Combining Independent Experiments for Structural Identifiability

Why Combine Experiments?

A model may produce ideal outputs that leave several parameter values indistinguishable in one experiment. A second experiment can resolve that ambiguity if its known initial condition or input makes the output depend on the shared parameters in a different way. The key is not simply to collect more outputs: the combined experiments must distinguish parameter values that each experiment alone cannot.

Structural identifiability concerns the model's ideal, exact output, not the precision of a finite noisy data set. The question is whether two admissible parameter values can produce exactly the same outputs under the specified experimental conditions. The argument therefore depends both on the model and on what is known about each experiment.

Definition: Let \(\Theta\) be the admissible parameter domain, and let experiment \(e\) have known design \(d_e\), such as its initial state or input profile. Write its ideal output map as \(F_e(\theta;d_e)\), where \(\theta\in\Theta\) is the parameter vector shared across experiments. The combined model is globally structurally identifiable on \(\Theta\) if \[ F_e(\theta;d_e)=F_e(\widetilde{\theta};d_e)\ \text{for every experiment }e \quad\Longrightarrow\quad \theta=\widetilde{\theta} \] for all \(\theta,\widetilde{\theta}\in\Theta\). If distinct admissible parameter values give the same full set of ideal outputs, the combined experiments do not globally identify the parameters.

This definition presumes that the model structure is specified, the initial conditions and inputs used in the output maps are known, and the same parameter vector applies across the experiments. The experiments may have separate trajectories and outputs; they need not share a state. Any unknown initial state, assay scale, or experiment-specific nuisance parameter must either be known, constrained, or included explicitly in the parameterization and identifiability test. Treating an unknown quantity as known can create apparent identifiability that the actual model does not have.

“Independent experiments” here means separate experimental runs whose conditions can be specified separately. In a structural analysis, independence does not itself imply any statistical assumption: the ideal outputs are compared exactly. The benefit comes from changing known conditions in a way that changes the parameter dependence of the output.

Stacking the Output Maps

For a finite set of measured outputs, the combined map is formed by placing the outputs from each experiment into one vector. For continuous-time outputs, the same idea applies to the collection of output functions: equality of the combined outputs means equality at all observed times in every experiment.

F_{\mathrm{stack}}(\theta) = \begin{bmatrix} F_1(\theta;d_1)\\ F_2(\theta;d_2)\\ \vdots\\ F_m(\theta;d_m) \end{bmatrix}.

The combined model is globally identifiable on \(\Theta\) exactly when \(F_{\mathrm{stack}}\) is one-to-one on \(\Theta\). Equivalently, for any specified ideal output vector, the set of admissible parameters that produce it must contain at most one value.

This also explains how experiments can resolve ambiguity. For a particular output from experiment \(e\), define its parameter fiber as the set of all values in \(\Theta\) that produce that output. The parameter values compatible with all experiments lie in the intersection of their fibers. A single experiment may have a large fiber, while a second experiment with a different design can shrink the intersection to one value. If the new experiment has the same parameter dependence, the intersection may remain large.

Worked Examples

Worked Example: Distinct Initial States Identify Two Linear Coefficients

Consider a known linear observation rule at the initial measurement time: \(y=a x_1+b x_2\). The states \(x_1\) and \(x_2\) are known and normalized; \(a\) and \(b\) are shared, unknown response coefficients. Suppose two separate hypothetical experiments use initial states \((x_1,x_2)=(1,1)\) and \((1,2)\), and produce ideal outputs of 5 and 8 response units, respectively. Assume \(a,b>0\).

The first experiment gives \(a+b=5\). By itself, it cannot distinguish, for example, \((a,b)=(2,3)\) from \((3,2)\); both pairs are positive and yield the same output. The second experiment gives \(a+2b=8\). Stacking the two equations gives

\begin{bmatrix} 1&1\\ 1&2 \end{bmatrix} \begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}5\\8\end{bmatrix}.

Subtract the first equation from the second: \(b=8-5=3\). Substituting into \(a+b=5\) gives \(a=5-3=2\). Thus the combined outputs uniquely determine \(a=2\) and \(b=3\) response units per normalized state unit. The coefficient matrix has determinant \(1(2)-1(1)=1\), so it has full column rank and the solution is unique.

The known initial states are important: they provide two different known weightings of the same coefficients. The conclusion applies to this observation model and these data; it does not establish identifiability for an unspecified dynamic model with additional unknown states or parameters.

Worked Example: Two Input Levels Resolve a Nonlinear Ambiguity

Suppose the steady-state response at a known input level \(u\) follows the hypothetical saturable model \(y=\theta_1u/(\theta_2+u)\), with \(\theta_1>0\), \(\theta_2>0\), and \(u>0\). Here \(\theta_1\) is a response scale and \(\theta_2\) has the same units as the input. Measurements are ideal, the same parameters apply in both runs, and the input levels are known. Consider outputs \(y=2\) at \(u=1\) input unit and \(y=3\) at \(u=2\) input units.

One experiment alone does not identify both parameters. At \(u=1\), \(y=2\) requires \(\theta_1=2(\theta_2+1)\) when numerical values are expressed in the stated units. Every positive choice of \(\theta_2\) gives a corresponding positive \(\theta_1\), so there is a continuum of compatible pairs.

For the combined analysis, rearrange the model:

\frac{\theta_1}{y}=1+\frac{\theta_2}{u}.

Substituting the two hypothetical observations yields \(\theta_1/2=1+\theta_2\) and \(\theta_1/3=1+\theta_2/2\). The first equation gives \(\theta_1=2+2\theta_2\); the second gives \(\theta_1=3+1.5\theta_2\). Equating these expressions gives \(0.5\theta_2=1\), hence \(\theta_2=2\) input units and \(\theta_1=6\) response units. Substitution returns \(6(1)/(2+1)=2\) and \(6(2)/(2+2)=3\), as required.

The algebra also verifies uniqueness for these outputs: the two equations for \(\theta_1,\theta_2\) have only the solution just calculated. The distinct input levels provide different parameter weightings; the conclusion relies on the stated model, positive domain, known inputs, and shared parameters.

Worked Example: Rank Deficiency and a Restricted Parameter Domain

Consider a linear output \(y=X\theta\), where \(X=\begin{bmatrix}1&0\end{bmatrix}\) and \(\theta=(\theta_1,\theta_2)^\mathsf{T}\). On the unconstrained domain \(\Theta=\mathbb{R}^2\), the output is \(y=\theta_1\), so \((4,2)\) and \((4,9)\) both give \(y=4\). The parameters are not uniquely determined on this domain.

The null space of \(X\) contains every vector \((0,c)^\mathsf{T}\), since \(X(0,c)^\mathsf{T}=0\). In particular, adding \((0,7)^\mathsf{T}\) to \((4,2)^\mathsf{T}\) leaves the output unchanged. But now restrict the admissible domain to \(\Theta=\{(q,0)^\mathsf{T}:0\leq q\leq 10\}\). On this domain \(y=q\), so each output has at most one admissible parameter value even though \(X\) still has a nontrivial null space.

Thus full column rank guarantees uniqueness for a linear map, but rank deficiency does not by itself rule out uniqueness on every restricted domain. For an even simpler case, if \(\Theta=\{(0,0)^\mathsf{T}\}\) and \(X=0\), the parameter is unique within \(\Theta\), despite the nontrivial null space. The domain must be included in the claim.

Linear Rank Tests and Their Scope

When stacked outputs are linear in the parameters, \(y=X\theta\), a null-space calculation gives a direct test. Over the unconstrained domain \(\mathbb{R}^p\), the map is one-to-one if and only if the only solution of \(Xv=0\) is \(v=0\), or, equivalently, \(X\) has full column rank. To see both directions, if \(Xv=0\) for a nonzero \(v\), then \(X(\theta+v)=X\theta\) for every \(\theta\), so two distinct parameter values have identical outputs. Conversely, if \(X\theta=X\widetilde{\theta}\), then

X(\theta-\widetilde{\theta})=0.

If the null space is trivial, this implies \(\theta-\widetilde{\theta}=0\), so \(\theta=\widetilde{\theta}\). For a restricted domain \(\Theta\), the exact condition is that no distinct pair \(\theta,\widetilde{\theta}\in\Theta\) have \(X(\theta-\widetilde{\theta})=0\). Trivial null space is still sufficient, but it need not be necessary.

For nonlinear models, a matrix rank test alone does not establish global uniqueness. The relevant question remains whether the stacked nonlinear map takes the same value at two distinct admissible parameter vectors. Algebraic rearrangement, solving the equality of two output maps, or a symmetry argument may expose such pairs. A local derivative calculation can be informative, but it is not by itself a proof of global injectivity.

Checking Whether a New Experiment Helps

A useful analysis proceeds by comparing what each design adds to the parameter constraints, rather than counting experiments or observations. In a linear model, the rows contributed by a new experiment must add independent information in parameter directions not already distinguished. In a nonlinear model, the new design must change the output equations so that parameter values previously equivalent can be separated.

Specify the common parameters and domain.
Write the shared parameter vector and state its admissible set. Keep known conditions separate from unknown quantities.

Write one ideal output map per experiment.
Insert that run's known initial state or input into the model. Do not assume different runs imply different parameter dependence.

Stack outputs and test equality.
For a linear map, analyze rank and the null space with the parameter-domain qualification. For a nonlinear map, solve or otherwise analyze equality of the stacked outputs for two admissible parameter values.

Check the experimental assumptions.
Verify that the conditions are known and that shared and experiment-specific quantities have been represented correctly. Then state the scope of the uniqueness conclusion.

Diagnostics, Alternatives, and Limitations

Several checks help reveal an uninformative combination. Identical input profiles or initial conditions may simply repeat an existing output map. In a linear calculation, a rank-deficient matrix on an unconstrained parameter space exposes a direction along which parameters can change without changing the stacked output. In a nonlinear calculation, explicitly substituting a proposed pair of distinct parameter values into every experiment's output map can confirm an ambiguity. Conversely, failure to find such a pair numerically is not a proof that none exists.

The main alternatives serve different purposes. An experiment-by-experiment analysis identifies which ambiguities remain in each run; stacking then tests whether their joint constraints remove those ambiguities. Rank and null-space methods are efficient for linear parameter dependence, while equality-of-maps arguments are more general for nonlinear models. Differential-algebra methods and symmetry analyses, covered in the earlier material on those subjects, can also expose structural ambiguity; whichever method is used, its final claim must refer to the combined map and the admissible domain.

More experiments do not guarantee more identifiable parameters. If every run repeats the same relationship among parameters, the combined map can remain non-injective. Nor does structural identifiability guarantee that a parameter can be estimated precisely from finite, noisy observations; that is a separate practical question. A structural conclusion is conditional on the assumed model, known designs, and parameter sharing. If any of these are misspecified, the algebra may answer a different question from the one the experiment actually poses.

Common mistake: “There are two unknown parameters and two experiments, so the parameters must be identifiable.” The number of experiments alone says nothing about whether their outputs provide distinct parameter constraints. Check the stacked map, or the rank in a linear model.
Common mistake: “The design matrix is rank deficient, so the parameters cannot be identifiable under any conditions.” That conclusion is valid for an unconstrained linear parameter space, not every restricted admissible domain. Test injectivity on the actual domain.
Common mistake: Treating a run-specific unknown initial state as a known design value. If it is unknown, include it in the model and assess whether it confounds the shared parameters.

Key takeaway: Combining experiments identifies shared parameters only when the stacked ideal output map is one-to-one on the stated admissible parameter domain. Distinct known conditions can shrink ambiguity, but the result must be demonstrated from the model rather than inferred from the number of runs.

Check Your Understanding

Use the stated model and parameter domain in each question. Short solutions follow the questions.

For \(y=a+ub\), two hypothetical experiments use \(u=1\) and \(u=3\), with ideal outputs 5 and 9. Solve for \(a\) and \(b\), and explain why the pair of designs identifies both coefficients.
For \(y=\theta_1+\theta_2\), two runs use the same design. Can stacking identify both parameters on \(\mathbb{R}^2\)? State the identifiable combination.
For \(X=\begin{bmatrix}1&1\\1&1\end{bmatrix}\) on \(\mathbb{R}^2\), give a nonzero null-space vector and explain what it implies.
If \(X=0\) but the admissible domain is \(\Theta=\{(2,3)^\mathsf{T}\}\), is the parameter unique on that domain? Does \(X\) have full column rank?

Solutions. (1) The equations are \(a+b=5\) and \(a+3b=9\). Subtracting gives \(2b=4\), so \(b=2\) and \(a=3\); the distinct design values yield independent equations. (2) No. Both outputs repeat the same equation, so only the sum \(\theta_1+\theta_2\) is determined; for example, \((1,4)\) and \((2,3)\) give the same outputs. (3) The vector \((1,-1)^\mathsf{T}\) is in the null space because \(X(1,-1)^\mathsf{T}=(0,0)^\mathsf{T}\). Adding it to a parameter vector leaves both outputs unchanged, so the map is not injective on \(\mathbb{R}^2\). (4) Yes, there is only one admissible parameter value, so it is unique on \(\Theta\). No: the zero matrix has rank zero, not full column rank. This illustrates why rank deficiency alone does not rule out uniqueness on a restricted domain.