Competing Destinations From One State

Introduction

A person or contract in one state may leave it through several different routes. For example, a hypothetical income-protection contract might leave its active state when the insured becomes disabled or dies. Each route has its own transition intensity, but the event of leaving the active state is governed by both intensities together. The key calculation is to find the total chance of an exit and then allocate that probability among its competing destinations.

This tutorial considers constant cause-specific intensities over a fixed time interval. It develops the probability of no exit, the probability that the first exit goes to each destination, and the distinction between those probabilities and occupancy probabilities when a destination permits further transitions. These methods are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Suppose a process starts in state \(0\), with possible first transitions to states \(1,\ldots,r\). If the transition intensity from state \(0\) to state \(j\) is constant at \(q_{0j}\) per unit time, then \(q_{0j}\) is the cause-specific intensity for exit to destination \(j\). The total exit intensity is \(\lambda=\sum_{j=1}^{r}q_{0j}\). A cause-specific intensity describes an instantaneous rate, not a finite-time probability.

Total Exit and Destination Probabilities

Assume the process begins in state \(0\), each cause-specific intensity is constant and nonnegative, and the only transitions considered during the interval are exits from state \(0\). Write \(p_{00}(t)\) for the probability of still being in state \(0\) at time \(t\). Over a short interval of length \(h\), the probability of an exit is approximately \(\lambda h\), so the probability of no exit is approximately \(1-\lambda h\). This gives the exponential holding-time survival probability:

p_{00}(t)=e^{-\lambda t}, \qquad P(\text{an exit by }t)=1-e^{-\lambda t}.

The probability that the first exit occurs by time \(t\) and goes to destination \(j\) is not generally \(1-e^{-q_{0j}t}\). All competing causes can end the time spent in state \(0\), so the probability of still being there until a possible exit must use the total intensity \(\lambda\). The first-exit probability for cause \(j\) is:

P(\text{first exit is to }j\text{ by }t) =\int_0^t e^{-\lambda u}q_{0j}\,du =\frac{q_{0j}}{\lambda}\left(1-e^{-\lambda t}\right), \qquad \lambda>0.

The fraction \(q_{0j}/\lambda\) is the probability that an exit, whenever it occurs, is to destination \(j\), under these constant competing intensities. It is a proportion of exits, not the probability of exiting by time \(t\). Multiplying that fraction by the probability of exiting by \(t\) gives the finite-time destination probability.

Why the Destination Formula Holds

The derivation uses the first-exit time. If the process has remained in state \(0\) until time \(u\), the probability of a transition to \(j\) in a short interval from \(u\) to \(u+du\) is approximately \(q_{0j}du\). Remaining in state \(0\) until \(u\) has probability \(e^{-\lambda u}\). Thus the probability of a first exit to \(j\) in that short interval is approximately \(e^{-\lambda u}q_{0j}du\). Integrating these disjoint possibilities over the interval gives:

Find the chance of still being in the origin state.
At time \(u\), no cause has occurred with probability \(e^{-\lambda u}\).

Apply the selected cause-specific rate.
Conditional on still being in state \(0\), the chance of a transition to \(j\) in a sufficiently short interval \(du\) is approximately \(q_{0j}du\).

Accumulate over possible exit times.
Integrate \(e^{-\lambda u}q_{0j}\) from \(u=0\) to \(u=t\). Since \(q_{0j}\) is constant, the integral is \(\frac{q_{0j}}{\lambda}(1-e^{-\lambda t})\).

The destination probabilities add to the total exit probability, which provides a useful check:

\sum_{j=1}^{r}\frac{q_{0j}}{\lambda}(1-e^{-\lambda t}) =\frac{\sum_{j=1}^{r}q_{0j}}{\lambda}(1-e^{-\lambda t}) =1-e^{-\lambda t}.

If \(\lambda=0\), all cause-specific intensities are zero. The process remains in state \(0\) with probability one, and every exit probability is zero; the displayed formulas with division by \(\lambda\) are not used in that case.

Worked Calculations

Worked Example: Allocating Five-Year Exits Between Two Causes

A hypothetical contract starts in its active state. Its constant annual transition intensities are \(0.12\) to disability and \(0.08\) to death. Treat disability and death as absorbing destinations for this calculation. Find the probability of remaining active for five years and the probability that the first exit is to each destination by then.

Method. Add the competing intensities to obtain the total exit intensity:

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

For \(t=5\) years, the probability of remaining active is:

p_{00}(5)=e^{-0.20(5)}=e^{-1}=0.367879\ldots

Thus the total probability of exit by five years is \(1-0.367879\ldots=0.632120\ldots\). The disability share of exits is \(0.12/0.20=0.6\), and the death share is \(0.08/0.20=0.4\). Therefore:

P(\text{disability first by 5}) =0.6(0.632120\ldots)=0.379272\ldots,

P(\text{death first by 5}) =0.4(0.632120\ldots)=0.252848\ldots.

Rounded to four decimal places, the probabilities are \(0.3679\) still active, \(0.3793\) first exit to disability, and \(0.2528\) first exit to death. The check is \(0.3679+0.3793+0.2528=1.0000\). The common mistake avoided here is treating each rate as if it acted alone when calculating the chance of remaining active.

Worked Example: Keeping Time Units and Event Meaning Straight

A hypothetical long-term care model begins in the independent state. The constant intensities are \(0.10\) per year for entering care and \(0.15\) per year for death directly from the independent state. Suppose care and death are absorbing for the five-year calculation. Find the five-year probability of each first exit, and the probability that an exit by five years is death.

Method. The rates are already annual, so use \(t=5\) years. The total intensity is \(0.10+0.15=0.25\) per year. The probability of no exit is \(e^{-0.25(5)}=e^{-1.25}=0.2865048\ldots\), and the total exit probability is \(1-e^{-1.25}=0.7134952\ldots\).

The shares of exits are \(0.10/0.25=0.4\) to care and \(0.15/0.25=0.6\) to death. Hence:

P(\text{care first by 5})=0.4(0.7134952\ldots)=0.2853981\ldots,

P(\text{death first by 5})=0.6(0.7134952\ldots)=0.4280971\ldots.

Conditional on an exit having occurred by five years, the probability that death was the destination is:

\frac{P(\text{death first by 5})}{P(\text{some exit by 5})} =\frac{0.4280971\ldots}{0.7134952\ldots}=0.6.

To four decimal places, the requested probabilities are \(0.2854\) for care, \(0.4281\) for death, and \(0.6000\) for death conditional on an exit. The check \(0.2865+0.2854+0.4281=1.0000\) includes the no-exit outcome. Do not divide the five-year probability by five or interpret an annual intensity as a one-year probability.

Worked Example: Recovering Cause-Specific Intensities

A hypothetical study follows individuals starting in state \(0\) for two years. Under a constant-intensity model with two absorbing destinations, the probability of still being in state \(0\) at two years is \(0.40\). The probability that the first exit is to destination 1 by then is \(0.36\). Infer both cause-specific annual intensities and the probability of first exit to destination 2 by two years.

Method. First infer the total intensity using the no-exit probability:

e^{-2\lambda}=0.40,\qquad \lambda=-\frac{\ln(0.40)}{2}=0.4581454\ldots\text{ per year}.

The total probability of exit by two years is \(1-0.40=0.60\). Destination 1 therefore accounts for \(0.36/0.60=0.60\) of exits. Its intensity is this share multiplied by the total intensity:

q_{01}=0.60(0.4581454\ldots)=0.2748872\ldots\text{ per year}.

The remaining exit share is \(1-0.60=0.40\), so:

q_{02}=0.40(0.4581454\ldots)=0.1832581\ldots\text{ per year}.

The probability of first exit to destination 2 is the remaining \(0.60-0.36=0.24\). To four significant figures, the intensities are \(0.2749\) and \(0.1833\) per year. As a check, they sum to \(0.4581\) per year, and \(e^{-2(0.4581454\ldots)}=0.40\). The common mistake avoided is to treat \(0.36\), a finite-time probability, as the intensity \(q_{01}\).

First-Exit Probabilities Versus State Occupancy

A destination-specific formula requires careful interpretation. The integral above gives the probability that the first exit from state \(0\) occurs by time \(t\) and goes to \(j\). If destination \(j\) is absorbing, this is also the probability of occupying \(j\) at time \(t\) after starting in state \(0\). But if transitions out of \(j\) are possible, a person may enter \(j\) and leave it again before time \(t\). The first-exit probability then need not equal the probability of being in \(j\) at time \(t\). Calculating that later occupancy requires accounting for those additional transitions, which are outside the one-origin-state competing-exit calculation here.

The formula also depends on the cause-specific intensities remaining constant throughout the interval. If rates change with time, the probability of no exit involves the accumulated total intensity over time, and the constant-rate expressions above cannot be applied by substituting a single instantaneous rate. Confirm the assumed time unit as well: when rates are per month, \(t\) must be measured in months.

Common Exam Traps

Using one cause's rate for survival in the origin state. With competing exits, no exit means avoiding every cause, so use \(\lambda=\sum_j q_{0j}\) in the exponential.
Reporting the total exit probability as a destination probability. The total \(1-e^{-\lambda t}\) must be allocated using \(q_{0j}/\lambda\).
Confusing a rate with a probability. An intensity has units of inverse time and can exceed one; a finite-time probability is between zero and one.
Ignoring the endpoint interpretation. A first-exit destination is not necessarily the state occupied at time \(t\) if the destination is not absorbing.
Mixing units. The product \(\lambda t\) must be dimensionless: annual rates go with years and monthly rates with months.

Key takeaway: Add all cause-specific intensities to obtain the total exit intensity. Calculate total exit probability as \(1-e^{-\lambda t}\), then multiply by \(q_{0j}/\lambda\) for the probability of first exit to destination \(j\). Check that the destination probabilities sum to the total exit probability.

Check Your Understanding

For each problem, distinguish the probability of no exit from the probabilities assigned to the competing destinations.

A state has constant exit intensities \(0.06\) and \(0.09\) per year to two absorbing destinations. Find the probability of remaining in the state and the probability of first exit to each destination over two years. (Answer: no exit \(0.7408\); destination probabilities \(0.1037\) and \(0.1555\).)
Two competing exit intensities are \(0.20\) and \(0.30\) per month. Find the probability of no exit and the probability of first exit to each destination over three months. (Answer: no exit \(0.2231\); destination probabilities \(0.3107\) and \(0.4661\).)
Two destinations have intensities \(0.04\) and \(0.06\) per year. Over ten years, what is the probability of exit to the second destination, and what fraction of all exits goes there? (Answer: probability \(0.3793\); fraction \(0.6000\).)
Under a constant model with two absorbing destinations, the no-exit probability after one year is \(0.64\), and the first-destination probability is \(0.18\). Find both annual cause-specific intensities and the second-destination probability by one year. (Answer: intensities \(0.2231\) and \(0.2231\) per year; second-destination probability \(0.18\).)
A process has constant intensities \(0.10\) and \(0.20\) per year. Given that it is still in the origin state at year 3, find the probability that it exits to the second destination during the following two years. (Answer: \(0.3008\).)

What You'll Learn