Remaining Probability Under Constant Intensities

Introduction

In a competing decrement model, a person may leave a defined group through any one of several causes. For example, an active policyholder might leave the active group through death, surrender, or transition into a benefit-eligible state. The probability of remaining in the group depends on the combined effect of all causes that count as exits.

When each cause-specific intensity is constant over the interval, the survival differential equation has a simple exponential solution. The key steps are to add the intensities, express the interval in the same time unit as the rates, and solve the differential equation. The result is a probability of remaining in the group, not a probability assigned to any particular exit cause.

Definition: Suppose a person is in a specified group at time 0 and can leave through one of $m$ mutually exclusive first-exit causes. Let $\mu^{(j)}$ be the constant cause-specific intensity of exit through cause $j$, measured per unit time while the person remains in the group. The total intensity is $\mu=\sum_{j=1}^{m}\mu^{(j)}$. Write $R(t)=P(\text{remain in the group through time }t)$, given membership at time 0.

The model concerns remaining in the group without an exit during the interval. Its usual assumptions are that the listed causes account for all exits relevant to the question, the cause-specific intensities remain constant during the interval, and there is no re-entry after an exit. Each cause-specific intensity is conditional on still being in the group immediately before the time considered. These are intensities, not probabilities for a whole year or other fixed period.

Solving the Survival Differential Equation

For competing exits, the total intensity is the sum of the cause-specific intensities. A short interval $dt$ therefore has an exit probability approximately $\mu\,dt$, conditional on being in the group at its start. This gives the survival differential equation

\frac{dR(t)}{dt}=-\mu R(t), \qquad R(0)=1.

The negative sign reflects that remaining probability decreases as time passes. Because $\mu$ is constant, the equation can be solved by separating variables and integrating.

Separate variables.
For $R(t)>0$, divide by $R(t)$ and multiply by $dt$: $\frac{dR}{R}=-\mu\,dt$.

Integrate from 0 to $t$.
The initial condition $R(0)=1$ gives $\int_1^{R(t)}\frac{dR}{R}=-\int_0^t\mu\,ds$, so $\ln R(t)=-\mu t$.

Exponentiate.
Thus $R(t)=e^{-\mu t}$. Substituting $\mu=\sum_{j=1}^{m}\mu^{(j)}$ gives the remaining probability under competing constant intensities.

Theorem: If all cause-specific exit intensities are constant over an interval of length $t$, measured in the same time unit as the intensities, then $$ R(t)=\exp\left(-t\sum_{j=1}^{m}\mu^{(j)}\right). $$ The result applies to the probability of no exit through any listed cause during that interval. It does not by itself give the probability of exit through an individual cause.

The same formula applies conditionally from a later time $s$, provided the intensities stay constant and the person is in the group at time $s$. Over an interval from $s$ to $t$, the relevant length is $t-s$, not $t$:

P(\text{remain in the group from }s\text{ to }t\mid\text{in the group at }s) =\exp\left(-\mu(t-s)\right).

This conditional form follows by starting the survival calculation at time $s$, where the conditional probability of remaining for zero additional time is 1. Using the total time since original entry instead would answer a different question.

Worked Calculations

Worked Example: Combining Two Constant Exit Intensities

A hypothetical active group has two possible exits: death, with constant intensity $0.08$ per year, and surrender, with constant intensity $0.12$ per year. Find the probability that a person in the group at entry remains in it for three years. Give the probability to four decimals.

Method. Add the cause-specific intensities to obtain the total intensity, then use the exponential solution. Both rates are stated per year, and the interval is three years.

Step 1: Find the total intensity.

\mu=0.08+0.12=0.20\text{ per year}.

Step 2: Substitute the interval length.

R(3)=e^{-(0.20)(3)}=e^{-0.60}=0.5488116\ldots

Rounded to four decimals, the probability is 0.5488. The result is between zero and one, and it is less than the probability of remaining for one year, as it should be for a longer interval. The common mistake to avoid is treating $0.20$ as a three-year probability rather than a rate per year; the exponential solution accounts for the accumulating exposure over all three years.

Worked Example: Using a Conditional Starting Time

In a hypothetical disability income model, a person leaves the active group through death at a constant intensity of $0.06$ per year, withdrawal at $0.10$ per year, or transition into disability at $0.04$ per year. Given that the person is still active at month 6, find the probability of remaining active through month 24. Give the answer to four decimals.

Method. Calculate the length of the interval conditional on being active at month 6, convert months to years, and apply the constant-intensity formula.

Step 1: Calculate the total intensity.

\mu=0.06+0.10+0.04=0.20\text{ per year}.

Step 2: Calculate the conditional interval length. The interval runs from month 6 to month 24, which is $24-6=18$ months, or $18/12=1.5$ years.

Step 3: Calculate the probability.

P(\text{remain active from month 6 to month 24}\mid\text{active at month 6}) =e^{-(0.20)(1.5)} =e^{-0.30} =0.7408182\ldots

Rounded to four decimals, the answer is 0.7408. The common mistake is to use 24 months as though the condition were “active at entry.” That would use a two-year interval and answer a different question. Here, the condition resets the starting point to month 6.

Worked Example: Expected Number Remaining in a Hypothetical Cohort

A hypothetical cohort contains 640 people, all in the same defined group at time 0. Each person faces constant exit intensities of $0.03$ per year for death, $0.05$ per year for withdrawal, and $0.02$ per year for a transition to another state. Find (a) the probability that an individual remains in the group for 2.5 years and (b) the expected number of the original cohort remaining at that time. Give the probability to four decimals and the expected number to the nearest cent.

Method. First find the individual remaining probability using the total intensity. Then multiply that probability by the initial cohort size to obtain the expected count. No independence assumption between different people is needed for this expected value.

Step 1: Add the intensities.

\mu=0.03+0.05+0.02=0.10\text{ per year}.

Step 2: Calculate the probability for one individual.

R(2.5)=e^{-(0.10)(2.5)}=e^{-0.25}=0.7788008\ldots

To four decimals, the probability is 0.7788.

Step 3: Calculate the expected number remaining. Let $I_k$ equal 1 if person $k$ remains and 0 otherwise. The number remaining is the sum of these indicators, so its expectation is the sum of their expectations:

E\left[\sum_{k=1}^{640} I_k\right] =\sum_{k=1}^{640} E[I_k] =640(0.7788008\ldots) =498.4325\ldots

Rounded to the nearest cent, the expected number is 498.43 people. This is an expected count, not a guarantee that exactly 498.43 people will be observed. The check is that the result is below the initial 640, consistent with a probability less than one. A common error is to report the probability $0.7788$ as the expected cohort count without multiplying by 640.

Interpretation and Limits

The exponential form is a direct consequence of the constant total intensity: over every instant that a person remains in the group, the same total rate governs the chance of exit. This produces a survival curve that starts at 1 and decreases smoothly. For $\mu>0$, $e^{-\mu t}$ is positive, decreases as $t$ increases, and tends toward zero as the interval becomes very long.

The formula is not appropriate if any cause-specific intensity changes during the interval and a constant-rate approximation is not part of the stated model. It also does not describe a model in which re-entry into the group counts as remaining in it. In either case, the assumptions used to obtain this solution do not match the question.

A frequent exam trap is to calculate $1-\mu t$ as the exact remaining probability. That expression is only a linear approximation for a short interval; in general it is not the solution of the differential equation. For example, with $\mu=0.20$ per year over three years, $1-(0.20)(3)=0.40$, whereas the exact result is $e^{-0.60}=0.5488$. The difference is material, so use the exponential formula when the model specifies constant intensities.

Before calculating, check three items: the cause-specific intensities all refer to exits from the same group; their time units agree; and the requested interval begins at the correct conditioning time. These checks prevent most errors in remaining-probability calculations.

Key takeaway: Under constant competing exit intensities, add the cause-specific intensities to obtain the total intensity, then calculate the probability of no exit over an interval of length $t$ as $e^{-\mu t}$. Match the time unit to the rates and use the interval since the stated starting condition.

Check Your Understanding

For each question, assume the stated intensities are constant while the person remains in the group. Probabilities are to four decimals unless otherwise stated.

A hypothetical group has cause-specific intensities $0.03$ and $0.07$ per year. What is the probability that a person in the group at entry remains for four years? (Answer: 0.6703)
In a hypothetical active group, death and withdrawal intensities are $0.025$ and $0.045$ per year. Given a person is active at month 6, what is the probability of remaining active through month 24? (Answer: 0.9003)
A hypothetical cohort of 1,200 people has three exit intensities, $0.04$, $0.02$, and $0.06$ per year. Find the probability of remaining for two years and the expected number remaining. (Answer: probability 0.7866; expected number 943.95)
A group has a single total exit intensity of $0.18$ per year. Given a person is in the group at year 2, calculate the probability of remaining through year 2.75. (Answer: 0.8737)
A hypothetical group has two constant exit intensities, $0.03$ and $0.05$ per year. How long does it take for the probability of remaining in the group to fall to $0.5000$? Give the time to four decimals in years. (Answer: 8.6643 years)

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