Identifiability of Branching Trajectory Models

Why Identifiability Matters

A branching trajectory model can fit observed cell states well without uniquely determining the biological story behind them. The same apparent spread of cells might arise from branches that are far apart with little measurement noise, or branches that are closer together with more noise. Before interpreting estimated branch positions or lineage proportions, ask a more basic question: could a different parameter setting produce exactly the same distribution of observations?

Identifiability concerns what the probability model makes possible in principle. It is distinct from whether a particular data set is large or informative enough to estimate parameters precisely. A parameter can be identifiable but estimated imprecisely when branches overlap or one lineage is rarely sampled. Conversely, collecting more observations cannot resolve structural non-identifiability: if two parameter settings imply exactly the same distribution, no amount of data from that distribution can distinguish them.

Definition: A parameter, or a specified combination of parameters, is identifiable under a model if equality of the observable probability distributions implies equality of that parameter, or combination, subject to stated symmetries such as relabeling branches. It is structurally non-identifiable if distinct parameter settings give the same observable distribution. It is weakly identified when distinct settings give different distributions, but those distributions are so similar that realistic data may not distinguish them reliably.

A Two-Branch Model at a Fixed Progression Stage

To make the issue precise, consider cells observed at a fixed, known progression value \(s\), represented by a vector \(\boldsymbol{Y}\) in a \(d\)-dimensional state space. A cell belongs to one of two branches, indexed by a latent label \(Z\). At this stage, suppose the model is

P(Z=1)=\pi,\qquad P(Z=2)=1-\pi,

and

\boldsymbol{Y}\mid Z=k,s \sim N\!\left(\boldsymbol{m}_k(s),\sigma^2 I_d\right),\qquad k\in\{1,2\}.

Here \(\boldsymbol{m}_k(s)\) is the mean state on branch \(k\), \(I_d\) is the \(d\)-dimensional identity matrix, and \(\sigma^2\) is the common isotropic observation-noise variance in each coordinate. The observable distribution is the mixture

f(\boldsymbol{y}\mid s) =\pi\,\phi_d\!\left(\boldsymbol{y};\boldsymbol{m}_1(s),\sigma^2I_d\right) +(1-\pi)\,\phi_d\!\left(\boldsymbol{y};\boldsymbol{m}_2(s),\sigma^2I_d\right),

where \(\phi_d(\boldsymbol{y};\boldsymbol{m},\Sigma)\) is the density of a \(d\)-variate normal distribution with mean \(\boldsymbol{m}\) and covariance \(\Sigma\). This model assumes the cells are sampled from the stated mixture, branch-specific deviations are absent beyond the specified noise, and the observation noise is Gaussian with the same variance in every coordinate and on both branches. The mixture weight \(\pi\) is the fraction of sampled cells assigned to branch 1 under this model. It is not automatically a biological fate probability: unequal sampling of branches can make the sampled fraction differ from the probability that a progenitor takes a fate.

A useful local trajectory parameterization is

\boldsymbol{m}_k(s)=\boldsymbol{b}+s\boldsymbol{d}_k,

where \(\boldsymbol{b}\) is the shared branchpoint state and \(\boldsymbol{d}_k\) is the direction and rate of change on branch \(k\) in the chosen coordinate and progression units. This is a linear-trajectory assumption, not a general property of cell differentiation. The identifiability conclusions below are conditional on the model and the observed coordinate system.

What the Distribution Reveals

Let \(\Delta=\boldsymbol{m}_1(s)-\boldsymbol{m}_2(s)\) denote the branch separation at the observed stage. The mixture mean and covariance follow by conditioning on the branch label. Writing \(\boldsymbol{\mu}=E(\boldsymbol{Y}\mid s)\), the mean is

\boldsymbol{\mu} =\pi\boldsymbol{m}_1(s)+(1-\pi)\boldsymbol{m}_2(s).

For the covariance, add the average within-branch covariance to the covariance of the branch means. The within-branch contribution is \(\sigma^2 I_d\). The branch-mean contribution is \[ \pi(\boldsymbol{m}_1-\boldsymbol{\mu})(\boldsymbol{m}_1-\boldsymbol{\mu})^\mathsf{T} +(1-\pi)(\boldsymbol{m}_2-\boldsymbol{\mu})(\boldsymbol{m}_2-\boldsymbol{\mu})^\mathsf{T}. \] Since \(\boldsymbol{m}_1-\boldsymbol{\mu}=(1-\pi)\Delta\) and \(\boldsymbol{m}_2-\boldsymbol{\mu}=-\pi\Delta\), substitution gives

\operatorname{Cov}(\boldsymbol{Y}\mid s) =\sigma^2I_d+\pi(1-\pi)\Delta\Delta^\mathsf{T}.

This identity is a central identifiability warning. The covariance alone combines observation noise, branch separation, and lineage proportion. It does not, in general, determine these quantities separately. The full mixture distribution contains more information than its covariance: its shape, including higher-order features, can distinguish some models with identical first two moments.

Theorem: A finite mixture of multivariate normal distributions is identifiable from its full probability distribution up to permutation of its component labels, provided zero-weight components and duplicate components are removed. Thus, in the two-component model above, if \(0<\pi<1\) and the two component distributions are distinct, the full distribution identifies the component means, covariances, and weights up to swapping labels. If a component has zero weight or the components coincide, that component distinction is not identifiable.

The label qualification matters. Replacing \((\pi,\boldsymbol{m}_1,\boldsymbol{m}_2)\) by \((1-\pi,\boldsymbol{m}_2,\boldsymbol{m}_1)\) leaves the distribution unchanged. Therefore the model can identify an unordered pair of branches and their weights, but not which branch deserves the name “1” without an external convention. The theorem concerns the full distribution under the stated finite Gaussian mixture model; it does not say that a finite sample will estimate the parameters precisely.

Worked Example: Different Noise and Separation, Same Covariance

Consider two hypothetical two-dimensional models with equal sampled branch proportions, \(\pi=0.5\), and centered overall mean. In Model A, let the branch means be \((1,0)\) and \((-1,0)\), so \(\Delta=(2,0)\), and let \(\sigma^2=1\). In Model B, let the means be \((0.5,0)\) and \((-0.5,0)\), so \(\Delta=(1,0)\), and let \(\sigma^2=1.75\). Find and compare their covariance matrices.

For Model A, \[ \pi(1-\pi)\Delta\Delta^\mathsf{T} =0.5(0.5) \begin{pmatrix}4&0\\0&0\end{pmatrix} =\begin{pmatrix}1&0\\0&0\end{pmatrix}. \] Adding the noise covariance gives \[ \operatorname{Cov}(\boldsymbol{Y})= \begin{pmatrix}1&0\\0&1\end{pmatrix} +\begin{pmatrix}1&0\\0&0\end{pmatrix} =\begin{pmatrix}2&0\\0&1\end{pmatrix}. \]

For Model B, the branch contribution is \[ 0.5(0.5) \begin{pmatrix}1&0\\0&0\end{pmatrix} =\begin{pmatrix}0.25&0\\0&0\end{pmatrix}. \] Thus \[ \operatorname{Cov}(\boldsymbol{Y})= \begin{pmatrix}1.75&0\\0&1.75\end{pmatrix} +\begin{pmatrix}0.25&0\\0&0\end{pmatrix} =\begin{pmatrix}2&0\\0&1.75\end{pmatrix}. \] These matrices are not equal: the models do not match in the second coordinate. To make a covariance-matching comparison, restrict attention to the branch-separation direction, the first coordinate. Its variance is \(2\) in both models: \(1+0.25(2^2)=2\) for Model A and \(1.75+0.25(1^2)=2\) for Model B. Consequently, a covariance analysis of that coordinate alone cannot separate the models. In the two-dimensional data, the second coordinate also informs \(\sigma^2\), so the full covariance distinguishes them. This illustrates why identifiability depends on which observations and summaries are available.

Branch Geometry and Progression Scale

Even if the component means at a given stage are identifiable, branchpoint position and trajectory directions need not be. At one observed value of \(s\), the data inform \(\boldsymbol{m}_k(s)\), not a unique decomposition into \(\boldsymbol{b}\) and \(\boldsymbol{d}_k\). If \(s\ne0\), any proposed branchpoint \(\boldsymbol{b}'\) can be paired with \[ \boldsymbol{d}'_k=\{\boldsymbol{m}_k(s)-\boldsymbol{b}'\}/s \] to give the same mean at that stage. Likewise, if the numerical scale of \(s\) is not fixed, replacing \(s\) by \(cs\) and each \(\boldsymbol{d}_k\) by \(\boldsymbol{d}_k/c\), for nonzero \(c\), leaves the trajectory means unchanged. A calibrated progression scale and observations across stages are needed to interpret slopes in their stated units.

Worked Example: One Stage Does Not Locate the Branchpoint

Suppose cells are observed only at \(s=1\), and the two branch means are \((2,0)\) and \((0,2)\). In Model A, choose \(\boldsymbol{b}=(0,0)\), \(\boldsymbol{d}_1=(2,0)\), and \(\boldsymbol{d}_2=(0,2)\). Then \(\boldsymbol{b}+s\boldsymbol{d}_1=(2,0)\) and \(\boldsymbol{b}+s\boldsymbol{d}_2=(0,2)\).

In Model B, choose \(\boldsymbol{b}'=(1,0)\), \(\boldsymbol{d}'_1=(1,0)\), and \(\boldsymbol{d}'_2=(-1,2)\). At the same observed stage, the means are \((1,0)+(1,0)=(2,0)\) and \((1,0)+(-1,2)=(0,2)\). With the same \(\pi\) and noise covariance, the two models give exactly the same observable distribution at \(s=1\), despite having different branchpoints and directions. Observations at additional known progression values can constrain the linear trajectories; extrapolating a branchpoint remains dependent on the linearity assumption and the quality of stage information.

Exact Equivalence and Weak Identification

Worked Example: An Unobserved Branch Can Vary Freely

Suppose \(\pi=1\), the observed branch has mean \((0,0)\), and its noise covariance is \(I_2\). The observable distribution is \(N((0,0),I_2)\). Set the unobserved branch mean first to \((1,0)\), then to \((100,-50)\), keeping its weight at zero. In either case, its density is multiplied by zero, so the observable distribution remains exactly \(N((0,0),I_2)\). The mean and noise parameters of the absent component cannot be learned from these observations. A model that requires both branches to have strictly positive weights rules out this exact boundary case, but does not prevent practical difficulty when one weight is very small.

A second exact degeneracy occurs if \(\boldsymbol{m}_1(s)=\boldsymbol{m}_2(s)\): the mixture reduces to one normal distribution, and every value of \(\pi\) gives the same distribution. Near this case, rather than exactly at it, branch weights and separation may be formally identifiable but weakly identified. Small changes in the data can then lead to large changes in estimated weights or branch positions.

Checking the Model and Choosing What to Interpret

Identifiability is conditional on assumptions, so diagnostics should check whether those assumptions are plausible. For example, a visibly heavy-tailed or multimodal residual pattern may indicate that common Gaussian noise is inadequate. Different residual spreads across branches contradict the common-\(\sigma^2\) assumption. A curved trajectory challenges the linear mean path. These checks do not prove identifiability; they assess whether the model used to make identifiability claims is a reasonable description of the data.

The main alternatives answer different questions. Moment-based analysis is transparent and gives the covariance decomposition, but covariance alone generally cannot separate branch spread from noise. Full-distribution mixture modeling can use component shape and higher-order information, under its distributional assumptions, but becomes practically fragile when components overlap or a weight is small. External information, such as a calibrated progression measure or a separately justified noise estimate, can constrain otherwise confounded parameters. Such information changes what is identifiable only when it provides a real restriction; simply fixing an uncertain quantity for convenience does not establish that the fixed value is correct.

A further limitation is that a fitted mixture proportion describes the sampled cells represented by the model. To interpret it as a lineage fate probability requires assumptions about how cells enter the sample and whether sampling differs by lineage. Likewise, inferred coordinates and distances depend on the representation of cell state: a branch separation in a particular embedding is not automatically a physical distance or a causal effect.

Key takeaway: State which object is identifiable: the full observable mixture, a covariance contribution, the branch means at an observed stage, or trajectory parameters such as a branchpoint and slopes. These are not interchangeable claims. Positive weights, distinct component distributions, a fixed progression scale, suitable stage coverage, and credible noise assumptions determine which interpretation is supported.

Common Mistakes

Treating equal covariance as equal distributions. Matching first and second moments does not establish equality of full mixture distributions. Compare the full model or state explicitly that the conclusion concerns covariance only.
Calling a branch label identifiable. Mixture component labels can be swapped without changing the distribution. Use an external biological convention if a named branch is required.
Interpreting a sampled mixture weight as a fate probability. The weight is a sampling proportion under the model; unequal lineage sampling can separate it from a biological fate probability.
Assuming more cells fix structural non-identifiability. More observations reduce sampling uncertainty, but cannot distinguish parameter settings that imply exactly the same distribution.
Reading a branchpoint from one post-branch stage. At a single stage, multiple intercept-and-slope combinations can yield the same branch means. A branchpoint interpretation requires additional stage information or explicit constraints.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

In the two-branch model, what condition makes the mixture proportion \(\pi\) unidentifiable because the components coincide?
Why does the covariance formula not, by itself, separate branch separation from observation noise?
If the progression coordinate is rescaled by a factor of four, how must the trajectory directions change to preserve the same mean paths?
A model has two distinct component means and positive weights. What ambiguity remains even if the full mixture distribution identifies its components?

Solutions. (1) If \(\boldsymbol{m}_1(s)=\boldsymbol{m}_2(s)\) and the component noise distributions are the same, changing \(\pi\) leaves the observable distribution unchanged. (2) The covariance is \(\sigma^2I_d+\pi(1-\pi)\Delta\Delta^\mathsf{T}\), so noise and the branch contribution are added; a covariance summary may admit multiple decompositions. (3) If \(s'=4s\), set \(\boldsymbol{d}'_k=\boldsymbol{d}_k/4\); then \(\boldsymbol{b}+s'\boldsymbol{d}'_k=\boldsymbol{b}+s\boldsymbol{d}_k\). (4) The two component labels can still be exchanged, replacing \(\pi\) with \(1-\pi\) and swapping the component parameters without changing the distribution.