Closed-Form Exit Probabilities Under Constant Intensities

From Constant Intensities to Interval Exit Probabilities

In a competing decrement model, an individual can leave an active group through one of several causes, such as death, withdrawal, or entry into a benefit state. A cause-specific intensity describes the instantaneous rate of exit through one cause among individuals still active. It is a rate per unit time, not itself the probability of exiting during an interval.

When the cause-specific intensities are constant, the probability of exit through each cause over a specified interval has a closed form. The calculation combines two features: the chance that the individual remains in the group until a given time, and the intensity of the particular cause at that time.

Definition: Suppose there are $r$ mutually exclusive exit causes, with constant cause-specific intensities $\lambda_1,\ldots,\lambda_r$ over an interval of length $h$. The total intensity is $\Lambda=\sum_{j=1}^{r}\lambda_j$. Let $Q_j(h)$ denote the probability, from the start of the interval, of exiting through cause $j$ before its end. For $\Lambda>0$, $$ Q_j(h)=\frac{\lambda_j}{\Lambda}\left(1-e^{-\Lambda h}\right). $$ The probability of remaining in the group throughout the interval is $e^{-\Lambda h}$. If $\Lambda=0$, then every $\lambda_j=0$, so $Q_j(h)=0$ for every cause and the probability of remaining is 1.

All intensities and $h$ must use the same time unit. For example, an intensity stated per month must be paired with an interval measured in months. The formula requires constant intensities over the interval; it does not directly apply when an intensity varies with time.

Why the Formula Holds

At elapsed time $s$, the probability of still being active is $e^{-\Lambda s}$, since the total intensity of leaving is $\Lambda$. Among individuals active at that time, the instantaneous exit rate through cause $j$ is $\lambda_j$. Thus, the unconditional density of an exit through cause $j$ at time $s$ is the product of the probability of still being active and the cause-specific intensity.

f_j(s)=e^{-\Lambda s}\lambda_j

To obtain the probability of an exit through cause $j$ during the interval, integrate this density from 0 to $h$. When $\Lambda>0$, the integral gives the stated expression.

Q_j(h)=\int_0^h \lambda_j e^{-\Lambda s}\,ds =\lambda_j\left[\frac{-e^{-\Lambda s}}{\Lambda}\right]_0^h =\frac{\lambda_j}{\Lambda}\left(1-e^{-\Lambda h}\right)

If $\Lambda=0$, dividing by $\Lambda$ is not valid. Because every intensity is nonnegative and their sum is zero, each intensity must be zero. The cause-specific exit density is then zero throughout the interval, so its integral is zero. This gives the separate all-zero case without using a zero denominator.

Check the time unit.
Express every cause intensity and the interval length using the same unit, then confirm the intensities are constant over that interval.

Find the total intensity.
Add the cause-specific intensities to obtain $\Lambda$.

Choose the appropriate case.
If $\Lambda>0$, use the closed-form formula. If $\Lambda=0$, every cause-specific exit probability is zero.

Calculate each cause probability.
For positive $\Lambda$, multiply the common interval exit factor $1-e^{-\Lambda h}$ by the cause's intensity share $\lambda_j/\Lambda$.

Worked Applications

Worked Example: Three Constant Causes Over Two Years

A hypothetical active group is subject to three mutually exclusive exits: death, withdrawal, and entry into a specified benefit state. Their constant intensities are respectively $0.04$, $0.03$, and $0.05$ per year. Find the probability of exit through each cause during the next two years and the probability of remaining active. Treat the stated causes as the complete set of exits; no interest rate or discounting is involved.

Method. Add the cause-specific intensities, calculate the probability of any exit, and allocate that probability according to each cause's share of the total intensity.

The total intensity is $$ \Lambda=0.04+0.03+0.05=0.12\text{ per year}. $$

The interval is $h=2$ years, so $$ \Lambda h=0.12(2)=0.24,\qquad e^{-0.24}=0.7866279. $$ The probability of any exit is therefore $$ 1-e^{-0.24}=1-0.7866279=0.2133721. $$

For death, the intensity share is $0.04/0.12=1/3$, giving $$ Q_{\text{death}}(2)=\frac{1}{3}(0.2133721)=0.0711240. $$ For withdrawal, the share is $0.03/0.12=1/4$, giving $$ Q_{\text{withdrawal}}(2)=\frac{1}{4}(0.2133721)=0.0533430. $$ For entry into the benefit state, the share is $0.05/0.12=5/12$, giving $$ Q_{\text{benefit}}(2)=\frac{5}{12}(0.2133721)=0.0889051. $$ The probability of remaining active is $e^{-0.24}=0.7866279$. Rounded to four decimal places, the exit probabilities are $0.0711$, $0.0533$, and $0.0889$, and the remaining probability is $0.7866$.

Check. Before rounding, the exit probabilities sum to $0.0711240+0.0533430+0.0889051=0.2133721$, the total exit probability. Adding the remaining probability gives $0.2133721+0.7866279=1$.

Common mistake avoided: The fraction $0.04/0.12$ is death’s share of exits under these constant intensities, not the two-year probability of death. The interval exit factor $1-e^{-0.24}$ must also be applied.

Worked Example: Monthly Intensities and Expected Exits

A hypothetical cohort of 800 independent individuals is active at the start of a three-month period. Each individual is subject to three mutually exclusive exits with constant monthly intensities: withdrawal $0.01$, entry into a care state $0.02$, and death $0.01$. These are the only exits. Find the probability of each type of exit and the expected number of individuals in each exit category and still active at the end of the period. No interest rate or discounting is relevant.

Method. Use months for both the intensities and the interval, calculate the individual probabilities, and multiply each probability by 800 to obtain expected counts. Under the stated independent, identical assumptions, the counts have a multinomial distribution with the calculated exit probabilities and the remaining probability.

The total intensity is $$ \Lambda=0.01+0.02+0.01=0.04\text{ per month}. $$ Over $h=3$ months, $\Lambda h=0.04(3)=0.12$, so $$ 1-e^{-\Lambda h}=1-e^{-0.12}=1-0.8869204=0.1130796. $$

The intensity shares are $0.01/0.04=0.25$ for withdrawal, $0.02/0.04=0.50$ for entry into care, and $0.01/0.04=0.25$ for death. Therefore, the respective exit probabilities are $$ Q_{\text{withdrawal}}(3)=0.25(0.1130796)=0.0282699, $$ $$ Q_{\text{care}}(3)=0.50(0.1130796)=0.0565398, $$ $$ Q_{\text{death}}(3)=0.25(0.1130796)=0.0282699. $$ The probability of remaining active is $e^{-0.12}=0.8869204$.

Multiplying each probability by 800 gives expected counts:

Withdrawal: $800(0.0282699)=22.616$.
Entry into care: $800(0.0565398)=45.232$.
Death: $800(0.0282699)=22.616$.
Still active: $800(0.8869204)=709.536$.

The expected counts are rounded to three decimal places. Their displayed values sum to $22.616+45.232+22.616+709.536=800.000$.

Common mistake avoided: Pairing a monthly rate with an interval of three years would give the wrong exponent. Here the exponent is $0.04$ per month multiplied by 3 months, not by 3 years or by 0.25 years.

Worked Example: A Zero Cause and the All-Zero Case

Consider a hypothetical one-year model with three possible exits: a specified event, withdrawal, and death. In Plan A, their constant intensities are $0.04$, $0$, and $0.06$ per year. In Plan B, all three intensities are zero. For each plan, calculate the cause-specific exit probabilities and the probability of remaining active. Assume the listed causes are the complete set of exits; there is no interest or discounting.

Method. In Plan A, apply the formula because the total intensity is positive, while retaining the zero intensity for withdrawal. In Plan B, apply the separate $\Lambda=0$ case rather than dividing by zero.

For Plan A, $$ \Lambda=0.04+0+0.06=0.10\text{ per year},\qquad 1-e^{-\Lambda h}=1-e^{-0.10}=0.0951626, $$ where $h=1$ year. The specified-event intensity share is $0.04/0.10=0.4$, so its exit probability is $0.4(0.0951626)=0.0380650$. Withdrawal has intensity share $0/0.10=0$, so its exit probability is $0$. Death has intensity share $0.06/0.10=0.6$, so its exit probability is $0.6(0.0951626)=0.0570976$. The probability of remaining active is $e^{-0.10}=0.9048374$.

For Plan B, $$ \Lambda=0+0+0=0. $$ Every cause-specific intensity is zero, so the probability of exiting through each cause is $0$, and the probability of remaining active for the year is $1$. The expression $\lambda_j/\Lambda$ is not used for Plan B because its denominator would be zero.

Check. In Plan A, the exits sum to $0.0380650+0+0.0570976=0.0951626$, and adding $0.9048374$ gives 1. In Plan B, the total probability is $0+0+0+1=1$.

Common mistake avoided: A zero intensity for one cause does not imply that no one exits through any cause. In Plan A, withdrawal has zero probability, but the other causes still produce exits. Only when the total intensity is zero are all cause-specific exit probabilities zero.

Interpretation and Exam Traps

The factor $\lambda_j/\Lambda$ allocates the total probability of exit over the interval among causes when the intensities are constant. It does not give the probability of exit from the active group by itself. The exponential factor accounts for the possibility of remaining active as time passes. This distinction matters especially over longer intervals, when omitting that factor can produce a cause probability that is much too large.

A second frequent error is to use the formula mechanically when $\Lambda=0$. The correct approach is to check the total before dividing: if the total is zero, nonnegativity of the cause intensities implies that each is zero, and no exit occurs. If just one cause intensity is zero while another is positive, the total remains positive and the formula gives probability zero for the zero-intensity cause.

The closed form also depends on constancy over the stated interval. If intensities change during follow-up, one cannot replace them with a single constant total and use this expression without further justification. In every application, check the cause set, the time units, the constancy assumption, and the value of the total intensity before calculating.

Key takeaway: For constant cause-specific intensities, each cause's interval exit probability is its share of the total intensity multiplied by the probability of any exit, $1-e^{-\Lambda h}$. If the total intensity is zero, use the separate result that every exit probability is zero and the remaining probability is one.

Check Your Understanding

For each problem, use the stated constant intensities and interval, and give probabilities to four decimal places unless expected counts are requested.

A hypothetical active group has two mutually exclusive exit causes with constant intensities $0.02$ and $0.03$ per year. Find the probability of exit through each cause and the probability of remaining active over one year. (Answer: cause probabilities $0.0195$ and $0.0293$; remaining probability $0.9512$.)
Over three months, three mutually exclusive causes have constant intensities $0.01$, $0.02$, and $0.01$ per month. Find each cause's exit probability and the remaining probability. (Answer: cause probabilities $0.0283$, $0.0565$, and $0.0283$; remaining probability $0.8869$.)
For a two-year interval, three constant cause-specific intensities are $0.05$, $0$, and $0.15$ per year. Find the probability for each cause and the probability of remaining active. (Answer: cause probabilities $0.0824$, $0.0000$, and $0.2473$; remaining probability $0.6703$.)
A cohort of 5,000 independent individuals is subject to two constant exit intensities of $0.04$ and $0.06$ per year for 1.5 years. Find the expected number exiting through each cause and the expected number remaining active, to one decimal place. (Answer: $278.6$ and $417.9$ exits; $4{,}303.5$ remaining.)
Three constant cause-specific intensities are all zero over an eight-month interval. State the probability of exit through each cause and the probability of remaining active. (Answer: each cause probability is $0$; remaining probability is $1$.)

What You'll Learn