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Exam ALTAM · State Models and Intensities · Tutorial 30 of 999

Building a Long-Term Care State Model

Learn to translate a hypothetical long-term care contract into clearly defined states, permitted transitions, and a valid intensity matrix.

Beginner 10 min read

What You'll Learn

  • Define mutually exclusive healthy, care-needing, and dead states for a simple long-term care model.
  • Identify permitted transitions from the model assumptions and distinguish them from forbidden transitions.
  • Construct a row-oriented intensity matrix with diagonal entries equal to negative total exit intensities.
  • Assemble and check matrices with constant or calendar-time-dependent intensities.
  • Avoid treating a permitted transition with a momentarily zero rate as structurally forbidden.

Why Define the States Before the Rates?

A long-term care model needs to represent more than whether a person is alive or dead. For example, a hypothetical contract may pay a benefit while a covered person needs care, but not while that person is healthy. The model must therefore distinguish health status as well as death. Its states and permitted transitions describe what changes are possible; the intensities quantify how quickly those changes occur.

The central construction is a state list followed by a transition list. Only after both are clear should rates be placed into a matrix. This prevents a common modeling error: putting a plausible-looking rate in the wrong position or allowing a change of status that the assumptions exclude. The method is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Consider a model with states \(H\) (healthy), \(C\) (care-needing), and \(D\) (dead). A person occupies exactly one state at a time. In the simple model developed first here, the permitted transitions are \(H\to C\), \(H\to D\), and \(C\to D\). There is no recovery transition \(C\to H\). State \(D\) is absorbing: once entered, it cannot be left.

These state descriptions are modeling assumptions, not universal descriptions of long-term care. “Care-needing” must be defined in a way that fits the hypothetical contract or population being modeled; for instance, it might mean meeting a specified care-eligibility criterion. The model also assumes that health and care-needing statuses do not overlap. If recovery is possible and relevant, it must be included as an allowed transition rather than silently treated as impossible.

From Permitted Transitions to the Intensity Matrix

Use the state order \(H,C,D\), with rows representing the state currently occupied and columns representing the destination state. Write \(q_{ij}(t)\) for the transition intensity from state \(i\) to a different state \(j\), at time \(t\). An intensity is a rate per unit time; the time unit must be stated. For example, an intensity expressed per year must be paired with time measured in years.

For each allowed transition, put its intensity in the corresponding off-diagonal position. Put zero in an off-diagonal position for a transition forbidden throughout the model. The diagonal entry in each row is the negative of the sum of the intensities of all permitted exits from that state. The resulting matrix \(Q(t)\) is called an intensity matrix or generator matrix.

$$ Q(t)= \begin{pmatrix} -\bigl(q_{HC}(t)+q_{HD}(t)\bigr) & q_{HC}(t) & q_{HD}(t)\\ 0 & -q_{CD}(t) & q_{CD}(t)\\ 0 & 0 & 0 \end{pmatrix}. $$

The first row has two possible exits from health: entering care-needing status or dying. The second row has only one permitted exit, death. The third row has none. Consequently, the diagonal is negative in the two transient states and zero in the absorbing state. Every row sums to zero, which provides a quick arithmetic check on the construction.

Important distinction: A forbidden transition has intensity zero throughout the model’s time interval, as specified by the model. A permitted transition may have intensity zero at particular times. Therefore, a zero off-diagonal value at one instant does not, by itself, establish that the transition is structurally forbidden.

Why the Diagonal Is the Negative Total Exit Intensity

The diagonal entries are not additional transitions. They are determined by the off-diagonal intensities. To see why, consider a short interval of length \(h\) in a state with outgoing intensities \(q_{ij}(t)\). Under the usual continuous-time intensity model, the probability of a particular permitted exit \(i\to j\) during that interval is \(q_{ij}(t)h+o(h)\), where \(o(h)\) denotes a term that becomes negligible compared with \(h\) as \(h\) tends to zero. The chance of remaining in state \(i\) must make the total probability one.

Adding the first-order probabilities of all permitted exits gives \(\bigl(\sum_{j\ne i}q_{ij}(t)\bigr)h+o(h)\). Thus the probability of remaining in \(i\) is \(1-\bigl(\sum_{j\ne i}q_{ij}(t)\bigr)h+o(h)\). The instantaneous rate of change of that stay probability at \(h=0\) is the negative sum of the outgoing intensities. This gives the diagonal rule.

$$ q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t), \qquad \sum_j q_{ij}(t)=0. $$

This construction assumes finite, nonnegative transition intensities and the stated state structure. If a model allows a transition not included in the simple long-term care structure, the permitted-transition list and matrix must be revised before applying the diagonal rule.

Worked Matrix Constructions

Worked Example: Constant Intensities With No Recovery

Problem. A hypothetical model uses the states \(H,C,D\) in that order. Time is measured in years. A healthy person enters care-needing status with constant intensity \(0.04\) per year and dies with constant intensity \(0.01\) per year. A care-needing person dies with constant intensity \(0.12\) per year. Recovery is excluded. Construct and check the intensity matrix.

Method. First map each stated rate to its permitted source-to-destination transition. Then set forbidden positions to zero and compute each diagonal as the negative total exit intensity. This avoids treating the rate of entering care as the only exit from health.

From \(H\), the permitted exits are \(H\to C\) at \(0.04\) and \(H\to D\) at \(0.01\). The total exit intensity from \(H\) is \(0.04+0.01=0.05\), so \(q_{HH}=-0.05\). From \(C\), the only permitted exit is \(C\to D\), with intensity \(0.12\), so \(q_{CC}=-0.12\). State \(D\) has no exit, so its row is all zeros.

$$ Q= \begin{pmatrix} -0.05 & 0.04 & 0.01\\ 0 & -0.12 & 0.12\\ 0 & 0 & 0 \end{pmatrix} \quad\text{per year}. $$

Check. The row sums are \(-0.05+0.04+0.01=0\), \(0-0.12+0.12=0\), and \(0\). The off-diagonal positions match the permitted transitions, and the diagonal entries are the negative total exit intensities. The matrix is consistent with all the stated assumptions.

Worked Example: A Permitted Rate That Is Zero at One Time

Problem. In a hypothetical model, the states are \(H,C,D\) and time \(t\) is measured in years from the start of observation. On \(0\le t\le4\), the permitted intensity from \(H\) to \(C\) is \(0.01t\) per year. The intensity from \(H\) to \(D\) is \(0.02\) per year, and the intensity from \(C\) to \(D\) is \(0.10\) per year. No recovery is allowed. Construct \(Q(t)\), then evaluate it at \(t=0\) and \(t=4\).

Method. Keep the allowed transition \(H\to C\) in the model because it is specified as permitted over the interval, even though its intensity is zero at \(t=0\). This avoids mistaking a momentary zero for a forbidden transition.

At general time \(t\), the total exit intensity from \(H\) is \(0.01t+0.02\), so \(q_{HH}(t)=-(0.01t+0.02)\). The \(C\)-state exit intensity is \(0.10\), giving \(q_{CC}(t)=-0.10\). Death remains absorbing.

$$ Q(t)= \begin{pmatrix} -(0.01t+0.02) & 0.01t & 0.02\\ 0 & -0.10 & 0.10\\ 0 & 0 & 0 \end{pmatrix} \quad\text{per year}. $$

At \(t=0\), \(q_{HC}(0)=0.01(0)=0\), and \(q_{HH}(0)=-(0+0.02)=-0.02\). At \(t=4\), \(q_{HC}(4)=0.01(4)=0.04\), and \(q_{HH}(4)=-(0.04+0.02)=-0.06\). Therefore:

$$ Q(0)= \begin{pmatrix} -0.02 & 0 & 0.02\\ 0 & -0.10 & 0.10\\ 0 & 0 & 0 \end{pmatrix}, \qquad Q(4)= \begin{pmatrix} -0.06 & 0.04 & 0.02\\ 0 & -0.10 & 0.10\\ 0 & 0 & 0 \end{pmatrix} \quad\text{per year}. $$

Check. Each row sums to zero at both times. In particular, the zero in the \(H,C\) position of \(Q(0)\) is only the value of a permitted time-varying intensity at the endpoint. It does not change the model’s stated transition structure.

Worked Example: Adding Recovery When the Model Allows It

Problem. A different hypothetical care model allows recovery. Its states are \(H\) (healthy), \(C\) (currently needing care), and \(D\) (dead), in that order. All intensities are constant and measured per year. From \(H\), the intensities to \(C\) and \(D\) are \(0.025\) and \(0.008\). From \(C\), the intensities to \(H\) and \(D\) are \(0.06\) and \(0.14\). Death is absorbing. Construct and check the matrix.

Method. Use only the transitions stated for this version of the model, including the recovery transition \(C\to H\). This avoids carrying over the no-recovery assumption from a different model.

The total exit intensity from \(H\) is \(0.025+0.008=0.033\), so \(q_{HH}=-0.033\). The total exit intensity from \(C\) is \(0.06+0.14=0.20\), so \(q_{CC}=-0.20\). Since death is absorbing, its row is zero. Placing the rates in row-source, column-destination positions gives:

$$ Q= \begin{pmatrix} -0.033 & 0.025 & 0.008\\ 0.06 & -0.20 & 0.14\\ 0 & 0 & 0 \end{pmatrix} \quad\text{per year}. $$

Check. The row sums are \(-0.033+0.025+0.008=0\), \(0.06-0.20+0.14=0\), and \(0\). The positive entry in row \(C\), column \(H\) represents recovery. The first model’s no-recovery restriction does not apply here because this example explicitly permits recovery.

Modeling Choices and Common Traps

The state labels alone do not determine the transition structure. In the simple no-recovery model, the allowed routes out of health are entering care-needing status and death, while the only allowed route out of care-needing status is death. A model that allows recovery has a different structure. Similarly, if the contract or population distinguishes different care levels, those levels may require separate states; combining them into one state is appropriate only if the distinction is unnecessary for the quantities being modeled.

When a rate depends on time, write the time argument explicitly. For example, \(q_{HC}(t)\) indicates that the intensity may change with calendar time \(t\). A matrix with constant entries describes a time-homogeneous intensity model; a matrix whose entries vary with \(t\) describes a time-inhomogeneous model. In either case, diagonal entries must use the rates at the same time as the off-diagonal entries.

A frequent exam trap is to interpret an observed zero as proof that a transition is forbidden. That conclusion is valid only when the model says the intensity is zero throughout the relevant interval. In a time-dependent model, an allowed rate can be zero at a particular time and positive at other times. The structural transition list, not a single matrix value, determines whether a route is forbidden.

  • Check the state order. State the order once and use it consistently for both rows and columns.
  • Check the direction. A rate for recovery belongs in row \(C\), column \(H\), not row \(H\), column \(C\).
  • Check total exits. The magnitude of each diagonal entry is the sum of all outgoing rates from that state.
  • Check units and time arguments. Rates per year require time in years; time-varying rates should be evaluated at the stated time.
  • Check the assumptions. Recovery and other routes appear only if the model permits them; death is absorbing only under the stated model assumption.
Key takeaway: Define the states and permitted routes first. Place each permitted intensity by source row and destination column, set structurally forbidden routes to zero, and make each diagonal the negative sum of that row’s outgoing intensities. A permitted transition can still have zero intensity at a particular time.

Check Your Understanding

For each problem, use the stated state order and time units. Check that every row of your proposed intensity matrix sums to zero.

  1. A no-recovery model has states \(H,C,D\). Per year, \(H\to C\) has intensity \(0.03\), \(H\to D\) has intensity \(0.02\), and \(C\to D\) has intensity \(0.09\). Write the matrix. (Answer: \(\begin{pmatrix}-0.05&0.03&0.02\\0&-0.09&0.09\\0&0&0\end{pmatrix}\) per year.)
  2. On \(0\le t\le6\), a permitted \(H\to C\) intensity is \(0.005t\) per year, \(H\to D\) is \(0.01\), and \(C\to D\) is \(0.08\). Give \(Q(t)\) and \(Q(0)\). (Answer: \(Q(t)=\begin{pmatrix}-(0.005t+0.01)&0.005t&0.01\\0&-0.08&0.08\\0&0&0\end{pmatrix}\); \(Q(0)=\begin{pmatrix}-0.01&0&0.01\\0&-0.08&0.08\\0&0&0\end{pmatrix}\), per year; \(H\to C\) is still permitted.)
  3. A model uses states \(H,C,D\) and permits \(H\to C\), \(H\to D\), \(C\to H\), and \(C\to D\). The corresponding constant intensities per year are \(0.04,0.01,0.03,0.12\). Construct the matrix. (Answer: \(\begin{pmatrix}-0.05&0.04&0.01\\0.03&-0.15&0.12\\0&0&0\end{pmatrix}\) per year.)
  4. A no-recovery model has \(q_{HC}(t)=0.02\), \(q_{HD}(t)=0.01t\), and \(q_{CD}(t)=0.06+0.01t\), with \(t\) in years. Find \(Q(3)\), and state why the \(H\to D\) route is not forbidden at \(t=0\). (Answer: \(Q(3)=\begin{pmatrix}-0.05&0.02&0.03\\0&-0.09&0.09\\0&0&0\end{pmatrix}\) per year; \(H\to D\) is permitted by the model, although its intensity is zero at \(t=0\).)
  5. A proposed matrix in state order \(H,C,D\) has first row \((-0.04,0.03,0.02)\), second row \((0,-0.10,0.10)\), and third row \((0,0,0)\), all per year. Identify the error and correct the first row if the stated \(H\to C\) and \(H\to D\) rates are \(0.03\) and \(0.02\). (Answer: The first row sums to \(0.01\), not zero; its diagonal must be \(-(0.03+0.02)=-0.05\), so the corrected first row is \((-0.05,0.03,0.02)\).)
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