Two Different Recovery Questions
A dynamic pharmacology model may contain hidden states, such as concentrations in an unmeasured compartment or the amount of a biomarker in a tissue. It may also contain fixed parameters, such as clearance rates or turnover constants. These pose different questions: can the state trajectory be recovered from the outputs, given the model? And can the fixed parameter values be recovered from ideal outputs? Answering the first question does not automatically answer the second.
State observability is a property of a state-space system with specified system matrices. Structural parameter identifiability is a property of the model’s mapping from parameters to ideal input-output behavior. The distinction matters in QSP: a model can have enough information in its outputs to reconstruct all its states while different parameter values still produce exactly the same model behavior.
In contrast, structural parameter identifiability asks whether distinct parameter values can give identical ideal outputs under the specified experimental conditions. To isolate this question, the examples below assume the model structure, inputs, and initial states are known unless stated otherwise. Structural identifiability concerns noiseless, ideal observations; it is not a claim that parameter estimates will be precise with finite, noisy measurements.
The Observability Matrix
For the autonomous linear system \(\dot{x}=Ax,\ y=Cx\), the output and its successive derivatives at an observation time depend on the initial state through \(C, CA, CA^2,\ldots\). The observability matrix for an \(n\)-state system is
For a known, constant \(A\) and \(C\), the system is observable if and only if \(\mathcal{O}\) has rank \(n\). This is a state-recovery test: it treats the matrices as known. It does not test whether the entries of \(A\), or any parameters used to construct \(A\), can be uniquely recovered.
Why the Rank Condition Works
The solution of \(\dot{x}=Ax\) is \(x(t)=e^{At}x(0)\), so \(y(t)=Ce^{At}x(0)\). Differentiating at \(t=0\) gives \(y^{(j)}(0)=CA^j x(0)\). Thus, when \(\mathcal{O}\) has rank \(n\), its rows contain enough independent information to solve for the \(n\) entries of \(x(0)\).
The rank condition is also necessary. If \(\mathcal{O}\) has rank less than \(n\), there is a nonzero vector \(v\) such that \(CA^jv=0\) for \(j=0,\ldots,n-1\). The Cayley–Hamilton theorem expresses every higher power of \(A\) as a combination of lower powers, so \(CA^jv=0\) for all \(j\geq 0\). Consequently, the output trajectories starting from \(x(0)\) and \(x(0)+v\) are identical: their difference is \(Ce^{At}v=0\). The initial state is not uniquely determined.
This argument concerns recovery when \(A\) and \(C\) are known. If they depend on unknown parameters, the rank test is conditional on particular parameter values; it is not a parameter-identifiability test.
Worked Example: Recovering Two Hidden States
Consider a hypothetical two-state system with known matrix \(A=\begin{bmatrix}-3&0\\0&-1\end{bmatrix}\), output matrix \(C=\begin{bmatrix}1&1\end{bmatrix}\), and initial state \(x(0)=\begin{bmatrix}2\\1\end{bmatrix}\). The measured output is the sum of the two states. Check observability and recover the initial state from the ideal output and its initial derivative.
The observability matrix is
Since the determinant is nonzero, the matrix has rank 2, so both initial states are observable. At time zero, \(y(0)=x_1(0)+x_2(0)=3\), and \(\dot{y}(0)=-3x_1(0)-x_2(0)=-7\). Solving \(x_1(0)+x_2(0)=3\) and \(-3x_1(0)-x_2(0)=-7\) gives \(x_1(0)=2\) and \(x_2(0)=1\).
This recovery assumes the system matrix is known and that the ideal output derivative is available. It establishes observability for this specified system; it says nothing yet about whether unknown parameters in \(A\) could be determined.
An Observable System With Non-Identifiable Parameters
Now let the state matrix depend on three positive rate parameters:
Assume the initial state is known and both states begin above zero. The first state decays at rate \(k_1+k_2\), and the second at rate \(k_3\). The distinction between \(k_1\) and \(k_2\) does not appear in the system matrix: they enter only through their sum. The following example checks both state observability and parameter identifiability for this model.
Worked Example: Observable States, Ambiguous Rate Parameters
For a hypothetical system, let \(x(0)=\begin{bmatrix}2\\1\end{bmatrix}\), \(k_1=1\) per hour, \(k_2=2\) per hour, and \(k_3=1\) per hour. Compare this parameter set with \(k_1=0.5\) per hour, \(k_2=2.5\) per hour, and \(k_3=1\) per hour.
For the first parameter set, \(k_1+k_2=3\) per hour. Its observability matrix is
The two states are observable. The second parameter set has the same sum, \(0.5+2.5=3\) per hour, and the same \(k_3\). It therefore gives exactly the same matrix \(A\), not merely the same output at a few selected times. With the stated initial state, both sets produce
where \(t\) is in hours. The rates \(k_1\) and \(k_2\) are different between the two sets, but every ideal output value is identical for all \(t\geq 0\). Thus the states are observable for these specified rates, while \(k_1\) and \(k_2\) are not separately structurally identifiable from these outputs and assumptions. The conclusion follows from the model equations, not from limited sampling or measurement noise.
The example shows why “all states can be recovered” is not equivalent to “all parameters can be recovered.” Even exact knowledge of the trajectory does not distinguish two parameter sets when the model uses them in a way that produces identical dynamics.
Worked Example: Measuring Both States Does Not Resolve the Split
Suppose the same hypothetical system is measured with two separate outputs, \(y_1=x_1\) and \(y_2=x_2\), and the initial state remains known as \((2,1)\). For the trajectories in the preceding example, the measured states are \(y_1(t)=2e^{-3t}\) and \(y_2(t)=e^{-t}\).
Their ideal logarithmic slopes are \(\frac{d}{dt}\log(y_1(t))=-3\) per hour and \(\frac{d}{dt}\log(y_2(t))=-1\) per hour. These measurements reveal \(k_1+k_2=3\) per hour and \(k_3=1\) per hour. They still do not determine whether \(k_1=1\) and \(k_2=2\), or \(k_1=0.5\) and \(k_2=2.5\), or another positive pair summing to 3 per hour. Directly measuring both states improves state information, but cannot distinguish parameters that enter the state equations only through the same sum.
Assumptions, Diagnostics, and Alternatives
An observability conclusion depends on its assumptions. The rank criterion above applies to a finite-dimensional linear system with constant, specified matrices and ideal continuous-time output. With a known input, the output contribution caused by that input can be accounted for; the criterion for recovering the initial state still concerns the system’s state and output matrices. Unknown initial conditions, unknown output scales, an uncertain model structure, or unmeasured inputs can change the recovery question and must be included in the analysis rather than silently treated as known.
For parameter identifiability, specify the model, which initial conditions and inputs are known, which outputs are measured, and the time interval or experimental conditions. Then ask whether two distinct parameter values can generate the same ideal outputs under those same conditions. In the example, the direct check is especially transparent: the matrix depends on \(k_1\) and \(k_2\) through \(k_1+k_2\), so any positive pair with the same sum yields the same state equations.
| Question | Object examined | Typical mathematical check |
|---|---|---|
| Are hidden states recoverable for a specified linear model? | Initial state \(x(0)\), with \(A\) and \(C\) treated as known | Rank of the observability matrix |
| Are fixed parameters recoverable under specified conditions? | Parameter values in the model-to-output mapping | Whether distinct values can produce identical ideal outputs |
A numerical observability matrix with full rank can support a state-recovery conclusion for the specified matrices, but it cannot establish that their unknown generating parameters are identifiable. Conversely, fitting a model successfully to noisy data does not prove structural identifiability: a dataset can favor one estimate because of noise, constraints, or an optimizer’s behavior even when exact output-equivalent parameter values exist.
In practice, check the model’s dependence on each parameter, calculate the observability matrix when the system is linear and its matrices are specified, and compare full output trajectories for proposed alternative parameter values. A rank that is nearly deficient can also warn that state reconstruction may be sensitive to measurement error, even when the exact rank is full. That is a practical estimation concern, distinct from structural identifiability.
Common Pitfall
Wrong reasoning: “The observability matrix has full rank, so all parameters can be estimated uniquely.” The rank test only shows that the states can be distinguished for the specified \(A\) and \(C\). It does not test whether different parameter values produce the same \(A\) or the same outputs.
Correction: State the result at the level actually tested. Full rank supports recovery of the hidden state under the specified linear model. Parameter identifiability requires a separate comparison of the parameter-to-output mapping. In the rate example, full state observability coexists with non-identifiability of \(k_1\) and \(k_2\).
Check Your Understanding
Answer each question using the stated assumptions. Brief solutions follow the questions.
- For a two-state system, what does full rank of the observability matrix establish, and what does it not establish?
- For the rate model, calculate the determinant of the observability matrix when \(k_1+k_2=4\) per hour and \(k_3=1\) per hour. Are the states observable?
- In the rate model, compare \((k_1,k_2,k_3)=(1,2,1)\) per hour with \((2,1,1)\) per hour. Are these distinguishable from ideal outputs if the initial state is the same and known?
- Why does measuring \(x_1\) and \(x_2\) separately still fail to determine \(k_1\) and \(k_2\) individually in this model?
Solutions. (1) Full rank establishes unique recovery of the initial state for the specified matrices and ideal outputs; it does not establish unique recovery of unknown parameters. (2) Here \(\mathcal{O}=\begin{bmatrix}1&1\\-4&-1\end{bmatrix}\), so its determinant is \(-1-(-4)=3\), which is nonzero; the states are observable. (3) Both parameter sets have \(k_1+k_2=3\) per hour and the same \(k_3\), so they produce the same state equations and ideal outputs; they are not distinguishable under these conditions. (4) Separate state measurements reveal the decay rate of \(x_1\), which is \(k_1+k_2\), but the equations contain no other separate dependence on those two parameters.