Competing Decrements With Piecewise-Constant Intensities

Why Piecewise-Constant Intensities Need Interval-by-Interval Updates

In a competing decrement model, an individual may leave a group through one of several causes. The cause-specific intensities can change over time: for example, a hypothetical benefit program might have different withdrawal and termination intensities before and after a contract anniversary. If each cause-specific intensity is constant within each of several intervals, the probabilities can be calculated exactly on each interval and then combined.

The key is to keep track of who is still exposed at the start of each interval. Cause-specific probabilities calculated from the start of an interval are conditional on being present then. To obtain probabilities for the original group, multiply by the probability of having remained through all earlier intervals.

This tutorial develops that update method for a finite sequence of intervals. It is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Let the observation period be divided at times \(0=t_0<t_1<\cdots<t_m\). On interval \(j\), from \(t_{j-1}\) to \(t_j\), let the cause-specific intensity for cause \(r\) be the constant \(\mu_{r,j}\), and let the interval length be \(h_j=t_j-t_{j-1}\). The total intensity on that interval is \(\mu_{\bullet,j}=\sum_r\mu_{r,j}\). The probability of remaining through interval \(j\), conditional on being present at its start, is \(e^{-\mu_{\bullet,j}h_j}\).

One Interval: Conditional Exits and Remaining Probability

Suppose an individual is present at the start of an interval of length \(h\). If the cause-specific intensities are constant and equal to \(\mu_1,\ldots,\mu_R\), the total intensity is \(\mu_\bullet=\sum_{r=1}^R\mu_r\). The probability of remaining to the interval end is \(e^{-\mu_\bullet h}\). Therefore, the probability of exiting by any cause during the interval is \(1-e^{-\mu_\bullet h}\).

The probability of exit through a particular cause is not generally that cause’s intensity multiplied by the interval length. Individuals must first remain in the group up to the time of exit, and the competing causes operate at the same time. With constant intensities, the conditional probability of exit through cause \(r\) is

\frac{\mu_r}{\mu_\bullet}\left(1-e^{-\mu_\bullet h}\right), \qquad \mu_\bullet>0.

If the starting group has a probability \(s\) of still being present at the beginning of the interval, the corresponding probability of exiting through cause \(r\), measured from the original starting point, is \(s\) times this conditional probability. The probability still present at the interval end is \(s e^{-\mu_\bullet h}\). If \(\mu_\bullet=0\), all cause intensities are zero, so there are no exits and the remaining probability is unchanged.

Worked Example: Two Causes Over One Year

A hypothetical insured group is observed for one year. Throughout the year, cause A has constant intensity \(0.15\) per year and cause B has constant intensity \(0.05\) per year. Find the probability of remaining in the group and the probability of exit from each cause by the end of the year.

Method. Add the cause intensities to obtain the total intensity, then use the constant-intensity competing-decrement formulas. This avoids treating an intensity as if it were an interval probability.

The total intensity is \(0.15+0.05=0.20\) per year. The interval length is one year, so the remaining probability is

e^{-(0.20)(1)}=e^{-0.20}=0.8187308.

The probability of exit through A is its share of the total intensity times the probability of any exit:

\frac{0.15}{0.20}\left(1-e^{-0.20}\right) =0.75(0.1812692)=0.1359519.

Similarly, for B:

\frac{0.05}{0.20}\left(1-e^{-0.20}\right) =0.25(0.1812692)=0.0453173.

To four decimal places, the remaining probability is 0.8187, the probability of exit through A is 0.1360, and the probability of exit through B is 0.0453. The check is \(0.8187308+0.1359519+0.0453173=1.0000000\), up to rounding. The error avoided is using \(0.15+0.05\) as an exit probability; it is the total intensity, not the probability of exit over one year.

Updating Across Successive Intervals

For piecewise-constant intensities, apply the one-interval calculation separately on every interval. Write \(s_0=1\) for the probability of being present at time zero. At the end of interval \(j\), the probability of remaining from the original starting point is

s_j=s_{j-1}e^{-\mu_{\bullet,j}h_j}.

The probability of exiting through cause \(r\) during interval \(j\), measured from the original starting point, is

q_{r,j} =s_{j-1}\frac{\mu_{r,j}}{\mu_{\bullet,j}} \left(1-e^{-\mu_{\bullet,j}h_j}\right), \qquad \mu_{\bullet,j}>0.

The interval probability conditional on being present at \(t_{j-1}\) is the same formula without the factor \(s_{j-1}\). The distinction matters: conditional probabilities for separate intervals do not have the same starting population. To obtain the probability of exit by cause \(r\) anywhere in the full observation period, add its unconditional interval probabilities:

q_r=\sum_{j=1}^{m}q_{r,j}.

Set the interval boundaries.
Record each interval length and the cause-specific intensities that apply throughout it.

Calculate the total intensity.
For each interval, add the intensities of all competing causes.

Calculate interval exits and survival.
Use the total intensity and interval length to find the conditional interval probabilities.

Apply the current remaining probability.
Multiply each conditional cause exit by the probability present at that interval’s start; update the remaining probability for the next interval.

Accumulate by cause and check.
Add unconditional exits for each cause and verify that all cause exits plus the final remaining probability sum to one.

Why the Interval Formula Holds

The constant-intensity cause-specific exit density at elapsed time \(u\) within an interval is the probability of still being present at \(u\), multiplied by the intensity of cause \(r\) at that time. Conditional on being present at the interval start, the probability of still being present at elapsed time \(u\) is \(e^{-\mu_\bullet u}\). Thus the cause-\(r\) exit probability during an interval of length \(h\) is

\int_0^h e^{-\mu_\bullet u}\mu_r\,du =\mu_r\frac{1-e^{-\mu_\bullet h}}{\mu_\bullet} =\frac{\mu_r}{\mu_\bullet}\left(1-e^{-\mu_\bullet h}\right), \qquad \mu_\bullet>0.

If the probability present at the interval start is \(s\), the density for an exit measured from the original starting point is multiplied by \(s\), so the integral is multiplied by \(s\) as well. This gives \(q_{r,j}\). The derivation also explains why the relevant share is \(\mu_r/\mu_\bullet\): with constant competing intensities, the cause-specific exit density has the same survival factor for every cause, leaving the intensity ratio constant throughout the interval.

Worked Example: Two Intervals With Changing Cause Shares

A hypothetical population starts with probability one in a group. During year 1, cause A has intensity \(0.30\) and cause B has intensity \(0.10\), per year. During year 2, the intensities change to \(0.05\) for A and \(0.15\) for B, per year. Calculate the probability remaining after two years and the cumulative exit probability for each cause.

Method. Calculate each interval’s results conditional on being present at its start, then multiply the second-year exits by the probability remaining after year 1. This avoids applying year 2 intensities to the original population.

In year 1, the total intensity is \(0.30+0.10=0.40\). The remaining probability is

s_1=e^{-0.40}=0.6703200.

The probability of any exit in year 1 is \(1-e^{-0.40}=0.3296800\). Cause A’s share is \(0.30/0.40=0.75\), and cause B’s share is \(0.10/0.40=0.25\). Their year 1 exits are therefore

q_{A,1}=0.75(0.3296800)=0.2472600, \qquad q_{B,1}=0.25(0.3296800)=0.0824200.

In year 2, the total intensity is \(0.05+0.15=0.20\). Conditional on surviving to year 2, the probability of any exit during that year is \(1-e^{-0.20}=0.1812692\). Cause A has share \(0.05/0.20=0.25\), and cause B has share \(0.15/0.20=0.75\). Applying the year 1 remaining probability gives

q_{A,2}=0.6703200(0.25)(0.1812692)=0.0303926,

q_{B,2}=0.6703200(0.75)(0.1812692)=0.0911777.

The cumulative cause probabilities are \(q_A=0.2472600+0.0303926=0.2776526\) and \(q_B=0.0824200+0.0911777=0.1735977\). The final remaining probability is

s_2=s_1e^{-0.20}=e^{-0.40}e^{-0.20}=e^{-0.60}=0.5488116.

To four decimal places, the answers are \(q_A=0.2777\), \(q_B=0.1736\), and \(s_2=0.5488\). Their sum is \(1.0000619\) when the intermediate values above are rounded as displayed; using unrounded values gives \(1.0000000\). Do not round intermediate probabilities when checking conservation.

Different Cause Mixes Across Intervals

The total intensity determines the probability of remaining within each interval, while the relative intensities determine how interval exits are divided among causes. Both can change at an interval boundary. Consequently, one cannot generally calculate a cause’s total exit probability by applying its intensity share from the first interval to the probability of any exit over the whole period.

Worked Example: Three Causes With a Changed Mix

A hypothetical group starts at time zero and is observed for two years. In year 1, the constant annual intensities for causes A, B, and C are \(0.10\), \(0.15\), and \(0.05\). In year 2, they are \(0.20\), \(0.05\), and \(0.05\). Find the cumulative exit probability for each cause and the probability remaining after two years.

Method. Use the interval update and calculate each cause’s share separately in each year. The common total intensity is \(0.30\) per year in both years, but the cause shares change.

In year 1, the exit probability is \(1-e^{-0.30}=0.2591818\). The cause shares are \(0.10/0.30=1/3\), \(0.15/0.30=1/2\), and \(0.05/0.30=1/6\). Thus

q_{A,1}=0.0863939,\quad q_{B,1}=0.1295909,\quad q_{C,1}=0.0431970,\quad s_1=e^{-0.30}=0.7408182.

In year 2, the total remains \(0.30\), but the cause shares are \(2/3\), \(1/6\), and \(1/6\). The unconditional probability of any year 2 exit is \(s_1(1-e^{-0.30})=0.7408182(0.2591818)=0.1920066\). Therefore,

q_{A,2}=\frac{2}{3}(0.1920066)=0.1280044,\qquad q_{B,2}=\frac{1}{6}(0.1920066)=0.0320011,\qquad q_{C,2}=\frac{1}{6}(0.1920066)=0.0320011.

Adding the exits by cause gives \(q_A=0.0863939+0.1280044=0.2143983\), \(q_B=0.1295909+0.0320011=0.1615920\), and \(q_C=0.0431970+0.0320011=0.0751981\). The final remaining probability is \(e^{-0.60}=0.5488116\). To four decimal places, these are \(0.2144\), \(0.1616\), \(0.0752\), and \(0.5488\). The unrounded probabilities sum to one. Keeping the year 1 and year 2 cause shares separate is essential; applying a single share to both years would give incorrect cumulative cause probabilities.

Common Exam Traps and Checks

A useful conservation check is that the final remaining probability plus the accumulated exit probabilities for all causes must equal one, assuming the causes are exhaustive and mutually exclusive and there are no other exits. In symbols, for \(R\) causes,

s_m+\sum_{r=1}^{R}q_r=1.

This follows interval by interval: the probability present at the start of an interval is divided among remaining through that interval and exits by the competing causes during it. Summing over intervals makes each intermediate remaining probability cancel between one interval’s endpoint and the next interval’s start.

A common error is to use an intensity as an interval probability. For example, with annual intensities \(0.12\) and \(0.08\) over one year, the incorrect calculation might assign cause exit probabilities \(0.12\) and \(0.08\), and remaining probability \(1-0.12-0.08=0.80\). The total intensity is \(0.20\), so the correct remaining probability is \(e^{-0.20}=0.8187\). The correct cause exits are \(0.12/0.20(1-e^{-0.20})=0.1088\) and \(0.08/0.20(1-e^{-0.20})=0.0725\). Together with the remaining probability, these sum to one.

Do not restart with probability one at each interval. The start probability for a later interval is the probability that the original group survived all preceding intervals.
Do not reuse a cause share after intensities change. Recalculate each cause’s share using the intensities for the current interval.
Keep units consistent. Multiply an intensity per year by an interval length in years, or convert one of those quantities before using the exponential.
Distinguish conditional from unconditional interval exits. The conditional probability applies only to those present at the interval start; the unconditional probability includes the earlier survival factor.

Key takeaway: For each interval, use its own total intensity and length to calculate the conditional survival and cause-specific exit probabilities. Multiply by the probability remaining at that interval’s start, update the remaining probability, and add unconditional exits by cause across intervals.

Check Your Understanding

For each problem, use piecewise-constant cause-specific intensities and retain adequate precision until the final answer.

A group has two causes with constant intensities \(0.06\) and \(0.04\) per year for two years. Find the probability remaining and the exit probability for each cause. (Answer: remaining \(0.8187\); cause 1 \(0.1088\); cause 2 \(0.0725\).)
During the first half-year, two causes have intensities \(0.20\) and \(0.10\) per year. During the following year, both intensities are \(0.10\) per year. Find the final remaining probability and each cause’s cumulative exit probability. (Answer: remaining \(0.7047\); cause 1 \(0.1709\); cause 2 \(0.1244\).)
For years 1, 2, and 3, respectively, the cause 1 and cause 2 intensities are \((0.10,0.10)\), \((0.20,0.10)\), and \((0.05,0.15)\), all per year. Find the cumulative exit probability for cause 2. (Answer: \(0.2245\).)
A hypothetical cohort of 800 starts in a group. For the first half-year, cause A and B intensities are \(0.20\) and \(0.20\) per year; for the next year they are \(0.05\) and \(0.15\) per year. Find the expected number remaining after the full 1.5 years and the expected number of exits from each cause. (Answer: remaining \(536.26\); cause A \(102.19\); cause B \(161.55\).)
Over each of two consecutive one-year intervals, the total intensity is \(0.20\) per year. Cause Y has intensity \(0.08\) in year 1 and \(0.16\) in year 2; cause X has intensity \(0.12\) and \(0.04\), respectively. Find the cumulative probability of exit through Y and the probability remaining after two years. (Answer: cause Y \(0.1912\); remaining \(0.6703\).)

What You'll Learn