Constructing the Disability Model Generator

Why the Generator Matters

A disability model needs to describe not only which states a person can occupy, but also how quickly the model allows movement between them. The generator matrix collects those instantaneous transition rates in one object. Its off-diagonal entries represent possible transitions; its diagonal entries ensure that probability is conserved across all states.

This tutorial uses three states: active, disabled, and dead. It focuses on constructing the generator from incidence, recovery, and mortality intensities. The generator describes local transition behavior; by itself, it is not a table of probabilities over a finite period. This construction is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: For a continuous-time Markov model with states $1,\ldots,n$, the generator entry $q_{ij}(t)$, for $i\ne j$, is the instantaneous intensity of transition from state $i$ to state $j$ at time $t$: $$ q_{ij}(t)=\lim_{h\downarrow 0}\frac{P(X(t+h)=j\mid X(t)=i)}{h}. $$ The diagonal entry is defined so that each row sums to zero: $$ q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t). $$

States, Allowed Transitions, and Matrix Layout

Use the state order active, disabled, dead for both rows and columns. Assume the only permitted transitions are active to disabled, active to dead, disabled to active, and disabled to dead. Death is absorbing: once dead, the model has no transition to another state.

Write $\alpha(t)$ for the active-to-disabled incidence intensity, $\mu_A(t)$ for mortality while active, $\rho(t)$ for the disabled-to-active recovery intensity, and $\mu_D(t)$ for mortality while disabled. With this state order, the generator is

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

An entry in row $i$, column $j$ describes movement from state $i$ to state $j$. Thus the incidence intensity belongs in the active row and disabled column. The disabled-state mortality intensity belongs in the disabled row and dead column. The zeros in the dead row express absorption, not an estimate that is merely small.

The quantities on the diagonal are not additional transition intensities. They are negative sums of the rates for leaving the row's state. They are generally negative, while off-diagonal generator entries are nonnegative. If a transition is structurally impossible under the model, its off-diagonal entry is zero. If the model allows another transition, such as disabled directly to a different living state, the state space and generator must be changed rather than silently omitting that transition.

Why the Diagonal Is the Negative Outflow

The diagonal rule follows from the fact that, conditional on occupying a state at time $t$, the probabilities of being in all possible states at time $t+h$ sum to one. For a fixed starting state $i$, the probability of remaining in $i$ over a short interval has first-order expansion $1+q_{ii}(t)h+o(h)$. The probability of moving to each other state $j$ has expansion $q_{ij}(t)h+o(h)$. Adding these terms gives

1+\left(q_{ii}(t)+\sum_{j\ne i}q_{ij}(t)\right)h+o(h)=1.

Subtract 1, divide by $h$, and let $h$ decrease to zero. The terms represented by $o(h)$, divided by $h$, tend to zero. Therefore

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

For example, a person who is active can leave that state through incidence or active mortality. The total first-order rate of leaving active is therefore $\alpha(t)+\mu_A(t)$, and the active diagonal is its negative. Recovery is not included in that sum because recovery leaves the disabled state, not the active state.

Worked Example: Constant Annual Intensities

A hypothetical income-protection model has constant continuous-time intensities, measured per year: active-to-disabled incidence $0.080$, active mortality $0.012$, disabled-to-active recovery $0.300$, and disabled mortality $0.045$. Assume a time-homogeneous Markov model with no other transitions, and use exponential holding times implied by these constant intensities. No interest or benefit valuation is involved. Construct and check the generator.

Step 1: Fix the state order and place the off-diagonal rates. Use active, disabled, dead for both rows and columns. The active-to-disabled entry is $0.080$, and the active-to-dead entry is $0.012$. The disabled-to-active entry is $0.300$, and the disabled-to-dead entry is $0.045$. The dead row is zero because death is absorbing.

Step 2: Derive the active diagonal. The total intensity of leaving active is $0.080+0.012=0.092$ per year. Thus $q_{AA}=-0.092$.

Step 3: Derive the disabled diagonal. The total intensity of leaving disabled is $0.300+0.045=0.345$ per year. Thus $q_{DD}=-0.345$.

Step 4: Assemble and check. The generator, with all entries per year, is

Q= \begin{pmatrix} -0.092 & 0.080 & 0.012\\ 0.300 & -0.345 & 0.045\\ 0 & 0 & 0 \end{pmatrix}.

The active row sums to $-0.092+0.080+0.012=0$. The disabled row sums to $0.300-0.345+0.045=0$, and the dead row sums to zero. Every off-diagonal entry is nonnegative, as required.

Method and mistake avoided. Place the given intensities first, then derive each diagonal from that row's outgoing rates. A common error is to put the recovery intensity into the active diagonal; recovery is an outflow from disabled, so it belongs in the disabled row's sum.

Time-Varying Intensities

The generator need not be constant. If intensities vary with age, calendar time, or time since entry into the model, the same construction applies at each time $t$. Each off-diagonal entry is the intensity at that time, and each diagonal is the negative sum of the other entries in its row at that same time. The rates must use a common unit of time.

For a time-varying model, the Markov assumption means that the current state and the specified time-dependent intensities contain the information needed to describe transitions from that point onward. If transition rates also depend on unmodeled past history, such as the number of earlier disability spells, the three-state model may not be sufficient without adding states or otherwise representing that history.

Worked Example: Generator at a Specified Duration

A hypothetical model measures time $t$ in years since coverage began. For $t\ge 0$, the incidence intensity is $\alpha(t)=0.060+0.010t$ per year, active mortality is $\mu_A(t)=0.010$ per year, recovery is $\rho(t)=0.240$ per year, and disabled mortality is $\mu_D(t)=0.060+0.010t$ per year. Assume these are the complete transition intensities in the active-disabled-dead model, with death absorbing. The rates are deterministic functions of duration; no interest or cash-flow distribution is involved. Find the generator at $t=2$ years.

Step 1: Evaluate each intensity at $t=2$. Incidence is $0.060+0.010(2)=0.080$ per year. Active mortality remains $0.010$ per year. Recovery remains $0.240$ per year. Disabled mortality is $0.060+0.010(2)=0.080$ per year.

Step 2: Sum outgoing rates by starting state. The active-state outflow is $0.080+0.010=0.090$ per year, so its diagonal is $-0.090$. The disabled-state outflow is $0.240+0.080=0.320$ per year, so its diagonal is $-0.320$. The dead-state outflow is zero.

Step 3: Write the generator and verify it.

Q(2)= \begin{pmatrix} -0.090 & 0.080 & 0.010\\ 0.240 & -0.320 & 0.080\\ 0 & 0 & 0 \end{pmatrix}.

The row sums are $-0.090+0.080+0.010=0$, $0.240-0.320+0.080=0$, and $0$. All off-diagonal entries are nonnegative. Each entry is evaluated at the same duration, $t=2$; mixing rates from different durations would not give the generator at a single point in time.

Method and mistake avoided. Evaluate the time-dependent rates before taking row sums. A frequent error is to leave the diagonal at its value at $t=0$ while updating the off-diagonal intensities at $t=2$. That would violate the zero-row-sum condition.

Units and the Meaning of an Intensity

An intensity has units of inverse time. A rate of $0.006$ per month is not $0.006$ per year. To express a monthly intensity on an annual scale, multiply by 12. This conversion is valid for an intensity expressed per unit time; do not treat a finite-period transition probability as if it were an intensity. In particular, a one-month probability of disability is not automatically a monthly intensity.

Under constant intensities, the holding time in a state has an exponential distribution with parameter equal to the sum of the intensities for leaving that state. For an active person in this model, that parameter is $\alpha+\mu_A$; for a disabled person, it is $\rho+\mu_D$. The destination of the transition is selected in proportion to the corresponding outgoing intensity. These facts explain why the diagonal uses the total outflow, but do not make an intensity itself a finite-period probability.

Worked Example: Converting Monthly Intensities to Annual Units

A hypothetical disability model is supplied in continuous-time intensities per month: incidence $0.006$, active mortality $0.001$, recovery $0.025$, and disabled mortality $0.004$. Assume constant intensities, exponential holding times, no other transitions, and an absorbing dead state. Convert the model to a per-year generator. No interest rate or benefit payments are specified or needed.

Step 1: Convert every intensity using the same time scale. Multiply each monthly rate by 12. Incidence is $12(0.006)=0.072$ per year, and active mortality is $12(0.001)=0.012$ per year. Recovery is $12(0.025)=0.300$ per year, and disabled mortality is $12(0.004)=0.048$ per year.

Step 2: Find the row outflows. From active, the total outflow is $0.072+0.012=0.084$ per year, so the active diagonal is $-0.084$. From disabled, the total outflow is $0.300+0.048=0.348$ per year, so the disabled diagonal is $-0.348$. The dead state has no outflow.

Step 3: Assemble and check the per-year generator.

Q_{\text{year}}= \begin{pmatrix} -0.084 & 0.072 & 0.012\\ 0.300 & -0.348 & 0.048\\ 0 & 0 & 0 \end{pmatrix}.

The row sums are $-0.084+0.072+0.012=0$, $0.300-0.348+0.048=0$, and $0$. The units in every entry are now per year.

Method and mistake avoided. Convert all the off-diagonal intensities first, then derive the diagonals in the converted unit. Mixing monthly off-diagonal rates with an annual diagonal would produce inconsistent units even if the numerical row sum appeared plausible.

Model Checks and Common Traps

Before using a proposed generator, check its structure rather than relying only on arithmetic. For the three-state model, the following checks are useful:

Every off-diagonal entry is nonnegative.
Each row sums to zero, including the absorbing dead row.
Each nonzero off-diagonal entry matches a permitted transition and has the correct direction.
All intensities in a given generator use the same time unit.
The diagonal is the negative sum of the off-diagonal entries in that row, not a separate mortality or transition intensity.

A common exam trap is confusing a transition intensity with a probability. The quantity $\alpha(t)$ is a rate of active-to-disabled transition per unit time, not the probability of becoming disabled over a year. The generator alone does not justify replacing that intensity by a one-year probability. Another trap is reversing the row and column interpretation. With the convention used here, rows are starting states and columns are destination states; changing that convention requires transposing the layout and being consistent throughout.

Key takeaway: Put each permitted transition intensity in its starting-state row and destination-state column. Then make each diagonal the negative sum of the other entries in its row. Confirm state order, units, and row sums before using the generator.

Check Your Understanding

For each question, use the active-disabled-dead state order and assume death is absorbing unless the problem says otherwise.

A constant model has incidence $0.050$, active mortality $0.010$, recovery $0.200$, and disabled mortality $0.030$, all per year. State the generator. (Answer: $\begin{pmatrix}-0.060&0.050&0.010\\0.200&-0.230&0.030\\0&0&0\end{pmatrix}$, per year.)
A model has active-to-disabled intensity $0.004$ per month and active mortality intensity $0.001$ per month. What is the active diagonal in a per-year generator? (Answer: $-0.060$ per year.)
At a specified time, the disabled row has recovery intensity $0.180$ per year and disabled mortality intensity $0.070$ per year. What is its diagonal, and what is the row sum after insertion? (Answer: diagonal $-0.250$ per year; row sum $0$.)
At duration $t$ years, $\alpha(t)=0.040+0.005t$, $\mu_A(t)=0.010$, $\rho(t)=0.160$, and $\mu_D(t)=0.050$, all per year. Give the generator at $t=4$. (Answer: $\begin{pmatrix}-0.070&0.060&0.010\\0.160&-0.210&0.050\\0&0&0\end{pmatrix}$, per year.)
A proposed disabled row is $(0.150,-0.190,0.050)$, with entries per year and the stated state order. Is it a valid generator row? If not, correct the diagonal. (Answer: No; the row sums to $0.010$, so the diagonal must be $-0.200$, giving $(0.150,-0.200,0.050)$.)

What You'll Learn