Competing Exit Probabilities Under Constant Forces

Introduction

When several events can end a policy, pension membership, or other state of interest, each event competes to be the first exit. A cause-specific force describes the instantaneous rate of one type of exit among those still in the state. To find probabilities over a full interval, the calculation must account for the fact that a person can experience only the first exit: after any cause occurs, the person is no longer exposed to the others.

Under constant cause-specific forces, the total force is constant as well. This gives a direct way to calculate both the probability of surviving all competing causes and the probability of exiting by each particular cause. The formulas apply over any specified duration, provided the forces remain constant throughout that interval.

Definition: Suppose a person is in a state at time 0 and can leave it through causes $1,\ldots,m$. The cause-specific force for cause $j$, denoted $\lambda_j$, is constant if, for a person still in the state at time $t$, the probability of an exit by cause $j$ during the next short interval of length $h$ is $\lambda_j h+o(h)$, as $h$ tends to zero. Here $o(h)/h$ tends to zero. The total force is $\Lambda=\sum_{j=1}^{m}\lambda_j$.

The $\lambda_j$ are rates per unit time, not probabilities over a full year or other interval. For example, a force of $0.10$ per year does not mean there is exactly a 10% probability of exit by that cause in a year. The probability depends on the chance of remaining exposed to the competing causes throughout the year.

Survival and Cause-Specific Exit Probabilities

Let $S(t)$ be the probability of no exit from the state by any cause during the first $t$ time units. With constant forces, the probability of remaining in the state for a further short period is reduced by the sum of the cause-specific forces. Thus $S(t)$ satisfies $S'(t)=-\Lambda S(t)$, with $S(0)=1$. Its solution is $S(t)=e^{-\Lambda t}$.

Theorem: If the $m$ causes have constant nonnegative forces $\lambda_1,\ldots,\lambda_m$ over an interval of length $t$, and $\Lambda=\sum_{j=1}^{m}\lambda_j$, then $$ {}_tp^{(\mathrm{all})}=e^{-\Lambda t} $$ is the probability of no exit by any cause. The probability of exit by cause $j$ during the interval is $$ {}_tq^{(j)}=\frac{\lambda_j}{\Lambda}\left(1-e^{-\Lambda t}\right), \qquad \Lambda>0. $$ If $\Lambda=0$, all the forces are zero, the probability of survival is 1, and every cause-specific exit probability is 0.

In this notation, the superscript $(\mathrm{all})$ indicates survival from every listed cause, while $(j)$ identifies an exit by cause $j$. The probabilities refer to the first exit during the interval. They do not describe the probability that cause $j$ would act if the other causes were removed.

Why the Cause-Specific Formula Holds

At time $s$, the probability of still being in the state is $e^{-\Lambda s}$. Conditional on still being there, the instantaneous rate of exit by cause $j$ is $\lambda_j$. Therefore, the density of an exit by cause $j$ at time $s$ is $\lambda_j e^{-\Lambda s}$. Integrating this density over the interval gives the cause-specific exit probability.

{}_tq^{(j)} =\int_0^t \lambda_j e^{-\Lambda s}\,ds =\frac{\lambda_j}{\Lambda}\left(1-e^{-\Lambda t}\right).

This derivation shows why simply multiplying $\lambda_j$ by $t$ is not generally correct: that multiplication does not allow for the reduction in the number of people still exposed as exits occur. It also shows that, when $\Lambda>0$, the cause-specific share of all exits is

\frac{{}_tq^{(j)}}{1-{}_tp^{(\mathrm{all})}}=\frac{\lambda_j}{\Lambda}.

That share is independent of the interval length when all forces are constant. The absolute probability of exit by each cause still depends on $t$, because a longer interval gives more opportunity for any cause to occur.

Worked Example: Two Competing Causes Over Two Years

A hypothetical long-term benefit remains in force until a member either enters a specified care state (cause 1) or dies before entering that state (cause 2). Assume the constant forces are $\lambda_1=0.10$ per year and $\lambda_2=0.20$ per year. Calculate the probability of remaining in force for two years and the probability of exit by each cause within those two years. Give probabilities to four decimals.

Method. Add the cause-specific forces to obtain the total force, calculate survival under that total force, and allocate the total exit probability to causes in proportion to their forces.

Step 1: Total force.

\Lambda=0.10+0.20=0.30 \text{ per year}.

Step 2: Survival for two years.

{}_2p^{(\mathrm{all})}=e^{-(0.30)(2)}=e^{-0.60}=0.5488116\ldots

Thus the probability of remaining in force is $0.5488$ to four decimals. The total probability of exit is

1-0.5488116\ldots=0.4511884\ldots.

Step 3: Allocate exits to the two causes. The force shares are $0.10/0.30=1/3$ for cause 1 and $0.20/0.30=2/3$ for cause 2.

{}_2q^{(1)}=\frac{1}{3}(0.4511884\ldots)=0.1503961\ldots, \qquad {}_2q^{(2)}=\frac{2}{3}(0.4511884\ldots)=0.3007923\ldots.

The four-decimal answers are $0.1504$ for cause 1 and $0.3008$ for cause 2. Check: $0.5488+0.1504+0.3008=1.0000$. The common mistake avoided here is reporting $\lambda_j t$ as the cause-specific probability; for cause 1 that would give $0.10(2)=0.20$, rather than the correct $0.1504$.

Using the Formula for Several Causes

With more than two causes, the procedure is unchanged: sum every force to obtain $\Lambda$, calculate the probability of any exit, and multiply that probability by each cause's fraction of the total force. Each cause must be included exactly once. This provides a useful arithmetic check, because the cause-specific exit probabilities must sum to the total exit probability.

Worked Example: Three Causes Over a Fraction of a Year

For a hypothetical membership benefit, three constant exit forces are $0.08$, $0.12$, and $0.20$ per year. Calculate survival and the probability of exit by each cause over 1.5 years. Give probabilities to four decimals.

Method. Treat 1.5 as the interval length in years, use the sum of the forces for survival, and then apply each cause's share of the total force.

Step 1: Total force and survival.

\Lambda=0.08+0.12+0.20=0.40 \text{ per year}, \qquad {}_{{1.5}}p^{(\mathrm{all})}=e^{-(0.40)(1.5)}=e^{-0.60}=0.5488116\ldots.

The probability of any exit is $1-0.5488116\ldots=0.4511884\ldots$.

Step 2: Cause shares. The three shares are $0.08/0.40=0.20$, $0.12/0.40=0.30$, and $0.20/0.40=0.50$.

{}_{1.5}q^{(1)}=0.20(0.4511884\ldots)=0.0902377\ldots,

{}_{1.5}q^{(2)}=0.30(0.4511884\ldots)=0.1353565\ldots, \qquad {}_{1.5}q^{(3)}=0.50(0.4511884\ldots)=0.2255942\ldots.

To four decimals, the answers are $0.0902$, $0.1354$, and $0.2256$, respectively, with survival $0.5488$. Check: the unrounded cause probabilities sum to $0.4511884\ldots$, and adding survival gives 1. A common error is to use a one-year interval because the forces are quoted per year; the forces are annual rates, while the stated duration is 1.5 years.

Worked Example: Recovering Forces From Survival and a Cause Ratio

Over a two-year interval, a hypothetical group has probability $e^{-0.8}$ of remaining in the original state. Two causes compete to end that state, and their constant forces are in the ratio $1:3$. Find the two forces per year and the probability of exit by each cause. Give forces to four significant figures and probabilities to four decimals.

Method. First recover the total force from the survival probability. Then split that total using the stated force ratio and allocate the total exit probability.

Step 1: Recover the total force. Since $e^{-\Lambda(2)}=e^{-0.8}$, equating exponents gives

2\Lambda=0.8,\qquad \Lambda=0.4 \text{ per year}.

Step 2: Split the total force. The ratio $1:3$ has four total parts, so the forces are

\lambda_1=\frac{1}{4}(0.4)=0.10, \qquad \lambda_2=\frac{3}{4}(0.4)=0.30 \quad\text{per year}.

Step 3: Calculate exit probabilities. Survival is $e^{-0.8}=0.4493289\ldots$, so the total exit probability is $1-e^{-0.8}=0.5506710\ldots$. The cause shares are $0.10/0.40=0.25$ and $0.30/0.40=0.75$. Therefore,

{}_2q^{(1)}=0.25(0.5506710\ldots)=0.1376678\ldots, \qquad {}_2q^{(2)}=0.75(0.5506710\ldots)=0.4130032\ldots.

The forces are $0.1000$ and $0.3000$ per year to four significant figures. The exit probabilities are $0.1377$ and $0.4130$ to four decimals. Check: $0.4493+0.1377+0.4130=1.0000$. The ratio $1:3$ applies to the forces, not directly to survival; it determines the shares of the total exits.

Assumptions, Checks, and Exam Traps

The formulas rely on a specific model. The person starts in the state at the beginning of the interval, the listed causes are the only ways to leave it, each force is constant over the full interval, and an exit removes the person from exposure to the other causes. The calculation concerns the first exit. It does not model later transitions after that exit, and it does not apply unchanged if the forces vary with time.

Under continuous-time constant forces, simultaneous exits have probability zero. This is why the cause-specific probabilities partition the probability of any exit. In every calculation, verify

{}_tp^{(\mathrm{all})}+\sum_{j=1}^{m}{}_tq^{(j)}=1.

The most common errors are unit and interpretation errors. If forces are annual, $t$ must be measured in years; if a duration is given in months, convert it before multiplying by an annual force. Do not add the forces and then treat the sum as an exit probability. The sum is a force, and the total exit probability is $1-e^{-\Lambda t}$. Likewise, a cause-specific force is not itself a cause-specific probability: the probability must reflect survival from all competing causes until the time of that exit.

These calculations are useful whenever a decrement model assumes constant cause-specific rates over a defined interval. They provide probabilities suitable for decrement-table calculations while preserving the distinction between rates among those still exposed and probabilities over the whole interval. The final partition check is especially valuable under exam conditions: if survival plus the cause-specific exits does not equal one, revisit the interval length, total force, and allocation of exits.

Check Your Understanding

For each problem, assume the stated cause-specific forces are constant over the interval and that the listed causes are the only exits. Give probabilities to four decimals unless another precision is requested.

Two causes have forces $0.05$ and $0.15$ per year. Find survival and the probability of exit by each cause over one year. (Answer: survival $0.8187$; cause 1 $0.0453$; cause 2 $0.1360$.)
For two causes with forces $0.12$ and $0.08$ per year, calculate survival and both cause-specific exit probabilities over two years. (Answer: survival $0.6703$; cause 1 $0.1978$; cause 2 $0.1319$.)
Three causes have forces $0.04$, $0.06$, and $0.10$ per year. Find the probability of no exit and the probability of each cause-specific exit over three years. (Answer: survival $0.5488$; cause 1 $0.0902$; cause 2 $0.1354$; cause 3 $0.2256$.)
Two causes have constant forces in the ratio $3:4$. Survival over two years is $0.70$. Find the total force per year and the probability of exit by each cause. (Answer: total force $0.1783$ per year; cause 1 $0.1286$; cause 2 $0.1714$.)
Two causes have forces $0.09$ and $0.21$ per year. Find the probability of any exit over six months and the probability that an exit during those six months is due to cause 2. (Answer: probability of any exit $0.1393$; conditional probability of cause 2 given an exit $0.7000$.)

What You'll Learn