Deriving Exposure From the Remaining Probability

Why Remaining Probability Determines Exposure

In a decrement model, a member contributes exposure only while still in force. A person who exits early contributes less person-time than one who remains for the full observation period. The number still in force at the end of an interval therefore does not, by itself, tell us the cohort’s total exposure during that interval. To calculate exposure, we need to account for the probability of remaining in force at each time.

This tutorial derives expected exposure from the remaining probability and then specializes the result to a constant total decrement intensity. The same calculation applies whether the decrement is mortality, withdrawal, disability, or another competing exit, provided the model describes a member’s time to first exit and there is no re-entry during the interval.

Definition: Let \(T\) be the time from the start of observation to the member’s first decrement, and let \(n\) be the length of the observation interval. The member’s exposure during that interval is \(Z=\min(T,n)\). It is the time spent in force before exit or the end of the interval, whichever comes first. For a cohort of \(N\) members, total person-time exposure is the sum of the \(N\) individual exposures.

For example, a member who exits at time \(1.4\) during a two-year interval contributes \(1.4\) years of exposure. A member who remains in force for all two years contributes two years. Exposure is measured in person-years if time is measured in years.

Deriving the Exposure Integral

Write \({}_tp\) for the probability that a member remains in force from the start through time \(t\). In a continuous-time decrement model with no instantaneous exits, this is the probability that \(T>t\). The key identity is that an individual’s exposure can be expressed as the accumulated time during which that individual is still in force:

Z=\int_0^n \mathbf{1}_{\{T>t\}}\,dt

Here, \(\mathbf{1}_{\{T>t\}}\) equals 1 while the member is in force at time \(t\), and 0 after exit. If exit occurs at time \(T<n\), the integral is 1 over the time interval from 0 to \(T\), giving exposure \(T\). If the member remains in force through \(n\), the integral is 1 throughout the observation interval, giving exposure \(n\).

Taking expectations and interchanging expectation and integration gives the exposure formula. This interchange is valid because the indicator is nonnegative and bounded on the finite interval:

E[Z] = E\left[\int_0^n \mathbf{1}_{\{T>t\}}\,dt\right] = \int_0^n E[\mathbf{1}_{\{T>t\}}]\,dt = \int_0^n P(T>t)\,dt = \int_0^n {}_tp\,dt

This is the survival-integral identity for expected time to exit, capped at \(n\). The integrand is the probability of still being exposed at time \(t\), not the probability of having exited by then. For a cohort of \(N\) identical members, linearity of expectation gives expected total exposure \(N E[Z]\); independence among members is not required for this expectation.

Result: If \({}_tp\) is the probability that a member remains in force at time \(t\), expected exposure over the interval \([0,n]\) is \(\int_0^n {}_tp\,dt\). For \(N\) homogeneous members, expected total exposure is \(N\int_0^n {}_tp\,dt\).

Constant Total Decrement Intensity

Suppose there are \(J\) competing decrement causes. Let \(\mu^{(j)}(t)\) denote the cause-specific intensity of cause \(j\) for a member who is still in force at time \(t\). In the continuous-time competing-decrement model, the total intensity is the sum of the cause-specific intensities:

\mu^{(\tau)}(t)=\sum_{j=1}^{J}\mu^{(j)}(t)

The remaining probability through time \(t\) is the exponential of the negative accumulated total intensity. Thus, the exposure integral can be written directly in terms of the total intensity:

{}_tp = \exp\left(-\int_0^t \mu^{(\tau)}(s)\,ds\right), \qquad E[Z] = \int_0^n \exp\left(-\int_0^t \mu^{(\tau)}(s)\,ds\right)dt

If the total intensity is constant, \(\mu^{(\tau)}(t)=\lambda\) for all \(t\) in the interval, then \({}_tp=e^{-\lambda t}\). Integrating this remaining probability gives a closed form:

E[Z] = \int_0^n e^{-\lambda t}\,dt = \frac{1-e^{-\lambda n}}{\lambda}, \qquad \lambda>0

The parameter \(\lambda\) is a rate per unit of time, so \(\lambda n\) must be dimensionless. The answer has units of time. If \(\lambda=0\), no decrement occurs and exposure is \(n\). This agrees with the limiting value of the formula as \(\lambda\) approaches zero.

Worked Example: Full-Interval Exposure for a Cohort

A hypothetical block of 800 policies is observed for five years. Each policy is subject to a single decrement with constant intensity \(0.10\) per year. Find the expected total exposure in policy-years. Assume all policies are in force at the start, each follows the same decrement model, and there is no re-entry.

Method. First calculate expected exposure for one policy using the remaining-probability integral. Then multiply by the initial number of policies.

The remaining probability at time \(t\) is \(e^{-0.10t}\). The one-policy expected exposure over five years is:

E[Z] = \int_0^5 e^{-0.10t}\,dt = \frac{1-e^{-0.10(5)}}{0.10} = \frac{1-e^{-0.5}}{0.10}

Using \(e^{-0.5}=0.60653066\), the numerator is \(1-0.60653066=0.39346934\). Therefore:

E[Z]=\frac{0.39346934}{0.10}=3.9346934\text{ years per policy}

The expected cohort exposure is:

800(3.9346934)=3147.75472\text{ policy-years}

Rounded to two decimal places, the expected total exposure is 3,147.75 policy-years. As a check, the result is less than the maximum possible exposure of \(800(5)=4,000\) policy-years, as it must be because some policies exit before the interval ends. The common mistake avoided here is multiplying the initial cohort by five years without allowing for decrements.

Exposure When There Are Competing Causes

For exposure, competing causes enter through the total intensity because any decrement ends a member’s time in force. If all cause-specific intensities are constant, their sum is constant as well. The cause labels are relevant to cause-specific exit calculations, but the probability of remaining exposed depends on whether any decrement has occurred.

Worked Example: Combining Two Constant Intensities

A hypothetical group of 250 members is exposed for four years to two competing decrements. The constant intensity of withdrawal is \(0.08\) per year, and the constant intensity of another decrement is \(0.12\) per year. Find the expected total exposure in member-years. Assume every member is in force at time zero and there is no re-entry.

Method. Add the cause-specific intensities to obtain the total intensity, then integrate the remaining probability. The calculation uses the total intensity, not either cause-specific intensity alone.

The total intensity is:

\lambda=0.08+0.12=0.20\text{ per year}

For one member, the four-year expected exposure is:

E[Z] = \frac{1-e^{-0.20(4)}}{0.20} = \frac{1-e^{-0.8}}{0.20}

Using \(e^{-0.8}=0.44932896\), the numerator is \(1-0.44932896=0.55067104\). Hence:

E[Z]=\frac{0.55067104}{0.20}=2.7533552\text{ years per member}

Multiplying by 250 gives:

250(2.7533552)=688.3388\text{ member-years}

Rounded to two decimal places, expected total exposure is 688.34 member-years. It is below the maximum of \(250(4)=1,000\) member-years. Using \(0.08\) alone would overstate exposure because it would ignore the competing decrement with intensity \(0.12\).

Exposure Over a Later Subinterval

Sometimes exposure is required only over \([a,b]\), where \(0\leq a<b\). For a member who entered the model at time zero, the member contributes exposure at time \(t\) only if still in force at that time. Therefore, the expected exposure over the subinterval is:

E[Z_{[a,b]}]=\int_a^b {}_tp\,dt

For a cohort of \(N\) members present at time zero, expected total exposure over \([a,b]\) is \(N\int_a^b {}_tp\,dt\). This is different from the conditional expected exposure for a member known to be in force at time \(a\). Under a constant intensity \(\lambda\), that conditional exposure over the next \(b-a\) years is \(\frac{1-e^{-\lambda(b-a)}}{\lambda}\). To obtain the original cohort’s expected exposure over \([a,b]\), that conditional value must also be weighted by the probability of surviving to \(a\).

Worked Example: Cohort Exposure Between Two Times

A hypothetical cohort of 600 members is in force at time zero. The total decrement intensity is constant at \(0.15\) per year. Find the expected cohort exposure between years 2 and 5. Assume no re-entry.

Method. Integrate the original cohort’s remaining probability over the requested interval. Do not treat all 600 members as if they were still in force at year 2.

E[Z_{[2,5]}] = 600\int_2^5 e^{-0.15t}\,dt = 600\frac{e^{-0.15(2)}-e^{-0.15(5)}}{0.15}

The endpoint exponents are \(0.15(2)=0.30\) and \(0.15(5)=0.75\). Using \(e^{-0.30}=0.74081822\) and \(e^{-0.75}=0.47236655\), the exposure per initial member over this interval is:

\frac{0.74081822-0.47236655}{0.15} = \frac{0.26845167}{0.15} = 1.7896778\text{ years}

Thus, expected cohort exposure is:

600(1.7896778)=1073.80668\text{ member-years}

Rounded to two decimal places, the answer is 1,073.81 member-years. As a check by conditioning, the probability of remaining in force to year 2 is \(0.74081822\), so the expected number in force then is \(600(0.74081822)=444.49093\). Their conditional expected exposure over the next three years is \((1-e^{-0.15(3)})/0.15=(1-0.63762815)/0.15=2.4158123\) years each. Their product is approximately \(444.49093(2.4158123)=1,073.81\) member-years, consistent with the direct integral.

The common mistake avoided is using the conditional three-year exposure for all 600 original members. Only the members who remain in force at year 2 can contribute exposure after that time.

Common Mistakes and Checks

The integral form is useful because it makes the exposure interpretation explicit: at each time, weight that instant by the probability that a member is still exposed. Two errors are particularly easy to make.

Using endpoint survivors for the whole interval. Endpoint remaining probability describes who is still in force at the end, not the average number in force throughout the interval.
Using one cause’s intensity instead of the total. Every competing decrement removes the member from exposure, so constant cause-specific intensities must be summed before applying the constant-intensity formula.
Omitting the time units. If intensity is per year, the interval length must be in years. The product of intensity and time is dimensionless; exposure is then reported in years.
Confusing unconditional and conditional exposure. Exposure over \([a,b]\) for the original cohort includes the chance of reaching \(a\). Conditional exposure for someone already in force at \(a\) does not.

For a numerical check, expected exposure over \([0,n]\) must lie between 0 and \(n\) per member. For \(N\) members, it must lie between 0 and \(Nn\). If the total intensity increases, remaining probability at each positive time decreases, so expected exposure over a fixed interval cannot increase.

A wrong endpoint calculation illustrates the first error. In the first worked example, using \(800(5)e^{-0.10(5)}\) gives \(4,000(0.60653066)=2,426.12\) policy-years. That calculation treats the endpoint remaining proportion as though it were the proportion in force throughout all five years. The integral instead accounts for the larger proportions still in force earlier in the interval and gives the correct expected exposure of 3,147.75 policy-years.

Key takeaway: Expected person-time exposure is the area under the remaining-probability curve. With constant total intensity \(\lambda>0\), one member’s exposure over \([0,n]\) is \((1-e^{-\lambda n})/\lambda\); multiply by the initial cohort size for expected cohort exposure.

Check Your Understanding

For each problem, use the remaining-probability integral and report exposure to two decimal places.

A cohort of 120 members faces a constant total decrement intensity of \(0.25\) per year for two years. Find expected total exposure. (Answer: 188.87 member-years)
A group of 500 members faces three competing constant intensities, \(0.04\), \(0.06\), and \(0.10\) per year, over 2.5 years. Find expected total exposure. (Answer: 983.67 member-years)
A cohort of 300 members has a constant total intensity of \(0.12\) per year. Find its expected exposure between years 1 and 4, measured from the original cohort at time zero. (Answer: 670.34 member-years)
A hypothetical cohort of 100 members has total intensity \(0.10\) per year for the first two years and \(0.30\) per year for the next year. Find expected total exposure over all three years. Use the remaining probability accumulated from time zero. (Answer: 252.00 member-years)
A cohort of 80 members faces a constant total intensity of \(0.05\) per year for six years. Find expected total exposure and verify that it does not exceed the maximum possible exposure. (Answer: 414.69 member-years, below the maximum of 480.00 member-years)

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