Parameter Reparameterization and Identifiability

Why Reparameterize?

A model can be written with parameters that have a direct biological interpretation yet still fail to let its observations distinguish those parameters individually. In that situation, fitting the original parameters may return a numerical answer for each one, but the data can determine only particular combinations. Reparameterizing expresses the model in terms of those combinations and makes the distinction visible.

Here the focus is a two-rate turnover system with a production rate and a first-order loss rate. We will include an unknown measurement scale so that the model has both identifiable quantities and a redundant parameter pair. This is a structural question: assume an ideal, noise-free, continuously observed output and ask whether distinct parameter values can produce exactly the same output trajectory. Real data add sampling and noise limitations, which are separate from structural identifiability.

A Two-Rate Turnover Model

Let \(x(t)\) be an amount in the system. A known, dimensionless input \(u(t)\) controls production; material is removed in proportion to the current amount. Suppose the measured output is proportional to the amount, with unknown scale \(q\):

\frac{dx}{dt}=k_{\mathrm{in}}u(t)-k_{\mathrm{out}}x(t), \qquad y(t)=q x(t).

The parameters \(k_{\mathrm{in}}>0\) and \(k_{\mathrm{out}}>0\) are the production and loss rates. Their units are amount per time and inverse time, respectively. The scale \(q>0\) has units of output per amount, so \(y(t)\) is in output units. For the main result, assume \(u(t)=1\), that the initial amount is known to be \(x(0)=0\), and that \(q\), \(k_{\mathrm{in}}\), and \(k_{\mathrm{out}}\) are constant. Assume also that the output is measured in absolute units, without an additional unknown offset.

Definition: A parameter, or a function of parameters, is structurally identifiable under specified model assumptions if ideal observations uniquely determine its value. A parameter combination is identifiable when every parameter set producing the same ideal output has the same value of that combination. A redundant parameterization contains distinct parameter values that produce the same output trajectory.

The unknown measurement scale and production rate enter the output equation as a product. Define the transformed quantities

\alpha=qk_{\mathrm{in}}, \qquad \beta=k_{\mathrm{out}}.

Here \(\alpha\) has units of output per time and \(\beta\) has units of inverse time. Since \(y=qx\) and \(q\) is constant, differentiating the output gives

\frac{dy}{dt} =q\frac{dx}{dt} =qk_{\mathrm{in}}-k_{\mathrm{out}}qx =\alpha-\beta y.

The output model is therefore expressed directly in terms of the transformed quantities. With \(y(0)=0\) and a constant unit input, its solution is

y(t)=\frac{\alpha}{\beta}\left(1-e^{-\beta t}\right).

This form shows the roles of the combinations: \(\alpha/\beta\) is the steady-state output, while \(\beta\) controls the rate of approach to that steady state. It does not, by itself, establish that both are uniquely recoverable; that requires checking whether two different transformed parameter pairs could generate the same entire trajectory.

Why the Transformed Quantities Are Identifiable

Under the assumptions above, \(y(0)=0\), and \(\alpha>0\), the ideal trajectory initially rises. Its first and second derivatives at the initial time are

y'(0)=\alpha, \qquad y''(0)=-\beta y'(0)=-\alpha\beta.

Thus the trajectory determines \(\alpha\) from its initial slope and determines \(\beta\) from the ratio of its initial curvature to its initial slope:

\alpha=y'(0), \qquad \beta=-\frac{y''(0)}{y'(0)}.

This is a short uniqueness argument: if two parameter pairs produce the same ideal trajectory, they have the same derivatives at zero, and hence the same \(\alpha\) and \(\beta\). Both transformed quantities are globally identifiable for this model and these assumptions. This derivative argument describes what is determined by an ideal continuous trajectory; estimating derivatives from noisy, sparse observations can be difficult and is not required as a practical fitting method.

The original parameters are not individually recoverable when \(q\) is unknown. For any constant \(c>0\), define \(q^\ast=cq\) and \(k_{\mathrm{in}}^\ast=k_{\mathrm{in}}/c\). Then \(q^\ast k_{\mathrm{in}}^\ast=qk_{\mathrm{in}}=\alpha\), while \(k_{\mathrm{out}}\) is unchanged. The output equation is therefore unchanged. The data determine \(qk_{\mathrm{in}}\) and \(k_{\mathrm{out}}\), but not \(q\) and \(k_{\mathrm{in}}\) separately. If \(q\) is known independently, \(k_{\mathrm{in}}=\alpha/q\) becomes recoverable.

Key result: For a zero initial amount, known constant unit input, positive rates and scale, and an absolute output with no unknown offset, the ideal output trajectory uniquely determines \(\alpha=qk_{\mathrm{in}}\) and \(\beta=k_{\mathrm{out}}\). If \(q\) is unknown, \(q\) and \(k_{\mathrm{in}}\) remain redundant: only their product is determined.

Worked Example: Recovering the Transformed Rates

Suppose a hypothetical ideal trajectory has a steady-state output of \(0.800\) output units. At \(t=2\) time units its output is \(0.506\), and at \(t=4\) time units its output is \(0.692\). These rounded values are consistent with the exact trajectory \(y(t)=0.8(1-e^{-0.5t})\). Use the trajectory to recover the transformed rates.

The remaining distance to the steady state is \(y_{\infty}-y(t)=(\alpha/\beta)e^{-\beta t}\). Using the two observations, the ratio of those distances is approximately

\frac{0.800-0.506}{0.800-0.692} \approx \frac{0.294}{0.108} \approx 2.72.

The time difference is \(4-2=2\) time units, so the exponential rate is

\beta \approx \frac{\ln(2.72)}{2} \approx 0.500\ \text{time}^{-1}.

The steady-state output is \(\alpha/\beta=0.800\) output units. Therefore,

\alpha =\beta(0.800) \approx 0.500\times0.800 =0.400\ \text{output units per time}.

The transformed production-scale combination is approximately \(0.400\) output units per time, and the loss rate is approximately \(0.500\) per time. The steady-state value used here is the limit of the hypothetical ideal trajectory; it should not be treated as an exactly observed value merely because a finite experiment has several late measurements.

Worked Example: Two Different Parameter Sets, One Output

Consider two hypothetical parameter sets, using output units per amount for \(q\), amount per time for \(k_{\mathrm{in}}\), and inverse time for \(k_{\mathrm{out}}\):

(q,k_{\mathrm{in}},k_{\mathrm{out}})=(2.00,0.200,0.500) \quad\text{and}\quad (q,k_{\mathrm{in}},k_{\mathrm{out}})=(5.00,0.0800,0.500).

For the first set, \(qk_{\mathrm{in}}=2.00\times0.200=0.400\) output units per time. For the second, \(qk_{\mathrm{in}}=5.00\times0.0800=0.400\) output units per time. Both also have \(\beta=0.500\) per time. Consequently, both predict

y(t)=\frac{0.400}{0.500}\left(1-e^{-0.500t}\right) =0.800\left(1-e^{-0.500t}\right).

At \(t=2\) time units, each predicts \(0.800(1-e^{-1})\approx0.506\) output units. No ideal observation of this output trajectory can distinguish the two sets. The transformed quantities agree even though the measurement scales and production rates differ.

Conditions That Matter

The uniqueness result depends on the assumptions, not just on the equation’s appearance. In particular, a transient must be present: if the system begins at steady state under constant input, the output is flat. A constant trajectory does not reveal how quickly the system would return after a perturbation.

Worked Example: A Steady-State Trajectory Does Not Identify the Rate

Suppose the observed output is constantly \(1.20\) output units with \(q=3.00\) output units per amount. One possible system has \(\beta=0.500\) per time and \(\alpha=\beta y=0.600\) output units per time. It corresponds to \(k_{\mathrm{in}}=\alpha/q=0.200\) amount per time. Its steady-state amount is \(x=k_{\mathrm{in}}/\beta=0.400\) amount, giving \(y=qx=1.20\) output units.

A second system can have \(\beta=1.50\) per time, \(\alpha=1.50\times1.20=1.80\) output units per time, and \(k_{\mathrm{in}}=1.80/3.00=0.600\) amount per time. Its steady-state amount is again \(0.600/1.50=0.400\) amount, so it produces the same constant output. The observed trajectory cannot distinguish the loss rates \(0.500\) and \(1.50\) per time. The transformed quantities are not both identifiable from this particular trajectory because the required transient is absent.

For this turnover model, check the following conditions before interpreting a fit in terms of identifiable quantities:

A known input: the derivation assumes the input \(u(t)\) is known. An unknown input can be confounded with production.
A known initial condition: the main uniqueness argument assumes \(x(0)=0\). If the initial amount is not known, the initial output may add another unknown quantity, and the result must be reassessed under that model.
A visible transient: an initial condition at steady state, or a trajectory that is effectively flat over the observation window, provides little or no information about the rate of approach.
A calibrated output form: an unknown additive measurement offset or a time-varying measurement scale changes the output equation; the result above does not automatically apply.

A practical diagnostic is to plot the output over time and inspect whether it contains a discernible rise or decay. If a fitted curve is nearly flat, or if the estimated rate changes substantially when a few time points are removed, the data may not constrain the rate well. Such behavior can signal practical estimation difficulty even when the parameters are structurally identifiable under the ideal model. Conversely, a stable numerical estimate does not prove structural identifiability: an optimizer may select one point from a family of equivalent parameter sets.

Choosing a Parameterization

One approach is to fit \(q\), \(k_{\mathrm{in}}\), and \(k_{\mathrm{out}}\) directly. This preserves familiar biological labels, but with unknown \(q\) the objective has equivalent solutions along curves satisfying \(qk_{\mathrm{in}}=\alpha\). Reported estimates of \(q\) and \(k_{\mathrm{in}}\) can then depend on starting values, bounds, or other arbitrary numerical choices.

A second approach is to fit \(\alpha\) and \(\beta\) directly, then report that the output identifies the scale-weighted production rate and the loss rate. This avoids trying to estimate two quantities separately when the output only determines their product. If independent information fixes \(q\), the fitted \(\alpha\) can be converted to \(k_{\mathrm{in}}\). Additional experiments or measurements can also change what is recoverable, but their effect must be assessed under the expanded model and data assumptions.

Reparameterization does not create information or change the model’s predictions. It makes the identifiable quantities explicit and avoids presenting a redundant parameterization as though every parameter had a unique data-supported value. Its limits matter: the conclusion is specific to the stated model, input, initial condition, and observation scale. Structural identifiability describes uniqueness with ideal observations; it does not guarantee precise estimates from a finite, noisy time course.

Common pitfall: “The fitting software returned separate values for \(q\) and \(k_{\mathrm{in}}\), so the data identified both.” This is incorrect when the predictions depend on them only through \(qk_{\mathrm{in}}\). The correction is to check whether different parameter values can yield the same output; here, every positive rescaling \(q^\ast=cq\), \(k_{\mathrm{in}}^\ast=k_{\mathrm{in}}/c\) does so. Report the identifiable product unless an additional assumption or measurement fixes the scale.

Check Your Understanding

Answer each question using the model and assumptions in this tutorial. Brief solutions follow.

If the steady-state output is \(1.50\) output units and \(\beta=0.400\) per time, what is \(\alpha\), including its units?
If \(q=3.00\) output units per amount and \(\alpha=0.600\) output units per time, what is \(k_{\mathrm{in}}\)?
For \(\alpha=0.900\) output units per time and \(\beta=0.300\) per time, what are the steady-state output and the time constant \(1/\beta\)?
Why can a constant output at steady state fail to determine \(\beta\), even if the output is observed without noise?

Solutions. (1) Since \(y_{\infty}=\alpha/\beta\), \(\alpha=\beta y_{\infty}=0.400\times1.50=0.600\) output units per time. (2) Since \(\alpha=qk_{\mathrm{in}}\), \(k_{\mathrm{in}}=\alpha/q=0.600/3.00=0.200\) amount per time. (3) The steady-state output is \(\alpha/\beta=0.900/0.300=3.00\) output units, and the time constant is \(1/0.300=3.33\) time units. (4) At steady state, the derivative is zero and \(\alpha=\beta y\); different values of \(\beta\), paired with corresponding values of \(\alpha\), can produce the same constant output. There is no transient to reveal the rate of approach.