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Structural Identifiability · Tutorial 540 of 799

Proving Nonidentifiability With Symmetry Groups

You will construct and verify a continuous symmetry of a signaling model, then use it to prove that distinct parameter values can produce identical measured outputs.

Advanced 12 min read

What You'll Learn

  • Define output-preserving transformations of states and parameters.
  • Verify a continuous scaling symmetry in a signaling pathway.
  • Use uniqueness of ODE solutions to prove trajectory invariance.
  • Identify which parameters and initial conditions vary along a symmetry.
  • Assess how known initial states or calibrated measurements can break a symmetry.
  • Distinguish a symmetry-based proof from numerical or local diagnostics.

Why an Exact Symmetry Proves Nonidentifiability

A model may fit measured signaling data perfectly without determining every parameter that generated them. In structural identifiability, the question is whether ideal, noise-free observations over the specified experiment uniquely determine parameters. An exact symmetry gives a direct way to answer “no”: if a continuous change to the states and parameters leaves every measured output unchanged, then the data cannot distinguish the points along that transformation.

The key is to verify invariance of the complete model, not merely a similar-looking expression or a single time point. For an ordinary differential equation (ODE), this means showing that transformed states satisfy the transformed equations, with transformed initial conditions, and that the observation equations return the same outputs. Under the usual uniqueness condition for ODE solutions, those checks establish equality of the entire output trajectories.

Definition: Let a model have parameter vector \(\theta\), initial state \(x_0\), known input \(u(t)\), and measured output \(z(t)\). A transformation indexed by \(\lambda\) is output-preserving if it maps \((x,\theta,x_0)\) to \((x_\lambda,\theta_\lambda,x_{0,\lambda})\) while leaving the input and every measured output unchanged for all observation times. If a continuous family of such transformations produces distinct admissible parameter values, the affected parameters are structurally nonidentifiable for that experiment.

A one-parameter symmetry group has an identity transformation, an inverse, and a composition rule: applying the transformation for \(\lambda_1\) and then \(\lambda_2\) gives the transformation for \(\lambda_1\lambda_2\). A positive scaling, with \(\lambda>0\), is a common example. The group terminology is useful, but the proof rests on the model equations and the output invariance, not on the name.

A Signaling Model With a Hidden Intermediate

Consider a simple signaling cascade. The unmeasured state \(x(t)\) represents an intermediate signaling species, and \(y(t)\) is a downstream signal that is measured. The normalized input \(u(t)\) is known. The parameters \(k_{\mathrm{in}}, k_x, k_{xy}, k_y\) are positive rate or gain constants. The initial value \(x_0\) is unknown; \(y_0\) may be known or unknown.

$$ \begin{aligned} \frac{dx}{dt} &= k_{\mathrm{in}}u(t)-k_xx,\\ \frac{dy}{dt} &= k_{xy}x-k_yy,\\ z(t)&=y(t),\\ x(0)&=x_0,\qquad y(0)=y_0. \end{aligned} $$

Assume the input is fixed and known, the parameters and states are in an admissible positive range, and the ODE has a unique solution for each admissible parameter and initial-state choice. For example, its right-hand side is linear in the states and continuous in time when \(u(t)\) is continuous. Only \(y\) is measured: there is no direct measurement of \(x\) and no independently known scale for that hidden state.

For any \(\lambda>0\), define transformed states, parameters, and initial conditions by

$$ x_\lambda(t)=\lambda x(t),\qquad y_\lambda(t)=y(t), $$
$$ k_{\mathrm{in},\lambda}=\lambda k_{\mathrm{in}},\qquad k_{x,\lambda}=k_x,\qquad k_{xy,\lambda}=\frac{k_{xy}}{\lambda},\qquad k_{y,\lambda}=k_y, \qquad x_{0,\lambda}=\lambda x_0,\qquad y_{0,\lambda}=y_0. $$

The input \(u(t)\) is not transformed. The scale \(\lambda\) is dimensionless, so the transformations preserve the units of the states and parameters. When \(\lambda=1\), nothing changes; composing two transformations multiplies their scale factors, and the inverse of the transformation for \(\lambda\) has scale \(1/\lambda\).

Deriving the Invariance

To prove that this is a symmetry, verify both transformed state equations and the observation. Differentiating the transformed intermediate gives

$$ \frac{dx_\lambda}{dt} =\lambda\frac{dx}{dt} =\lambda k_{\mathrm{in}}u(t)-\lambda k_xx =k_{\mathrm{in},\lambda}u(t)-k_{x,\lambda}x_\lambda. $$

The transformed downstream derivative is

$$ \frac{dy_\lambda}{dt} =\frac{dy}{dt} =k_{xy}x-k_yy =\frac{k_{xy}}{\lambda}(\lambda x)-k_{y,\lambda}y_\lambda =k_{xy,\lambda}x_\lambda-k_{y,\lambda}y_\lambda. $$

The initial conditions also match the transformed model, because \(x_\lambda(0)=\lambda x_0=x_{0,\lambda}\) and \(y_\lambda(0)=y_0=y_{0,\lambda}\). Finally, the measured output is unchanged: \(z_\lambda(t)=y_\lambda(t)=y(t)=z(t)\). Thus the transformed state trajectory solves the transformed model with the transformed initial conditions and produces exactly the original output at every time. By uniqueness of solutions, this is the output trajectory of that transformed model.

Theorem: If a continuous transformation maps an admissible solution of an ODE model to a solution of the transformed model, maps the initial conditions accordingly, and leaves every measured output invariant, then every point on the transformation’s admissible orbit produces the same output trajectory. If the orbit contains distinct parameter values, those parameters are structurally nonidentifiable from those outputs under that experiment.

Here, any \(\lambda\ne1\) changes \(k_{\mathrm{in}}\) and \(k_{xy}\), and also changes \(x_0\) when \(x_0\ne0\). Those distinct choices produce the same \(z(t)\). The individual quantities \(k_{\mathrm{in}}\), \(k_{xy}\), and (when unknown and nonzero) \(x_0\) therefore cannot all be uniquely recovered from this output under the stated conditions. The transformation leaves combinations such as \(k_{xy}k_{\mathrm{in}}\) and \(k_{xy}x_0\) unchanged. This observation identifies invariants of this symmetry; it does not, by itself, prove that those combinations or the other parameters are globally identifiable.

Worked Example: Checking the Symmetry With Numerical Parameters

Suppose \(u(t)=1\), and take the hypothetical values \(k_{\mathrm{in}}=2\), \(k_x=1\), \(k_{xy}=3\), \(k_y=0.5\), \(x_0=1\), and \(y_0=0.2\). For this constant input, the intermediate trajectory is \(x(t)=2-e^{-t}\). Apply the symmetry with \(\lambda=4\).

The transformed values are \(k_{\mathrm{in},\lambda}=8\), \(k_{x,\lambda}=1\), \(k_{xy,\lambda}=0.75\), \(k_{y,\lambda}=0.5\), \(x_{0,\lambda}=4\), and \(y_{0,\lambda}=0.2\). The transformed intermediate is \(x_\lambda(t)=4x(t)=8-4e^{-t}\). Directly solving its transformed equation, \(dx_\lambda/dt=8-x_\lambda\), with \(x_\lambda(0)=4\), gives that same trajectory.

The downstream production term is identical at every time: \(k_{xy,\lambda}x_\lambda(t)=0.75(8-4e^{-t})=3(2-e^{-t})=k_{xy}x(t)\). At time zero, the original downstream derivative is \(3(1)-0.5(0.2)=2.90\) signal units per time unit. The transformed derivative is \(0.75(4)-0.5(0.2)=2.90\) signal units per time unit. Since the downstream initial state and the entire forcing term are unchanged, its complete trajectory and measured output are unchanged, not just its initial derivative.

What the Proof Does—and Does Not—Establish

The proof establishes nonuniqueness constructively: one parameter set is accompanied by infinitely many alternatives indexed by \(\lambda>0\), all with exactly the same output. It does not rely on noisy data, a particular fitting algorithm, or a weak signal. More observations of \(y(t)\) at additional times under the same model and input cannot distinguish points on this orbit, because the output functions agree for every time.

The conclusion depends on the experimental specification and what is known. If \(x_0\) is known to be a fixed, nonzero value, the proposed transformation generally violates that constraint by replacing it with \(\lambda x_0\). That removes this particular symmetry from the admissible model, though it does not prove that all parameters are identifiable. If \(x_0=0\) is known, the transformed initial value remains zero, and this aspect of the symmetry is not broken.

An independently calibrated measurement of the hidden state can also break this specific scaling. Suppose the model additionally measures \(w(t)=q x(t)\), where \(q\) is a known, fixed calibration factor. Then the transformed measurement would be \(w_\lambda(t)=q\lambda x(t)\), which is not generally equal to \(q x(t)\). If \(q\) is itself unknown and is transformed to \(q/\lambda\), however, that added measurement does not break the symmetry. What is known, what is estimated, and which outputs are measured must therefore be stated precisely.

Worked Example: Testing Whether a Calibrated Measurement Breaks the Symmetry

Suppose at a particular time \(x=2\) in hypothetical signal units, and an additional assay measures \(w=qx\) with known \(q=5\). The measured value is \(w=5(2)=10\). Under the transformation \(\lambda=3\), the hidden state becomes \(x_\lambda=6\). If the calibration is fixed, the predicted additional measurement is \(w_\lambda=5(6)=30\), not 10. Therefore this added output is not invariant under the earlier scaling, and that scaling is not an admissible symmetry of the augmented measurement model.

If instead \(q\) were unknown and transformed to \(q_\lambda=q/3=5/3\), then \(w_\lambda=(5/3)(6)=10\). The extra output would remain invariant, so it would not break the scaling symmetry in that model. In neither case does this one check prove that the full model is identifiable; it tests whether this particular transformation remains output-preserving.

Diagnostics, Alternatives, and Limitations

A useful diagnostic is to substitute the proposed transformation into every part of the model, rather than checking only the observation equation. Confirm that it preserves the state equations, the initial conditions, known inputs, known parameter constraints, and all measured outputs. A mismatch in any one of these means the transformation does not prove nonidentifiability for the stated experiment. Also check that transformed parameters remain within their admissible domain: a positive scaling preserves strict positivity, but bounds or fixed values may exclude part or all of its orbit.

A symmetry proof is a sufficient, explicit certificate of nonidentifiability, not a complete catalogue of all possible ambiguities. Failure to find a symmetry is not evidence that the parameters are identifiable. Generic rank tests and augmented-state observability tests, covered in the earlier material on structural identifiability, offer complementary ways to investigate local parameter distinguishability. They do not replace checking the assumptions and constraints of a proposed transformation. Changing inputs or adding outputs may remove a particular ambiguity, but each modified design requires its own analysis.

This argument concerns structural rather than practical identifiability. Structural nonidentifiability persists even with ideal continuous observations; practical difficulty can also arise when parameters are theoretically distinguishable but finite, noisy data provide little information about them. A symmetry may be global over a range of parameter values or may apply only under particular constraints. Establishing the latter requires care: a transformation that works only at a special parameter value does not automatically prove generic nonidentifiability.

Common pitfall: “The measured equation contains the product \(k_{xy}x\), so \(k_{xy}\) and \(x\) are confounded.” That observation alone is not a proof: the dynamics for \(x\), its initial condition, and any other measurements might distinguish their scales. The correction is to transform the states, parameters, and initial conditions together, then verify all equations and outputs. In this model, the full substitution confirms the ambiguity; in another model, it may not.

A symmetry proof is especially valuable because it says exactly which changes are invisible to the specified measurements. It can guide model interpretation and experimental design: a direct measurement with known calibration, a known nonzero initial state, or another independently scaled output may constrain the hidden scale. But those additions break a particular symmetry only when their values and constraints are genuinely known. Identifiability of the resulting model must still be checked rather than inferred from the design change alone.

Check Your Understanding

Answer each question using the model and assumptions stated above. Brief solutions follow the questions.

  1. For \(\lambda=2\), write the transformed values of \(k_{\mathrm{in}}\), \(k_{xy}\), and \(x_0\). Does the measured output change?
  2. Why does checking the transformed equations and initial conditions establish equality of the complete output trajectories rather than equality at one time?
  3. If \(x_0\) is known to be a fixed positive value, does the scaling transformation as defined remain admissible?
  4. If the added measurement is \(w=qx\), with both \(q\) and \(x\) unknown, how must \(q\) transform for \(w\) to remain unchanged?

Solutions. (1) The transformed quantities are \(2k_{\mathrm{in}}\), \(k_{xy}/2\), and \(2x_0\). The output does not change because \(x\) doubles while \(k_{xy}\) halves, leaving \(k_{xy}x\) unchanged, and \(y\) is unchanged. (2) The transformed states satisfy the transformed ODE and initial conditions; uniqueness then makes them the model’s solution, and the observation equation gives the same output at every time. (3) No. The transformation changes a fixed positive \(x_0\) to \(2x_0\), violating the known initial condition. This excludes this symmetry but does not prove identifiability. (4) Set \(q_\lambda=q/\lambda\); then \(q_\lambda x_\lambda=(q/\lambda)(\lambda x)=qx\).

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