Cause-Specific Exits From a Multiple-Decrement Table

From Multiple-Decrement Probabilities to Exit Counts

A multiple-decrement table records how many lives leave a group during an interval and the cause of each exit. To calculate those counts from probabilities, first identify the lives exposed at the start of the interval. Each cause-specific probability is applied to that same starting group; the causes are not applied one after another to successively smaller groups.

The calculation is useful both when constructing a table from probability assumptions and when checking whether a supplied table is internally consistent. The central check is straightforward: if the listed causes account for every exit, their counts must sum to the total number of exits.

Definition: Let \(l_x\) be the number of lives exposed at the start of the interval from age or duration \(x\) to \(x+1\). For cause \(j\), let \(q_x^{(j)}\) be the probability that a life exposed at \(x\) exits from cause \(j\) during that interval, before the interval ends. The number of exits from cause \(j\) is \(d_x^{(j)}\). The total number of exits is \(d_x\), and the probability of any exit is \(q_x\). When the listed causes are mutually exclusive and include every possible exit, \(d_x=\sum_j d_x^{(j)}\) and \(q_x=\sum_j q_x^{(j)}\).

The denominator for every \(q_x^{(j)}\) is the same: \(l_x\), the lives exposed at the interval's start. In a theoretical table, multiplying a probability by an exposure can produce a fractional expected count. In an observed table of actual lives, counts are integers, and the corresponding probability is calculated from the observed count and exposure.

The Cause-Specific Count Formula

A cause-specific probability is a conditional probability for one life drawn from the exposed group. If \(q_x^{(j)}\) is the probability of an exit from cause \(j\), then the expected proportion of the \(l_x\) exposed lives exiting from that cause is \(q_x^{(j)}\). Multiplying the proportion by the number exposed gives the cause count.

d_x^{(j)}=l_x q_x^{(j)}

If all causes are included, adding the cause counts gives the total. The number remaining in the table at the end of the interval is the starting exposure less those exits.

d_x=\sum_j d_x^{(j)} =l_x\sum_j q_x^{(j)} =l_xq_x, \qquad l_{x+1}=l_x-d_x

These equations also work in reverse: divide a cause count by the starting exposure to obtain its probability, or subtract known cause counts from the total to find an unknown count. A cause total can be recovered this way only if the total and the other causes use the same interval and exposure basis.

Worked Example: Calculate Counts for Three Causes

A hypothetical table has 1,200 lives exposed at the start of a one-year interval. The probabilities of exit during the interval are 0.040 from disability, 0.025 from withdrawal, and 0.015 from death. Calculate the exits by cause, the total exits, and the number remaining. Check the result against the total exit probability.

Method. Multiply each cause-specific probability by the starting exposure of 1,200. This treats each probability as conditional on entering the interval, not on surviving a previous cause-specific event.

Disability exits: \(d_x^{(1)}=1{,}200(0.040)=48\).

Withdrawal exits: \(d_x^{(2)}=1{,}200(0.025)=30\).

Death exits: \(d_x^{(3)}=1{,}200(0.015)=18\).

The cause counts sum to \(48+30+18=96\). The total probability is \(0.040+0.025+0.015=0.080\), so the total calculated directly is \(1{,}200(0.080)=96\). The two calculations agree. The number remaining is \(1{,}200-96=1{,}104\).

Check and trap avoided. The total exit count is 96, or 8% of the 1,200 exposed lives. A common mistake is to apply the second and third probabilities to a reduced exposure after calculating earlier exits. That would treat the given probabilities as sequential probabilities rather than multiple-decrement probabilities based on the starting exposure.

Recovering a Missing Cause

Sometimes the total probability or total exit count is known, while one cause-specific value is missing. When the listed causes are exhaustive, the missing probability is the total probability minus the known cause probabilities. Equivalently, the missing count is the total exit count minus the known cause counts. These are equivalent because each probability and count uses the same exposure.

Worked Example: Find the Unspecified Cause

A hypothetical group of 2,000 lives is exposed for one year. The total exit probability is 0.120. The known probabilities are 0.035 for a health-related exit and 0.045 for withdrawal. A third cause accounts for all other exits. Find the third cause's probability and exit count, the counts for the known causes, and the number remaining.

Step 1: Find the missing probability. The known probabilities sum to \(0.035+0.045=0.080\). Therefore, the third probability is \(0.120-0.080=0.040\).

Step 2: Calculate the cause counts from the exposure. Health-related exits are \(2{,}000(0.035)=70\). Withdrawal exits are \(2{,}000(0.045)=90\). Exits from the third cause are \(2{,}000(0.040)=80\).

Step 3: Check total exits. The cause counts sum to \(70+90+80=240\). Independently, the total exit probability gives \(2{,}000(0.120)=240\). The counts reconcile.

Step 4: Calculate the number remaining. The remaining number is \(2{,}000-240=1{,}760\).

Check and trap avoided. The three cause probabilities sum to \(0.035+0.045+0.040=0.120\), the stated total probability. Do not assign the entire difference between the total and one known cause to the unspecified cause: both known causes must be subtracted before finding the residual.

Working Backward from Observed Counts

When a table gives actual exits rather than probabilities, the cause-specific probability is the cause count divided by the lives exposed at the beginning of the interval. The total probability is the total count divided by that same exposure. This calculation is useful for checking a table or translating counts into probability form.

q_x^{(j)}=\frac{d_x^{(j)}}{l_x}, \qquad q_x=\frac{d_x}{l_x}

Worked Example: Check a Table of Observed Exits

A hypothetical one-year table begins with 1,500 lives exposed. It records 36 disability exits, 24 withdrawals, and 30 death exits. Calculate each cause-specific probability, the total number and probability of exits, and the number remaining. Then verify that the cause probabilities reconcile with the total.

Step 1: Calculate the total count. The three cause counts give \(d_x=36+24+30=90\).

Step 2: Calculate the cause probabilities. The disability probability is \(36/1{,}500=0.024\). The withdrawal probability is \(24/1{,}500=0.016\). The death probability is \(30/1{,}500=0.020\).

Step 3: Calculate and check the total probability. From the total count, \(q_x=90/1{,}500=0.060\). From the cause probabilities, \(0.024+0.016+0.020=0.060\). They agree.

Step 4: Find the number remaining. The number remaining is \(1{,}500-90=1{,}410\).

Check and trap avoided. Each cause probability uses 1,500 as its denominator, so the cause probabilities can be added directly. Dividing a cause count by 1,410 instead would calculate a ratio to the survivors, not the probability of that cause-specific exit among lives exposed to risk.

Why the Reconciliation Check Works

For each exposed life, the mutually exclusive outcomes over the interval are exit from cause 1, exit from cause 2, and so on, or no exit. Because one life cannot leave from two different causes in the same interval, the cause-specific exit groups do not overlap. If the causes listed are exhaustive, every exit belongs to exactly one group. Therefore, counting exits cause by cause counts every exit once and only once.

In probability terms, the events “exit from cause \(j\)” are disjoint. Their union is the event “exit from any cause,” so the probability of any exit is the sum of the cause-specific probabilities. Multiplication by the shared exposure \(l_x\) gives the corresponding count identity.

Fix the interval and exposure.
Identify the lives exposed at its start and confirm that every probability or count refers to that same interval.

Calculate each cause.
Use \(d_x^{(j)}=l_xq_x^{(j)}\), or \(q_x^{(j)}=d_x^{(j)}/l_x\), according to the information given.

Reconcile the totals.
Sum the cause counts and compare them with \(l_xq_x\), or sum the cause probabilities and compare them with \(q_x\).

Find the remaining lives.
Subtract total exits from the starting exposure. Check that exits plus lives remaining equal the original exposure.

Common Exam Traps

Using the wrong denominator. Cause-specific multiple-decrement probabilities are based on lives exposed at the start of the interval, not the survivors after exits.
Applying probabilities sequentially. Multiplying a second cause probability by the lives left after a first cause changes the meaning of the supplied probabilities. Use the common starting exposure unless the problem explicitly defines conditional probabilities on a changed group.
Mixing intervals. A one-year probability cannot be multiplied by an exposure covering a different duration without a justified conversion.
Omitting a cause or counting it twice. The sum of listed causes equals total exits only if those causes are exhaustive and mutually exclusive.
Confusing exits with survivors. Total exits are \(l_x-l_{x+1}\); the remaining lives are not another cause of exit.

Key takeaway: For a multiple-decrement interval, calculate each cause count using the same starting exposure: \(d_x^{(j)}=l_xq_x^{(j)}\). Then check that the cause counts sum to total exits and that total exits plus lives remaining equal the starting exposure.

Check Your Understanding

For each problem, use the exposure at the start of the stated interval and check cause totals against total exits.

A hypothetical table has 900 lives exposed for one year. The cause-specific probabilities are 0.020, 0.030, and 0.010, and these are all possible exit causes. Find each cause count, total exits, and lives remaining. (Answer: Cause counts 18, 27, and 9; total exits 54; 846 remain.)
There are 2,400 lives exposed. The total exit probability is 0.125, and three cause probabilities are 0.040, 0.035, and 0.020. Find the missing cause probability and the count for each cause. (Answer: Missing probability 0.030; cause counts 96, 84, 48, and 72; total exits 300.)
A table starts with 1,250 lives and records 25, 50, and 75 exits from three exhaustive causes. Find each cause probability, the total exit probability, and the number remaining. (Answer: Cause probabilities 0.020, 0.040, and 0.060; total probability 0.120; 1,100 remain.)
A hypothetical group of 1,800 lives has total exit probability 0.090. Two causes have observed counts of 63 and 27, and a third cause accounts for the remaining exits. Find the third count and all three cause probabilities. (Answer: Third count 72; cause probabilities 0.035, 0.015, and 0.040; total exits 162.)
There are 3,200 lives exposed and the total exit probability is 0.120. Three causes have counts 128, 64, and 160; a fourth cause accounts for every other exit. Find the fourth count and all four cause probabilities. (Answer: Fourth count 32; cause probabilities 0.040, 0.020, 0.050, and 0.010; total exits 384.)

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