Building an Active-Disabled-Dead Intensity Matrix

Why State Order Matters

A model of a hypothetical disability income contract may distinguish three conditions: active, disabled, and dead. An active person may become disabled or die; a disabled person may recover to the active state or die. The rates are not interchangeable: incidence starts in the active state, recovery starts in the disabled state, and each mortality intensity is attached to the state occupied immediately before death.

A matrix makes these directions explicit, but only if its state order is stated. In this tutorial, rows identify the state being left and columns identify the state being entered. We use the order \(A,D,\partial\), where \(A\) denotes active, \(D\) denotes disabled, and \(\partial\) denotes dead. All intensities below are per year unless stated otherwise. This three-state construction is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Let \(\alpha(t)\) be the incidence intensity from active to disabled, \(\rho(t)\) the recovery intensity from disabled to active, \(\mu_A(t)\) the mortality intensity while active, and \(\mu_D(t)\) the mortality intensity while disabled. With row states and column destinations ordered \(A,D,\partial\), the intensity matrix is

\mathbf{Q}(t)= \begin{pmatrix} -\bigl(\alpha(t)+\mu_A(t)\bigr) & \alpha(t) & \mu_A(t)\\ \rho(t) & -\bigl(\rho(t)+\mu_D(t)\bigr) & \mu_D(t)\\ 0 & 0 & 0 \end{pmatrix}.

The dead state is absorbing: once entered, no further transitions occur in this model.

The off-diagonal entries describe possible transitions. Thus \(q_{AD}(t)=\alpha(t)\) is incidence, \(q_{DA}(t)=\rho(t)\) is recovery, \(q_{A\partial}(t)=\mu_A(t)\) is active-state mortality, and \(q_{D\partial}(t)=\mu_D(t)\) is disabled-state mortality. The zero entries encode transitions excluded by the model: for example, there is no direct transition from active to dead that is separate from the active mortality transition, and no transition out of death.

The diagonal entries are not additional transition rates. Each is the negative of the total intensity of leaving that row's state. From active, the available exits are incidence and active-state death; from disabled, they are recovery and disabled-state death. Death has no exits, so its diagonal entry is zero.

Deriving the Diagonal Entries

The diagonal rule follows from the definition of an intensity matrix. For distinct states \(i\) and \(j\), the transition intensity at time \(t\) is the instantaneous rate of moving from \(i\) to \(j\), conditional on being in \(i\) immediately before \(t\). For a small positive interval \(h\), the probability of that transition is \(q_{ij}(t)h+o(h)\). The probability of remaining in state \(i\) is therefore one minus the sum of the probabilities of all exits, to first order in \(h\).

P(\text{remain in }i\text{ over }[t,t+h)\mid X(t)=i) =1-\left(\sum_{j\ne i}q_{ij}(t)\right)h+o(h).

The coefficient of \(h\) in the rate of leaving \(i\) is the total exit intensity. Consequently, the diagonal element must be the negative of that total. This gives zero row sums, including for an absorbing state whose total exit intensity is zero.

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

This argument assumes a continuous-time Markov model with finite, nonnegative transition intensities at the time under consideration. In particular, the next-state transition rates depend on the current state and the specified time, not on omitted information such as unrecorded duration in the state. If duration affects a rate and is not otherwise represented, this three-state intensity description may not capture the intended model.

Worked Example: Building a Constant-Intensity Matrix

A hypothetical contract model has incidence intensity \(0.03\) per year, active mortality intensity \(0.01\) per year, recovery intensity \(0.20\) per year, and disabled mortality intensity \(0.05\) per year. Construct the matrix in the order \(A,D,\partial\), and check its rows.

Method. First place each rate in the off-diagonal cell named by its source and destination. Then make each diagonal the negative sum of that row's off-diagonal entries. This avoids treating recovery as an exit from the active state.

The active row has exits to disabled and dead. Its total exit intensity is \(0.03+0.01=0.04\), so its diagonal entry is \(-0.04\). The disabled row has exits to active and dead, with total \(0.20+0.05=0.25\), so its diagonal entry is \(-0.25\). The dead state is absorbing.

\mathbf{Q}= \begin{pmatrix} -0.04 & 0.03 & 0.01\\ 0.20 & -0.25 & 0.05\\ 0 & 0 & 0 \end{pmatrix} \quad\text{per year}.

Check. The row sums are \(-0.04+0.03+0.01=0\), \(0.20-0.25+0.05=0\), and \(0+0+0=0\). Every off-diagonal entry is nonnegative, and each stated rate appears in the cell with the correct source and destination.

Reading and Evaluating Time-Dependent Rates

The same state structure applies when an intensity varies with calendar time. In that case, \(\mathbf{Q}(t)\) is evaluated at the specified time by first evaluating each rate function and then forming the diagonal entries from the rates at that same time. A rate measured in years must be evaluated using \(t\) in years; changing the time unit without changing the rate functions would change the model.

Worked Example: Evaluating a Matrix at a Specified Time

In a hypothetical model, time \(t\) is measured in years from the start of coverage. The intensities are \(\alpha(t)=0.02+0.003t\), \(\mu_A(t)=0.005+0.001t\), \(\rho(t)=0.12\), and \(\mu_D(t)=0.03+0.002t\), all per year. Find \(\mathbf{Q}(5)\).

Method. Evaluate the four functions at \(t=5\), retain the state-specific destinations, and calculate the diagonals as negative total exit intensities.

The rates at year 5 are

\begin{aligned} \alpha(5)&=0.02+0.003(5)=0.035,\\ \mu_A(5)&=0.005+0.001(5)=0.010,\\ \rho(5)&=0.12,\\ \mu_D(5)&=0.03+0.002(5)=0.040. \end{aligned}

The active exit intensity is \(0.035+0.010=0.045\). The disabled exit intensity is \(0.12+0.040=0.160\). Therefore,

\mathbf{Q}(5)= \begin{pmatrix} -0.045 & 0.035 & 0.010\\ 0.120 & -0.160 & 0.040\\ 0 & 0 & 0 \end{pmatrix} \quad\text{per year}.

Check. The active row sums to \(-0.045+0.035+0.010=0\), and the disabled row sums to \(0.120-0.160+0.040=0\). The mortality entries match the functions for the states from which death occurs.

Matching Each Rate to Its Source State

A reliable construction starts with a verbal transition description, not with the diagonal. Ask, for each rate: from which state does the transition begin, and which state does it enter? Incidence is \(A\to D\), recovery is \(D\to A\), and death can be \(A\to\partial\) or \(D\to\partial\), with a potentially different mortality intensity for each source state.

Fix the state order.
Write the same ordered list for the rows and columns, such as \(A,D,\partial\).

Place allowed transitions.
Enter incidence at row \(A\), column \(D\); recovery at row \(D\), column \(A\); and each mortality intensity in the death column of its source-state row.

Encode excluded transitions.
Use zero for a transition that the stated model does not allow, including every exit from the absorbing dead state.

Complete and check the diagonals.
For each row, take the negative sum of its off-diagonal intensities and verify that the full row sums to zero.

Worked Example: Auditing a Proposed Matrix

A hypothetical model specifies incidence \(0.025\) per year, active mortality \(0.005\) per year, recovery \(0.18\) per year, and disabled mortality \(0.04\) per year. An analyst proposes

\widetilde{\mathbf{Q}}= \begin{pmatrix} -0.030 & 0.025 & 0.005\\ 0.180 & -0.220 & 0.040\\ 0 & 0 & 0 \end{pmatrix}.

Determine whether the proposal is consistent with the stated rates and, if not, correct it.

Method. Read the off-diagonal entries by row and column, then compare each row's diagonal with the negative total of its exits. This distinguishes a correct-looking row sum from rates that actually agree with the description.

The active row is consistent: its incidence entry is \(0.025\), its active mortality entry is \(0.005\), and its diagonal is \(-(0.025+0.005)=-0.030\). In the disabled row, recovery \(0.18\) and disabled mortality \(0.04\) give total exits \(0.18+0.04=0.22\), so the diagonal \(-0.220\) is correct. The proposed matrix is consistent.

Check. The active row sums to \(-0.030+0.025+0.005=0\), and the disabled row sums to \(0.180-0.220+0.040=0\). The entry \(0.040\) is disabled mortality because it is in the disabled row and dead column; \(0.005\) is active mortality because it is in the active row and dead column.

Common Exam Trap: A Correct Row Sum Is Not Enough

A matrix can have zero row sums and still encode the wrong model. For example, using the data in the audit example, suppose an analyst swaps the mortality intensities between the two source states and writes

\begin{pmatrix} -0.065 & 0.025 & 0.040\\ 0.180 & -0.185 & 0.005\\ 0 & 0 & 0 \end{pmatrix}.

Both transient rows still sum to zero: \(-0.065+0.025+0.040=0\) and \(0.180-0.185+0.005=0\). But the matrix assigns active mortality \(0.040\) and disabled mortality \(0.005\), contrary to the given rates. The correction is to place \(0.005\) in the active-to-dead cell and \(0.040\) in the disabled-to-dead cell, then set the diagonals to \(-0.030\) and \(-0.220\), respectively.

A second frequent error is to include only mortality in a transient state's diagonal. That would give an active diagonal of \(-0.005\) even though incidence is also an exit. The correct active diagonal is \(-(0.025+0.005)=-0.030\). Always sum every allowed exit from the state, not only exits that are benefits or events of primary interest.

Key takeaway: In the order \(A,D,\partial\), incidence and active mortality belong in the active row; recovery and disabled mortality belong in the disabled row; and the dead row is zero. A valid matrix also has nonnegative off-diagonal intensities and each diagonal equal to the negative sum of its row's other entries.

Check Your Understanding

For each problem, use the stated order \(A,D,\partial\). All rates are per year unless stated otherwise.

A model has incidence \(0.04\), active mortality \(0.01\), recovery \(0.25\), and disabled mortality \(0.06\). Write its intensity matrix. (Answer: \(\begin{pmatrix}-0.05&0.04&0.01\\0.25&-0.31&0.06\\0&0&0\end{pmatrix}\).)
Let \(\alpha(t)=0.01+0.002t\), \(\mu_A(t)=0.004\), \(\rho(t)=0.10\), and \(\mu_D(t)=0.02+0.001t\). Find \(\mathbf{Q}(3)\). (Answer: \(\begin{pmatrix}-0.020&0.016&0.004\\0.100&-0.123&0.023\\0&0&0\end{pmatrix}\).)
A proposed matrix is \(\begin{pmatrix}-0.07&0.05&0.02\\0.12&-0.16&0.04\\0&0&0\end{pmatrix}\). Identify incidence, active mortality, recovery, and disabled mortality, and verify the row sums. (Answer: incidence \(0.05\), active mortality \(0.02\), recovery \(0.12\), disabled mortality \(0.04\); all row sums are zero.)
A model specifies incidence \(0.03\), active mortality \(0.02\), recovery \(0.14\), and disabled mortality \(0.03\). An analyst uses diagonal entries \(-0.02\) and \(-0.17\). Identify the error and give the correct diagonals. (Answer: the active diagonal omits incidence; the correct diagonals are \(-0.05\) and \(-0.17\).)
At time \(t=4\), suppose \(\alpha(t)=0.02+0.005t\), \(\mu_A(t)=0.006+0.001t\), \(\rho(t)=0.16\), and \(\mu_D(t)=0.04+0.005t\). Construct \(\mathbf{Q}(4)\) and check its transient row sums. (Answer: \(\begin{pmatrix}-0.052&0.040&0.012\\0.160&-0.220&0.060\\0&0&0\end{pmatrix}\); the transient row sums are both zero.)

What You'll Learn