Exact State Aggregation Through Lumpability

Why Exact Aggregation Requires a Closure Test

A compartment model may contain several states that are not distinguished by the question being asked. For example, a model may track two cell subtypes separately even when the reported outcome is their total abundance. Adding those states can simplify reporting, but it does not necessarily produce a valid lower-dimensional dynamical model. The total may change at a rate that depends on how its abundance is divided among the hidden subtypes.

Lumpability is the condition that makes aggregation exact: the aggregate state has a closed equation whose future can be calculated from the aggregate state itself, without needing the fine-scale state composition. This is stronger than matching an output at one time or for one initial condition. The condition must hold for every fine-scale state in the model's domain.

Definition: Consider the linear continuous-time model \( \dot{x}=Ax+Gu(t) \), where \(x\in\mathbb{R}^m\) is the fine-scale state, \(A\) is a fixed \(m\times m\) matrix, \(u(t)\) is an input, and \(G\) is a fixed input matrix. Let \(C\in\mathbb{R}^{r\times m}\) map fine states to aggregate states \(z=Cx\). The model is exactly lumpable through \(C\) if there is an \(r\times r\) matrix \(B\) such that \(CA=BC\). Then \(z\) obeys a closed aggregate equation for every initial state and admissible input.

For ordinary grouping, each row of \(C\) is an indicator row: it contains 1 for the fine states in one group and 0 elsewhere. Thus each component of \(z\) is the sum of the fine states in one group. More general linear maps are possible, but the same matrix condition applies.

Deriving the Aggregate Equation

Differentiate \(z=Cx\). Since \(C\) is fixed, \(\dot{z}=C\dot{x}\). Substituting the fine-scale model gives

\dot{z}=CAx+CG u(t).

If \(CA=BC\), then \(CAx=BCx=Bz\), so the aggregate equation is

\dot{z}=Bz+CG u(t).

For a fixed \(G\), the aggregate input matrix is simply \(H=CG\). It is always defined; no additional input-compatibility condition is needed. The input can enter different fine states, and its aggregate contribution is obtained by applying the same aggregation map to that input contribution.

The closure condition is also necessary for a closed linear autonomous part that works for every fine state. If \(CAx=BCx\) is required for all \(x\in\mathbb{R}^m\), then \((CA-BC)x=0\) for every \(x\), which implies \(CA=BC\). Conversely, the matrix equality directly gives closure. This proves the exact criterion, rather than merely providing a sufficient test.

Theorem: For fixed matrices \(A\), \(G\), and \(C\), the aggregate \(z=Cx\) of \( \dot{x}=Ax+Gu(t) \) has exact dynamics \( \dot{z}=Bz+Hu(t) \) for every fine state and input if and only if \(CA=BC\) for some \(B\). When this condition holds, \(H=CG\). If \(C\) has full row rank, \(B\) is unique.

The derivation assumes a fixed linear state equation and a fixed aggregation map. The input may vary with time; it must be sufficiently regular for the differential equation to have a solution, such as being locally integrable. If \(A\) depends on time or parameters, the equality must hold for the relevant values of those quantities, with the corresponding aggregate matrix allowed to depend on them. A condition verified at only one parameter setting does not establish exactness elsewhere.

What the Matrix Condition Means for Groups

Suppose \(C\) represents a partition into groups. Write \(I_a\) for the fine states in aggregate group \(a\). The element \((CA)_{aj}\) is the sum of the entries of column \(j\) of \(A\) over rows in group \(a\):

(CA)_{aj}=\sum_{i\in I_a} A_{ij}.

If fine state \(j\) belongs to group \(b\), then \((BC)_{aj}=B_{ab}\). Therefore, \(CA=BC\) holds precisely when, for every destination group \(a\), the summed effect on group \(a\) is the same for every fine state \(j\) within a given source group \(b\). In a compartment model, this means the net rate of change in each aggregate group, per unit amount in a source state, cannot vary with which fine state within that source group contains the amount.

For a transfer-only compartment model using the convention \(\dot{x}=Ax\), off-diagonal \(A_{ij}\) is the rate of transfer from state \(j\) to state \(i\). The diagonal entries account for departures from state \(j\), including transfers out and any elimination. The condition compares summed net effects, not just one selected pathway. Internal transfers within a group can cancel in that group's total, while different loss rates can prevent closure.

Worked Example: Aggregating Two Compartments Exactly

Consider three compartments with \(x_1\) and \(x_2\) grouped together, and \(x_3\) kept as the second aggregate state. Transfer rates are hypothetical. Use the column-state convention and the matrix

A= \begin{pmatrix} -3&3&1\\ 1&-5&1\\ 2&2&-2 \end{pmatrix}, \qquad C= \begin{pmatrix} 1&1&0\\ 0&0&1 \end{pmatrix}.

The columns of \(A\) sum to zero, so this example has transfers among the three compartments and no elimination. Calculate \(CA\) by adding the first two rows of \(A\) for the first aggregate and retaining the third row for the second:

CA= \begin{pmatrix} -2&-2&2\\ 2&2&-2 \end{pmatrix}.

A candidate aggregate matrix is \(B=\begin{pmatrix}-2&2\\2&-2\end{pmatrix}\). Multiplication gives

BC= \begin{pmatrix} -2&-2&2\\ 2&2&-2 \end{pmatrix} =CA.

Thus the condition holds. Define \(z_1=x_1+x_2\) and \(z_2=x_3\). The exact aggregate dynamics are

\dot{z}_1=-2z_1+2z_2,\qquad \dot{z}_2=2z_1-2z_2.

For instance, the rate of change in \(z_1\) is the same whether its amount is all in \(x_1\), all in \(x_2\), or divided between them. The internal exchange rates differ, but they do not alter the aggregate closure.

Now let a hypothetical input enter only \(x_1\), with \(G=(1,0,0)^\mathsf{T}\). Then \(H=CG=(1,0)^\mathsf{T}\), and the aggregate equation becomes \(\dot{z}=Bz+Hu(t)\). The input contributes \(u(t)\) to \(z_1\); this requires no further condition.

Specified Outputs and Exact Preservation

If the reported outputs are the aggregate states \(z=Cx\), lumpability preserves their trajectories exactly: solving the aggregate equation from \(z(0)=Cx(0)\) gives the same \(z(t)\) as aggregating the fine-scale solution at every time. More generally, suppose a specified linear output is \(w=Lx\). It can be calculated from the aggregate state alone exactly when there is a matrix \(D\) such that \(L=DC\). In that case \(w=Dz\). If no such \(D\) exists, two fine states can have identical aggregates but different outputs, so the output is not determined by the aggregate state.

When both conditions hold, the aggregate model preserves the specified outputs for every initial state and admissible input. In fact, the aggregate equation has the same initial aggregate \(z(0)=Cx(0)\), and the same forcing \(CG u(t)\). Uniqueness of solutions to the linear differential equation then gives identical aggregate trajectories. Applying \(D\) gives identical output trajectories \(w(t)\).

Worked Example: Detecting Hidden Composition Dependence

Suppose states \(x_1\) and \(x_2\) are proposed as one group, with \(x_3\) as the other. Consider the hypothetical transfer matrix

A= \begin{pmatrix} -1&0&0\\ 0&-3&0\\ 1&3&0 \end{pmatrix}, \qquad C= \begin{pmatrix} 1&1&0\\ 0&0&1 \end{pmatrix}.

Here state \(x_1\) transfers to \(x_3\) at rate 1, while \(x_2\) transfers to \(x_3\) at rate 3. Compute

CA= \begin{pmatrix} -1&-3&0\\ 1&3&0 \end{pmatrix}.

For the first source group, the first column of \(CA\) gives \(-1\) in the first aggregate, while the second gives \(-3\). They cannot both equal the same entry of \(B\) for that source group, so no \(B\) satisfies \(CA=BC\).

The failure is visible without matrix algebra. The two fine states \(x=(1,0,0)^\mathsf{T}\) and \(x=(0,1,0)^\mathsf{T}\) both have \(z=(1,0)^\mathsf{T}\). But their aggregate derivatives are respectively \((-1,1)^\mathsf{T}\) and \((-3,3)^\mathsf{T}\). The same aggregate state therefore has two different derivatives. Its future cannot be determined from \(z\) alone.

Discrete-Time Transition Models

The same idea applies to a discrete-time linear transition model \(x_{k+1}=Px_k\). With \(z_k=Cx_k\), exact aggregation requires a matrix \(Q\) satisfying \(CP=QC\), which then gives \(z_{k+1}=Qz_k\). For a Markov transition model represented with column probability vectors, \(P_{ij}\) is the probability of moving from state \(j\) to state \(i\), and the columns of \(P\) sum to 1. Under a partition, the summed probability of entering each destination group must be identical for all fine states in the same source group.

Worked Example: Lumpability of a Transition Matrix

Consider three hypothetical states with states 1 and 2 grouped, and state 3 alone. Let

P= \begin{pmatrix} 0.5&0.1&0.2\\ 0.2&0.6&0.3\\ 0.3&0.3&0.5 \end{pmatrix}, \qquad C= \begin{pmatrix} 1&1&0\\ 0&0&1 \end{pmatrix}.

Each column of \(P\) sums to 1. Summing the first two rows gives

CP= \begin{pmatrix} 0.7&0.7&0.5\\ 0.3&0.3&0.5 \end{pmatrix}.

The first two columns, corresponding to fine source states 1 and 2 in the same group, have the same aggregate probabilities. Thus \(Q=\begin{pmatrix}0.7&0.5\\0.3&0.5\end{pmatrix}\) satisfies \(CP=QC\), and \(z_{k+1}=Qz_k\). For example, from either state 1 or state 2, the probability of moving into the grouped destination \(\{1,2\}\) is 0.7, and the probability of moving to state 3 is 0.3. The aggregate transition probabilities do not depend on the hidden composition.

Diagnostics, Alternatives, and Limitations

A practical test is to compute \(CA\), then compare its columns within each source group. In each row of \(CA\), entries belonging to the same source group must be equal. Equivalently, seek \(B\) and verify the full matrix residual \(CA-BC\) is zero. For an exact mathematical model, any nonzero residual means the proposed aggregation is not exact. With numerical matrices, a small residual may reflect rounding, but a small residual alone establishes only approximate closure, not exact lumpability.

A useful diagnostic for failure is to construct two fine states with the same aggregate value but different aggregate derivatives. Such a pair demonstrates directly why no closed aggregate equation can work for all states. Also check output representation separately: a closed aggregate state does not preserve an output that distinguishes fine states within a group.

Exact lumpability differs from eliminating a state using a conserved total or restricting dynamics to a quasi-steady or invariant manifold. Those approaches rely on different structure or assumptions; they do not automatically imply \(CA=BC\) for a proposed grouping. Projection onto selected modes can also produce a useful reduced description without guaranteeing exact trajectories for all initial conditions. Lumpability's strength is its exactness under its matrix condition, while its limitation is that many biologically intuitive groupings fail that stringent test. The condition concerns the specified \(A\), aggregation \(C\), and outputs; changing rates or the grouping can change the result.

Common mistake: checking only the output definition. It may be easy to sum two states, but that does not show their sum has closed dynamics. Check \(CA=BC\), not just that \(z=Cx\) is well-defined.
Common mistake: requiring the input to satisfy an extra closure condition. For fixed \(G\), the aggregate input term is \(CG u(t)\). It is already a function of the input, and the aggregate input matrix is \(H=CG\).
Common mistake: treating a result for one trajectory as exact aggregation. A particular initial condition may conceal composition dependence. Exact lumpability requires the same aggregate derivative for every fine state with the same aggregate.

Key takeaway: Exact state aggregation is a closure property of the dynamics. For \(z=Cx\), verify \(CA=BC\); then the aggregate model is \(\dot{z}=Bz+CG u(t)\). Check separately that every specified linear output can be written as a function of \(z\).

Check Your Understanding

Answer each question, then compare with the solutions.

For a fixed input matrix \(G\), what is the aggregate input matrix when \(z=Cx\)? Is a further input-compatibility condition required?
In a partition-based compartment model, what must be equal for fine source states in the same group?
Can an output \(w=Lx\) be recovered from \(z=Cx\) if there is a matrix \(D\) with \(L=DC\)? Explain.
In the failed aggregation example, why do the two states \((1,0,0)^\mathsf{T}\) and \((0,1,0)^\mathsf{T}\) disprove closed aggregate dynamics?
For a discrete-time transition model \(x_{k+1}=Px_k\), state the exact lumpability condition.

Solutions. (1) The aggregate input matrix is \(H=CG\), and no additional condition is needed for fixed \(G\). (2) For every destination group, the sum of the corresponding entries in a fine source-state column of \(A\) must be the same across source states within that group. (3) Yes. Then \(w=Lx=DCx=Dz\), so it is determined by the aggregate. (4) They have the same \(z=(1,0)^\mathsf{T}\) but different aggregate derivatives, \((-1,1)^\mathsf{T}\) and \((-3,3)^\mathsf{T}\); one closed equation for \(z\) cannot give both. (5) There must be a matrix \(Q\) such that \(CP=QC\).