The Exit-Time Density for Each Cause

Why a Cause-Specific Exit Density Is Needed

In a competing decrement model, an individual begins in a specified group and may leave it for one of several mutually exclusive causes. For example, a hypothetical long-term care model might track transitions out of an active, non-claim state due to claim incidence or death. A cause-specific exit density describes how exits due to one chosen cause are distributed over time, accounting for the fact that other causes may remove individuals first.

The key distinction is between the intensity of a cause among those who are still in the group and the density of exits from the original starting cohort. The intensity applies only to individuals who have remained. The density for the starting cohort must therefore include the probability that an individual is still present when the time in question is reached.

Definition: Consider a cohort starting in the group at time 0. Let \(T\) be the time of exit and \(J\) the cause of exit. For cause \(j\), the cause-specific intensity \(\mu_j(t)\) is the instantaneous rate of exit due to cause \(j\), conditional on still being in the group at time \(t\). The cause-specific exit-time density \(f_j(t)\) is the density of exits from the original cohort at time \(t\) due to cause \(j\).

Assume there are finitely many competing causes, with nonnegative, deterministic intensities \(\mu_j(t)\) that are locally integrable. There is no re-entry to the starting group, and no separate point mass of exits at a fixed time. Write \(\mu_\bullet(t)=\sum_k\mu_k(t)\) for the total intensity and \(S(t)=P(T>t)\) for the probability of remaining in the group. Under these assumptions, the density for cause \(j\) is

f_j(t)=S(t)\mu_j(t) =\exp\left(-\int_0^t\sum_k\mu_k(s)\,ds\right)\mu_j(t).

The formula is used for \(t\geq 0\), at times where the intensities are defined. If intensities have jumps, it applies within each interval of continuity; the value assigned at a single jump time does not change probabilities over intervals. The density has units of inverse time. It is not itself a probability at a single instant.

Why Remaining Probability Multiplies the Intensity

The cause-specific intensity answers a conditional question: among individuals still present at time \(t\), how rapidly are exits due to cause \(j\) occurring? To obtain the rate of cause-\(j\) exits from the original cohort, both conditions must be met: an individual must remain until \(t\), and then exit due to \(j\) immediately after \(t\).

For a small positive interval \(h\), the intensity definition gives \(P(t\leq T<t+h,\ J=j\mid T\geq t)=\mu_j(t)h+o(h)\), where \(o(h)/h\) tends to zero as \(h\) tends to zero. Multiplying by the probability of reaching time \(t\) gives the unconditional probability for a member of the starting cohort:

P(t\leq T<t+h,\ J=j) =S(t)\{\mu_j(t)h+o(h)\}.

Divide by \(h\) and take the limit as \(h\) tends to zero. Since \(S(t)\) is the probability of still being in the group at the start of that interval, the limit is \(S(t)\mu_j(t)\), which is the cause-specific density. The survival factor is determined by all competing causes, not just cause \(j\).

Accumulate all competing intensities.
At each time, the total intensity is \(\mu_\bullet(t)=\sum_k\mu_k(t)\).

Find the probability of reaching time \(t\).
Use \(S(t)=\exp(-\int_0^t\mu_\bullet(s)\,ds)\).

Apply the chosen cause’s intensity at time \(t\).
Multiply \(S(t)\) by \(\mu_j(t)\) to obtain \(f_j(t)\).

As a consistency check, adding the cause-specific densities at the same time gives the density of exit due to any cause:

\sum_j f_j(t)=S(t)\sum_j\mu_j(t)=S(t)\mu_\bullet(t).

This agrees with the survival differential equation \(S'(t)=-S(t)\mu_\bullet(t)\): the instantaneous rate at which the remaining probability decreases is the total density of exits.

Worked Calculations

Worked Example: Constant Competing Intensities

A hypothetical group starts with 5,000 individuals at time 0. There are two possible exit causes, with constant intensities \(\mu_1(t)=0.03\) per year and \(\mu_2(t)=0.02\) per year. Assume a continuous-time competing decrement model with no re-entry. Find the cause-1 exit density at \(t=2\) years and estimate the number of cause-1 exits in the next 0.01 year, using the density at \(t=2\) as a local approximation.

Method. First calculate remaining probability using the total intensity; then multiply by the cause-1 intensity. This avoids treating cause 1 as if it were the only decrement.

The total intensity is constant: \(\mu_\bullet=0.03+0.02=0.05\) per year. Over two years, its accumulated value is \(0.05(2)=0.10\). Therefore, \(S(2)=e^{-0.10}=0.904837\), or 0.9048 to four decimal places.

The cause-1 density is \(f_1(2)=0.904837(0.03)=0.0271451\) per year. Thus, the density is approximately 0.027145 per year. For a short interval of length \(0.01\) year, the approximate cause-1 exit probability per starting individual is \(0.0271451(0.01)=0.000271451\). For 5,000 starting individuals, the estimated number is \(5{,}000(0.000271451)=1.3573\), or about 1.36 exits.

Check. The cause-2 density is \(0.904837(0.02)=0.0180967\) per year, and the total density is \(0.0271451+0.0180967=0.0452418\) per year. This equals \(S(2)\mu_\bullet=0.904837(0.05)=0.0452419\) per year, with the small difference due to rounding.

Worked Example: A Time-Varying Cause Intensity

In a hypothetical disability model, two causes remove individuals from an active state. Time \(t\) is measured in years since entry. Let \(\mu_1(t)=0.01+0.004t\) per year and \(\mu_2(t)=0.02\) per year for \(t\geq0\). Assume the cohort starts in the active state and the stated intensities apply continuously. Find the cause-1 density at \(t=3\) years.

Method. Integrate the sum of the intensities from entry to year 3 to obtain the remaining probability, and then use the instantaneous value of \(\mu_1\) at year 3. Integrating only the value of the cause-1 intensity at year 3 would not give the accumulated decrement rate.

The total intensity at time \(s\) is \(\mu_\bullet(s)=0.01+0.004s+0.02=0.03+0.004s\) per year. Its accumulated value through year 3 is \(\int_0^3(0.03+0.004s)\,ds=0.03(3)+0.002(3^2)=0.09+0.018=0.108\).

Consequently, \(S(3)=e^{-0.108}=0.897627\), or 0.8976 to four decimal places. The cause-1 intensity at the specific time \(t=3\) is \(\mu_1(3)=0.01+0.004(3)=0.022\) per year. Therefore, \(f_1(3)=0.897627(0.022)=0.0197478\) per year.

Check. At year 3, the cause-2 density is \(0.897627(0.02)=0.0179525\) per year. The two densities sum to \(0.0377003\) per year, which equals \(S(3)(0.022+0.02)=0.897627(0.042)=0.0377003\) per year.

Worked Example: Piecewise-Constant Intensities

A hypothetical care model has two competing exit causes, with time measured in years since the individual entered the modeled state. From time 0 up to year 1, the intensities are \(\mu_A=0.04\) and \(\mu_B=0.02\) per year. After year 1, they are \(\mu_A=0.02\) and \(\mu_B=0.03\) per year. Assume there are no point masses of exits at year 1. Find the cause-A density at \(t=1.5\) years.

Method. Accumulate the total intensity separately over the two time segments, then multiply the resulting remaining probability by the current cause-A intensity. This handles the fact that survival to year 1.5 depends on both rate regimes.

During the first year, the total intensity is \(0.04+0.02=0.06\) per year, contributing \(0.06(1)=0.06\) to the accumulated intensity. From year 1 to year 1.5, the total is \(0.02+0.03=0.05\) per year, contributing \(0.05(0.5)=0.025\). The total accumulated intensity is \(0.06+0.025=0.085\).

Thus \(S(1.5)=e^{-0.085}=0.918512\), or 0.9185 to four decimal places. At \(t=1.5\), the current cause-A intensity is \(0.02\) per year, so \(f_A(1.5)=0.918512(0.02)=0.0183702\) per year.

Check. The cause-B density at this time is \(0.918512(0.03)=0.0275554\) per year. Their sum, \(0.0459256\) per year, agrees with the total density \(S(1.5)(0.05)=0.0459256\) per year.

Interpreting the Density and Avoiding Common Errors

A density is a rate of probability per unit of time, not the probability of exit at exactly \(t\). For a short interval of length \(h\), the cause-\(j\) exit probability for one member of the starting cohort is approximately \(f_j(t)h\), provided the density does not change greatly over that interval. The approximation improves as \(h\) becomes smaller. For a larger interval, the density can vary, so using one value times the interval length may be inaccurate.

The units must be consistent. If time is measured in years, intensities must be expressed per year and the density is per year. If time is instead measured in months, both the intensity and the interval length must use months. Changing the time unit requires converting the intensity as well; the numerical value of a rate cannot be carried unchanged between years and months.

Common exam trap: Using \(e^{-\int_0^t\mu_j(s)\,ds}\mu_j(t)\) for the cause-\(j\) density treats cause \(j\) as the only cause that can remove individuals. The correct remaining probability uses the sum of all competing intensities. In the first worked example, this incorrect calculation gives \(e^{-0.03(2)}(0.03)=0.0282530\) per year. The correct result is \(e^{-(0.03+0.02)(2)}(0.03)=0.0271451\) per year because cause 2 also reduces the number who survive in the group to year 2.

The cause-specific density is useful whenever a model needs to distinguish the timing and cause of exits from a starting cohort. For example, a valuation may attach different cash flows to different exit causes. The density supplies the time-specific weighting for a particular cause, while the competing intensities determine how many individuals remain exposed to every cause. Keeping those roles separate is essential: use all causes to calculate remaining probability, then use only the selected cause’s intensity at the time of interest.

Check Your Understanding

For each problem, use the continuous-time competing decrement model and keep the stated time units consistent. Densities are given per unit of the stated time.

A cohort has two constant cause intensities of 0.012 and 0.008 per year. Find the density for the first cause at \(t=4\) years. (Answer: 0.0111 per year)
For \(t\) in years, \(\mu_1(t)=0.01+0.002t\) and \(\mu_2(t)=0.015\) per year. Find the cause-1 density at \(t=2\) years. (Answer: 0.0133 per year)
From year 0 to year 1, two cause intensities are 0.03 and 0.02 per year. After year 1, they are 0.01 and 0.04 per year. Find the second-cause density at \(t=1.5\) years. (Answer: 0.0371 per year)
At \(t=3\) years, a model gives a remaining probability of 0.90 and a cause-A intensity of 0.018 per year. Estimate cause-A exits during the next 0.01 year in a starting cohort of 10,000, using the density at year 3 as constant over that short interval. (Answer: 1.62 exits, approximately)
For \(t\) in years, let \(\mu_1(t)=0.02\) and \(\mu_2(t)=0.01+0.01t\) per year. Find the cause-2 density at \(t=2\) years. (Answer: 0.0277 per year)

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