The Probability of Remaining in the Group

Introduction

In a competing decrement model, a person or policy can leave a specified group through one of several causes. The chance of remaining in the group over a short interval is therefore linked to all the cause-specific exit intensities, not to any one of them in isolation. A useful local approximation follows directly from this structure.

The key is to distinguish an intensity, which is a rate per unit time, from an interval probability. For a sufficiently short interval, the total intensity times the interval length approximates the probability of exit. The probability of remaining is approximately one minus that quantity. The order of the approximation error depends on the regularity of the intensities.

Setup and Definitions

Suppose an individual is in a group at time $t$, and can leave through one of $m$ competing causes. Let $h>0$ be a short interval, measured in the same time unit used for the intensities. For cause $j$, write $q_j(t,h)$ for the conditional probability of exiting through that cause during $[t,t+h)$, given that the individual is in the group at time $t$. Write $p_0(t,h)$ for the conditional probability of remaining in the group throughout that interval.

Definition: The cause-specific exit intensity $\mu_j(t)$ is the instantaneous rate of exit through cause $j$, conditional on being in the group at time $t$. In particular, $q_j(t,h)=\mu_j(t)h+o(h)$ as $h$ tends to zero. The total exit intensity is $\lambda(t)=\sum_{j=1}^{m}\mu_j(t)$. The probability of remaining in the group is $p_0(t,h)$.

The intensities have units such as exits per year, while $q_j(t,h)$ and $p_0(t,h)$ are probabilities. Thus $\mu_j(t)h$ is dimensionless. The earlier result that total intensity is the sum of cause intensities gives $\lambda(t)$; the next step is to connect that rate to the probability of remaining.

Assume that exits through the listed causes are mutually exclusive outcomes for the individual over the interval, and that there are no other ways to leave the group. Then remaining and exiting through one of the causes exhaust the possibilities:

p_0(t,h)+\sum_{j=1}^{m}q_j(t,h)=1.

The Short-Interval Relationship

Using the definition of each cause-specific intensity, the cause probabilities have first-order expansions $q_j(t,h)=\mu_j(t)h+o(h)$. Add these expansions over the finite set of causes and use the fact that the probabilities sum to one. This gives the general first-order relationship:

p_0(t,h)=1-\lambda(t)h+o(h), \qquad \lambda(t)=\sum_{j=1}^{m}\mu_j(t).

This relationship needs only the instantaneous-intensity limits. To state a sharper error order, more regularity is needed. Under a standard continuous-time competing decrement model with cause intensities bounded and locally Lipschitz near $t$, the local probability of exit through cause $j$ has the expansion $q_j(t,h)=\mu_j(t)h+O(h^2)$. Summing the finitely many expansions gives:

Theorem: If the cause-specific intensities are bounded and locally Lipschitz near $t$, and exits through the causes exhaust the ways to leave the group, then $$ p_0(t,h)=1-\lambda(t)h+O(h^2) $$ as $h$ tends to zero. The $O(h^2)$ term means that, for sufficiently small $h$, the absolute approximation error is bounded by a constant times $h^2$. The constant is not determined by the instantaneous intensities alone.

Why the Error Is Second Order Under Regularity

The first-order term comes from the chance of one exit during the interval. The following argument explains why the remaining terms are of order $h^2$ when the intensities are locally Lipschitz. It is a local argument; it does not solve for the probability of remaining over a longer period.

Bound the chance of any exit.
Because the cause intensities are bounded near $t$, their sum is bounded there. Over a sufficiently short interval, the probability of an exit by time $t+u$, for $0\leq u\leq h$, is therefore $O(u)$. The probability of still being in the group at that point is $1+O(u)$.

Approximate the total intensity within the interval.
Local Lipschitz continuity implies $\lambda(t+u)=\lambda(t)+O(u)$. The rate of exit at $t+u$, conditional on still being in the group, is therefore close to $\lambda(t)$, with a difference of order $u$.

Accumulate the local exit probability.
The unconditional exit rate at $t+u$, conditional on being in the group at $t$, is the probability of still being in the group at $t+u$ multiplied by its total intensity then. By Steps 1 and 2, this rate is $\lambda(t)+O(u)$. Accumulating it over $0\leq u<h$ gives total exit probability $\lambda(t)h+O(h^2)$.

Take the complement.
Remaining and exiting are exhaustive outcomes. Hence $p_0(t,h)=1-\lambda(t)h+O(h^2)$, as claimed.

The second-order error reflects two local effects: the intensities may change slightly during the interval, and an individual who exits early is no longer exposed to the competing causes later in the interval. Both effects contribute terms of order $h^2$ under the stated regularity conditions.

If only the intensity limits are known, without the additional regularity, the justified statement is $1-\lambda(t)h+o(h)$. One should not claim an $O(h^2)$ error solely from the definition of intensity. In either case, the approximation is local: it does not provide an exact probability over a substantial time interval.

Worked Example: Combining Three Exit Intensities

A hypothetical membership group has three possible exit causes. At time $t=5$ years, their intensities are $\mu_1(t)=0.03+0.002t$, $\mu_2(t)=0.010$, and $\mu_3(t)=0.005$, all per year. Estimate the probability that a member remains in the group over the next $0.02$ years using the first-order approximation. Assume the intensities are locally Lipschitz.

Method. Evaluate each intensity at the start of the interval, add them to obtain the total intensity, and subtract the first-order exit probability from one.

First evaluate the time-varying cause intensity:

\mu_1(5)=0.03+0.002(5)=0.03+0.010=0.040 \text{ per year}.

The total intensity at time $5$ is:

\lambda(5)=0.040+0.010+0.005=0.055 \text{ per year}.

The interval is $h=0.02$ years, so the first-order probability of exit is:

\lambda(5)h=0.055(0.02)=0.0011.

Thus the estimated probability of remaining is:

p_0(5,0.02)\approx 1-0.0011=0.9989.

Check. The estimated exit probability $0.0011$ and remaining probability $0.9989$ sum to $1$. Under the stated regularity, the approximation error is $O((0.02)^2)$; its numerical size cannot be calculated from the instantaneous intensities alone. The common mistake avoided here is using the wrong value of the time-varying intensity: it must be evaluated at $t=5$ before the cause rates are added.

Worked Example: Evaluating Intensities at the Start of a Short Interval

In a hypothetical benefit program, the cause-specific intensities at time $t$, measured in years, are $\mu_1(t)=0.020+0.003t$, $\mu_2(t)=0.012$, and $\mu_3(t)=0.004$, all per year. For a member still in the group at $t=2$, estimate the probability of remaining over the next $0.05$ years. Assume the intensities are locally Lipschitz.

Method. Use the beginning-of-interval intensities for the local approximation. Do not average a rate at some other time or treat the interval length as a probability.

At $t=2$, the first intensity is:

\mu_1(2)=0.020+0.003(2)=0.026 \text{ per year}.

The total intensity is $0.026+0.012+0.004=0.042$ per year. Multiplying by the interval length gives the first-order exit estimate:

\lambda(2)h=0.042(0.05)=0.0021.

Therefore, the first-order remaining-probability estimate is:

p_0(2,0.05)\approx 1-0.0021=0.9979.

Check. The exit estimate $0.0021$ plus the remaining estimate $0.9979$ equals $1$. The $O(h^2)$ error here is of order $(0.05)^2=0.0025$, but that order statement does not specify an error bound of exactly $0.0025$. The common mistake avoided is claiming the linear estimate is exact simply because the cause intensities are known at the start of the interval.

Worked Example: Converting Days to the Intensity Time Unit

A hypothetical group has two exit causes with intensities $0.018$ and $0.006$ per month, and a third cause with intensity $0.003$ per month. Estimate the probability of remaining over the next 10 days. For this calculation, take one month to be 30 days and assume locally Lipschitz intensities.

Method. Convert the interval to months before multiplying it by the total intensity.

The interval in months is $h=10/30=1/3$. The total intensity is:

\lambda=0.018+0.006+0.003=0.027 \text{ per month}.

The first-order exit estimate is:

\lambda h=0.027\left(\frac{1}{3}\right)=0.009.

The estimated probability of remaining is $1-0.009=0.991$, or $0.9910$ to four decimal places.

Check. The exit and remaining estimates sum to $0.009+0.991=1$. A common wrong calculation is $1-0.027(10)=0.73$, which treats 10 days as 10 months. Converting the interval to the rate’s time unit corrects the error.

Interpretation and Common Traps

For a short interval, the total exit intensity controls the first-order decrease in the probability of remaining. Individual cause probabilities can also be approximated separately as $\mu_j(t)h$, and their first-order estimates sum to $\lambda(t)h$. The remaining estimate is their complement to one.

Keep the following distinctions clear:

Intensity versus probability: an intensity is a rate, not an interval probability. Multiply by an interval measured in the matching time unit to obtain a first-order probability estimate.
First-order result versus error order: the intensity definition alone gives an $o(h)$ remainder. An $O(h^2)$ error requires additional regularity, such as bounded, locally Lipschitz cause intensities in the standard continuous-time model.
Local versus long-interval use: $1-\lambda(t)h$ is a short-interval approximation using the intensity at the start. It is not an exact long-interval probability, nor does a single starting intensity describe how rates may change later.
Probability checks: for a sufficiently short interval, the approximation should be near one and should decrease as the nonnegative total intensity or interval length increases. If it is negative, the first-order approximation is being used outside a sensible short-interval range.

This local relationship is useful whenever a model needs a probability update over a small time step. It also provides a basic consistency check: the first-order probabilities of the competing exits add to the first-order total exit probability, and the probability of remaining is its complement. Later calculations over longer periods require more than this local approximation.

Key takeaway: For competing causes, $p_0(t,h)=1-\lambda(t)h+o(h)$, where $\lambda(t)$ is the sum of the cause-specific intensities. With bounded, locally Lipschitz intensities in the standard model, the error improves to $O(h^2)$. Always match the interval’s time unit to the intensity’s time unit.

Check Your Understanding

For each question, use the short-interval approximation and check the units and probability complement.

A group has exit intensities $0.014$ and $0.009$ per year. Estimate the probability of remaining over $0.04$ years. (Answer: $0.99908$)
At $t=3$ years, the cause intensities are $\mu_1(t)=0.010+0.002t$ and $\mu_2(t)=0.007$, per year. Estimate the probability of remaining over the next $0.025$ years. (Answer: $0.999425$)
Two cause intensities are $0.012$ and $0.008$ per month. Using a 30-day month, estimate the probability of remaining for 15 days. (Answer: $0.9900$)
For two causes, suppose $q_1(t,h)=0.040h+O(h^2)$ and $q_2(t,h)=0.015h+O(h^2)$. State the first-order approximation for the probability of remaining. (Answer: $p_0(t,h)=1-0.055h+O(h^2)$)
A total intensity is $0.08$ per year and an interval is $0.5$ years. What does the first-order approximation give, and can it be treated as an exact remaining probability from the information given? (Answer: $0.96$; no, it is only a first-order approximation, and the error size is not determined by the starting intensity alone.)

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