First Entry Is Not the Same as Being in Care
A long-term care model can describe an insured moving from healthy status to one or more care states, recovering, relapsing, or dying. These transitions make it important to state the event precisely. The probability of ever entering care by duration \(t\) counts an insured who enters care and later recovers. It is different from the probability of being in a care state at exactly duration \(t\), which can exclude someone who recovered and can include someone who entered care only shortly before \(t\).
For an insured who starts healthy, care entry and death are competing first events: death before any care entry prevents a later care entry. Recovery and relapse matter to later state occupancy and to the path through the model, but they do not reverse the event “has entered care at least once.” The calculation can therefore focus on paths that have not yet entered care.
Constant Intensities: Care Entry Versus Death
Suppose the insured starts in one healthy state. While still healthy and never previously in care, assume the transition intensities are constant: \(\lambda_C\) for transition into any care state and \(\lambda_D\) for death. If there are multiple care destinations, \(\lambda_C\) is the sum of the intensities from healthy into those destinations. Assume there are no other transitions out of this never-in-care healthy state.
The total rate of leaving that state is \(\lambda_C+\lambda_D\). The probability of still being healthy, alive, and never having entered care at duration \(s\) is \(e^{-(\lambda_C+\lambda_D)s}\). Thus, the probability of first care entry in a small interval around \(s\) is approximately this probability multiplied by \(\lambda_C\,ds\). Integrating the first-entry flow gives the result.
This expression has two components. The factor \(\lambda_C/(\lambda_C+\lambda_D)\) is the probability that care, rather than death, is the first transition if the insured is followed indefinitely under these constant intensities. The bracketed factor is the probability that some transition occurs by duration \(t\). Their product is the probability that the first transition by \(t\) is into care.
Worked Example: Two Care Destinations and Competing Death
A hypothetical insured starts healthy. The constant annual intensities from healthy are \(0.07\) to moderate care, \(0.05\) to severe care, and \(0.03\) to death. Find the probability of entering either care state before death and by duration 4 years.
Method. Add the two care-entry intensities to obtain the total intensity of first care entry. Use the competing-risk formula, with death as the other possible first transition.
The total care-entry intensity is \(\lambda_C=0.07+0.05=0.12\) per year. The total exit intensity from the never-in-care healthy state is \(\lambda_C+\lambda_D=0.12+0.03=0.15\) per year.
The probability is \(\frac{0.12}{0.15}(1-e^{-0.15(4)})\). The exponent is \(0.15(4)=0.60\), and \(e^{-0.60}=0.5488\) to four decimal places. Therefore, \(\frac{0.12}{0.15}(1-0.5488)=0.8(0.4512)=0.36096\).
Rounded to four decimal places, the probability is 0.3610. This is the probability of entering either care destination by year 4 before death; the calculation does not require the insured to remain in the first care state.
Check. The result is below the probability of any exit from healthy by year 4, \(1-e^{-0.60}=0.4512\), as it must be because some first exits are deaths. The common mistake to avoid is using only one care intensity instead of summing the rates for all care destinations.
Why Recovery and Relapse Do Not Undo First Entry
After a care entry, the insured may recover to a healthy state and later relapse into care. Those transitions change the probability of occupying each state at a later duration. They do not change whether \(T_C\leq t\) has already occurred. A useful way to see this is to distinguish two kinds of healthy status in the model: healthy with no prior care entry, and healthy after at least one care entry.
Only the first kind contributes to the probability of having never entered care. Recovery leads to the second kind. A relapse from that state is another care entry, but it does not create a new insured who has ever entered care. Therefore, for the probability of ever entering care, the relevant flow is the flow out of the never-in-care healthy state into care. It is not the total flow into care from every state in the full model.
Record whether care has ever been entered, not merely the insured’s current health state.
For an insured starting healthy, these paths remain alive and in the never-in-care healthy state under the assumptions above.
At each duration, multiply the probability of still being in the never-in-care state by its intensity of transition into any care state, then integrate over time.
Worked Example: Recovery and Relapse After Care Entry
A hypothetical policyholder starts healthy. While healthy and never previously in care, the annual intensity of entering care is \(0.10\), and the annual intensity of death is \(0.02\). Once in care, the insured can recover to healthy status at intensity \(0.40\); after recovery, the intensity of relapse into care is \(0.20\). Find the probability of having entered care at least once by duration 3 years.
Method. Calculate the first-care-entry probability from the never-in-care healthy state. Recovery and relapse are specified to clarify the model, but neither changes whether a first entry has already occurred.
For the first event, \(\lambda_C=0.10\), \(\lambda_D=0.02\), and the total exit intensity is \(0.12\). Thus, \(\frac{0.10}{0.12}(1-e^{-0.12(3)})\). The exponent is \(0.36\), and \(e^{-0.36}=0.6977\) to four decimal places. Therefore, \(\frac{0.10}{0.12}(1-0.6977)=\frac{5}{6}(0.3023)=0.2519\), to four decimal places.
The probability of ever entering care by year 3 is 0.2519. The recovery intensity \(0.40\) and relapse intensity \(0.20\) do not enter this first-entry calculation: both transitions occur only after an earlier care entry. They would be relevant to questions about the insured’s state at year 3.
Check. The result cannot exceed the probability of any first transition by year 3, \(1-e^{-0.36}=0.3023\), because death is a competing first transition. A common mistake is to let recovery “erase” an earlier care entry and count only insureds still in care at year 3; that calculates a different event.
Changing Intensities Over the Policy Duration
If the intensities vary with duration, let \(\lambda_C(s)\) and \(\lambda_D(s)\) denote the care-entry and death intensities from the never-in-care healthy state at duration \(s\). The probability of remaining in that state to duration \(s\) is the exponential of the negative accumulated exit intensity. Multiplying this probability by the instantaneous care-entry intensity gives the first-entry probability flow.
This is the time-varying counterpart of the constant-intensity formula. In practice, piecewise-constant intensities are handled by calculating the probability of first care entry in each interval conditional on still being in the never-in-care healthy state at its start. Multiply each conditional probability by the probability of surviving without any transition through the preceding intervals, then add the interval contributions.
Worked Example: Intensities Change After Year 2
A hypothetical insured starts healthy. During the first 2 years, the annual intensities from healthy are \(0.08\) to any care state and \(0.02\) to death. During the next 2 years, they are \(0.12\) to care and \(0.03\) to death. Intensities are constant within each interval. Find the probability of entering care by duration 4 years before death.
Method. Add the first-interval care-entry probability to the probability of reaching year 2 without any transition multiplied by the conditional care-entry probability in the second interval.
In years 0 to 2, the total exit intensity is \(0.08+0.02=0.10\). The probability of care entry in this interval is \(\frac{0.08}{0.10}(1-e^{-0.10(2)})=0.8(1-e^{-0.20})\). Since \(e^{-0.20}=0.8187\) to four decimal places, this is \(0.8(0.1813)=0.1450\), to four decimal places.
The probability of remaining in the never-in-care healthy state to year 2 is \(e^{-0.20}=0.8187\). In years 2 to 4, the total exit intensity is \(0.12+0.03=0.15\). Conditional on reaching year 2 without transition, the care-entry probability in the next 2 years is \(\frac{0.12}{0.15}(1-e^{-0.15(2)})=0.8(1-e^{-0.30})\). Using \(e^{-0.30}=0.7408\), this is \(0.8(0.2592)=0.2074\), to four decimal places.
The second-interval contribution from the original starting population is \(0.8187(0.2074)=0.1698\), to four decimal places. Adding the two interval contributions gives \(0.1450+0.1698=\mathbf{0.3148}\), to four decimal places.
Check. The probability of no transition in both intervals is \(e^{-0.20}e^{-0.30}=e^{-0.50}=0.6065\). The total probability of any transition by year 4 is therefore \(1-0.6065=0.3935\), which exceeds the care-entry probability \(0.3148\), as required. The common mistake is to apply the second-interval probability to the whole original population rather than only to those still in the never-in-care healthy state at year 2.
Common Exam Traps and Model Boundaries
- Confusing first entry with occupancy. “Has entered care by duration \(t\)” includes someone who has since recovered. “Is in care at duration \(t\)” does not necessarily include that person.
- Ignoring competing death. The care-entry intensity alone is not enough when death can occur before care. The probability of remaining in the no-event state must account for both care entry and death.
- Counting relapse as a new first entry. Relapse may create another care episode, but not another occurrence of the event “ever entered care.” Do not add repeated entry flows from recovered states to a first-entry probability.
- Using the formula outside its assumptions. If there are several states in which the insured may remain without having entered care, the probability of still being event-free is not generally a single exponential. Track the event-free states and their transition equations.
- Mixing time units. If intensities are annual, duration must be measured in years. If the duration is given in months, convert it or consistently express the intensities per month.
Check Your Understanding
For each problem, assume transition intensities are constant within each stated interval and round probabilities to four decimal places.
- A healthy insured has annual intensities \(0.09\) to any care state and \(0.03\) to death. What is the probability of entering care by duration 2 years before death? (Answer: 0.1600)
- A healthy insured can enter either of two care states, with annual intensities \(0.04\) and \(0.06\), and has annual death intensity \(0.02\). Find the probability of entering either care state by duration 5 years before death. (Answer: 0.3760)
- During year 0 to 1, care-entry and death intensities are \(0.05\) and \(0.05\). During year 1 to 3, they are \(0.10\) and \(0.025\). Find the probability of care entry by duration 3 years before death. (Answer: 0.2077)
- A healthy insured has annual intensities \(0.08\) to care and \(0.02\) to death before any care entry. After care entry, recovery and relapse can occur. What is the probability of having entered care by duration 3 years, regardless of subsequent recovery? (Answer: 0.2073)
- For the first 2 years, a healthy insured has annual care-entry and death intensities of \(0.07\) and \(0.03\). For the next year, they are \(0.11\) and \(0.04\). Find the probability of entering care by duration 3 years before death. (Answer: 0.2105)