Person-Time Exposure in a Decrement Model

Why Person-Time Exposure Matters

A cohort count describes how many people are in a group at a particular time. Person-time exposure describes how much time, in total, the cohort is expected to spend in that group over an interval. These quantities are related, but they are not interchangeable: as decrements remove members, the expected number remaining usually falls below the starting cohort size.

For a hypothetical disability-income portfolio, for example, members might leave the active group through disability or death. If the question concerns time spent active during the next three years, multiplying the starting number of members by three counts three years for everyone, including those expected to leave early. Expected exposure instead accumulates the probability of still being active at each point in the interval. The concept is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Let \(T\) be the time from the start of observation until an individual exits a group through any decrement. For an interval \([a,b]\), the individual's person-time in the group is the length of time spent in the group during that interval. If the individual is in the group at time 0 and cannot re-enter after exit, define \(R(t)=P(T>t)\), the probability of remaining in the group through time \(t\). Expected person-time per starting member over \([a,b]\) is \(\int_a^b R(t)\,dt\). For a cohort of \(N_0\) members with the same decrement probabilities, total expected person-time is \(N_0\int_a^b R(t)\,dt\).

Exposure is measured in person-years when time is measured in years, or in person-months when time is measured in months. It is an expectation, not a promise about the realized total for a particular cohort. The maximum possible exposure over an interval of length \(b-a\) is \(N_0(b-a)\), attained if everyone remains in the group throughout the interval. When decrements can occur, expected exposure is generally less.

In a competing-decrement model, let \(\lambda_j(t)\) be the intensity of decrement cause \(j\) at time \(t\), conditional on still being in the group immediately before \(t\). If there are \(J\) causes and no re-entry, the total intensity is \(\lambda(t)=\sum_{j=1}^{J}\lambda_j(t)\). Under the usual continuous-time model, the remaining probability from the start is \(R(t)=\exp(-\int_0^t \lambda(s)\,ds)\). Thus exposure depends on the combined rate of leaving the group, not on any one cause considered in isolation. The formula assumes that remaining in the group is exactly the event of having had no decrement by time \(t\); a model with re-entry or movement among several states requires the probability of occupying the specified state instead.

From Remaining Probability to Expected Exposure

The exposure formula has a direct interpretation. At each instant \(t\), a starting member contributes one unit of person-time per unit of time if still in the group, and contributes nothing if already exited. The probability of making that contribution is \(R(t)\). Accumulating these expected contributions over the interval gives expected person-time.

Represent time in the group.
For an interval \([0,h]\), let \(I(T>t)\) equal 1 if the member remains in the group at time \(t\), and 0 otherwise. Then the member's time in the group is \(\int_0^h I(T>t)\,dt\).

Take the expectation.
Expected time is \(E[\int_0^h I(T>t)\,dt]\). Interchanging expectation and integration for this bounded, nonnegative indicator gives \(\int_0^h E[I(T>t)]\,dt\).

Recognize the remaining probability.
Since \(E[I(T>t)]=P(T>t)=R(t)\), expected person-time per starting member is \(\int_0^h R(t)\,dt\). For \(N_0\) comparable members, linearity of expectation gives \(N_0\int_0^h R(t)\,dt\).

This justification does not require the members' decrement times to be independent: linearity of expectation applies even when members' outcomes are dependent, provided each has the stated remaining probability. Independence would be relevant to the distribution or variance of the cohort's realized exposure, not to this expected total.

Constant Total Intensity

If the total intensity is a constant \(\lambda>0\) per year, then the remaining probability is \(R(t)=e^{-\lambda t}\). For a starting cohort of \(N_0\), exposure over \([0,h]\) is therefore

E_{\text{cohort}}=N_0\int_0^h e^{-\lambda t}\,dt =N_0\frac{1-e^{-\lambda h}}{\lambda}.

If the decrement causes have constant intensities, \(\lambda\) is their sum. For example, a mortality intensity of \(0.07\) per year and a withdrawal intensity of \(0.13\) per year produce total intensity \(0.20\) per year. The individual cause intensities affect which exit occurs, but the total determines the probability of remaining in the group and hence the total exposure.

Worked Example: Constant Competing Intensities

A hypothetical cohort of 1,000 members is observed for three years. The constant cause-specific intensities are \(0.12\) per year for withdrawal and \(0.08\) per year for death. Calculate the expected person-years in the group over the three years. State the maximum possible exposure and explain why it differs from the expected exposure.

Method. Add the cause-specific intensities to obtain the total intensity, then integrate the remaining probability over the observation period. This method avoids counting all 1,000 members as present for all three years.

The total intensity is \(\lambda=0.12+0.08=0.20\) per year. Under a constant total intensity, the remaining probability is \(R(t)=e^{-0.20t}\). The expected exposure per starting member is

\int_0^3 e^{-0.20t}\,dt =\frac{1-e^{-0.60}}{0.20} =\frac{1-0.5488116361}{0.20} =2.2559418195\text{ years}.

Multiplying by the starting cohort size gives \(1{,}000(2.2559418195)=2{,}255.9418195\) person-years. Rounded to two decimals, expected exposure is 2,255.94 person-years.

The maximum possible exposure is \(1{,}000(3)=3{,}000\) person-years, which would occur if no member exited during the three years. The expected total is lower because the probability of remaining decreases over time. As a check, \(2{,}255.94\) lies between zero and the maximum of \(3{,}000\) person-years.

Exposure When Intensities Change

When the total intensity changes over time, use the remaining probability that applies in each interval. Exposure over the full period is the sum of the interval contributions. If an interval begins at time \(a\), its contribution per original cohort member is the remaining probability at \(a\) multiplied by the expected time in the group during that interval, conditional on still being there at \(a\). Equivalently, integrate the original cohort's remaining probability over that interval. Keeping track of this distinction prevents treating a later interval as if the original cohort were all still present.

Worked Example: Two Intervals With Different Intensities

A hypothetical cohort starts with 600 members. The total decrement intensity is \(0.10\) per year during year 1 and \(0.30\) per year during year 2. Calculate expected exposure over the two-year period. Assume no re-entry and constant intensity within each year.

Method. Calculate the expected exposure in each interval using the remaining probability from the start of the cohort, then add the two amounts. The key trap is to apply the second-year exposure to 600 rather than to the expected number still present at the start of year 2.

During year 1, \(R(t)=e^{-0.10t}\). Exposure per original member in that year is

\int_0^1 e^{-0.10t}\,dt =\frac{1-e^{-0.10}}{0.10} =\frac{1-0.9048374180}{0.10} =0.9516258196\text{ years}.

For 600 starting members, the first-year expected exposure is \(600(0.9516258196)=570.9754918\) person-years.

At time 1, the remaining probability is \(e^{-0.10}=0.9048374180\). During year 2, the additional total intensity is \(0.30\) per year, so the original cohort's remaining probability at time \(1+s\), for \(0\leq s\leq1\), is \(e^{-0.10}e^{-0.30s}\). The second-year exposure per original member is

e^{-0.10}\int_0^1e^{-0.30s}\,ds =\frac{e^{-0.10}-e^{-0.40}}{0.30} =\frac{0.9048374180-0.6703200460}{0.30} =0.7817245733\text{ years}.

For the original cohort, second-year expected exposure is \(600(0.7817245733)=469.0347440\) person-years. Adding both intervals gives \(570.9754918+469.0347440=1{,}040.0102358\) person-years, or 1,040.01 person-years rounded to two decimals. The maximum possible is \(600(2)=1{,}200\) person-years, so the result is within the required bound.

The same calculation can be expressed using expected numbers remaining. In the example, the expected number at the start of year 2 is \(600e^{-0.10}=542.9024508\). A member present then has an expected \(0.8639392643\) years in the group during the next year, since \((1-e^{-0.30})/0.30=0.8639392643\). Their product is \(542.9024508(0.8639392643)=469.0347440\) person-years, the same second-year result. The number remaining is a cohort count expectation; multiplying it by the conditional remaining time gives exposure for that interval.

Worked Example: A Longer Interval With a Rate Change

A hypothetical cohort of 500 people is followed for five years. The total intensity is \(0.12\) per year from time 0 to time 2, then \(0.08\) per year from time 2 to time 5. Find expected exposure across the full period and compare it with the maximum possible exposure.

Method. Calculate the first two years using the first intensity. For the final three years, account for the probability of surviving in the group to time 2 before applying the later interval's exposure. This avoids resetting the remaining probability to 1 at the change in intensity.

For the first interval, the remaining probability is \(e^{-0.12t}\). Exposure per original member is

\int_0^2 e^{-0.12t}\,dt =\frac{1-e^{-0.24}}{0.12} =\frac{1-0.7866278611}{0.12} =1.7781011578\text{ years}.

The first-interval cohort exposure is \(500(1.7781011578)=889.0505789\) person-years. At time 2, the remaining probability is \(e^{-0.24}=0.7866278611\). For the next three years, the exposure per original member is

e^{-0.24}\int_0^3e^{-0.08s}\,ds =\frac{e^{-0.24}-e^{-0.48}}{0.08} =\frac{0.7866278611-0.6187833918}{0.08} =2.0980558663\text{ years}.

The final-interval cohort exposure is \(500(2.0980558663)=1{,}049.0279332\) person-years. The total is \(889.0505789+1{,}049.0279332=1{,}938.0785121\) person-years, which rounds to 1,938.08 person-years. The maximum is \(500(5)=2{,}500\) person-years; the calculated expectation is positive and below that maximum.

Common Mistakes and Interpretation

A frequent error is to multiply the starting cohort size by the full interval length and call the result expected exposure. That calculation is the maximum exposure, not the expected exposure under a model with decrements. For the 1,000-member cohort followed for three years in the first example, the incorrect calculation is \(1{,}000(3)=3{,}000\) person-years. It assumes everyone stays for all three years. The correction is to integrate the remaining probability: \(1{,}000(1-e^{-0.60})/0.20=2{,}255.94\) person-years.

Other checks help catch mistakes before interpreting a result:

Check the time unit. An intensity stated per year must be combined with time measured in years. A three-year interval is \(h=3\), not \(h=36\), unless the intensity has first been converted to a monthly basis.
Use the total intensity for remaining in the group. With competing decrements, the probability of avoiding every exit cause depends on the sum of the cause-specific intensities.
Carry forward the remaining probability. At a change point, later-interval exposure is still measured for the original cohort. Do not assume all original members are present at the start of the new interval.
Check bounds and units. For \(N_0\) members over an interval of length \(h\), exposure must be between zero and \(N_0h\), and its units are member-years or the corresponding chosen time unit.

Person-time exposure is useful because it reflects both cohort size and how long members are expected to remain exposed to the possibility of a decrement. It is therefore different from a headcount at the start or end of an interval. A calculation should make clear whether it reports exposure per starting member, total cohort exposure, or the maximum possible cohort exposure.

Check Your Understanding

For each problem, state the time unit and calculate expected cohort exposure. Round final numerical answers to two decimals.

A cohort of 400 members is subject to a constant total decrement intensity of \(0.25\) per year for two years. Find expected exposure. (Answer: 629.55 person-years)
A cohort of 300 members has total intensity \(0.20\) per year during the first year and \(0.10\) per year during the second year. Find total expected exposure over both years. (Answer: 505.64 person-years)
A cohort of 250 members has constant total intensity \(0.15\) per year and is followed for 18 months. Find expected exposure. (Answer: 335.81 person-years)
A cohort of 1,000 members has total intensity \(0.10\) per year from time 0 to 1 and \(0.20\) per year from time 1 to 3. Find exposure over all three years, retaining the first-year remaining probability in the later interval. (Answer: 2,443.16 person-years)
A cohort of 750 members has constant total intensity \(0.40\) per year for six months. Find expected exposure and compare it with the maximum possible exposure. (Answer: 339.88 person-years; maximum 375.00 person-years)

What You'll Learn