Why the Closure Identity Matters
A multiple-decrement table separates exits from a population by cause: for example, a hypothetical insured population might leave active status because of death, disability, or withdrawal. The table’s cause-specific probabilities describe exits from the people who were present at the start of the period. Their sum must reconcile with the probability of any exit, and the probability of no exit must complete the total to one.
This reconciliation is called the multiple-decrement closure identity. It is a basic check on probability columns and table lives. It also gives a careful way to accumulate cause-specific exits over several periods: each period’s conditional probabilities must be applied only to the portion of the original group that has survived to that period.
These probabilities are conditional on being active at the start of the period. The cause categories must be mutually exclusive: one life cannot be counted as exiting from two causes during the same period. They must also be exhaustive: every exit in the model is assigned to one of the listed causes. If an exit type has been omitted, it must be added as another cause before applying the exhaustive version of the identity.
The One-Period Closure Identity
Let \(E_j\) be the event that a life exits from cause \(j\) during the period, and let \(S\) be the event that the life remains active through the period. The events \(E_1,\ldots,E_m,S\) form a partition of all possible outcomes: exactly one occurs. Since the cause-specific exit events are mutually exclusive, the probability of their union is the sum of their probabilities.
The first equality follows because “any exit” is the union of the cause-specific exit events. The second follows because survival through the period is the complement of any exit. Together, they give the closure identity: the probabilities of all cause-specific exits plus the probability of remaining active sum to one.
The identity also holds in a table of lives. If \(l_x\) lives are active at the start of the period, the expected number exiting from cause \(j\) is \(d_x^{(j)}=l_xq_x^{(j)}\). The expected number remaining active is \(l_{x+1}=l_xp_x^{(\tau)}\). Multiplying the probability identity by \(l_x\) gives the corresponding table reconciliation.
In a deterministic table, the counts may be whole numbers; in a probability model, expected counts need not be integers. The reconciliation concerns the quantities as defined by the model, not a requirement that every expected decrement count be a whole number.
Worked Example: Recovering a Missing Cause and Checking Table Lives
A hypothetical active population has four exhaustive exit causes during a year. For a life active at the start of the year, the probabilities for causes 1, 2, and 3 are \(0.021\), \(0.034\), and \(0.015\). The probability of any exit is \(0.080\). Find the probability for cause 4 and, for a starting table of 5,000 lives, the expected number exiting from each cause and the number remaining active.
Method. Apply closure to find the missing cause, then multiply each probability and the survival probability by the starting number of lives. The check this avoids is treating the total exit probability as separate from, rather than equal to, the sum of all exhaustive cause probabilities.
The sum of the three supplied cause probabilities is \(0.021+0.034+0.015=0.070\). Therefore, the missing probability is \(q_x^{(4)}=0.080-0.070=0.010\). The total is checked: \(0.021+0.034+0.015+0.010=0.080\).
The probability of remaining active is \(p_x^{(\tau)}=1-0.080=0.920\). The expected cause-specific counts are: cause 1, \(5{,}000(0.021)=105\); cause 2, \(5{,}000(0.034)=170\); cause 3, \(5{,}000(0.015)=75\); and cause 4, \(5{,}000(0.010)=50\).
The expected number remaining active is \(5{,}000(0.920)=4{,}600\). The table check is \(105+170+75+50+4{,}600=5{,}000\), so the result closes.
Closure Across Several Periods
For multiple periods, each \(q_{x+k}^{(j)}\) is conditional on being active at the start of period \(k+1\), not on belonging to the original cohort. Define \({}_k p_x^{(\tau)}\) as the probability of remaining active through the first \(k\) periods, with \({}_0p_x^{(\tau)}=1\). The probability of reaching the start of period \(k+1\) and then exiting from cause \(j\) is \({}_k p_x^{(\tau)}q_{x+k}^{(j)}\).
Consequently, the cumulative probability of exiting from cause \(j\) during the first \(n\) periods, measured from the original starting group, is \(\sum_{k=0}^{n-1}{}_kp_x^{(\tau)}q_{x+k}^{(j)}\). Summing over causes gives the cumulative probability of any exit. In each period, the total exit contribution is the probability of having reached that period multiplied by the conditional probability of exiting in it.
To justify the last equality, use the one-period identity \(q_{x+k}^{(\tau)}=1-p_{x+k}^{(\tau)}\), together with \({}_{k+1}p_x^{(\tau)}={}_kp_x^{(\tau)}p_{x+k}^{(\tau)}\). Thus each term in the middle sum is \({}_kp_x^{(\tau)}-{}_{k+1}p_x^{(\tau)}\). Adding these terms cancels the intermediate survival probabilities, leaving \(1-{}_np_x^{(\tau)}\). This telescoping argument proves that cumulative cause-specific exit probabilities add to the cumulative probability of any exit.
Worked Example: Accumulating Cause Probabilities Over Three Years
For a hypothetical group, there are two exhaustive exit causes. Conditional on being active at the start of each year, the cause probabilities are \(0.030\) and \(0.020\) in year 1, \(0.010\) and \(0.040\) in year 2, and \(0.020\) and \(0.030\) in year 3. Find the three-year survival probability and the cumulative probability, from the original cohort, of exiting from each cause.
Method. First find the total exit probability in each year and multiply the annual survival probabilities. Then weight each year’s cause-specific probability by the probability of reaching that year. This avoids adding probabilities that have different conditional starting populations.
The total exit probability is \(0.030+0.020=0.050\) in year 1, \(0.010+0.040=0.050\) in year 2, and \(0.020+0.030=0.050\) in year 3. Each annual survival probability is \(1-0.050=0.950\). Therefore, survival through all three years is \(0.950^3=0.9025(0.950)=0.857375\).
For cause 1, the year-by-year contributions from the original cohort are: \(1(0.030)=0.030\) in year 1; \(0.950(0.010)=0.0095\) in year 2; and \(0.9025(0.020)=0.01805\) in year 3. The cumulative probability is \(0.030+0.0095+0.01805=0.05755\).
For cause 2, the contributions are: \(1(0.020)=0.020\) in year 1; \(0.950(0.040)=0.038\) in year 2; and \(0.9025(0.030)=0.027075\) in year 3. The cumulative probability is \(0.020+0.038+0.027075=0.085075\).
The total cumulative exit probability is \(0.05755+0.085075=0.142625\), which equals \(1-0.857375=0.142625\). The calculation therefore closes.
Using the Identity to Check Table Lives
A table calculation provides another useful form of the same result. In each period, the number exposed to the decrement probabilities is the number active at the period’s start. The cause-specific expected counts for that period must sum to the total decrement count, and adding that count to the survivors must recover the starting number. Over several periods, the cumulative cause counts must account for the difference between the original starting group and the final survivors.
Worked Example: Reconciling Two Years of Expected Decrements
A hypothetical group starts with 10,000 active lives. In year 1, the conditional probabilities of three exhaustive exit causes are \(0.040\), \(0.030\), and \(0.010\). In year 2, conditional on still being active at its start, the probabilities are \(0.020\), \(0.050\), and \(0.030\). Calculate the expected exits by cause in each year, the survivors after each year, and verify the two-year closure.
Method. Calculate year 1 from 10,000 lives. Use year 1 survivors—not the original 10,000—as the year 2 exposure. The check this avoids is applying a conditional year 2 probability to lives that have already exited.
In year 1, the total exit probability is \(0.040+0.030+0.010=0.080\). Expected exits are \(10{,}000(0.040)=400\), \(10{,}000(0.030)=300\), and \(10{,}000(0.010)=100\). Total exits are \(400+300+100=800\), leaving \(10{,}000-800=9{,}200\) active lives.
In year 2, the total exit probability is \(0.020+0.050+0.030=0.100\). The year 2 expected exits are \(9{,}200(0.020)=184\), \(9{,}200(0.050)=460\), and \(9{,}200(0.030)=276\). They total \(184+460+276=920\), leaving \(9{,}200-920=8{,}280\) active lives.
Across both years, cause 1 accounts for \(400+184=584\) expected exits, cause 2 for \(300+460=760\), and cause 3 for \(100+276=376\). Total cumulative exits are \(584+760+376=1{,}720\). The final check is \(1{,}720+8{,}280=10{,}000\), as required.
Common Mistake: Adding Conditional Probabilities Directly
The annual cause-specific probabilities in the three-year example are conditional on being active at the start of their respective years. Adding the three annual probabilities for cause 1 without adjustment gives \(0.030+0.010+0.020=0.060\). For cause 2, the corresponding incorrect sum is \(0.020+0.040+0.030=0.090\). These are not cumulative probabilities from the original cohort: the year 2 and year 3 probabilities apply only to those who survive to those years.
The correction is to weight each annual cause probability by the probability of reaching that year. This gives \(0.05755\) for cause 1 and \(0.085075\) for cause 2. Their sum, \(0.142625\), agrees with one minus three-year survival. The unweighted totals sum to \(0.150\), which fails the closure check because it does not equal \(1-0.857375=0.142625\).
Check Your Understanding
For each problem, use the closure identity and show how conditional probabilities relate to the population exposed to each period.
- A one-year model has three exhaustive causes with probabilities \(0.018\), \(0.027\), and \(0.035\). Find total exit probability and survival probability. (Answer: total exit \(0.080\); survival \(0.920\).)
- Two exhaustive causes have conditional probabilities \((0.020,0.010)\) in year 1 and \((0.030,0.040)\) in year 2. Find two-year survival and the cumulative probability from the original cohort for each cause. (Answer: survival \(0.9021\); cause 1 \(0.0491\); cause 2 \(0.0488\).)
- A table begins with 2,400 lives and has three exhaustive one-year cause probabilities \(0.025\), \(0.015\), and \(0.010\). Find expected exits by cause and survivors. (Answer: exits \(60\), \(36\), and \(24\); survivors \(2{,}280\).)
- For two years, cause A has conditional probabilities \(0.040\) and \(0.030\), and cause B has conditional probabilities \(0.020\) and \(0.010\). Find two-year survival and both cumulative cause probabilities from the original cohort. (Answer: survival \(0.9024\); cause A \(0.0682\); cause B \(0.0294\).)
- A model records two listed causes with probabilities \(0.030\) and \(0.020\), but also allows an unclassified exit with probability \(0.010\). Find the probability of any exit and survival. Explain why the two listed causes alone do not close the model. (Answer: total exit \(0.060\); survival \(0.940\); the unclassified exit must be included as a third cause for the cause probabilities to sum to total exit.)