Reconstructing Lives From Multiple-Decrement Probabilities

Introduction

A multiple-decrement table tracks a group exposed to several mutually exclusive reasons for leaving an active state. To reconstruct the table, begin with the lives active at the start of an interval and apply each cause-specific probability to that same starting group. The survivors then become the starting lives for the next interval.

The order matters: annual probabilities apply to those active at the beginning of that year, not to the original cohort in every year. Keeping that exposure basis clear allows the exits by cause and the year-end survivors to be calculated consistently.

Definition: Suppose \(l_k\) lives are active at the start of year \(k\), and there are \(m\) possible decrement causes. The cause-specific annual decrement probability \(q_k^{(j)}\) is the probability that an active life exits from cause \(j\) during year \(k\), conditional on being active at its start. The causes are mutually exclusive within the year. The probability of remaining active through the year is \(1-\sum_{j=1}^{m}q_k^{(j)}\), which must be nonnegative.

Here, “year \(k\)” denotes one annual interval in the reconstruction. If the cohort is indexed by age, \(k\) can instead identify the age at the start of the interval. These probabilities are multiple-decrement probabilities: they describe the cause of exit when the competing causes operate together. They are not probabilities from a hypothetical table in which a cause operates alone.

A table may represent actual observed lives, in which case counts are integers, or expected table lives, in which case counts can be fractional. When probabilities are applied to a theoretical cohort, the calculated exits are expected counts; fractional results are therefore valid.

Reconstructing One Annual Interval

For each cause, multiply the number active at the start of the year by that cause’s probability. The survivors are the starting lives less all exits during the year.

d_k^{(j)}=l_k q_k^{(j)}, \qquad l_{k+1}=l_k-\sum_{j=1}^{m}d_k^{(j)} =l_k\left(1-\sum_{j=1}^{m}q_k^{(j)}\right).

The notation \(d_k^{(j)}\) denotes exits during year \(k\) from cause \(j\). The calculation uses one common exposure, \(l_k\), for every cause in that year. Since the causes are mutually exclusive, an exit from one cause is not also counted as an exit from another cause.

Annual check: For each year, the exits by cause plus the survivors must equal the starting lives: \( \sum_{j=1}^{m}d_k^{(j)}+l_{k+1}=l_k \). Also, the sum of the cause-specific probabilities cannot exceed 1.

Worked Example: Reconstructing a One-Year Cohort

A hypothetical benefit program has 10,000 active members at the start of a year. During the year, members can exit through retirement, disability, or withdrawal. The corresponding annual probabilities, conditional on being active at the start, are 0.04, 0.03, and 0.01. Calculate the exits by cause and the year-end active lives.

Method and assumptions. Treat the probabilities as annual competing-decrement probabilities for this one interval. The three causes are mutually exclusive, and there are no entrants during the year. The sum is \(0.04+0.03+0.01=0.08\), which is at most 1.

Retirement exits: \(10{,}000(0.04)=400\).

Disability exits: \(10{,}000(0.03)=300\).

Withdrawal exits: \(10{,}000(0.01)=100\).

The total exits are \(400+300+100=800\), so year-end active lives are \(10{,}000-800=9{,}200\).

Check and trap avoided. The reconstructed column totals \(400+300+100+9{,}200=10{,}000\), matching the starting lives. A common mistake is to subtract only one cause’s exits and treat that as the total decrement; the correction is to add all mutually exclusive cause-specific exits before finding survivors.

Following the Cohort Across Years

For multiple years, repeat the same calculation, but update the exposure at the beginning of every year. The end of one year is the beginning of the next, provided the model has no new entrants or other additions to the active group.

Set the starting lives.
Use the given cohort size as \(l_0\), or use the active lives carried forward from the prior year.

Calculate each cause’s exits.
For year \(k\), calculate \(d_k^{(j)}=l_kq_k^{(j)}\) for every cause using the same \(l_k\).

Calculate survivors and carry them forward.
Set \(l_{k+1}=l_k-\sum_jd_k^{(j)}\). Use this \(l_{k+1}\), not the original cohort, in the next year.

Check the reconstruction.
Verify each annual column and, over several years, confirm that cumulative exits plus final survivors equal the initial lives.

Worked Example: Two Years With Changing Probabilities

A hypothetical pension plan has 5,000 active members at the start of year 1. The annual probabilities of exit by causes A, B, and C are, respectively, 0.08, 0.04, and 0.02 in year 1; in year 2 they are 0.10, 0.05, and 0.01. Find the exits by cause in each year and the number active after year 2.

Method and assumptions. Apply each year’s cause-specific probabilities to that year’s beginning active lives. Assume no entrants and no other changes to the cohort. The sums are 0.14 in year 1 and 0.16 in year 2, both valid probabilities.

Year 1. Starting lives are \(l_0=5{,}000\).

Cause A exits: \(5{,}000(0.08)=400\). Cause B exits: \(5{,}000(0.04)=200\). Cause C exits: \(5{,}000(0.02)=100\). Total exits are \(400+200+100=700\), so \(l_1=5{,}000-700=4{,}300\).

Year 2. The starting lives are now \(l_1=4{,}300\).

Cause A exits: \(4{,}300(0.10)=430\). Cause B exits: \(4{,}300(0.05)=215\). Cause C exits: \(4{,}300(0.01)=43\). Total exits are \(430+215+43=688\), so \(l_2=4{,}300-688=3{,}612\).

Check and trap avoided. Across both years, exits are \(400+430=830\) for A, \(200+215=415\) for B, and \(100+43=143\) for C. Cumulative exits total \(830+415+143=1{,}388\), and \(5{,}000-1{,}388=3{,}612\), as required. Applying year 2’s probabilities to the original 5,000 would incorrectly give 500, 250, and 50 exits. The correction is to use the 4,300 members who actually remain active at the start of year 2.

Why the Recursion Holds

The formulas follow directly from the conditional meaning of the probabilities. At the start of year \(k\), there are \(l_k\) lives exposed to the year’s decrement risks. For any one cause \(j\), the fraction of that starting group expected to leave through \(j\) is \(q_k^{(j)}\). Thus its expected exits are \(l_kq_k^{(j)}\). Because the causes are mutually exclusive, the probabilities of the different exits add. The remaining fraction is one minus their sum, giving the survivor count:

l_k\left(1-\sum_{j=1}^{m}q_k^{(j)}\right) =l_k-\sum_{j=1}^{m}l_kq_k^{(j)} =l_k-\sum_{j=1}^{m}d_k^{(j)}.

This reasoning treats the probability basis as applicable to each life active at the start of that interval. It does not require the actual realized count for each cause to equal its expected count exactly. In a finite observed group, actual exits can differ because of random variation; a decrement table reconstructed from probabilities commonly reports expected lives.

Worked Example: Fractional Expected Lives Over Three Years

A hypothetical long-term care portfolio is represented by 2,500 active policyholders at the start of year 1. Two causes, A and B, may end active status. The annual probabilities are: year 1, 0.06 for A and 0.04 for B; year 2, 0.05 for A and 0.025 for B; year 3, 0.04 for A and 0.03 for B. Calculate the exits by cause in each year and the expected active lives after year 3.

Method and assumptions. Use the conditional annual probabilities as given, with no entrants. Keep fractional expected lives through the calculations and round only the final reported values to one decimal place. The annual probability sums are 0.10, 0.075, and 0.07, each no greater than 1.

Year 1: Starting lives \(l_0=2{,}500\). Cause A exits are \(2{,}500(0.06)=150\), and cause B exits are \(2{,}500(0.04)=100\). Therefore \(l_1=2{,}500-150-100=2{,}250\).

Year 2: Starting lives \(l_1=2{,}250\). Cause A exits are \(2{,}250(0.05)=112.5\), and cause B exits are \(2{,}250(0.025)=56.25\). Therefore \(l_2=2{,}250-112.5-56.25=2{,}081.25\).

Year 3: Starting lives \(l_2=2{,}081.25\). Cause A exits are \(2{,}081.25(0.04)=83.25\), and cause B exits are \(2{,}081.25(0.03)=62.4375\). Therefore \(l_3=2{,}081.25-83.25-62.4375=1{,}935.5625\).

Check and trap avoided. Cumulative exits are \(150+112.5+83.25=345.75\) for A and \(100+56.25+62.4375=218.6875\) for B. Their total is \(564.4375\), and \(2{,}500-564.4375=1{,}935.5625\). To one decimal place, the expected year-3 survivors are 1,935.6. Rounding intermediate expected lives to whole people would change later exits; preserve the unrounded values until the final step.

Common Errors and Practical Checks

A reliable reconstruction is a sequence of annual columns, not a collection of unrelated calculations. For each year, show the beginning lives, each cause’s probability and exits, the total exits, and the survivors. The checks below help identify both arithmetic and interpretation errors.

Using the wrong exposure. Each year’s probabilities apply to the lives active at that year’s start. Carry forward the prior year’s survivors.
Applying the probability sum to each cause separately. Calculate \(d_k^{(j)}=l_kq_k^{(j)}\) for each cause; sum those exits only when finding total exits or survivors.
Ignoring a probability check. If the cause-specific probabilities sum to more than 1, they cannot describe mutually exclusive exits during the same interval on the stated basis.
Rounding too early. Expected lives need not be integers. Retain full precision during a multiyear reconstruction and round only when reporting the final result.
Forgetting the cohort boundary. Under the no-entrants assumption, cumulative exits over all causes plus the final survivors must equal the initial lives.

Key takeaway: In each year, multiply every cause-specific probability by the same beginning-of-year active lives. Subtract the combined exits to obtain survivors, then use those survivors as the next year’s exposure.

Check Your Understanding

For each problem, treat the listed values as annual cause-specific probabilities conditional on being active at the start of the year. Assume no entrants and retain fractional expected lives where they arise.

A cohort begins with 3,200 active members. During one year, the probabilities of exit from causes A, B, and C are 0.03, 0.02, and 0.05. Find each cause’s exits and the year-end survivors. (Answer: A: 96; B: 64; C: 160; survivors: 2,880.)
A cohort begins with 6,000 lives. In year 1, cause A and B probabilities are 0.04 and 0.03; in year 2 they are 0.02 and 0.05. Find exits by cause in each year and survivors after year 2. (Answer: Year 1: A 240, B 180, survivors 5,580; year 2: A 111.6, B 279, survivors 5,189.4.)
A cohort begins with 4,000 lives. For three successive years, the probabilities for causes A, B, and C are (0.05, 0.02, 0.03), (0.04, 0.04, 0.02), and (0.025, 0.015, 0.01). Find the annual exits by cause and final survivors. (Answer: Year 1: 200, 80, 120; year 2: 144, 144, 72; year 3: 81, 48.6, 32.4; final survivors: 3,078.)
A cohort begins with 2,400 active members. In year 1, cause probabilities are 0.10, 0.05, and 0.05; in year 2 they are 0.0625, 0.03125, and 0.03125. Find the exits by cause for each year and the final active count. (Answer: Year 1: 240, 120, 120; year 2: 120, 60, 60; final active count: 1,680.)
A cohort begins with 7,500 lives. The probabilities for causes A, B, and C are (0.08, 0.04, 0.02) in year 1 and (0.06, 0.03, 0.01) in year 2. Find cumulative exits by cause and year-2 survivors. (Answer: Cumulative A: 987; B: 493.5; C: 214.5; year-2 survivors: 5,805.)

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