Identifiable Parameter Combinations

Why an Output May Identify a Combination, Not Each Parameter

A model can reproduce an output trajectory exactly while leaving uncertainty about how its individual parameters contribute. This is not necessarily a failure of numerical estimation: the model equations themselves may make different parameter values observationally indistinguishable. In such cases, the output can still determine a function of the parameters uniquely. Identifying that function is more informative than reporting separate parameter estimates that the model cannot support.

The distinction is structural. Assume an exact model, specified inputs and initial conditions, and an ideal observation of the model output over the stated time interval. Structural identifiability asks what the output determines under those assumptions, before considering measurement noise or the amount of data collected. The earlier material on global and local structural identifiability provides the broader framework; here the focus is on identifiable combinations.

Definition: Let \(\theta\) denote the model parameters and let \(y(t;\theta)\) be the output under fixed model assumptions, inputs, and initial conditions. A parameter combination \(g(\theta)\) is globally structurally identifiable if equality of the ideal output trajectories implies equality of that combination: if \(y(t;\theta_1)=y(t;\theta_2)\) for all observed times, then \(g(\theta_1)=g(\theta_2)\). An individual parameter is the special case \(g(\theta)=\theta_j\). If distinct parameter values produce the same output but have different values of \(g\), the combination is not globally identifiable.

This definition depends on the specified parameter domain. For example, requiring reaction rates to be nonnegative can rule out some parameter pairs that would be allowed if rates could take any real value. Conclusions about whether individual parameters can be separated must therefore state their domain and respect its boundaries.

A Single Exponential Decay Identifies a Rate Sum

Suppose an observed amount \(x(t)\) is depleted by two parallel first-order processes with nonnegative rate constants \(k_1\) and \(k_2\). Assume the amount itself is observed without an offset, the initial amount \(x_0\) is known and positive, and the model is exact:

\frac{dx}{dt}=-(k_1+k_2)x,\qquad x(0)=x_0,\qquad k_1,k_2\geq 0.

Solving this differential equation gives an exponential with a single decay rate. Because \(x_0>0\), the trajectory is positive, so its logarithm is defined. Taking logarithms turns the trajectory into a straight line whose slope is the negative of the rate sum:

x(t)=x_0 e^{-(k_1+k_2)t},\qquad \log\!\left(\frac{x(t)}{x_0}\right)=-(k_1+k_2)t.

Thus the output determines \(\lambda=k_1+k_2\). To see why, suppose two parameter pairs produce the same trajectory for all \(t\). For any \(t>0\), equality of their positive outputs implies equality of their logarithms, and therefore \(\lambda_1t=\lambda_2t\). Since \(t>0\), \(\lambda_1=\lambda_2\). The sum is identifiable even when its components are not.

If the identified sum is positive, every value of \(k_1\) between zero and \(\lambda\), paired with \(k_2=\lambda-k_1\), is nonnegative and gives the same trajectory. There is then a continuum of parameter pairs, so the output does not identify either rate separately. The boundary case matters: if \(\lambda=0\), nonnegativity and \(k_1+k_2=0\) force \(k_1=k_2=0\). In that case, the constant trajectory does identify both rates separately within the nonnegative domain. It would be incorrect to claim that this model never identifies the individual rates. With an unrestricted real-valued domain, by contrast, a zero sum could arise from many pairs, such as \(k_1=1\) and \(k_2=-1\).

Worked Example: Deriving the Identifiable Combination

Consider a hypothetical exact trajectory from the model above, with \(x_0=80\) mg/L and \(x(4)=29.43\) mg/L after 4 h. Assume both rates are nonnegative and the initial amount and measurement scale are known. What combination does the trajectory determine, and can it determine the two rates separately?

The solution gives \(x(4)/x_0=e^{-4(k_1+k_2)}\). Substituting the hypothetical values,

k_1+k_2 =-\frac{1}{4\ \mathrm{h}}\log\!\left(\frac{29.43\ \mathrm{mg/L}}{80\ \mathrm{mg/L}}\right) \approx 0.2500\ \mathrm{h}^{-1}.

The ratio is dimensionless, so the resulting sum has units of \(\mathrm{h}^{-1}\). For example, \((k_1,k_2)=(0.1000,0.1500)\ \mathrm{h}^{-1}\) and \((0.2000,0.0500)\ \mathrm{h}^{-1}\) have the same sum and produce the same trajectory. Since the sum is positive, the observed trajectory does not select between these pairs or identify either rate separately. It does identify the sum under the stated assumptions.

Reaction Kinetics Can Identify a Parameter Product

A similar issue appears when a reaction law contains two unknown quantities only through their product. Consider a simplified hypothetical reaction in which substrate \(S\) is consumed at a rate proportional to both its current concentration and a constant enzyme concentration \(E_0\). Assume \(E_0\) remains constant, the substrate concentration is observed directly, \(S(0)=S_0>0\) is known, and both \(k_{\mathrm{cat}}\) and \(E_0\) are nonnegative:

\frac{dS}{dt}=-k_{\mathrm{cat}}E_0S.

Here \(k_{\mathrm{cat}}\) has units of \(\mathrm{\mu M}^{-1}\mathrm{h}^{-1}\) and \(E_0\) has units of \(\mathrm{\mu M}\), so their product has units of \(\mathrm{h}^{-1}\). The solution is

S(t)=S_0e^{-k_{\mathrm{cat}}E_0t}.

Under these assumptions the trajectory determines the product \(q=k_{\mathrm{cat}}E_0\), not necessarily either factor. If \(q>0\), for each positive choice of \(E_0\), the value \(k_{\mathrm{cat}}=q/E_0\) gives the same output. If both quantities may be zero, \(q=0\) means at least one factor is zero; it does not, by itself, show which one. Further model restrictions could alter these conclusions, so they are statements about this reaction model and parameter domain.

Worked Example: Estimating an Identifiable Reaction Product

In a hypothetical experiment, the model assumptions above hold, the initial substrate concentration is \(50.0\ \mathrm{\mu M}\), and the concentration after \(5.00\) h is \(15.06\ \mathrm{\mu M}\). Suppose \(k_{\mathrm{cat}}\) and \(E_0\) are both unknown and positive. Find the product determined by the trajectory, and show why the individual values are not determined.

From the model solution, \(S(5)/S_0=e^{-5q}\), where \(q=k_{\mathrm{cat}}E_0\). Therefore,

q=-\frac{1}{5.00\ \mathrm{h}} \log\!\left(\frac{15.06\ \mathrm{\mu M}}{50.0\ \mathrm{\mu M}}\right) \approx 0.2400\ \mathrm{h}^{-1}.

For example, \(E_0=2.00\ \mathrm{\mu M}\) and \(k_{\mathrm{cat}}=0.1200\ \mathrm{\mu M}^{-1}\mathrm{h}^{-1}\) have product \(0.2400\ \mathrm{h}^{-1}\). So do \(E_0=3.00\ \mathrm{\mu M}\) and \(k_{\mathrm{cat}}=0.08000\ \mathrm{\mu M}^{-1}\mathrm{h}^{-1}\). Both pairs predict \(S(5)=50.0e^{-0.2400(5.00)}\ \mathrm{\mu M}\approx15.06\ \mathrm{\mu M}\). The product is identified; the two positive factors are not identified separately from this output.

Worked Example: Checking the Zero-Product Boundary

Consider the same reaction model with \(S_0=40.0\ \mathrm{\mu M}\), and suppose an ideal trajectory is constant at \(40.0\ \mathrm{\mu M}\) throughout the observation interval. What can be concluded if \(k_{\mathrm{cat}}\geq0\) and \(E_0\geq0\)?

A constant positive trajectory implies \(q=0\), because \(S(t)=S_0e^{-qt}\) can equal \(S_0\) for all \(t\) only when \(e^{-qt}=1\), hence \(q=0\). Since \(q=k_{\mathrm{cat}}E_0\), this says \(k_{\mathrm{cat}}E_0=0\). It does not identify both factors: \((k_{\mathrm{cat}},E_0)=(0,2.00)\) and \((0.1200,0)\), using compatible concentration and rate units, both give zero product and the same constant trajectory. The product is identified as zero, but the factors remain ambiguous. This differs from the decay-rate sum, where a zero sum of two nonnegative rates forces both rates to be zero.

Justification, Diagnostics, and Alternatives

The general check in these examples is direct: write the output as a function of the parameters, then ask what equality of two complete output trajectories forces. If the output depends on \((k_1,k_2)\) only through \(k_1+k_2\), no feature of that output can distinguish pairs with the same sum. If it depends on \((k_{\mathrm{cat}},E_0)\) only through their product, pairs with the same product likewise remain indistinguishable. This is a consequence of the equations, not a claim that a particular fitting algorithm has failed.

Several assumptions make these conclusions precise. The model form and conditions such as constant \(E_0\) are fixed; the initial state is positive; the measured output corresponds directly to the modeled concentration or amount; and the ideal trajectory is available over time. If the output is zero throughout, its logarithm cannot be used, and a positive known initial state is essential to the arguments given here. An unknown observation offset or an unmodeled process can also invalidate the simple exponential representation. The identifiability conclusion applies to the specified model, not automatically to a more complicated system.

In practice, a diagnostic is to inspect whether the equations and fitted predictions depend on parameters separately or only through a repeated sum or product. If different parameter pairs give exactly the same predicted trajectory, a narrow numerical estimate for each parameter does not establish that the output identifies them: a fitting procedure may have imposed bounds, fixed a parameter, or selected one point among equivalent possibilities. Conversely, noisy and sparse observations can make even an identifiable combination imprecise. Structural identifiability and practical precision are different questions.

One alternative is to report the identifiable combination, with its units, rather than present unsupported component values. Another is to impose a justified fixed value or constraint based on external knowledge; that can yield estimates conditional on the constraint, but it does not mean the original output independently identified both parameters. Independent measurements or a changed experimental design may provide information that separates components, but whether a particular design does so must be established for that model.

These combinations are useful in QSP because they show what a calibration can support at the level of the model output. A rate sum can predict an overall decay even when the individual removal routes are unresolved; a reaction-rate product can describe effective depletion even when enzyme abundance and catalytic activity are not separated. The interpretation must remain at that level: the combination is identified under the stated assumptions, while its biological components may require additional evidence.

Common Pitfalls and Limitations

Claiming that no individual parameters are identified in every case. For the nonnegative decay rates, a positive sum leaves a continuum of pairs, but a zero sum forces both rates to zero. Check the parameter domain and boundary values.
Confusing a precise estimate with structural identifiability. A numerical optimizer can return separate values even when an entire family predicts the same output. A reported point estimate does not remove that ambiguity.
Interpreting a product as evidence for each factor. Knowing \(k_{\mathrm{cat}}E_0\) does not determine \(k_{\mathrm{cat}}\) and \(E_0\) separately unless further information or restrictions are supplied.
Ignoring units. A sum of first-order rates has units of inverse time; a product of parameters has the units implied by its factors. In the reaction example, \(k_{\mathrm{cat}}E_0\) is an inverse-time rate.

Key takeaway: State exactly which function of the parameters the ideal output determines, and verify the conclusion over the stated parameter domain. An identifiable sum or product is a valid model result; it is not automatically evidence that its component parameters have been separately identified.

Check Your Understanding

Answer each question using the model assumptions stated in the tutorial. Brief solutions follow the questions.

For \(x(t)=x_0e^{-(k_1+k_2)t}\) with \(k_1,k_2\geq0\), a trajectory identifies \(k_1+k_2=0.60\ \mathrm{h}^{-1}\). Give two distinct rate pairs consistent with it, and state what is not identified.
In the same model, the identified sum is \(0\). What do nonnegative rates imply?
For \(S(t)=S_0e^{-k_{\mathrm{cat}}E_0t}\), what combination is identified if \(S_0>0\) and the ideal trajectory is observed? Give its units when \(k_{\mathrm{cat}}\) is in \(\mathrm{\mu M}^{-1}\mathrm{h}^{-1}\) and \(E_0\) is in \(\mathrm{\mu M}\).
If the reaction product \(k_{\mathrm{cat}}E_0=0\) and both factors are nonnegative, are both factors necessarily identified? Explain.
Why does a precise numerical fit for both factors not, by itself, prove that they are structurally identifiable?

Solutions. (1) For example, \((0.20,0.40)\) and \((0.50,0.10)\ \mathrm{h}^{-1}\) both sum to \(0.60\ \mathrm{h}^{-1}\); the trajectory does not determine either rate separately. (2) Both rates must equal zero, since two nonnegative numbers can sum to zero only if each is zero. (3) The identified combination is \(k_{\mathrm{cat}}E_0\), with units \(\mathrm{h}^{-1}\). (4) No. A zero product requires at least one factor to be zero, but does not show which factor is zero. (5) Equivalent parameter pairs may generate exactly the same output; a fitting procedure can select a particular pair without the output distinguishing it from the others.