Exceptional Parameter Values in Identifiability Proofs

Generic Results and Special Cases

A structural-identifiability proof often solves for parameters by manipulating output equations. Those manipulations can involve dividing by a rate, an initial condition, or a difference between two rates. If the quantity in a denominator is zero, the derivation no longer applies. That does not, by itself, show that the parameter is unidentifiable: the special case may have other informative equations, or it may genuinely produce the same output for multiple parameter values.

The distinction matters in QSP models because rates can be zero at a boundary, modes can coincide, and a state may start at zero under a particular experiment. A useful analysis asks two questions separately: what can be recovered for general parameter values, and what can be recovered at the special values where the generic argument breaks down?

Definition: For a specified model, input, initial condition, and observed output, two parameter vectors \(\theta\) and \(\tilde{\theta}\) are input-output equivalent if they produce the same output for all observed times. A parameter vector is globally identifiable at a specified point if no other admissible vector is input-output equivalent to it. It is locally identifiable if no distinct equivalent vectors occur sufficiently near it. A parameter is generically identifiable if it is identifiable for general parameter values, apart from a special set where the input-output map loses the relevant information. The exact meaning of “identifiable” (global or local) should be stated when giving a classification.

Here, “general” does not mean “most likely” under a probability distribution. In algebraic identifiability calculations, the exceptional set often consists of parameter values satisfying one or more equations, such as \(k_1=0\) or \(a-b=0\). A generic result applies away from those equations. If the parameter domain excludes a boundary value, such as \(k_1=0\) when rates are required to be strictly positive, that boundary is not part of the domain being classified. If it is included, it must be checked separately.

Structural conclusions assume an exact model, known inputs and specified initial conditions, and ideal, noiseless output information over the stated observation interval. They concern whether the output uniquely determines parameters in principle, not how reliably finite noisy observations estimate them.

A Zero Rate Can Hide Another Rate

Consider two states in sequence, with the first state feeding the second. The measured output is the second state:

\begin{aligned} \dot{x}_1(t)&=-k_1x_1(t),\\ \dot{x}_2(t)&=k_1x_1(t)-k_2x_2(t),\\ x_1(0)&=1,\qquad x_2(0)=0,\qquad y(t)=x_2(t). \end{aligned}

Assume \(k_1,k_2\geq 0\), with time measured in days, and suppose the output is known continuously and exactly. The first state is \(x_1(t)=e^{-k_1t}\). From the second-state equation, \(y'(0)=k_1\). Differentiating that equation once gives \(y''(0)=-k_1^2-k_1k_2\). Thus, if \(k_1>0\), both rates can be recovered:

k_1=y'(0),\qquad k_2=-\frac{y''(0)}{y'(0)}-y'(0).

The division by \(y'(0)\) is legitimate only when \(k_1>0\). If \(k_1=0\), then \(x_2(t)=0\) for all \(t\), regardless of \(k_2\). The output determines that there is no transfer into the measured state, but it contains no information about that state’s elimination rate \(k_2\). Therefore \(k_2\) is identifiable for \(k_1>0\) but not at \(k_1=0\), when the zero-rate boundary is included in the domain.

Worked Example: Recovering Rates Away From a Zero-Transfer Boundary

Suppose the exact derivatives of the hypothetical output at time zero are \(y'(0)=0.30\ \text{day}^{-1}\) and \(y''(0)=-0.21\ \text{day}^{-2}\). Use the model above to recover the rates.

Because \(y'(0)=k_1\), \(k_1=0.30\ \text{day}^{-1}\). This is nonzero, so the recovery formula for \(k_2\) applies:

k_2=-\frac{-0.21\ \text{day}^{-2}}{0.30\ \text{day}^{-1}} -0.30\ \text{day}^{-1} =0.70\ \text{day}^{-1}-0.30\ \text{day}^{-1} =0.40\ \text{day}^{-1}.

The rates are \(k_1=0.30\ \text{day}^{-1}\) and \(k_2=0.40\ \text{day}^{-1}\). The calculation is consistent with the assumed domain \(k_1,k_2\geq 0\). If instead \(y'(0)=0\) and \(y(t)\) is identically zero, this calculation cannot recover \(k_2\): the model output is independent of \(k_2\) when \(k_1=0\).

This example also illustrates a useful discipline in identifiability proofs: record the conditions attached to every rearrangement. Cross-multiplying an equation may remove a denominator from the written expression, but it does not automatically make the original solution valid at values where the denominator vanished. Those values require their own analysis.

Repeated Eigenvalues: Singular Formula or Real Loss?

Repeated eigenvalues are another common source of apparent exceptions. Consider two independent exponential modes contributing to one output:

y(t)=A e^{-at}+B e^{-bt},

where \(A\) and \(B\) are unknown positive initial amplitudes and \(a,b\) are positive rates, all expressed using consistent time units. When \(a\neq b\), the output contains two distinct exponential modes. When \(a=b=r\), the modes coincide and the output becomes \(y(t)=(A+B)e^{-rt}\). At the repeated-rate value, the output depends on the amplitudes only through their sum. No exact observation of this output can distinguish different positive splits of that sum.

Worked Example: Amplitudes at a Repeated Rate

Suppose a hypothetical model has \(a=b=0.50\ \text{day}^{-1}\), \(A=2\) concentration units, and \(B=3\) concentration units. Its output is

y(t)=2e^{-0.50t}+3e^{-0.50t} =5e^{-0.50t} \quad\text{concentration units, with \(t\) in days.}

The alternative amplitudes \(A=1\) concentration unit and \(B=4\) concentration units, with the same rates, produce exactly the same output at every time. More generally, any positive \(A'\) and \(B'\) satisfying \(A'+B'=5\) concentration units give the same trajectory. The common rate \(r=0.50\ \text{day}^{-1}\) and the sum \(A+B=5\) concentration units are determined by this trajectory, but \(A\) and \(B\) separately are not. The amplitudes are therefore not identifiable at this repeated-rate point.

The loss of amplitude information can also be seen directly from derivatives. If \(a=b=r\), then for every nonnegative integer \(m\),

y^{(m)}(0)=(-r)^m(A+B).

Every derivative involves \(A\) and \(B\) only through \(A+B\). Adding further exact derivatives cannot separate the two amplitudes. This is a genuine loss in the input-output map at the repeated-rate value, not merely a bad choice of algebraic formula.

By contrast, a formula for distinct rates may contain the factor \(a-b\) in a denominator and fail at \(a=b\), even when the output still identifies the quantities of interest at the repeated value. A failed formula is a warning to reanalyze the case, not a verdict about identifiability. The amplitude example shows one possible outcome of that reanalysis: some quantities remain determined, while others do not.

A Vanishing Initial Condition Can Erase Rate Information

Now consider a single observed state obeying \(\dot{x}=-kx\), with known initial condition \(x(0)=x_0\) and output \(y(t)=x(t)\). Its solution is \(y(t)=x_0e^{-kt}\). If \(x_0\neq 0\), the initial output and its derivative give \(k=-y'(0)/y(0)\). If \(x_0=0\), however, the output is zero for every time and every \(k\). The rate cannot be recovered from this experiment.

Worked Example: A Rate With a Zero Initial State

For a hypothetical assay, suppose the measured state satisfies \(\dot{x}=-kx\), with \(x(0)=0\) concentration units and output \(y=x\). Compare \(k=0.20\ \text{day}^{-1}\) with \(k=0.80\ \text{day}^{-1}\).

For either rate, the solution is \(y(t)=0e^{-kt}=0\) concentration units at every time. The two distinct parameter values therefore produce identical outputs, so \(k\) is not identifiable under this initial condition. If the known initial amount were instead \(x_0=4\) concentration units and the initial derivative were \(-0.80\) concentration units per day, then

k=-\frac{y'(0)}{y(0)} =-\frac{-0.80\ \text{concentration units per day}} {4\ \text{concentration units}} =0.20\ \text{day}^{-1}.

The initial condition is part of the experiment specification, not necessarily a parameter. Thus \(x_0=0\) is an exceptional experimental configuration for estimating \(k\); it is not automatically an exceptional value of the model parameter \(k\).

How to Analyze an Exceptional Value

A short case-by-case argument prevents generic classifications from being applied outside their conditions. In particular, distinguish an exceptional value that creates input-output equivalence from one that merely makes a chosen expression inconvenient. The following steps are useful when a derivation includes divisions or assumes distinct modes.

State the model and domain.
Specify the parameters allowed, which quantities are known, the initial conditions, the input, and the measured output. A value outside the domain is not an exceptional point of that model’s classification.

List the conditions used in the generic argument.
For example, note every requirement such as \(k_1\neq 0\), \(a\neq b\), or \(x_0\neq 0\). These conditions identify where that argument may fail.

Substitute the exceptional value into the model itself.
Derive or simplify the output in that case rather than relying on a formula obtained by dividing by a quantity that is now zero.

Compare equivalent parameter values.
Determine whether distinct admissible parameter vectors produce the same output. State which parameters or combinations remain recoverable, and whether the conclusion is local or global.

This procedure supports a precise conclusion. For example, in the two-state transfer model, \(k_2\) is recoverable when \(k_1>0\), but not when \(k_1=0\); the exceptional case is a genuine loss of information about \(k_2\). In the repeated-rate amplitude model, the sum of amplitudes remains recoverable while the individual amplitudes do not. In the single-state model, a zero initial condition removes all information about the rate from that experiment.

Common Pitfalls and Limits

Calling a failed division proof of nonidentifiability. If a formula divides by \(a-b\), the formula does not apply at \(a=b\). Analyze the output at equal rates before classifying the parameters.
Assuming a zero rate always causes loss of identifiability. A zero rate may still be determined from an informative output. The consequence depends on how that rate enters the model and on what is observed.
Reporting only a generic result. “Identifiable for general parameter values” does not imply identifiable at every admissible value. State important exceptional conditions and their consequences.
Confusing a parameter boundary with an experiment condition. A zero initial state can make a rate unidentifiable without making the rate itself zero or exceptional.
Ignoring parameter labels or symmetries. Distinct parameter vectors can sometimes differ only by exchanging two modes. Say whether the claim concerns labeled parameters or an unordered set of quantities.

Exceptional-value analysis is limited to the model and experiment being studied. Changing the observed outputs, inputs, or initial conditions can change which quantities are identifiable. Moreover, a parameter that is structurally identifiable in the nonexceptional case may still be difficult to estimate from finite noisy data; that is a different question from exact input-output uniqueness.

Key takeaway: A generic identifiability proof applies only under the conditions used in its derivation. At each important exceptional value, simplify the model’s output and check directly whether distinct admissible parameters remain input-output equivalent.

Check Your Understanding

For each question, give a conclusion and a brief justification.

In the two-state transfer model, which rate becomes unidentifiable when \(k_1=0\), and why?
For \(y(t)=A e^{-rt}+B e^{-rt}\), which amplitude quantity is determined by the output, and which amplitude information is lost?
In \(y(t)=x_0e^{-kt}\), is \(k=0\) necessarily unidentifiable when \(x_0\neq 0\)? Explain using the output.
A derivation divides by \(a-b\) and concludes that parameters are identifiable when \(a\neq b\). What does this establish about the case \(a=b\)?

Solutions. (1) \(k_2\) is unidentifiable: when \(k_1=0\), no material enters \(x_2\), so \(y(t)=0\) for every \(k_2\). (2) The sum \(A+B\) is determined, but the separate amplitudes are not, since every split with the same sum gives the same output. (3) No. With \(x_0\neq 0\), \(y(t)=x_0\) when \(k=0\), and a different \(k\) gives a different exponential trajectory, so \(k=0\) is distinguishable in this model. (4) It establishes nothing by itself about \(a=b\); the output must be analyzed at that value without using the division by \(a-b\).