Why Transition Intensities Need a Time Scale
In a long-term care model, a transition rate may change as a person gets older, as a policy remains in force, or for both reasons. These are distinct modelling assumptions. A care-entry rate that rises with attained age reflects the insured’s age at the time of transition. A rate that rises with policy duration reflects how long the contract has been in force. The two time scales move together for an individual policy, but they do not have the same meaning.
Let \(x\) be age at policy issue and \(t\) be time since issue, measured in years. Attained age at time \(t\) is \(x+t\). For states \(r\) and \(s\), write \(\lambda_{rs}(x,t)\) for the instantaneous transition intensity from \(r\) to \(s\) at policy duration \(t\). An age-only model has the form \(\lambda_{rs}(x+t)\); a duration-only model has the form \(\lambda_{rs}(t)\). A model allowing both effects uses both arguments.
Assume a continuous-time Markov model with a specified set of states. Conditional on the current state and the known age-duration schedule, future transition behaviour does not depend on the earlier path. At each age and duration, intensities must be nonnegative. The intensity matrix \(Q_x(t)\) has off-diagonal entry \(\lambda_{rs}(x,t)\), and each diagonal entry is the negative sum of the other entries in its row. Thus every row sums to zero.
If transition rates depend on how long the person has been in the current care state, rather than only on age and policy duration, the stated Markov assumption may not hold with the existing state set. One approach is to add time in state as a model variable; another is to use a semi-Markov formulation. Do not silently treat duration since issue as duration in a care state.
Age Effects, Duration Effects, and Their Combination
An attained-age model assigns rates by current age. For example, an incidence intensity might be lower before age 65 and higher from age 65 onward. A duration model assigns rates by time since issue: a rate might change after the first few policy years even if age is not explicitly used. In a combined model, a transition may depend on both variables, such as increasing with age and receiving an additional increase after a duration threshold.
The distinction matters in both assumptions and calculations. In one policy, age is \(x+t\), so age and duration advance together. A rate specified as a function of age must be evaluated at \(x+t\), whereas a duration-only rate must be evaluated at \(t\). Across policies issued at different ages, the same duration corresponds to different attained ages. This difference can help distinguish the intended assumption and is a reason to state the time scale explicitly.
Worked Example: Age-Banded Care Entry with Competing Mortality
A hypothetical insured is age 60 at issue and begins in the active state. During policy year 1, the intensity of transition from active to care is 0.020 per year and the active-state mortality intensity is 0.010 per year. During policy year 2, the care-entry intensity is 0.040 per year and mortality remains 0.010. Once care or death occurs, the insured cannot return to active during this calculation. Find the probabilities of remaining active, entering care, and dying by duration 2.
Method. Treat each one-year interval as having constant competing intensities. If the two rates are \(\lambda_C\) for care and \(\lambda_D\) for death, the probability of no transition for one year is \(e^{-(\lambda_C+\lambda_D)}\). The probability that care is the first transition in that year is \(\frac{\lambda_C}{\lambda_C+\lambda_D}(1-e^{-(\lambda_C+\lambda_D)})\).
In year 1, the total active-state exit intensity is \(0.020+0.010=0.030\). The probability of remaining active through the year is \(e^{-0.030}=0.9704455\). The probability of entering care during year 1 is \( (0.020/0.030)(1-e^{-0.030})=0.0197030\). The probability of dying during year 1 is \( (0.010/0.030)(1-e^{-0.030})=0.0098515\). These three probabilities sum to 1.
An insured still active at duration 1 faces the year-2 rates. The probability of remaining active for the full second year, conditional on being active at its start, is \(e^{-0.050}=0.9512294\). Therefore the probability of remaining active to duration 2 from issue is \(0.9704455e^{-0.050}=e^{-0.080}=0.9231163\).
The conditional probability of care entry during year 2 is \( (0.040/0.050)(1-e^{-0.050})=0.0390165\). Allowing for survival in the active state through year 1, the unconditional contribution from year 2 is \(0.9704455(0.0390165)=0.0378634\). Total care-entry probability is \(0.0197030+0.0378634=0.0575664\).
Similarly, the conditional probability of death during year 2 is \( (0.010/0.050)(1-e^{-0.050})=0.0097541\). Its unconditional contribution is \(0.9704455(0.0097541)=0.0094658\). Total death probability is \(0.0098515+0.0094658=0.0193173\). The check is \(0.9231163+0.0575664+0.0193173=1.0000000\), up to rounding. Rounded to four decimals, the probabilities are 0.9231, 0.0576, and 0.0193.
The age bands are the source of the rate change: the insured ages from 60 to 62, and the care-entry intensity rises at attained age 61. The common mistake is to use the year-1 rate for both years; the second-year care intensity is 0.040, not 0.020.
Competing Intensities: A Short Derivation
The calculation above follows from the probability of staying in the starting state. Suppose the possible first transitions out of state \(r\) have intensities \(\lambda_{r1}(x,u),\ldots,\lambda_{rm}(x,u)\). Define the total exit intensity by \(\Lambda_r(x,u)=\sum_{j=1}^{m}\lambda_{rj}(x,u)\). Conditional on remaining in state \(r\) to time \(u\), the probability of a transition in the next small interval \(du\) is approximately \(\Lambda_r(x,u)du\). The resulting probability of no exit through time \(t\) is
For a particular cause \(j\), the probability of first exiting to state \(j\) in the small interval around \(u\) is the probability of still occupying \(r\) at \(u\), multiplied by the cause-\(j\) intensity and \(du\). Integrating these disjoint first-exit possibilities gives
This formula is for the first exit from a specified state. It does not by itself give the probability of occupying a later care state when paths can pass through intermediate states. Those probabilities require accounting for subsequent transitions as well.
Worked Example: A Policy-Duration Effect
A hypothetical policyholder begins in the active state. The active-to-care intensity depends on policy duration: it is 0.010 per year for durations \(0\) to \(1\), then 0.030 per year for durations \(1\) to \(2\). The active-state mortality intensity is 0.010 per year throughout. Assume constant rates within each interval and no return to active. Calculate the probabilities of care entry, death, and remaining active by duration 2.
Method. Use the competing-intensity formula separately in each duration interval, multiplying the second-interval probabilities by the probability of still being active at duration 1.
In the first interval, the total exit intensity is \(0.010+0.010=0.020\). The probability of remaining active to duration 1 is \(e^{-0.020}=0.9801987\). The probability of care entry in that interval is \( (0.010/0.020)(1-e^{-0.020})=0.0099007\), and the probability of death is also 0.0099007.
In the second interval, the total intensity is \(0.030+0.010=0.040\). Conditional on being active at duration 1, the probability of care entry is \( (0.030/0.040)(1-e^{-0.040})=0.0294079\). Thus the unconditional second-interval contribution to care is \(0.9801987(0.0294079)=0.0288256\). The total care-entry probability is \(0.0099007+0.0288256=0.0387263\).
The conditional probability of death in the second interval is \( (0.010/0.040)(1-e^{-0.040})=0.0098026\). Its unconditional contribution is \(0.9801987(0.0098026)=0.0096085\). Total death probability is \(0.0099007+0.0096085=0.0195092\). Remaining active at duration 2 is \(e^{-0.020}e^{-0.040}=e^{-0.060}=0.9417645\). The check is \(0.0387263+0.0195092+0.9417645=1.0000000\). Rounded to four decimals, the probabilities are 0.0387, 0.0195, and 0.9418.
This is a duration-only assumption: the change occurs at policy duration 1, not at a specified attained age. A common mistake is to describe the later rate as an age-65 rate without knowing the issue age. The duration schedule gives the same rate at duration 1 for policies issued at different ages; an age schedule need not.
Building a Combined Age-Duration Intensity Matrix
A combined model makes the dependence explicit. In the following example, \(x\) is issue age and \(t\) is duration, both in years. Let the states be active (0), moderate care (1), severe care (2), and dead (3). All intensities are per year. The functions below are hypothetical assumptions, not empirical estimates.
Worked Example: Evaluating Rates at a Given Age and Duration
Suppose the permitted transitions are \(0\to1\), \(0\to3\), \(1\to2\), \(1\to3\), and \(2\to3\). Let \(\lambda_{01}(x,t)=0.010+0.001(x+t-65)+0.002\max(t-3,0)\), \(\lambda_{03}(x,t)=0.004+0.0002(x+t-65)\), \(\lambda_{12}(x,t)=0.030+0.004\max(t-4,0)\), \(\lambda_{13}(x,t)=0.020\), and \(\lambda_{23}(x,t)=0.050\). Evaluate the intensity matrix for a policy issued at age \(x=65\), at duration \(t=5\), and estimate transition probabilities over the next 0.01 years using the small-interval definition.
Attained age is \(65+5=70\). Therefore, \(\lambda_{01}=0.010+0.001(70-65)+0.002(5-3)=0.019\), \(\lambda_{03}=0.004+0.0002(70-65)=0.005\), \(\lambda_{12}=0.030+0.004(5-4)=0.034\), \(\lambda_{13}=0.020\), and \(\lambda_{23}=0.050\).
The row for active has total exit intensity \(0.019+0.005=0.024\); the row for moderate care has total \(0.034+0.020=0.054\). The valid intensity matrix is
For an insured active at that instant, over \(h=0.01\) years the first-order probabilities are \(0.019(0.01)=0.00019\) for transition to moderate care and \(0.005(0.01)=0.00005\) for death. The probability of remaining active is \(1-0.024(0.01)=0.99976\) to first order. These add to 1. For someone in moderate care, the corresponding first-order probabilities are 0.00034 to severe care, 0.00020 to death, and 0.99946 to remain in moderate care. The approximation omits terms of order smaller than \(h\); it is not an exact one-year transition calculation.
At the same duration \(t=5\), a policy issued at age 67 has attained age 72. The same assumptions give \(\lambda_{01}=0.021\), \(\lambda_{03}=0.0054\), and \(\lambda_{12}=0.034\). The active-state total exit intensity is then \(0.0264\), compared with \(0.024\) for the age-65 issue. This difference comes from the age-dependent terms, while the duration components are unchanged.
Common Traps and Model Checks
The most frequent errors arise from using the correct rate on the wrong time scale or treating an intensity as a probability. A rate of 0.03 per year is not automatically a 3% probability over a year when other transitions compete. Under constant competing intensities, the probability of a specified cause is its share of the total intensity multiplied by the probability of any exit during the interval.
- Confusing \(x\) with \(x+t\). If a function is defined by attained age, evaluate it at the insured’s current age \(x+t\), not at issue age alone.
- Using age bands as duration bands. A threshold at age 70 and a threshold at policy duration 5 are different rules unless the issue age makes them coincide.
- Ignoring competing transitions. Calculate the cause-specific probability using all exit intensities from the starting state, not only the intensity for the cause of interest.
- Using a single matrix when rates vary. A matrix evaluated at one age and duration is local to that point. If rates change over the period, transition probabilities must reflect that change.
- Calling policy duration “time in care.” These are different clocks. If the model requires a rate based on time since entering care, that dependence must be represented explicitly.
A practical check is to write the units and arguments beside every rate before calculation: for example, “per year, evaluated at age \(x+t\), duration \(t\).” Then confirm that off-diagonal intensities are nonnegative and each diagonal entry is the negative row sum. These checks help keep the model aligned with its assumptions and with the long-term care valuation being developed. This treatment is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
Check Your Understanding
Work each problem by identifying the time scale and the competing intensities before calculating.
- A person is active at an instant when the active-to-care intensity is 0.025 per year and the active-to-death intensity is 0.005 per year. Using the first-order small-interval approximation, find the probabilities of care, death, and remaining active over the next 0.1 years. (Answer: 0.0025, 0.0005, and 0.9970, respectively.)
- Under constant intensities, active-to-care is 0.030 per year and active-to-death is 0.010 per year. Find the probability that care is the first exit from active by two years. (Answer: \(\frac{0.030}{0.040}(1-e^{-0.080})=0.0577\), to four decimals.)
- For a duration-only model, care intensity is 0.020 per year for the first two policy years and 0.040 per year for the next year; active-state mortality is 0.010 per year throughout. Starting active, find the probability of care entry by duration 3. (Answer: \(\frac{0.020}{0.030}(1-e^{-0.060})+e^{-0.060}\frac{0.040}{0.050}(1-e^{-0.050})=0.0756\), to four decimals.)
- A model defines \(\lambda_{01}(x,t)=0.012+0.001(x+t-60)\) per year. Find the intensity at duration 4 for someone issued at age 62. State whether the evaluation uses issue age or attained age. (Answer: attained age \(66\), so \(\lambda_{01}(62,4)=0.018\) per year.)
- At a given age and duration, a state has intensities 0.018 to moderate care, 0.007 to severe care, and 0.005 to death. Give the diagonal intensity and the first-order probability of remaining in the state over the next 0.02 years. (Answer: diagonal intensity \(-0.030\) per year; remaining probability \(1-0.030(0.02)=0.9994\), to first order.)