Ask whether different parameters can look identical

Try

A stock disappears through two channels with rates a and b. Its total follows M′=−(a+b)M, so M(t)=M₀e^(−(a+b)t). If only the remaining stock is observed, can even a perfectly measured time series distinguish the two loss channels?

Compare rate pairs (1,3) and (2,2). Their sum is the same, so their total-stock curves coincide at every time. This is structural ambiguity, not a failure of a numerical optimizer.

Predict

Would measuring total stock twice as often resolve the ambiguity? Predict whether measuring the accumulated loss through just one named channel would supply genuinely new information instead.

Experiment

Live experiment: different mechanisms, the same observed curve

A reservoir starts with mass 10 and loses material through two parallel first-order channels with rates a and b. Only remaining total mass is initially measured: M′=−(a+b)M. Time uses a consistent teaching unit, rates its reciprocal. Predict whether more exact observations of total mass can distinguish the two channels.

Channel rate a 0.511.52Channel rate b 0.511.52Observation time 00.5123Reset experiment

Solid: M(t)=10exp(−2t). Dashed reference: a=b=1, so total rate 2. At t=1, remaining mass is 1.353353. Curves coincide whenever the sum is two, even if the individual mechanisms differ. The graph samples the exact ODE solution; it contains no measured data or fitted noise.

Accumulated loss through A: 2.161662; through B: 6.484985. Remaining plus both losses is 10. Channel A accounts for fraction a/(a+b)=0.25 of any positive accumulated loss.

For a positive observation time, a single exact total mass determines only r=a+b through r=−log(M/10)/t. Both sensitivities ∂M/∂a and ∂M/∂b equal −tM=-1.353353, so total-mass sensitivity columns coincide. Across the entire time curve, the displayed rate grid contains 3 indistinguishable pairs: (0.5, 1.5); (1, 1); (1.5, 0.5). With unrestricted positive rates, there is a continuous family a+b=2, not just this finite grid.

If accumulated loss through A is independently measured as well as total mass, q=lossA/(10−M) identifies the channel fraction. Then a=rq and b=r(1−q). This additional positive-time observation separates mechanisms under the stated model. Noisy short-time measurements may still be poorly conditioned; structural identifiability is not a guarantee of precise estimation.

Compare (a,b)=(0.5,1.5) with (1,1). Decide which sensor you would add and justify it from the equations. A smaller residual cannot distinguish models that predict the same observed quantity. Settings are temporary.

The widget’s settings are temporary synthetic exploration. Compare equal-sum parameter pairs using the total-stock curve, then inspect the channel-specific integrated loss. Do not interpret a parameter fraction displayed at time zero as a measured loss ratio: both accumulated losses are zero there, so that observed ratio is undefined.

For positive a+b, channel A accumulated loss is M₀·(a/(a+b))·(1−e^(−(a+b)t)).

Observation	Identified combination in ideal noiseless data
Total remaining stock	a+b
Total plus channel A loss at positive time	Sum and fraction a/(a+b)

The sum and fraction together recover both rates, provided total loss is nonzero and the model assumptions hold.

Understand

Structural identifiability concerns injectivity of the parameter-to-observation map over a stated parameter domain. If two distinct admissible parameter vectors give the same observations, no estimator using only those observations can distinguish them. More accurate measurement of the same map cannot repair its noninjectivity.

For a linear observation y=Xθ, full column rank gives unique θ in an unrestricted real parameter space. If Xv=0 for a nonzero v, then θ and θ+v produce identical y. Conversely, a zero kernel makes equality of outputs imply equality of parameters. Parameter constraints can alter this conclusion at special boundary points, so state the domain when applying the rank argument.

Practical identifiability is different. Two columns can be independent but nearly parallel, making parameter estimates highly sensitive to small noise. A unique exact solution may still have wide uncertainty. Local sensitivity rank can diagnose nearby ambiguity, but local rank alone need not establish global injectivity for a nonlinear map. For example, observing θ² does not distinguish positive and negative θ unless the domain is restricted.

An informative redesign changes the observation map: measure another compartment, distinguish a channel or intervene under a justified model. Replication reduces noise for existing outputs; it does not add a missing parameter direction.

Apply

Find two indistinguishable parameter settings in a model you propose. Then design one additional measurement and show algebraically why it separates them. State when the new formula fails, including zero-time or zero-loss cases. If no ambiguity is found numerically, do not call that a proof of global identifiability.

Continue

Only after checking what can be identified should estimation attempt to choose parameter values. Residuals then reveal what the fitted model still fails to explain.