Cumulative Probability of Exit by Cause

From Exit-Time Density to Cumulative Probability

A cause-specific intensity describes the instantaneous rate of exit among people who are still in the group. It is not itself the probability of exit over an interval. To find that probability, account for both the cause-specific intensity and the probability that a person has remained in the group long enough to be exposed to it. Integrating their product gives the cumulative probability of exit by that cause.

This tutorial considers a first exit from a group, such as a hypothetical insured population that can leave through disability or withdrawal. The causes are mutually exclusive: once a person exits through one cause, that person cannot subsequently exit through another cause in this model.

Definition: Let $T$ be the time of first exit and $J$ the cause of that exit. For cause $j$, the cumulative incidence by time $t$ is $$ F_j(t)=P(T\le t,\ J=j). $$ It is the probability, measured from the initial group at time 0, of having exited by cause $j$ on or before time $t$. It is not the probability of cause $j$ conditional on an exit having occurred.

Write $\mu_j(t)$ for the cause-specific intensity at time $t$, conditional on still being in the group immediately before $t$. Assume the intensities are nonnegative and integrable over the interval considered, and that the listed causes account for all possible first exits. The total intensity is $\mu(t)=\sum_j\mu_j(t)$. If $S(t)=P(T>t)$ is the probability of remaining in the group, then the cause-specific exit-time density is $f_j(t)=S(t)\mu_j(t)$. This is the exit-time density for each cause, expressed here to establish the cumulative probability that follows.

Theorem: Under the stated first-exit model, $$ F_j(t)=\int_0^t f_j(s)\,ds =\int_0^t S(s)\mu_j(s)\,ds. $$ For an interval $(a,b]$, the probability of first exiting by cause $j$ during that interval, measured from the initial group, is $$ P(a<T\le b,\ J=j)=F_j(b)-F_j(a) =\int_a^b S(s)\mu_j(s)\,ds. $$

The factor $S(s)$ matters. A person who has already exited before time $s$ cannot exit at $s$ through another cause. The intensity $\mu_j(s)$ applies only to those still exposed to exit, so the density for the full initial group is the intensity multiplied by the probability of still being exposed.

Why the Integral Gives the Cumulative Incidence

Consider a short interval from $s$ to $s+\Delta t$. Among members still in the group at time $s$, the probability of cause-$j$ exit in that short interval is approximately $\mu_j(s)\Delta t$. The probability that a member of the initial group is still present at $s$ is $S(s)$. Therefore, the probability of a cause-$j$ exit in the short interval, measured from the initial group, is approximately $S(s)\mu_j(s)\Delta t$.

Adding these contributions over short intervals from 0 to $t$, then taking the limit as their widths approach zero, gives the integral. Equivalently, the cause-specific density is $f_j(s)=S(s)\mu_j(s)$, and integrating a density over a time interval gives the probability that the event occurs there.

P(s<T\le s+\Delta t,\ J=j) \approx S(s)\mu_j(s)\Delta t \quad\Longrightarrow\quad F_j(t)=\int_0^t S(s)\mu_j(s)\,ds.

When all causes are exhaustive, the cumulative incidences and the remaining probability partition the initial group:

S(t)+\sum_j F_j(t)=1.

This identity is a useful check on a calculation. It also clarifies the interpretation: $F_j(t)$ is an unconditional probability from time 0, whereas $F_j(t)/(1-S(t))$, when $S(t)<1$, is the share of exits by time $t$ attributed to cause $j$. Those quantities answer different questions.

Worked Example: Two Competing Causes With a Changing Share

A hypothetical benefit program begins with a group of members at time 0. Over the next year, members can first exit through either disability, cause 1, or withdrawal, cause 2. Time is measured in years, and the intensities for $0\le t\le1$ are $\mu_1(t)=t$ and $\mu_2(t)=1-t$. Find each cumulative incidence at $t=0.5$, and check the remaining probability. No interest assumption is needed because this is a probability calculation.

Method. First find the total intensity, then the remaining probability, then integrate each cause-specific density. The cause intensities are nonnegative on the stated interval, and their sum is $\mu(t)=t+(1-t)=1$.

The remaining probability solves $S'(t)=-\mu(t)S(t)$, with $S(0)=1$. Thus $S(t)=e^{-t}$. The two exit-time densities are $$ f_1(t)=te^{-t},\qquad f_2(t)=(1-t)e^{-t}. $$

For cause 1, $$ F_1(0.5)=\int_0^{0.5}te^{-t}\,dt =\left[-(t+1)e^{-t}\right]_0^{0.5} =1-1.5e^{-0.5} =0.090204. $$ Here $e^{-0.5}=0.60653066$, so $1-1.5(0.60653066)=0.09020401$.

For cause 2, $$ F_2(0.5)=\int_0^{0.5}(1-t)e^{-t}\,dt =\left[te^{-t}\right]_0^{0.5} =0.5e^{-0.5} =0.303265. $$ The remaining probability is $S(0.5)=e^{-0.5}=0.606531$. The check is $0.090204+0.303265+0.606531=1.000000$, up to rounding to six decimal places.

The cumulative incidence of withdrawal is larger here, even though the total intensity is constant: the cause-specific intensities change over time, and the members who remain are exposed to those changing cause intensities. A common mistake is to integrate $\mu_j(t)$ alone; that does not account for members who have already exited.

Calculating Cumulative Incidence With Time-Varying Causes

The total intensity determines the probability of remaining in the group, while the separate cause intensities determine how the exits are allocated across causes. In particular, a simple share of the total intensity need not be the share of cumulative exits over an interval. The cause intensities may change with time, and each time point is weighted by the probability $S(t)$ of still being in the group at that time.

Worked Example: Cause-Specific Probabilities and Expected Counts

In a hypothetical one-year contract, members can first leave through either cause 1 or cause 2. For $0\le t\le1$, measured in years, suppose the intensities are $\mu_1(t)=0.2+0.6t$ and $\mu_2(t)=0.8-0.6t$. Find the two cumulative incidences at one year and the expected numbers of exits by cause in an initial group of 1,500 members. State the expected number remaining. The rates are per year; no interest assumption is required.

Method. Sum the cause intensities to obtain the total intensity. It is $1$ per year throughout the interval, so $S(t)=e^{-t}$. Both cause intensities are nonnegative on $[0,1]$, as required.

For cause 1, $$ \begin{aligned} F_1(1) &=\int_0^1 (0.2+0.6t)e^{-t}\,dt\\ &=0.2(1-e^{-1})+0.6\left[1-2e^{-1}\right]\\ &=0.2(0.63212056)+0.6(0.26424112)\\ &=0.12642411+0.15854467\\ &=0.28496878. \end{aligned} $$ The integral $\int_0^1te^{-t}\,dt=1-2e^{-1}$ follows from the antiderivative $-(t+1)e^{-t}$.

The total probability of any exit by one year is $1-e^{-1}=0.63212056$. Therefore, $$ F_2(1)=0.63212056-0.28496878=0.34715178, \qquad S(1)=e^{-1}=0.36787944. $$ The probability check is $0.28496878+0.34715178+0.36787944=1.00000000$.

Multiply each probability by 1,500. The expected number exiting by cause 1 is $1{,}500(0.28496878)=427.45317$, or 427.5 members to one decimal place. By cause 2 it is $1{,}500(0.34715178)=520.72767$, or 520.7 members. The expected number remaining is $1{,}500(0.36787944)=551.81916$, or 551.8 members. The rounded counts sum to $427.5+520.7+551.8=1{,}500.0$.

The method avoids treating the end-of-year intensity values or their unweighted averages as probabilities. It uses the density at each time, $S(t)\mu_j(t)$, before integrating.

Probabilities During a Later Part of the Interval

A cumulative incidence is measured from the original group at time 0. Consequently, the probability of cause-$j$ exit during a later interval $(a,b]$, still measured as a fraction of that original group, is $F_j(b)-F_j(a)$. This difference removes cause-$j$ exits that had already occurred by $a$. It does not change the denominator to “members still present at $a$.”

Worked Example: Isolating Cause Exits Between Two Times

Use the hypothetical program in the first example, with $\mu_1(t)=t$, $\mu_2(t)=1-t$, and $S(t)=e^{-t}$ over the first year. Find the probabilities, as fractions of the initial group, of exiting through each cause between 0.5 and 1 year. Check their sum against the probability of being present at 0.5 but not at 1 year. Time is in years, and no interest assumption is needed.

Method. Find each cause’s cumulative incidence at both endpoints and subtract. From the first example, at $t=0.5$, $F_1(0.5)=0.09020401$ and $F_2(0.5)=0.30326533$.

At $t=1$, the cause 1 cumulative incidence is $$ F_1(1)=\int_0^1 te^{-t}\,dt=1-2e^{-1}=0.26424112. $$ The cause 2 cumulative incidence and remaining probability are $$ F_2(1)=\int_0^1(1-t)e^{-t}\,dt=e^{-1}=0.36787944, \qquad S(1)=e^{-1}=0.36787944. $$

Thus, the initial-group probability of a cause 1 exit during $(0.5,1]$ is $0.26424112-0.09020401=0.17403711$. For cause 2, it is $0.36787944-0.30326533=0.06461411$. Their sum is $0.23865122$.

The probability of remaining at 0.5 but not remaining at 1 is $S(0.5)-S(1)=0.60653066-0.36787944=0.23865122$, which agrees. This check confirms that the two cause-specific interval probabilities account for all exits during that interval. A common mistake would be to subtract the cumulative probabilities and then describe the results as conditional on surviving to 0.5; they are not conditional probabilities.

Common Mistakes and Checks

The most frequent error is omitting $S(t)$ from the cause-specific density. The intensity is a rate among those still at risk, not a rate applied to the entire initial group at every time. Another error is confusing cumulative incidence with the share of exits attributed to a cause. The latter uses the total exit probability as its denominator; cumulative incidence uses the initial group.

For example, in the first worked model, the cause 2 intensity is $1-t$. Integrating it alone over $[0,1]$ gives $$ \int_0^1(1-t)\,dt=0.5. $$ That is not the cause 2 cumulative incidence. The correct calculation is $$ F_2(1)=\int_0^1(1-t)e^{-t}\,dt=e^{-1}=0.367879. $$ The missing factor $e^{-t}$ is the probability of remaining in the group up to time $t$. The incorrect value also fails the probability check: combining it with $F_1(1)=0.264241$ and $S(1)=0.367879$ gives a total greater than 1.

Before finalizing a calculation, verify the time units used for the intensities and the integration limits, check that all cause intensities are nonnegative on the interval, and confirm that the cumulative incidences plus the remaining probability equal 1 when the causes are exhaustive. If the causes do not cover every possible exit, the remaining probability plus the listed cumulative incidences will not generally equal 1; an additional cause must be included to use the exhaustive-cause check.

Key takeaway: For each cause, multiply its cause-specific intensity at time $t$ by the probability of still being in the group at $t$, then integrate. The result is a cumulative incidence from the initial group, not an intensity and not a conditional cause share.

Check Your Understanding

Use the first-exit model stated in each problem. Probabilities are measured from the initial group unless the question says otherwise.

In a hypothetical model over $0\le t\le0.5$ years, $\mu_1(t)=t$ and $\mu_2(t)=1-t$. Find the cumulative incidence for cause 1 by 0.5 years. (Answer: 0.0902)
For $0\le t\le1$, let $\mu_1(t)=0.2+0.6t$ and $\mu_2(t)=0.8-0.6t$. Find the cause 1 cumulative incidence by $t=0.5$. (Answer: 0.1328)
For $0\le t\le0.75$, let $\mu_1(t)=t$ and $\mu_2(t)=1-t$. Find $F_1(0.75)$, $F_2(0.75)$, and $S(0.75)$. (Answer: $F_1=0.1734,\ F_2=0.3543,\ S=0.4724$)
A hypothetical initial group contains 800 members, with $\mu_1(t)=t$, $\mu_2(t)=1-t$, and time measured in years on $[0,1]$. Find the expected numbers exiting by each cause and remaining at one year, to one decimal place. (Answer: 211.4 by cause 1, 294.3 by cause 2, and 294.3 remaining)
For $0\le t\le1$, let $\mu_1(t)=0.2+0.6t$ and $\mu_2(t)=0.8-0.6t$. Find the fraction of the initial group exiting by cause 1 during $(0.25,0.75]$. (Answer: 0.1494)

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