How Multiple Outputs Resolve Parameter Ambiguity

Why the Choice of Measured State Matters

A model can contain distinct rate constants while a particular measurement reports only their combined effect. In that case, collecting an ideal, continuous trajectory of that output does not distinguish every parameter value: different rate pairs can generate exactly the same curve. Measuring another state can supply a different equation for the same rates and remove that ambiguity.

The result depends on which states are measured, not simply on how many. A single output may be uninformative about a parameter split, while a different single output can sometimes identify both rates through its time course. The example below makes these distinctions explicit.

Definition: Under stated model assumptions and known initial conditions, parameters are structurally identifiable from an output if ideal, continuous, noise-free knowledge of that output uniquely determines them, at least for generic parameter values. If distinct parameter values give exactly the same output trajectory, they are not uniquely identifiable from that output. In this tutorial, identifiability always refers to the specified measured states and assumptions.

A Coupled Competing-Reaction Model

Consider a closed network in which state \(A\) can transition to either \(B\) or \(C\). Both transitions remove material from \(A\), and each has its own first-order rate constant. Assume the states are measured in nM, time is measured in hours, and \(k_1\) and \(k_2\) have units of h\(^{-1}\).

\begin{aligned} \frac{dA}{dt} &= -(k_1+k_2)A,\\ \frac{dB}{dt} &= k_1A,\\ \frac{dC}{dt} &= k_2A, \end{aligned} \qquad A(0)=A_0>0,\quad B(0)=C(0)=0.

Assume \(A_0\) is known, both rate constants are constant and nonnegative, the model is correct, and there are no unmodelled inputs, losses, or measurement-scale factors. Define \(q=k_1+k_2\). Solving the first equation gives \(A(t)=A_0e^{-qt}\). Integrating the other two equations gives, for \(q>0\),

B(t)=A_0\frac{k_1}{q}\left(1-e^{-qt}\right), \qquad C(t)=A_0\frac{k_2}{q}\left(1-e^{-qt}\right).

These equations show two kinds of information in the outputs. The decay of \(A\) depends on the sum \(q\), whereas the amounts in \(B\) and \(C\) also depend on how that sum is divided between \(k_1\) and \(k_2\). This observation is specific to the stated network and assumptions.

One Output: The Remaining State Reveals Only the Sum

Suppose the only measured state is \(A\). Its ideal trajectory is \(A(t)=A_0e^{-qt}\). If \(A_0\) is known and \(A(t)>0\), the trajectory determines \(q\), for example by \(q=-\log(A(t)/A_0)/t\) at any \(t>0\). But it does not determine how \(q\) is divided between the two rates.

More formally, for any \(r\) satisfying \(0\leq r\leq q\), the parameter pair \(k_1=r,\ k_2=q-r\) gives the same \(A(t)\). Thus, when \(q>0\), there is a continuum of nonnegative parameter pairs with identical \(A\) output. This is an exact structural ambiguity, not a consequence of measurement noise or too few sampling times.

Worked Example: Identical A Trajectories From Different Rates

Suppose a hypothetical system has \(A_0=100\) nM and the measured \(A\) trajectory includes \(A(1)=74.0818\) nM and \(A(2)=54.8812\) nM. Under the model, the one-state output gives

q=-\frac{\log(A(1)/A_0)}{1\ \mathrm{h}} =-\frac{\log(74.0818/100)}{1\ \mathrm{h}} \approx 0.3000\ \mathrm{h}^{-1}.

The corresponding prediction at two hours is \(100e^{-0.3000(2)}=54.8812\) nM, consistent with the stated trajectory. The pair \(k_1=0.1000\) h\(^{-1}\), \(k_2=0.2000\) h\(^{-1}\) gives \(q=0.3000\) h\(^{-1}\). So does \(k_1=0.1800\) h\(^{-1}\), \(k_2=0.1200\) h\(^{-1}\). Both pairs predict the same \(A(t)\) at every time, not just at the two listed times. Measuring \(A\) alone therefore does not distinguish these rate pairs.

Adding an Output: Separating the Rates

Now suppose both \(A\) and \(B\) are measured. The \(A\) trajectory determines \(q\), while the \(B\) trajectory supplies information about \(k_1\). For any \(t>0\) and \(q>0\), rearrange the expression for \(B(t)\):

k_1 =\frac{qB(t)}{A_0(1-e^{-qt})}, \qquad k_2=q-k_1.

The denominator is positive for \(t>0\) and \(q>0\). Consequently, under the stated assumptions, the two output equations determine the two rates. The same separation is visible at the initial time: \(A'(0)=-qA_0\) and \(B'(0)=k_1A_0\). Since \(A_0>0\), the initial slopes give \(q=-A'(0)/A_0\) and \(k_1=B'(0)/A_0\), followed by \(k_2=q-k_1\).

Worked Example: Recovering Both Rates From A and B

Consider hypothetical measurements at \(t=1\) h with \(A_0=100\) nM, \(A(1)=74.0818\) nM, and \(B(1)=8.63939\) nM. First recover the shared total rate:

q=-\frac{\log(74.0818/100)}{1\ \mathrm{h}} \approx 0.3000\ \mathrm{h}^{-1}.

Then use the \(B\) output to calculate \(k_1\):

\begin{aligned} k_1 &=\frac{(0.3000\ \mathrm{h}^{-1})(8.63939\ \mathrm{nM})} {(100\ \mathrm{nM})(1-e^{-0.3000})}\\ &=\frac{2.59182\ \mathrm{nM\,h}^{-1}}{25.9182\ \mathrm{nM}} \approx 0.1000\ \mathrm{h}^{-1}. \end{aligned}

Therefore \(k_2=q-k_1=0.3000-0.1000=0.2000\) h\(^{-1}\). Substitution checks the result: \(B(1)=100(0.1000/0.3000)(1-e^{-0.3000})\approx8.63939\) nM. The second output has resolved the split that was ambiguous from \(A\) alone.

Justification and an Important Output-Dependent Qualification

The separation can also be stated as an invertible mapping. The two quantities obtained from the outputs are \(q=k_1+k_2\) and \(k_1\). The Jacobian of this mapping with respect to the rates is

\frac{\partial(q,k_1)}{\partial(k_1,k_2)} = \begin{pmatrix} 1&1\\ 1&0 \end{pmatrix}, \qquad \det\!\left( \frac{\partial(q,k_1)}{\partial(k_1,k_2)} \right)=-1\ne 0.

Because its determinant is nonzero, this mapping has a unique inverse: \(k_1\) is the second quantity and \(k_2=q-k_1\). This establishes the separation for the model and measurements considered here; it is not a general theorem that any two outputs identify any two parameters.

In fact, a single output \(B\) can also identify both rates under the ideal assumptions. When \(k_1>0\), differentiating \(B'(t)=k_1A(t)\) once gives \(B''(t)=-qk_1A(t)\). Hence \(B'(0)=k_1A_0\) and \(B''(0)=-qk_1A_0\), so that \(k_1=B'(0)/A_0\) and \(q=-B''(0)/B'(0)\). Then \(k_2=q-k_1\). The distinction is therefore not simply “one output versus two”: it is whether the chosen output equations contain enough independent information about the parameters.

Worked Example: Checking Identifiability From B Alone

Suppose a hypothetical, ideal \(B\) trajectory has \(A_0=100\) nM, initial slope \(B'(0)=10.0\) nM/h, and second derivative \(B''(0)=-3.00\) nM/h\(^{2}\). Since \(k_1>0\), the derivative formulas apply:

k_1=\frac{10.0\ \mathrm{nM\,h}^{-1}}{100\ \mathrm{nM}} =0.1000\ \mathrm{h}^{-1}, \qquad q=-\frac{-3.00\ \mathrm{nM\,h}^{-2}}{10.0\ \mathrm{nM\,h}^{-1}} =0.3000\ \mathrm{h}^{-1}.

Thus \(k_2=q-k_1=0.2000\) h\(^{-1}\). This example does not contradict the ambiguity from \(A\) alone: \(B\) has a different output equation and, through its curvature, carries information that \(A\) by itself lacks.

Assumptions, Diagnostics, and Limitations

The identifiability conclusions above rely on a known positive initial amount \(A_0\), known zero initial amounts for \(B\) and \(C\), constant nonnegative rates, the specified reaction structure, and ideal observations of the stated output or outputs. If these assumptions change, the output equations and the identifiability conclusion must be reconsidered. For example, an unknown output scale or unknown initial condition can change which combinations of parameters the observations determine.

The exact ambiguity in \(A\) alone can be diagnosed by checking whether the fitted output equation depends on \(k_1\) and \(k_2\) only through \(q=k_1+k_2\). If so, any pair with the same sum has the same prediction. For multiple outputs, a useful check is to derive their equations and ask whether the inferred quantities can be inverted to recover the individual rates. A full-rank local sensitivity calculation can support that check, but local rank alone does not establish global uniqueness in every model; the parameter mapping still needs to be examined.

Structural identifiability is an ideal-data property. It does not guarantee that rates can be estimated precisely from sparse, noisy observations. In particular, estimating initial slopes or curvature from real measurements may be unstable, and the denominator in the expression for \(k_1\) can be small at very early times. Those practical difficulties do not restore the exact ambiguity in \(A\) alone, nor do they invalidate the ideal identifiability result for \(A\) and \(B\).

A common mistake is to conclude that observing \(A\) precisely, or at many time points, must separate \(k_1\) and \(k_2\). More observations of the same ideal \(A\) trajectory can estimate \(q\) more reliably in practice, but every such prediction still depends only on \(k_1+k_2\). Another mistake is to claim that adding any second output automatically solves the problem. The equations must actually supply independent information, and exceptional parameter values can remove that information. For instance, if \(k_1=0\), then \(B(t)=0\) and the \(B\)-only derivative argument cannot recover \(q\).

Measured state(s)	Information under the stated assumptions	Conclusion
\(A\) only	The trajectory determines \(q=k_1+k_2\).	The individual rates are not separated when \(q>0\).
\(B\) only	For \(k_1>0\), its initial slope and curvature determine \(k_1\) and \(q\).	Both rates are identifiable from an ideal continuous trajectory.
\(A\) and \(B\)	\(A\) determines \(q\), and \(B\) determines \(k_1\).	Both rates are separated for \(q>0\).

Key takeaway: Multiple outputs resolve parameter ambiguity when their output equations contribute complementary information. In this network, \(A\) alone reveals only the total loss rate, while \(A\) together with \(B\) directly separates the two transition rates. The conclusion depends on the selected outputs and stated assumptions.

Check Your Understanding

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

If \(A_0=80\) nM and \(A(2)=80e^{-0.4}\) nM, what rate combination does the \(A\)-only output determine?
For \(q=0.4\) h\(^{-1}\), give two distinct nonnegative rate pairs that produce the same \(A\) trajectory.
Why does observing \(A\) and \(B\) identify \(k_2\) once \(q\) and \(k_1\) are known?
Under what condition does the \(B\)-only derivative argument in the tutorial fail to recover \(q\)?

Solutions. (1) \(q=-\log(A(2)/80)/2=0.2000\) h\(^{-1}\); only the sum is determined. (2) For example, \((k_1,k_2)=(0.1000,0.3000)\) h\(^{-1}\) and \((0.2500,0.1500)\) h\(^{-1}\) both sum to \(0.4000\) h\(^{-1}\). (3) The \(B\) output determines \(k_1\), and the model gives \(k_2=q-k_1\). (4) If \(k_1=0\), then \(B'(0)=0\), so the ratio \(-B''(0)/B'(0)\) is undefined and \(B\) is constant at its known initial value.