The Competing Decrement Framework

Introduction

A multiple decrement model begins by specifying who is exposed to a set of possible exits and the period over which those exits are observed. For example, a hypothetical long-term care contract may be held by an active policyholder who is not receiving care. During the next year, that policyholder may enter care, die before entering care, lapse the contract, or remain active and outside care at year-end.

The central discipline is to define the group and the interval before assigning probabilities. Each probability is conditional on being in the group at the start of the interval. The exit causes must be distinct outcomes for that interval, and remaining in the group is a separate outcome.

Definition: An at-risk group is the set of individuals who satisfy a specified condition at the beginning of an observation interval. A decrement is an event that takes an individual out of that group during the interval. The remaining-in-group outcome occurs when no defined decrement takes the individual out of the group by the end of the interval.

Defining the Competing Outcomes

Suppose the group at time 0 contains individuals who are active, insured, and not in long-term care. Consider an interval from time 0 to time \(t\). Possible decrements from this group might be care entry, death while still outside care, and lapse while still outside care. These events compete in the sense that the first event taking an individual out of the group determines that individual's outcome for this interval.

The cause definitions must be precise enough to make the outcomes mutually exclusive. If an individual enters care and later dies before the interval ends, that person has already left the at-risk group through care entry. For this interval, the decrement is care entry, not death from the active, non-care group. Death may be a later event in the person's history, but it is not a second exit from the original group.

Let there be \(m\) defined causes. Write \(q_j(0,t)\) for the probability that an individual in the at-risk group at time 0 leaves it during the interval through cause \(j\), and write \(p(0,t)\) for the probability of being in the group at time \(t\). All of these probabilities use the same starting group and the same interval.

Definition: For a specified at-risk group and interval \((0,t]\), \(q_j(0,t)\) is the probability of exit from the group by cause \(j\), conditional on membership in the group at time 0. The probability \(p(0,t)\) is the probability of remaining in the group through time \(t\). If the \(m\) exit causes are mutually exclusive and exhaustive of exits from the group, then \(p(0,t)+\sum_{j=1}^{m}q_j(0,t)=1\).

The word exhaustive matters: every person who leaves the group during the interval must be assigned to one of the causes included in the model. If a relevant exit has been omitted, the probabilities as defined will not add to 1. The remaining outcome is not an additional decrement; it is the outcome for anyone who experiences none of the defined exits during the interval.

Why the Probabilities Form a Complete Partition

The probability identity follows directly from how outcomes are defined, not from any assumption that different causes occur independently. Each person who was in the group at time 0 has exactly one interval outcome: remain in the group, or leave it through one of the mutually exclusive causes. These outcomes are disjoint and together include the whole starting group.

\{\text{remain}\},\ \{\text{exit by cause 1}\},\ldots,\{\text{exit by cause }m\}

For disjoint, exhaustive events, the probability of their union is the sum of their probabilities. Their union is the event that the person was in the starting group, which has conditional probability 1 given membership at time 0. Therefore,

p(0,t)+q_1(0,t)+q_2(0,t)+\cdots+q_m(0,t)=1.

This identity is a useful check on a calculation. It does not supply missing probabilities unless the causes are complete: if only some exits are listed, the difference from 1 may represent an omitted cause rather than an arithmetic error.

Worked Examples

Worked Example: Building the Interval Probability Partition

A hypothetical insurer observes 12,000 active policyholders who are not receiving care at the start of a one-year interval. By year-end, 240 have entered care, 360 have died before care entry, 600 have lapsed while still outside care, and all others remain active and outside care. Calculate the interval probability for each outcome and check the partition.

Method. Use the number in the at-risk group at the start, 12,000, as the denominator for every outcome. The four outcomes are mutually exclusive under the stated first-exit definitions.

Care entry has probability \(q_{\text{care}}(0,1)=240/12{,}000=0.0200\). Death before care entry has probability \(q_{\text{death}}(0,1)=360/12{,}000=0.0300\). Lapse while outside care has probability \(q_{\text{lapse}}(0,1)=600/12{,}000=0.0500\).

The number remaining is \(12{,}000-240-360-600=10{,}800\). Therefore, \(p(0,1)=10{,}800/12{,}000=0.9000\). The total exit probability is \(0.0200+0.0300+0.0500=0.1000\), and the partition check gives \(0.9000+0.1000=1.0000\).

Check and common mistake avoided. The event counts sum to \(240+360+600+10{,}800=12{,}000\), matching the starting group. Dividing by the year-end remaining count would give the wrong denominator; these are probabilities conditional on being at risk at the start.

Worked Example: Combining Two Successive Intervals

A hypothetical group of 8,000 active, non-care policyholders is followed for two one-year intervals. In year 1, the conditional probabilities of care entry, death before care entry, and lapse are 0.040, 0.025, and 0.035. Among those who remain in the group at the end of year 1, the corresponding conditional probabilities in year 2 are 0.050, 0.030, and 0.020. Calculate the numbers experiencing each outcome in each year, then find the two-year probabilities of care entry, death before care entry, lapse, and remaining in the group.

Method. Apply the first year's probabilities to the original 8,000. Apply the second year's probabilities only to those who remain at risk after year 1. The year-2 probabilities are conditional on that smaller starting group, not on all 8,000.

In year 1, care entries are \(8{,}000(0.040)=320\), deaths are \(8{,}000(0.025)=200\), and lapses are \(8{,}000(0.035)=280\). The remaining count is \(8{,}000-320-200-280=7{,}200\), consistent with a year-1 remaining probability of \(1-0.040-0.025-0.035=0.900\).

In year 2, the at-risk starting count is 7,200. Care entries are \(7{,}200(0.050)=360\), deaths are \(7{,}200(0.030)=216\), and lapses are \(7{,}200(0.020)=144\). The year-2 remaining count is \(7{,}200-360-216-144=6{,}480\).

Across both years, care entry occurs for \(320+360=680\) people, so its probability from the original group is \(680/8{,}000=0.0850\). Death before care entry totals \(200+216=416\), giving \(416/8{,}000=0.0520\). Lapse totals \(280+144=424\), giving \(424/8{,}000=0.0530\). The two-year remaining probability is \(6{,}480/8{,}000=0.8100\).

The check is \(0.0850+0.0520+0.0530+0.8100=1.0000\). Equivalently, the outcome counts sum to \(680+416+424+6{,}480=8{,}000\). The common mistake is to apply the year-2 percentages to the original 8,000; that ignores that only 7,200 remain at risk at the start of year 2.

Worked Example: Combining Two At-Risk Subgroups

At the start of a one-year interval, a hypothetical portfolio has 4,000 members in subgroup A and 6,000 in subgroup B. In subgroup A, the probabilities of care entry, death before care entry, and lapse are 0.030, 0.020, and 0.050. In subgroup B, they are 0.060, 0.040, and 0.100. Calculate the outcome counts and overall interval probabilities for the full 10,000-person group.

Method. Calculate counts within each subgroup using that subgroup's starting size, add counts by outcome, and divide by the full starting size. The subgroup probabilities are not averaged equally because the subgroup sizes differ.

For subgroup A, care entry is \(4{,}000(0.030)=120\), death is \(4{,}000(0.020)=80\), and lapse is \(4{,}000(0.050)=200\). The remaining count is \(4{,}000-120-80-200=3{,}600\).

For subgroup B, care entry is \(6{,}000(0.060)=360\), death is \(6{,}000(0.040)=240\), and lapse is \(6{,}000(0.100)=600\). The remaining count is \(6{,}000-360-240-600=4{,}800\).

Adding the subgroup counts gives 480 care entries, 320 deaths, 800 lapses, and 8,400 remaining. The overall probabilities are \(480/10{,}000=0.0480\), \(320/10{,}000=0.0320\), \(800/10{,}000=0.0800\), and \(8{,}400/10{,}000=0.8400\), respectively. They sum to \(1.0000\), and the counts sum to 10,000.

A simple unweighted average of subgroup care probabilities would be \((0.030+0.060)/2=0.045\), which is not the portfolio probability. The correct weighted value is \(0.4(0.030)+0.6(0.060)=0.048\), because 40% of the starting group is in A and 60% is in B.

Common Traps and Model Checks

Competing decrement calculations are often straightforward arithmetic, but the event definitions and denominators require care. Use the following checks before relying on a result:

State the starting condition. Identify exactly who is at risk at the beginning, such as active policyholders who have not entered care.
Fix the interval. A probability for one year and a probability for a two-year period have different conditioning and interpretation.
Make causes exclusive. Classify an individual by the first event that takes them out of the defined group during the interval.
Use the correct denominator. For an interval probability, use the number at risk at that interval's start.
Separate conditional intervals. When calculating a second interval, use the survivors of the first interval as its starting group.
Check completeness. The remaining probability plus all defined cause probabilities must equal 1 when the causes cover every possible exit.

The interval probabilities summarize which outcome occurs by the end of the interval. By themselves, they do not specify when within the interval an exit happens. They also do not imply that hypothetical causes are independent; the probability partition follows from mutually exclusive outcome definitions. More detailed descriptions of event timing require additional model information.

Key takeaway: A competing decrement framework is built from a clearly defined group at risk, a fixed observation interval, mutually exclusive exit causes, and the complementary outcome of remaining in the group. Keep the conditioning group consistent, then verify that all outcome probabilities total 1.

Check Your Understanding

For each problem, define the interval outcomes using the stated at-risk group and check the denominator carefully.

A hypothetical group of 5,000 active members who are not receiving care is followed for one year. There are 150 care entries, 100 deaths before care entry, and 250 lapses while outside care. Find each cause probability and the remaining probability. (Answer: care 0.0300, death 0.0200, lapse 0.0500, remain 0.9000.)
A hypothetical insurer starts with 6,000 active, non-care contracts. Over one year, 240 enter care, 180 die before care entry, and 180 lapse. Find the remaining count and all four interval probabilities. (Answer: remain 5,400; care 0.0400, death 0.0300, lapse 0.0300, remain 0.9000.)
A group of 10,000 begins year 1 at risk. In year 1, the probabilities of exit by cause A and cause B are 0.020 and 0.030. Among those remaining, the year-2 conditional probabilities are 0.040 for A and 0.010 for B. Find the two-year probabilities of exit by A, exit by B, and remaining at the end of year 2. (Answer: cause A 0.0580, cause B 0.0395, remain 0.9025.)
A hypothetical portfolio has 4,000 people in subgroup A and 6,000 in subgroup B. The one-year cause-1, cause-2, and remaining probabilities are respectively (0.020, 0.030, 0.950) in A and (0.050, 0.010, 0.940) in B. Find the overall probabilities. (Answer: cause 1 0.0380, cause 2 0.0180, remain 0.9440.)
Among 1,000 active, non-care members at the start of a year, 60 enter care, 20 die before care entry, 20 lapse while outside care, and 900 remain in the group. Another 5 people die after entering care before year-end. For the original active, non-care group, identify the interval decrement probabilities and explain how the five later deaths are classified. (Answer: care entry 0.0600, death before care 0.0200, lapse 0.0200, remain 0.9000; the five deaths after care entry are subsequent events, not exits by death from the original group.)

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