment conclusion. Finite-sample values close to the boundary do not establish exact equality or a uniquely observed class. Numerical rounding can also make an interior value appear to coincide with a boundary.

Report the distinction explicitly: an equality condition in a mathematical model is not a generic identification guarantee for noisy exposure experiments. The proof is useful because it tells us which additional evidence would matter and why, while preserving the conditions required for its stronger conclusion.

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For every finite Λ>0, exp(−Λt) tends to zero as t tends to infinity. For Λ=0 it remains one. Since the exponential is bounded between zero and one, the population zero probability tends to π under the fixed-rate model.

Thus π is the asymptotic plateau of q(t) in this idealised setting. A finite-duration observation does not automatically reach that limit, especially when positive rates can be very small.

A visibly flat segment over a limited range can therefore be compatible with slow decay rather than a true endpoint mass. Long-term real observations may also violate the constant-rate assumption through changing conditions, dropout or detection.

The mathematical limit identifies the model quantity, not an instruction to extend every study indefinitely. Decide whether the necessary stationarity, exposure and observation conditions are credible before interpreting an apparent plateau as a permanent population state.

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Suppose all nonzero rates are known to satisfy Λ≥λ_min>0. At exposure t, their X values are at most b=exp(−λ_min t)<1.

Then m≤π+(1−π)b, so π≥(m−b)/(1−b), with zero as the lower bound when the numerator is negative. This is extra information beyond the two zero fractions.

For the original m=1/2, if independent evidence justifies b=2/5, the lower bound is(1/2−2/5)/(3/5)=1/6. Combined with the moment upper bound, it identifies π=1/6 under that more restricted class.

The positive-rate construction with X≈0.7236 violates this floor, which is why it is no longer a competing admissible model. Do not choose the floor merely because it produces the preferred answer. Its evidential role is precisely that it supplies a constraint not contained in the two original summaries.

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A single Poisson component in a ZIP model has one supplied positive rate. That is a distributional restriction, not independent evidence that every possible positive-rate alternative lies above a known floor.

Using the fitted component rate as a universal floor for competing models would build the desired structure into the comparison. A heterogeneous population can contain rates smaller or larger than the fitted active mean.

A defensible floor might follow from a separately validated mechanism or a design constraint, but its meaning and uncertainty must be stated. If it is uncertain, the inferred lower endpoint-mass bound must reflect that uncertainty rather than treat the floor as exact.

This distinction helps separate data information from structural assumptions. The two exposure moments constrain an admissible class. A floor narrows that class. Identification after narrowing can be mathematically valid while still depending on the credibility of the added restriction.

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Let q(t)=3/5 and q(2t)=2/5. Verify the moment inequalities, calculate the endpoint-mass upper bound and construct its extremal two-support distribution.

Then find a no-endpoint distribution with the same moments. Compute q(3t) under both proposals and explain whether a third-exposure observation could distinguish them.

Before calculating, identify the quantity being bounded: a population atom at rate zero under the fixed-rate Poisson mixture. Do not interpret the result as the fraction of observed zeros, the active Poisson mean or a probability for a particular individual’s biological state.

The changed moments require rebuilding both the numerator and denominator of the bound. Reusing the original one-sixth or the support value two-fifths without checking the equations would conceal the information structure the exercise is intended to test.

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Now m=3/5 and r=2/5, with variance r−m²=1/25. The endpoint upper bound is(1/25)/(1−6/5+2/5)=1/5.

The nonendpoint support is c=(3/5−2/5)/(1−3/5)=1/2. Thus mass 1/5 at one and 4/5 at one-half reproduces the moments.

A no-endpoint alternative puts equal mass at X=2/5 and 4/5. Its mean is 3/5 and its second moment is 2/5. All corresponding rates are positive and finite.

At3t, the endpoint model predicts 1/5+(4/5)(1/2)³=3/10. The alternative predicts[(2/5)³+(4/5)³]/2=36/125=0.288. The difference is another exact model prediction, not an observed effect size or a statement about statistical power in a particular study.

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Observed zero fractions are estimates of m and r. Their uncertainty depends on sample sizes, dependence, exposure assignment, missing observations and whether the same units contribute to both durations.

Substituting estimates into a sharp nonlinear population bound does not automatically produce a confidence bound for π. A principled inferential procedure must account for uncertainty in the moments and the feasible model region.

For example, one can conceptually consider all compatible moment pairs in an appropriately justified uncertainty set, then evaluate the possible endpoint-mass bounds across that set. The construction of that uncertainty set is an additional statistical problem, not supplied by the algebra here.

Report whether a number is an exact model limit, a plug-in calculation or an uncertainty statement about data. A narrow displayed interval cannot be justified solely by the exactness of an inequality whose inputs were estimated imprecisely.

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The same zero fraction can describe different observation frames. A hundred short-duration records and a selected group of long-duration survivors need not represent the same distribution of latent rates.

If high-rate units leave a cohort or low-rate units are preferentially observed for longer, comparing the two reported fractions as moments of one X can be invalid. Exposure units must also match the assumed rate scaling.

Record eligibility, censoring, exclusions and the number of units actually observed at each duration. A zero from a record with no opportunity for an event may reflect the design rather than a latent rate class. A missing record is not automatically a zero.

The moment proof begins after those observation conditions have been justified. It can expose the information limit of a coherent model, but it cannot establish that the dataset satisfies the model’s common-population and comparable-exposure premises.

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If every event is independently recorded with a fixed positive probability p, a Poisson rate Λ is thinned to observed rate pΛ. The exposure-zero function then concerns that observed rate.

A fixed p>0 preserves which rates are exactly zero but changes the scale of positive rates and any proposed lower bound on them. If some units have p=0, an observation-zero-rate class can appear even when their underlying event rates are positive.

Therefore a zero-rate atom in the recorded process need not prove biological or physical impossibility of an event. It may reflect detection or data capture under the specified observation channel.

This is a stipulated thinning example, not a universal detection model. Its purpose is to show why a latent endpoint interpretation must name the process being modelled. Event generation, recording and eligibility can each introduce different zero mechanisms, and count summaries alone need not separate them.

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Our two moments concern q(t) and q(2t), the zero probabilities at two exposures. A paired record can contain more information, such as counts within separate windows or transitions after an event.

To use that information, specify the joint process and how the latent rate persists. A model with a fixed rate per unit and conditionally independent increments predicts a different joint structure from one that redraws its rate between windows.

Do not infer that joint structure merely because a marginal zero curve fits. The ZIP–hurdle equivalence guide explains a related distinction between matched marginal probabilities and added dependence assumptions.

Additional data are useful when the competing complete models make different predictions for them. Calling a second record longitudinal or paired does not by itself identify an unobserved class; the model and observation relationship still determine what the new record can distinguish.

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The two constructed models were matched only on zero probabilities at two durations. Their positive-count masses and other moments can differ.

If full counts were recorded, discarding them can remove evidence relevant to the comparison. The mean-and-zero count-mixture guide derives an adjacent-probability check showing how positive shape can distinguish models that share selected summaries.

This does not mean full counts always identify a scientifically unique mechanism. Different latent representations can still agree on complete observable laws under particular assumptions. It means the information set matters. State whether the analysis used zero indicators, complete counts, known exposure histories or independent state measurements before claiming a particular identification limit or an identified latent class.

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If m=1 under the finite nonnegative-rate model, E(X)=1 with X≤1 forces X=1 almost surely, so π=1. The main formula has a zero denominator at m=r=1 and should not be used there.

If r=m² with 0<m<1, X is constant below one, so π=0. If r=m for 0<m<1, equality of X² and X in expectation forces endpoint values zero or one. A positive mass at X=0 would require an infinite rate, outside the finite-rate model.

Consequently some formal boundaries of the bounded-moment region are limits of finite-rate mixtures rather than attainable nondegenerate finite-rate populations. Check those domain details before inserting a boundary estimate into an interpretation. The interior examples avoided them deliberately so both competing constructions remained valid ordinary Poisson mixtures with finite rates.

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The Stan User's Guide on finite mixtures documents mixture and zero-inflated probability constructions. It anchors the count-model notation; the bounded-moment endpoint proof and numerical constructions here are independently derived teaching examples.

The existing Cox-process guide explains why heterogeneous intensities affect count distributions and why time structure matters. Our fixed-rate exposure construction is narrower than a general time-varying Cox process.

Keep the complete equations beside any numerical result. A reader should be able to reconstruct q(t), q(2t), the support values and their weights without trusting a latent-class label returned by software. The proof’s value lies in its explicit information boundary: two exact zero summaries permit particular endpoint masses under a named class of models, not every interpretation of real zeros.

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A nonzero long-looking zero fraction proves a permanent inactive class: finite positive rates can also produce many zeros. A sharp upper bound gives the actual class fraction: the no-endpoint construction disproves that conclusion.

Two exposures fit a ZIP model, so all other rate distributions are ruled out: the paired moment constructions show otherwise. An additional exposure always identifies the whole distribution: it adds information but need not remove every alternative.

A fitted active rate is an independently known lower rate bound: it is a model parameter, not automatically evidence restricting every competing population. A plug-in algebraic bound is a confidence bound: the uncertainty in estimated moments requires separate treatment.

The repair is to name the observed summaries, write their common latent-variable representation, construct alternatives and state what extra assumption or observation would exclude one of them. This is a reusable way to distinguish mathematical identification from a convenient fitted story.

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Does the bound prove anyone is permanently inactive?

No. In the original example, both zero endpoint mass and mass one-sixth are compatible with the two exact zero probabilities. The upper limit is conditional on the model.

Why transform the rate to X?

It turns the two exposure probabilities into first and second moments of a variable in(0,1], making the endpoint mass and its constraints explicit.

Can a third exposure help?

Yes. The two original proposals predict 0.20 and 0.22 at 3t. That distinguishes those proposals in principle, while statistical power and broader identification remain separate questions.

What would a known rate floor add?

It restricts the positive-rate support and can supply a lower bound on π. It must come from independent evidence or a justified mechanism, not be chosen to force a desired answer.

Does a model-zero rate mean a true biological impossibility?

Not automatically. The model may describe recorded events, and observation or detection processes can create their own zero-rate groups.

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Calculate the two original X support values and verify both moments directly. Check the endpoint model with mass 1/6 at one and 5/6 at 2/5. Convert every nonendpoint support value into its finite positive rate −log(X)/t.

Derive the Cauchy–Schwarz bound and compare it with the simple upper limit π≤m. Then compute the two third-exposure probabilities and verify the lower and upper third-moment inequalities.

Repeat the changed m=3/5,r=2/5 exercise. Finally explain which conclusion would change if a rate floor were independently known, if rates varied over time, or if the two cohorts differed. This sequence tests the model, arithmetic, information limit and transfer conditions together rather than treating a matched zero curve as a certificate for an unobserved permanent state.

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Suppose a unit keeps its rate Λ across two adjacent windows of length t, with conditionally independent Poisson increments. The probability of zero events in both windows is E[exp(−2Λt)]=r.

Each individual window has zero probability m. The covariance of their zero indicators is therefore r−m², which is 1/20 in the original example. Two independent units drawn from the population would instead have joint-zero probability m².

This gives an operational reading of the moment inequality: a shared latent rate can make zero histories positively associated even without a zero-rate class. It does not prove that this was the source of an observed association; other dependence mechanisms or observation rules can also matter.

The both-zero event across those adjacent windows is exactly the zero event over their combined exposure in this model. Counting it as an additional independent third summary would double-count the same probability. Retain the observation and dependence structure before deciding how much new information a repeated record supplies.

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The support values X are dimensionless, while Λ has inverse-exposure units. Converting a support point back to a rate requires the actual duration: Λ=−log(X)/t.

If t is expressed in hours rather than minutes, the numerical rate changes inversely so that Λt remains unchanged. The endpoint mass π and all model zero probabilities are invariant under that consistent unit conversion.

An independently proposed rate floor must be converted too. Combining a per-hour lower rate with an exposure stated in minutes without conversion changes b=exp(−λ_min t) and can create a spurious lower bound on π.

A useful transfer check is to repeat the original construction with t=2 in a chosen exposure unit. Every nonzero rate halves relative to t=1, while the moments m and r at t and 2t stay exactly the same. The zero-rate support remains zero. This simple rescaling tests the distinction between a latent probability, a dimensionless transformed rate and a dimensional event intensity.

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For a finite mixture with fixed component rates, differentiation gives q′(t)=−Σ_i w_i λ_i exp(−λ_i t)≤0 and q″(t)=Σ_i w_i λ_i² exp(−λ_i t)≥0. The zero curve decreases and is convex.

A Cauchy–Schwarz argument also gives q((s+t)/2)²≤q(s)q(t), a log-convexity constraint. Setting s=0 and using q(0)=1 recovers r≥m² for the two-duration comparison.

These shape checks can assess compatibility with the declared class, but they do not determine an endpoint mass from a few curve values. Both original constructions satisfy them while having different π.

Do not infer that every decreasing convex curve must be a valid fixed-rate Poisson-mixture zero curve; we have stated necessary properties for the finite mixtures considered, not a complete characterisation of all possible exposure functions. Use the inequalities to check the model and the explicit constructions to assess what its supplied summaries identify.

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Return to the Zero-Inflated Models subject guide for count distributions, exposure and model choice. Follow the full ZIP–hurdle mapping and the mean-and-zero mixture comparison for two different levels of observable agreement.

Use The Gold Standard of Education: Learning Transfers into Life for the wider practice of worked examples and changed-assumption checks. The How X Works hub connects the same reasoning to other subjects.

A careful final statement names the exposure model, the moments supplied, the endpoint bound and its attainability. It then identifies what additional data or independently justified restriction would be needed before a stronger latent-state claim could follow.

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