From a Care-State Graph to an Intensity Matrix
A long-term care model begins by specifying the states a person can occupy and which transitions are permitted. The graph records that structure, but calculations require a numerical representation that is consistent across every state. The intensity matrix, also called a generator matrix, combines the transition rates with the restrictions in the graph.
This tutorial focuses on constructing and checking that matrix. It does not calculate future state probabilities or benefit values. Those require further steps after the model’s states and intensities have been specified.
The row index identifies the state the person is leaving; the column index identifies the state the person may enter. Thus, rows—not columns—sum to zero. A transition prohibited by the specified graph has intensity zero. Every off-diagonal entry must be nonnegative. The diagonal entry is generally negative, or zero if the state has no possible exits.
A matrix can be constant over time when all the specified intensities are constant and use the same time unit. If rates vary with age or duration, the entries can instead depend on time. Here, each example specifies a fixed set of rates; the construction itself applies at a given time even when rates are not constant.
Why the Diagonal Is Determined by the Other Rates
The diagonal is not an additional transition rate. It makes the matrix consistent with the probabilities of remaining in the current state. To see why, consider a person in state \(i\) at time \(t\), and suppose the rates are defined over a common time unit. For each distinct state \(j\), the probability of moving from \(i\) to \(j\) during a small interval of length \(h\) is \(q_{ij}(t)h+o(h)\). The probability of remaining in \(i\) is therefore
The coefficient of \(h\) in the probability of remaining in \(i\) is the diagonal intensity. Consequently, it is the negative sum of the outgoing intensities. The resulting row sum is zero:
This argument is also a practical check: total probability over the possible destinations, including remaining in the state, must be one to first order in \(h\). The row-sum condition does not mean that the diagonal is a probability. It is a rate, with the same time unit as the off-diagonal entries.
Worked Example: A Four-State Care Model
Method: list the graph’s outgoing edges by state, then complete each row. Consider a hypothetical model with states \(H\) (healthy and living independently), \(M\) (moderate care), \(S\) (severe care), and \(D\) (dead). Time is measured in years. The only permitted transitions and their constant annual intensities are \(H\to M:0.08\), \(M\to H:0.20\), \(M\to S:0.30\), \(M\to D:0.04\), \(S\to M:0.10\), and \(S\to D:0.12\). Death is absorbing.
Use the state order \(H,M,S,D\). In row \(H\), the only outgoing transition is to \(M\), so the off-diagonal entries are \(q_{HM}=0.08\), \(q_{HS}=0\), and \(q_{HD}=0\). The diagonal is \(q_{HH}=-(0.08+0+0)=-0.08\).
In row \(M\), the outgoing rates are \(0.20\), \(0.30\), and \(0.04\). Their sum is \(0.54\), so \(q_{MM}=-0.54\). In row \(S\), the outgoing rates are \(0.10\) to \(M\), zero to \(H\), and \(0.12\) to \(D\). Their sum is \(0.22\), so \(q_{SS}=-0.22\). Since \(D\) has no exits, every off-diagonal entry in its row is zero and \(q_{DD}=0\).
Check. The row sums are \(-0.08+0.08=0\), \(0.20-0.54+0.30+0.04=0\), \(0+0.10-0.22+0.12=0\), and \(0\). All off-diagonal entries are nonnegative, and the zeros outside the listed edges agree with the graph. The common mistake avoided is to put a positive total exit rate on the diagonal; the diagonal must be negative.
Encoding the Graph, Not Just the Rates
A list of rates is not enough to specify a matrix unless the states and permitted edges are clear. In particular, do not infer an edge simply because two states sound clinically related. If a transition is not permitted in the stated model, enter zero. If an edge is permitted but its rate has not been supplied, the model is incomplete; a rate cannot be invented from the other entries.
The state order is a bookkeeping choice rather than a model assumption. Changing the order changes the positions of rows and columns, but not the underlying transitions. State the order before presenting the matrix so that each entry can be interpreted unambiguously.
Worked Example: Checking a Home-Care and Facility Graph
Method: construct one row at a time from the specified edges. In a hypothetical annual model, let \(A\) denote independent living, \(C\) home care, \(F\) facility care, and \(D\) death. The graph permits \(A\to C:0.06\), \(A\to D:0.01\), \(C\to A:0.15\), \(C\to F:0.25\), \(C\to D:0.05\), \(F\to C:0.08\), and \(F\to D:0.18\). There are no other transitions, and death is absorbing.
Use the order \(A,C,F,D\). From \(A\), the total outgoing rate is \(0.06+0.01=0.07\), giving diagonal \(-0.07\). From \(C\), the total is \(0.15+0.25+0.05=0.45\), giving diagonal \(-0.45\). From \(F\), it is \(0.08+0.18=0.26\), giving diagonal \(-0.26\). The death row is all zero.
Check. The row sums are \(-0.07+0.06+0.01=0\), \(0.15-0.45+0.25+0.05=0\), \(0.08-0.26+0.18=0\), and \(0\). In particular, \(q_{AF}=0\): the graph does not permit a direct move from independent living to facility care. The common mistake avoided is to fill every off-diagonal cell with a rate just because the destination state exists.
Time Units and Absorbing States
Every entry in an intensity matrix must use the same unit of time. If all rates are supplied per month and the model is to be expressed per year, multiply each rate by 12. This is a change of units for an intensity, not a conversion from a transition probability. A rate over a period should not be treated as though it were the probability of making the transition during that period.
An absorbing state is one from which the model permits no departure. Its entire row consists of zeros, including its diagonal. Death is commonly represented this way when the model ends at death. This modeling convention should not be confused with a living state that happens to have no specified transition in a partial list: a row of zeros for such a state would assert that it has no possible exits.
Worked Example: Converting Monthly Intensities to Annual Intensities
Method: convert the time unit first, then calculate the diagonals. Consider a hypothetical model with states \(H\) (independent living), \(L\) (limited care), \(S\) (intensive care), and \(D\) (dead). Its supplied rates are constant monthly intensities: \(H\to L:0.006\), \(H\to D:0.0005\), \(L\to H:0.012\), \(L\to S:0.018\), \(L\to D:0.002\), \(S\to L:0.004\), and \(S\to D:0.010\). Death is absorbing. Express the matrix in annual units.
There are 12 months in a year, so multiply each off-diagonal monthly intensity by 12. The annual rates are \(H\to L:0.006(12)=0.072\) and \(H\to D:0.0005(12)=0.006\). From \(L\), they are \(L\to H:0.012(12)=0.144\), \(L\to S:0.018(12)=0.216\), and \(L\to D:0.002(12)=0.024\). From \(S\), they are \(S\to L:0.004(12)=0.048\) and \(S\to D:0.010(12)=0.120\).
The annual total outgoing rates are \(0.072+0.006=0.078\) from \(H\), \(0.144+0.216+0.024=0.384\) from \(L\), and \(0.048+0.120=0.168\) from \(S\). Thus the diagonal entries are \(-0.078\), \(-0.384\), and \(-0.168\), respectively.
Check. The first three row sums are \(-0.078+0.072+0.006=0\), \(0.144-0.384+0.216+0.024=0\), and \(0.048-0.168+0.120=0\); the death row sums to zero. The common mistake avoided is to use monthly off-diagonal entries with annual diagonal entries, or to multiply only some of the rates by 12. All entries must have the same time unit.
A Construction and Validation Checklist
A systematic construction reduces errors when a graph has several care states or recovery routes. Use the following sequence before relying on the matrix in a valuation or probability calculation:
Use the same order for both rows and columns, and label it next to the matrix.
Enter its supplied intensity in the row of the origin state and the column of the destination state. Direction matters: a recovery edge does not imply a relapse edge.
Do not assume that every pair of states is connected or that transitions occur in both directions.
Convert the off-diagonal rates consistently before calculating totals.
Set each diagonal entry equal to minus the sum of the other entries in its row; verify zero row sums and nonnegative off-diagonal entries.
These checks are structural, not merely arithmetic. A matrix can have row sums of zero and still represent the wrong graph—for example, if a recovery rate is placed in the wrong column. Conversely, a matrix that faithfully copies the listed edges but has a nonzero row sum has an incorrectly completed diagonal or a missing rate. Both the graph match and the row-sum check are necessary.
Common Exam Traps
A particularly tempting error is to treat the diagonal as another outgoing transition rate. For example, if a state has a single outgoing intensity of \(0.08\) per year, entering \(q_{ii}=+0.08\) gives row sum \(0.08+0.08=0.16\), not zero. The correct entry is \(q_{ii}=-0.08\), so that \(-0.08+0.08=0\).
Other frequent errors include reversing the row and column interpretation, omitting the death intensity from a living state’s outgoing total, and assigning a positive exit rate to an absorbing state. Another subtle error is to infer a direct transition from an initial state to a severe care state when the specified graph allows only progression through an intermediate care state. The matrix must encode the graph as given, not a different pathway that seems plausible.
Correct construction matters because later calculations use the matrix as the mathematical representation of the care-state model. A misplaced rate changes which movements are allowed; a wrong diagonal changes the total rate out of a state; and an inconsistent time unit changes the scale of every transition. For a model aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus, show the state order, the edge rates, the diagonal calculation, and the row-sum checks rather than presenting an unexplained matrix.
Check Your Understanding
For each problem, construct or interpret the matrix using the stated state order and time unit.
- A hypothetical annual model has states \(A\) (independent), \(B\) (care), and \(D\) (dead). The only transitions are \(A\to B:0.11\), \(A\to D:0.03\), \(B\to A:0.07\), and \(B\to D:0.09\); \(D\) is absorbing. Give the diagonal entries in order \(A,B,D\). (Answer: \(-0.14,\,-0.16,\,0\))
- In a hypothetical annual model ordered \(H,C,F,D\), the permitted rates are \(H\to C:0.04\), \(H\to D:0.02\), \(C\to H:0.12\), \(C\to F:0.16\), \(C\to D:0.03\), \(F\to C:0.05\), and \(F\to D:0.20\), with death absorbing. Give the complete matrix. (Answer: \(\begin{pmatrix}-0.06&0.04&0&0.02\\0.12&-0.31&0.16&0.03\\0&0.05&-0.25&0.20\\0&0&0&0\end{pmatrix}\))
- A two-state living-care model has states \(R\) (residential care) and \(H\) (home care). In annual units, \(R\to H=0.06\) and \(H\to R=0.10\), with no other transitions. What is the matrix, and what is the annual total exit intensity from \(H\)? (Answer: \(Q=\begin{pmatrix}-0.06&0.06\\0.10&-0.10\end{pmatrix}\); the total exit intensity from \(H\) is \(0.10\) per year.)
- A hypothetical model has states \(X,Y,Z,D\), ordered as written. Its annual transitions are \(X\to Y:0.09\), \(X\to D:0.01\), \(Y\to Z:0.13\), \(Y\to D:0.04\), \(Z\to Y:0.06\), and \(Z\to D:0.15\), with death absorbing. Give the three living-state diagonal entries. (Answer: \(-0.10,\,-0.17,\,-0.21\))
- Monthly intensities in a hypothetical model are \(P\to C:0.005\), \(P\to D:0.001\), and \(C\to D:0.020\), where \(P\) is independent living, \(C\) is care, and \(D\) is absorbing. Give the full annual intensity matrix in order \(P,C,D\). (Answer: \(\begin{pmatrix}-0.072&0.060&0.012\\0&-0.240&0.240\\0&0&0\end{pmatrix}\))