Purpose and Interpretation
An associated single-decrement table turns a sequence of probabilities for one selected cause into a corresponding sequence of survivor and decrement counts. Once the probabilities are specified, the calculation is recursive: apply each year's probability to the number alive at the start of that year, then carry the survivors forward. The resulting table gives a consistent count basis for the selected cause.
The counts in this table are not necessarily the observed numbers of exits from a real group exposed to several causes. They describe a hypothetical cohort following the associated single-decrement probabilities: only the selected decrement is represented in the table. The probabilities must therefore be understood as annual conditional probabilities on that associated basis, not as unconditional shares of the initial cohort.
This construction is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. The focus here is on building and checking the table when the associated probabilities have already been specified.
The initial count \(l'_{x_0}\) is the table's starting radix, or starting cohort size. It may be a convenient value such as 1,000 or 10,000; it need not equal the size of an observed population. Because the table tracks expected counts, its entries may be fractional even when its initial count is an integer. Use the supplied probabilities for the correct attained ages and keep the survivor count from one row as the starting count for the next.
The Recursion That Builds the Columns
For each one-year interval, the number decrementing is the number exposed at the beginning of the interval multiplied by the conditional probability of decrement. The remainder survive to the next row:
Thus, once the initial count and the sequence of \(q'_x\) values are known, every later survivor and decrement count is determined. The table is usually arranged with one row per starting age and columns for \(l'_x\), \(q'_x\), and \(d'_x\). The final survivor count in one interval is \(l'_{x+1}\), which becomes the exposure count in the next interval.
Write the given cohort size as \(l'_{x_0}\), at the starting age \(x_0\).
Pair \(l'_x\) with the supplied \(q'_x\) for the interval beginning at age \(x\).
Find \(d'_x=l'_xq'_x\), then \(l'_{x+1}=l'_x-d'_x\).
Use \(l'_{x+1}\), not the original cohort size, as the next interval's exposure count.
For every row, verify \(l'_x=d'_x+l'_{x+1}\); across the full table, verify that total decrements equal the initial count minus the final survivor count.
Why the Product Gives the Survivor Count
The survivor recursion follows directly from the definition of a conditional probability. Among \(l'_x\) lives present at age \(x\), the expected number decrementing in the next year is \(l'_xq'_x\). Subtracting that number leaves \(l'_x(1-q'_x)\) survivors. Applying this argument successively gives the product formula:
Each factor is the probability of surviving one particular year in the associated model, conditional on reaching that year's start. Multiplying the factors gives the probability of surviving all \(n\) consecutive years; multiplying that probability by the initial count gives the expected number remaining. The total decrement count across those years is therefore
This identity is also a useful check: decrements in successive rows must account for exactly the difference between the starting and ending cohort counts. The recursion assumes that each supplied \(q'_x\) applies to the associated cohort alive at the start of that interval. It does not apply if the entries are instead unconditional percentages of the initial cohort or if probabilities for a different basis are substituted without conversion.
Worked Example: Constructing Three Successive Rows
A hypothetical associated single-decrement cohort starts with 10,000 lives at age 50. Its annual associated probabilities for the selected cause are \(q'_{50}=0.06\), \(q'_{51}=0.08\), and \(q'_{52}=0.05\). Construct the survivor and decrement columns through age 53, and find the total decrement count.
Method. Apply each conditional probability to the survivor count at the start of its own year. This avoids applying all three probabilities to the original 10,000 lives.
At age 50, \(l'_{50}=10{,}000\). The first-year decrement is \(d'_{50}=10{,}000(0.06)=600\), so \(l'_{51}=10{,}000-600=9{,}400\).
At age 51, \(d'_{51}=9{,}400(0.08)=752\), so \(l'_{52}=9{,}400-752=8{,}648\).
At age 52, \(d'_{52}=8{,}648(0.05)=432.4\), so \(l'_{53}=8{,}648-432.4=8{,}215.6\).
| Starting age | \(l'_x\) | \(q'_x\) | \(d'_x\) | \(l'_{x+1}\) |
|---|---|---|---|---|
| 50 | 10,000 | 0.06 | 600 | 9,400 |
| 51 | 9,400 | 0.08 | 752 | 8,648 |
| 52 | 8,648 | 0.05 | 432.4 | 8,215.6 |
The total decrement count is \(600+752+432.4=1{,}784.4\). The endpoint check gives \(10{,}000-8{,}215.6=1{,}784.4\), as required. To two decimal places, the final survivor count is 8,215.60 and the total decrement count is 1,784.40.
Common mistake avoided. Adding the annual probabilities gives \(0.06+0.08+0.05=0.19\), but this is not the three-year probability of decrement. The associated probability of surviving all three years is \(0.94(0.92)(0.95)=0.82156\), so the cumulative decrement probability is \(1-0.82156=0.17844\), consistent with \(1{,}784.4/10{,}000\).
Worked Example: Continuing a Table With Fractional Counts
A hypothetical cohort begins with 2,400 lives at age 60. The associated probabilities for a selected decrement are 0.10, 0.05, 0.12, and 0.04 at ages 60, 61, 62, and 63, respectively. Build all four rows, then determine the survivors and total decrements at age 64.
Method. Iterate the row equations without rounding intermediate counts. Fractional values are retained because these are expected counts.
At age 60, \(d'_{60}=2{,}400(0.10)=240\), and \(l'_{61}=2{,}400-240=2{,}160\).
At age 61, \(d'_{61}=2{,}160(0.05)=108\), and \(l'_{62}=2{,}160-108=2{,}052\).
At age 62, \(d'_{62}=2{,}052(0.12)=246.24\), and \(l'_{63}=2{,}052-246.24=1{,}805.76\).
At age 63, \(d'_{63}=1{,}805.76(0.04)=72.2304\), and \(l'_{64}=1{,}805.76-72.2304=1{,}733.5296\).
The total decrements are \(240+108+246.24+72.2304=666.4704\). Independently, the endpoint difference is \(2{,}400-1{,}733.5296=666.4704\). Reporting to two decimal places, \(l'_{64}=1{,}733.53\) and the total decrements are 666.47.
Check. Each row satisfies \(l'_x=d'_x+l'_{x+1}\). In particular, the last row gives \(1{,}805.76=72.2304+1{,}733.5296\). Rounding the age-63 survivor count before calculating \(d'_{63}\) would introduce avoidable error.
Worked Example: Recovering a Probability and Completing the Table
A hypothetical selected-cause table starts with \(l'_{70}=3{,}200\). The decrement count in the first year is \(d'_{70}=224\). The next two associated probabilities are \(q'_{71}=0.06\) and \(q'_{72}=0.08\). Find \(q'_{70}\), complete the table to age 73, and verify the cumulative decrement count.
Method. Rearrange the defining equation \(d'_x=l'_xq'_x\) to recover the missing probability, then continue the usual survivor recursion.
The first-year probability is \(q'_{70}=224/3{,}200=0.07\). Therefore \(l'_{71}=3{,}200-224=2{,}976\).
For age 71, \(d'_{71}=2{,}976(0.06)=178.56\), so \(l'_{72}=2{,}976-178.56=2{,}797.44\).
For age 72, \(d'_{72}=2{,}797.44(0.08)=223.7952\), so \(l'_{73}=2{,}797.44-223.7952=2{,}573.6448\).
Total decrements are \(224+178.56+223.7952=626.3552\). The endpoint check gives \(3{,}200-2{,}573.6448=626.3552\). To two decimal places, \(l'_{73}=2{,}573.64\) and total decrements are 626.36.
Common mistake avoided. The first decrement count is not itself the probability. The probability is \(224/3{,}200=0.07\), because the denominator must be the count exposed at age 70.
Practical Checks and Interpretation
A correctly constructed table should meet both local and cumulative checks. Locally, each row must satisfy \(d'_x=l'_xq'_x\) and \(l'_{x+1}=l'_x-d'_x\). Cumulatively, summing all decrement counts must give the difference between the initial and final survivor counts. These checks are particularly valuable when probabilities vary by age or when the table contains fractional counts.
- Confirm that every \(q'_x\) is between 0 and 1, inclusive.
- Match each probability to the age and interval shown in its row.
- Use the current survivor count as exposure for the next year's probability.
- Retain full precision in intermediate calculations unless the question specifies a rounding rule.
- Check that total decrements equal initial lives minus final survivors.
If a supplied probability is 0, the row has no decrements and all lives survive that year. If it is 1, all lives present at the start decrement in that interval and the next survivor count is zero. A later row with zero exposure consequently has zero decrements under the recursion. These boundary values do not change the construction, though probabilities for years with no lives exposed do not affect the resulting counts.
Check Your Understanding
Construct the requested associated-table quantities using the stated annual probabilities. Keep intermediate counts unrounded.
- A cohort begins with 8,000 lives and has associated probabilities 0.03 and 0.04 in successive years. Find the survivor count after two years and each year's decrement count. (Answer: \(d'_0=240\), \(l'_1=7{,}760\), \(d'_1=310.4\), \(l'_2=7{,}449.6\).)
- A cohort starts with 1,250 lives. Its three successive associated probabilities are 0.20, 0.10, and 0.05. Find the three decrement counts, final survivor count, and total decrements. (Answer: \(d'_0=250\), \(d'_1=100\), \(d'_2=45\), \(l'_3=855\), total decrements \(=395\).)
- A table starts with 4,000 lives and records 240 decrements in the first year. The next year's associated probability is 0.10. Find the first-year probability, next-year survivor count, and next-year decrement count. (Answer: \(q'_0=0.06\), \(l'_1=3{,}760\), \(d'_1=376\).)
- A cohort of 3,600 has successive associated probabilities 0.08, 0.12, and 0.05. Find the survivor count after three years and the total decrement count. (Answer: \(l'_3=2{,}768.832\), total decrements \(=831.168\).)
- A cohort begins with 5,000 lives. Its first two associated probabilities are 0.10 and 0.20. Find the two annual decrement counts and the overall three-year survival probability if the third-year probability is 0.25. (Answer: \(d'_0=500\), \(d'_1=900\), overall survival probability \(=0.9(0.8)(0.75)=0.54\).)