Deriving Multiple-Decrement Exit Probabilities

Why Competing Exit Probabilities Need Integration

When several causes can remove a member from the same group, each cause acts only while the member is still exposed to all the competing causes. A disability claim, for example, might end through recovery or death; an active pension member might leave through retirement, withdrawal, or death. The probability of an exit through a particular cause therefore depends both on that cause’s intensity and on the probability of remaining in the group until the exit time.

This tutorial derives that relationship for a first exit from one group. It focuses on the probability of each cause of exit, not on valuing the benefits paid at those exits. The method applies to a hypothetical group whose members begin in the same state and are subject to specified cause-specific intensities.

Definition: Suppose a member is in an initial group at time 0 and can leave it through one of $J$ mutually exclusive causes. The cause-specific intensity $\mu_j(t)$ is the instantaneous rate of exit through cause $j$ at time $t$, conditional on still being in the initial group immediately before $t$. The total exit intensity is $$ \mu_{\mathrm{tot}}(t)=\sum_{j=1}^{J}\mu_j(t). $$ The cause-specific exit probability by time $T$ is the probability that the member’s first exit from the group occurs by $T$ and is through cause $j$.

Assume the intensities are nonnegative and measurable, and their sum is integrable on every finite interval considered. The group is occupied continuously until the first exit; the causes are mutually exclusive at that exit; and the model has no probability mass for simultaneous exits or for an exit at a fixed instant. Rates may change over time and may be discontinuous. Under these assumptions, survival is absolutely continuous, and its differential equation is understood to hold almost everywhere, not necessarily at every time.

Survival and Cause-Specific Exit Probabilities

Write $S(t)$ for the probability of remaining in the initial group through time $t$, with $S(0)=1$. The total intensity governs whether the member remains in the group, while each individual intensity determines the type of exit. In this model, the survival probability and the probability of exit through cause $j$ by time $T$ are

S(t)=\exp\!\left(-\int_0^t \mu_{\mathrm{tot}}(u)\,du\right), \qquad q_j(T)=\int_0^T S(t)\mu_j(t)\,dt.

The integrand in the second formula is the probability of still being exposed just before $t$, multiplied by the cause-$j$ intensity at that time. Thus, an exit through cause $j$ at a later time contributes only if no cause has already removed the member. The probability of no exit by time $T$ is $S(T)$.

Theorem: Under the assumptions above, $S$ is absolutely continuous and satisfies $$ S'(t)=-S(t)\mu_{\mathrm{tot}}(t) $$ for almost every $t$. The probability $q_j(T)$ of a first exit through cause $j$ by $T$ is $\int_0^T S(t)\mu_j(t)\,dt$, and the probabilities balance: $$ S(T)+\sum_{j=1}^{J}q_j(T)=1. $$

Why the Formula Holds

The survival formula follows by solving the almost-everywhere differential equation, with the initial condition $S(0)=1$. Since $S(t)>0$ on a finite interval with integrable total intensity, the chain rule for absolutely continuous functions gives $(\log S(t))'=-\mu_{\mathrm{tot}}(t)$ almost everywhere. Integration from 0 to $t$ yields $\log S(t)=-\int_0^t\mu_{\mathrm{tot}}(u)\,du$, and exponentiating gives the stated expression for $S(t)$.

For cause $j$, the probability of exiting through $j$ during a short interval after $t$, conditional on being in the group at $t$, is governed by $\mu_j(t)$. Multiplying by the probability $S(t)$ of being in the group gives the cause-$j$ exit probability per unit time, $S(t)\mu_j(t)$, almost everywhere. Integrating this exit density over $[0,T]$ gives $q_j(T)$. Finally,

\sum_{j=1}^{J}q_j(T) =\int_0^T S(t)\sum_{j=1}^{J}\mu_j(t)\,dt =\int_0^T S(t)\mu_{\mathrm{tot}}(t)\,dt =1-S(T).

The last equality follows by integrating $S'(t)=-S(t)\mu_{\mathrm{tot}}(t)$ and using $S(0)=1$. This proves the probability balance. If the intensities are discontinuous, the integral identities still apply; the differential equation need not hold at the discontinuity itself.

Worked Example: Constant Competing Intensities

In a hypothetical active-member group, withdrawal has constant intensity $0.02$ per year and death has constant intensity $0.03$ per year. Assuming these are the only competing exits, find the probability of each type of first exit within 5 years and the probability of no exit. Give probabilities to four decimal places.

Method. Add the competing intensities to obtain the total intensity, calculate survival, then integrate each cause-specific intensity against survival. This avoids treating each cause as if it acted alone.

The total intensity is $0.02+0.03=0.05$ per year. For $0\le t\le5$, survival is $S(t)=e^{-0.05t}$. Therefore,

q_{\mathrm{withdrawal}}(5) =\int_0^5 0.02e^{-0.05t}\,dt =\frac{0.02}{0.05}(1-e^{-0.25}) =0.4(0.2211992169) =0.0884796868.

For death,

q_{\mathrm{death}}(5) =\int_0^5 0.03e^{-0.05t}\,dt =\frac{0.03}{0.05}(1-e^{-0.25}) =0.6(0.2211992169) =0.1327195301.

The probability of no exit is $S(5)=e^{-0.25}=0.7788007831$. Rounded to four decimals, the withdrawal, death, and no-exit probabilities are $0.0885$, $0.1327$, and $0.7788$. They sum to $1.0000$. Using the unrounded values, the probability balance is exactly 1.

Check and trap avoided. The cause probabilities sum to $0.2211992169=1-e^{-0.25}$, the total probability of an exit by year 5. Calculating $1-e^{-0.02(5)}=0.0952$ for withdrawal would instead treat death as absent; it is not the probability of withdrawal in the competing-exit model.

Rates That Change Over Time

The cause-specific intensities need not remain constant or keep the same relative proportions. If the rates change, the total accumulated intensity determines survival up to each time, and each cause is weighted by that survival at its own times of exposure. In general, one cannot find a cause probability by multiplying the total exit probability by an unweighted average share for that cause.

Worked Example: Piecewise-Constant Cause Intensities

In a hypothetical group, the intensity of exit through cause A is $0.10$ per year from time 0 to 1 and $0.20$ per year from time 1 to 2. The intensity of exit through cause B is $0.20$ per year from time 0 to 1 and $0.10$ per year from time 1 to 2. Find the probability of each first exit by time 2 and the probability of remaining in the group. Give probabilities to four decimal places.

Method. Calculate survival from the total intensity, then integrate each cause’s intensity over both intervals. The rates are discontinuous at time 1, so use the integral formulas rather than assuming a pointwise differential equation there.

The total intensity is $0.30$ per year on both intervals. Hence $S(t)=e^{-0.30t}$, including $S(1)=e^{-0.30}=0.7408182207$, and $S(2)=e^{-0.60}=0.5488116361$. Define the common interval integral

\int_0^1 e^{-0.30t}\,dt =\frac{1-e^{-0.30}}{0.30} =0.8639392644.

Cause A has intensity $0.10$ in the first year and $0.20$ in the second. On the second interval, write $t=1+u$, so the survival factor is $e^{-0.30}e^{-0.30u}$. Thus,

\begin{aligned} q_A(2) &=0.10\int_0^1e^{-0.30t}\,dt +0.20e^{-0.30}\int_0^1e^{-0.30u}\,du\\ &=0.8639392644(0.10+0.20e^{-0.30})\\ &=0.8639392644(0.2481636441) =0.2143983162. \end{aligned}

For cause B, the two cause-specific rates are reversed:

\begin{aligned} q_B(2) &=0.20\int_0^1e^{-0.30t}\,dt +0.10e^{-0.30}\int_0^1e^{-0.30u}\,du\\ &=0.8639392644(0.20+0.10e^{-0.30})\\ &=0.8639392644(0.2740818221) =0.2367900478. \end{aligned}

Rounded to four decimals, $q_A(2)=0.2144$, $q_B(2)=0.2368$, and $S(2)=0.5488$. These displayed values sum to $1.0000$; the unrounded probabilities also sum to 1. The cause probabilities differ because more cause-A intensity occurs later, when fewer members remain exposed.

Check and trap avoided. The total exit probability is $1-e^{-0.60}=0.4511883639$, and $0.2143983162+0.2367900478=0.4511883640$ at the displayed intermediate precision. Assigning half of the total exit probability to each cause would be wrong: the cause rates are not in a constant 1:1 ratio over time.

Worked Example: Proportional Intensities That Vary With Time

For a hypothetical care-state group, let the cause-A intensity be $0.10(1+t)$ per year and the cause-B intensity be $0.20(1+t)$ per year for $0\le t\le1$, where $t$ is measured in years since entry. Find the probability of each first exit by year 1 and the probability of no exit. Give probabilities to four decimal places.

Method. Integrate the total time-varying intensity to obtain survival, and then integrate each cause-specific intensity against that survival. The common time-varying factor means the cause shares are constant, but survival still has to be included.

The total intensity is $0.30(1+t)$. Its integral from 0 to $t$ is $0.30t+0.15t^2$, so

S(t)=e^{-0.30t-0.15t^2}, \qquad S(1)=e^{-0.45}=0.6376281516.

The total exit probability is $1-S(1)=0.3623718484$. Since the cause-A and cause-B intensities are respectively one-third and two-thirds of the total intensity at every time, their exit probabilities have those same shares of the total exit probability. Equivalently, integrating each intensity against $S(t)$ gives

\begin{aligned} q_A(1)&=\frac{1}{3}(1-e^{-0.45})=0.1207906161,\\ q_B(1)&=\frac{2}{3}(1-e^{-0.45})=0.2415812323. \end{aligned}

To see why the shares follow, the integrands satisfy $\mu_A(t)=\frac13\mu_{\mathrm{tot}}(t)$ and $\mu_B(t)=\frac23\mu_{\mathrm{tot}}(t)$ at every time. Taking these constant factors outside the integrals gives $q_A(1)=\frac13\int_0^1S(t)\mu_{\mathrm{tot}}(t)\,dt=\frac13(1-S(1))$, and similarly for cause B.

Rounded to four decimals, the probabilities are $0.1208$ for A, $0.2416$ for B, and $0.6376$ for no exit. They sum to $1.0000$. The constant cause shares work here because the intensity ratio is constant; they would not generally work if that ratio changed with time.

Conditional Probabilities and Exam Checks

The same reasoning applies if a member is known to be in the group at some later time $s$. Conditional on still being there at $s$, survival to $t\ge s$ is

{}_sS(t)=\exp\!\left(-\int_s^t\mu_{\mathrm{tot}}(u)\,du\right), \qquad P(\text{first exit through }j\text{ in }[s,T]\mid\text{in group at }s) =\int_s^T {}_sS(t)\mu_j(t)\,dt.

This is the same formula with the starting time reset to $s$. In exam calculations, identify the interval over which exposure is being measured, use the total intensity for survival over that interval, and use the relevant cause-specific intensity in the exit integral. A final check should include all causes and the probability of no exit.

Use total intensity for survival. Omitting a competing cause overstates the time members remain exposed.
Use cause-specific intensity for the exit being counted. The total intensity alone does not identify which cause produces an exit.
Respect time variation. If cause shares change, integrate each cause separately rather than applying a single share to total exits.
Handle discontinuities with integrals. The survival equation holds almost everywhere under the stated assumptions; it need not hold pointwise at a rate change.

Check Your Understanding

For each problem, assume the stated causes are mutually exclusive first exits and no other exits are possible. Give probabilities to four decimal places.

A group has constant cause-A intensity $0.04$ per year and cause-B intensity $0.01$ per year. Find the probability of each exit and of no exit by year 3. (Answer: $q_A=0.1114,\ q_B=0.0279,\ S(3)=0.8607$.)
On $0\le t\le2$, two causes have intensities $\mu_A(t)=0.06(1+t)$ and $\mu_B(t)=0.04(1+t)$. Find each cause probability and the no-exit probability by time 2. (Answer: $q_A=0.1978,\ q_B=0.1319,\ S(2)=0.6703$.)
From time 0 to 1, two causes each have intensity $0.10$ per year. From time 1 to 2, their intensities are $0.30$ and $0.10$ per year, respectively. Find the probability of each cause and of no exit by time 2. (Answer: $q_A=0.2931,\ q_B=0.1581,\ S(2)=0.5488$.)
Two causes have constant intensities $0.02$ and $0.08$ per year. Find the probability that cause B is the first exit by year 4, the probability of no exit by year 4, and the conditional probability that B caused the exit given that an exit occurred by year 4. (Answer: $q_B=0.2637,\ S(4)=0.6703,\ P(B\mid\text{exit by }4)=0.8000$.)

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