Why a Proof Needs a Checklist
A structural identifiability argument asks what the model and observation design can determine in an ideal setting: outputs are known exactly, without measurement error, under the stipulated inputs and observation conditions. The central task is not to find a parameter estimate. It is to compare parameter values that would generate exactly the same outputs and determine which quantities must therefore agree.
A conclusion such as “the model is identifiable” is incomplete unless it says whether it concerns an individual parameter, the full parameter vector, or a combination of parameters. It must also specify the assumptions and experiment under which equivalence was tested. The checklist below makes those parts explicit.
“Generic” means that the conclusion holds away from exceptional parameter sets, commonly sets of lower dimension such as a parameter being exactly zero. State any exceptions relevant to a proof. A parameter combination, such as a product or ratio, can be identifiable even when its component parameters are not.
The Proof Checklist
Begin with the model in a form that makes the assumptions visible. For a dynamical model this commonly includes the state equations, observation equation, known input \(u(t)\), initial conditions, parameter domain, and any known quantities. The protocol matters: equivalence for one fixed input need not be equivalence across several input profiles or outputs.
Write the model and define which quantities are unknown, which are known, which initial conditions are fixed or unknown, and what input and output records constitute the experiment. State parameter constraints, such as positivity.
Introduce a second admissible parameter vector \(\theta'\). Require its predicted outputs to equal those from \(\theta\) under the same stipulated protocol. Do not replace equality of outputs with similarity of fitted curves.
Use direct output equations, equality of trajectories, or input-output equations to obtain necessary constraints on \(\theta\) and \(\theta'\). Keep track of assumptions used in differentiating, dividing, or eliminating variables.
Determine whether each parameter must agree, whether only a finite set of values is possible, or whether a continuous family remains. Check that proposed equivalent values satisfy the original model and parameter domain.
List globally identifiable parameters, locally identifiable but not globally identifiable parameters, and nonidentifiable parameters or combinations. State the experiment and any exceptional cases to which the conclusion applies.
An input-output equation can be useful when the internal states are not observed. It expresses a relation among measured outputs, known inputs, and parameters after eliminating unobserved states. Such an equation must retain the conditions under which it was derived. For example, multiplying by a denominator may add roots that did not satisfy the original equation, while dividing by a quantity that can be zero may discard a valid exceptional case.
Worked Example: An Exponential Decay Output
Model and assumptions. Suppose a hypothetical concentration output is \(y(t)=A e^{-kt}\), with unknown \(A>0\) in mg/L and \(k>0\) in h\(^{-1}\). Time is observed continuously from \(t=0\), and the model is assumed to describe the ideal output exactly. There is no additional unknown assay scale.
Equivalence proof. Let \((A',k')\) be another admissible pair with the same output at all observed times. At \(t=0\), equality gives
For any \(t>0\), equality then gives \(e^{-kt}=e^{-k't}\). Since the exponential function is one-to-one, \(-kt=-k't\), and therefore \(k=k'\). Thus both \(A\) and \(k\) are globally structurally identifiable under these assumptions.
Numerical check. Take hypothetical exact outputs generated by \(A=80\) mg/L and \(k=0.500\) h\(^{-1}\). At \(t=0\), \(y(0)=80\) mg/L. At \(t=2\) h,
Using these exact values, \(A=y(0)=80\) mg/L, and
The rounded output 29.43 mg/L gives a rounded numerical recovery; the proof of structural identifiability used equality of the ideal trajectories, not the rounded calculation. If the initial amplitude were not observed at \(t=0\), but the same continuous trajectory were observed over an interval, equality at two distinct times would still force \(k=k'\), and then equality at either time would force \(A=A'\).
Identifiable Combinations and Equivalent Parameter Families
A proof should not stop after showing that two named parameters cannot be separately recovered. Solve the equivalence relations far enough to state exactly what is preserved. This identifies the combinations the output does determine and makes nonidentifiability concrete.
Worked Example: Equilibrium Binding Identifies an Affinity Ratio
Model and assumptions. Consider a hypothetical equilibrium-binding experiment with known free ligand concentration \(C\), measured bound concentration \(Y\), and model
Here \(B>0\) is the total binding capacity in nM, \(k_{\mathrm{on}}>0\) has units nM\(^{-1}\) min\(^{-1}\), and \(k_{\mathrm{off}}>0\) has units min\(^{-1}\). The equilibrium model is assumed valid, concentrations are known, and outputs are observed without error at the stipulated ligand concentrations. It contains the rates only through their ratio \(K\), which has units nM.
Deriving identifiable quantities. Suppose two positive parameter sets give the same output for all positive \(C\). Cancelling \(C\) and cross-multiplying gives
Since this equality holds for all \(C\), the coefficients of \(C\) agree, so \(B=B'\). The constant terms then give \(BK'=B'K\), and positivity of \(B\) implies \(K'=K\). Thus \(B\) and \(K=k_{\mathrm{off}}/k_{\mathrm{on}}\) are globally identifiable under a sufficiently informative concentration protocol. The individual rate constants are not: any positive pair with the same ratio produces the same equilibrium outputs.
Numerical demonstration. Suppose hypothetical exact observations are \(Y=50\) nM at \(C=2\) nM and \(Y=75\) nM at \(C=6\) nM. Substitution gives
Equating the two expressions for \(B\) gives \(25K+50=12.5K+75\), so \(12.5K=25\) and \(K=2\) nM. Then \(B=25(2+2)=100\) nM. For example, \(k_{\mathrm{on}}=0.400\) nM\(^{-1}\) min\(^{-1}\) and \(k_{\mathrm{off}}=0.800\) min\(^{-1}\) give \(K=2\) nM. So do \(k_{\mathrm{on}}=0.200\) nM\(^{-1}\) min\(^{-1}\) and \(k_{\mathrm{off}}=0.400\) min\(^{-1}\). The equilibrium data distinguish neither pair of rates; they identify the ratio.
Individual Parameters, Full Vectors, and Exceptional Cases
The distinction between a parameter and a parameter vector is essential. Suppose the output is \(y(t)=\theta_1\), with \(\theta_1>0\) and an unrestricted positive \(\theta_2\). Output equivalence forces \(\theta_1'=\theta_1\) but places no restriction on \(\theta_2'\). Thus \(\theta_1\) is globally identifiable, while \(\theta_2\) is nonidentifiable. The full vector is not locally identifiable: infinitely many equivalent vectors \((\theta_1,\theta_2')\) remain. It would be incorrect to infer that \(\theta_1\) is not locally identifiable merely because the full-vector equivalence class is infinite.
Global and local identifiability also differ when a parameter has a finite ambiguity. For \(y=\theta^2\) with \(\theta\in\mathbb{R}\), output equality implies \((\theta')^2=\theta^2\), so \(\theta'=\theta\) or \(\theta'=-\theta\). Generically there are two possible parameter values: \(\theta\) is locally identifiable but not globally identifiable. At the exceptional value \(\theta=0\), the only solution is \(\theta'=0\). This exceptional case does not change the generic classification.
When proving a conclusion, inspect points where a coefficient vanishes, a denominator is zero, or the number of solutions changes. The earlier material on exceptional parameter values emphasizes why such cases should be recorded rather than silently removed. If an assumption excludes a case, say so; if it does not, check it in the original model.
Proof Routes, Diagnostics, and Limitations
Several standard routes can implement the checklist. Directly equating output expressions is often the shortest route for simple observation models. An input-output equation can eliminate unobserved states and expose parameter combinations. A symmetry argument can exhibit a transformation of parameters and states that leaves every stipulated output unchanged; it proves nonuniqueness when the transformation generates distinct admissible parameter values. These are complementary proof strategies, not substitutes for defining the protocol and solving the resulting equivalence problem.
| Proof route | Useful when | Check carefully |
|---|---|---|
| Direct output equality | The output has a simple closed form. | Use equality over the specified times or inputs, not just at one selected observation. |
| Input-output equation | States are hidden but can be eliminated. | Track denominators, differentiability assumptions, and exceptional cases introduced or lost by algebra. |
| Symmetry or invariance | A transformation preserves the model outputs. | Verify that transformed parameters remain admissible and that the transformation changes the parameter being called nonidentifiable. |
Common warning signs include claiming identifiability from a single numerical simulation, treating a successful parameter fit as a proof, or claiming nonidentifiability from a parameter count alone. A symmetry proves ambiguity only if it preserves the stipulated inputs, initial-condition assumptions, and observed outputs. Conversely, failure to find a symmetry does not prove identifiability; the equivalence equations may still have multiple solutions.
Structural conclusions are conditional on the mathematical model and experimental protocol. A parameter may be structurally identifiable under ideal continuous observations yet difficult to estimate from finite, noisy measurements. That is a different question from the one addressed here. Likewise, a conclusion under one input profile cannot automatically be carried over to multiple profiles, multiple outputs, or a different initial-condition design. The proof should name the conditions it actually used.
Check Your Understanding
Answer each question, then compare with the solutions below.
- In the equilibrium-binding example, why do the data identify \(K\) but not \(k_{\mathrm{on}}\) and \(k_{\mathrm{off}}\) separately?
- If an individual parameter has exactly three possible values among equivalent vectors for generic true values, is it globally or locally identifiable?
- For output \(y=\theta_1\) with unrestricted positive \(\theta_2\), classify \(\theta_1\), \(\theta_2\), and the full vector.
- Why can multiplying an input-output equation by a denominator create a proof problem?
- What must be held fixed when checking whether a parameter transformation is an output-preserving symmetry?
Solutions. (1) The equilibrium output depends on the rates only through \(K=k_{\mathrm{off}}/k_{\mathrm{on}}\). Every positive pair with the same ratio gives the same equilibrium curve, so the two rates cannot be separated by these outputs. (2) It is locally identifiable because three is finite, but not globally identifiable because the value is not unique. (3) \(\theta_1\) is globally identifiable, \(\theta_2\) is nonidentifiable, and the full vector is not locally identifiable because infinitely many equivalent vectors remain. (4) Multiplication can admit values that fail the original equation when the denominator is zero; check candidate solutions against the original model and treat zero-denominator cases separately. (5) Hold fixed the stipulated input, initial-condition assumptions, model structure, and observation protocol, and verify that the transformed parameter vector remains in the admissible domain.