Why the Choice of Decrement Basis Matters
A pension valuation may involve retirement, withdrawal, and death as competing ways to leave active service. These causes cannot all occur first for the same member in the same year. The probability of a particular exit therefore depends on the other exits operating alongside it. A single-decrement calculation, by contrast, asks what would happen if one specified cause operated on its own.
The distinction is important whenever a benefit is conditional on a specific event. A retirement pension may be paid only to members who retire before withdrawing or dying; an active-service pension may be paid only while the member remains in service. Such questions generally require the actual probabilities under all relevant competing decrements. A hypothetical question about the isolated retirement or mortality experience may instead use an associated single-decrement basis.
The prime is a reminder that \(q'_{x,j}\) is not the observed probability of cause \(j\) in the multiple-decrement population. In particular, one cannot generally obtain an associated single-decrement probability merely by renaming \(q_{x,j}\). A conversion from actual multiple-decrement probabilities to an associated basis requires assumptions about the decrement process within the year. If those assumptions are not given, the conversion is not determined.
Choose the Basis From the Benefit Condition
Start by translating the cash flow into an event statement. If it is payable only on an actual retirement, withdrawal, or death from the active population, use the relevant multiple-decrement probability. If the question explicitly describes a hypothetical population exposed to only one decrement, use the corresponding single-decrement basis. If a payment depends on remaining active, use the probability of avoiding all competing exits, not merely mortality.
| Valuation question | Relevant probability | Interpretation |
|---|---|---|
| Benefit paid when an active member actually retires | \(q_{x,R}\) | Actual retirement probability with all specified exits operating |
| Benefit paid if an active member actually dies | \(q_{x,D}\) | Actual death probability before another exit |
| Payment made only if a member remains active | \(p_x=1-\sum_jq_{x,j}\) | Probability of avoiding every listed exit for the year |
| Hypothetical experience with only mortality operating | \(q'_{x,D}\) | Isolated mortality probability on the stated single-decrement basis |
An associated single-decrement basis is not inherently more or less conservative than a multiple-decrement basis. It answers a different question. It can be useful for an isolated-risk comparison, but it does not automatically describe the probabilities that govern actual benefit payments in a population exposed to competing exits.
Worked Example: One-Year Benefits on Actual Exits
A hypothetical pension plan covers active members aged \(x\). During the next year, the actual multiple-decrement probabilities are \(q_{x,R}=0.12\) for retirement, \(q_{x,W}=0.08\) for withdrawal, and \(q_{x,D}=0.02\) for death. At the end of the year, the plan pays \( \$20{,}000\) on retirement, \( \$4{,}000\) on withdrawal, and \( \$12{,}000\) on death. The effective annual interest rate is \(i=5\%\). Find the expected present value (EPV) per active member at the start of the year.
Method. Each payment is triggered by an actual exit, so use the multiple-decrement probabilities. The exits are mutually exclusive. The remaining-active probability is \(1-0.12-0.08-0.02=0.78\), although no benefit is specified for remaining active.
The expected payment at the end of the year is \(0.12(\$20{,}000)+0.08(\$4{,}000)+0.02(\$12{,}000)=\$2{,}400+\$320+\$240=\$2{,}960\). The discount factor is \(v=1/(1.05)=0.95238095\). Therefore,
Rounded to the nearest cent, the EPV is \(\$2{,}819.05\).
Suppose a separate hypothetical mortality-only basis gives \(q'_{x,D}=0.025\). That basis would give an isolated mortality-only payment value of \(0.025(\$12{,}000)/1.05=\$285.71\). It does not replace the actual death component of this plan’s multiple-decrement EPV, which is \(0.02(\$12{,}000)/1.05=\$228.57\). Common mistake avoided: substituting an isolated mortality probability for the actual death probability when valuing a benefit on death in a population with competing exits.
Why Actual Cause-Specific Probabilities Add to the EPV
For a single year, suppose the benefit paid at year-end on cause \(j\) is \(B_j\), and the effective annual interest rate is \(i\). Let \(I_j\) equal 1 if cause \(j\) occurs during the year and 0 otherwise. Because only one first exit can occur, the year-end payment is \(\sum_j B_j I_j\). Taking expectations gives \(\sum_j B_jP(I_j=1)=\sum_j B_jq_{x,j}\). Discounting this expected payment by \(v=1/(1+i)\) gives the EPV.
The same reasoning extends to multiple years. To receive a benefit on cause \(j\) in year \(k+1\), the member must first remain exposed to all decrements through the previous \(k\) years and then exit by cause \(j\). If the annual probabilities vary by age or year, denote the probability of remaining active in year \(r\) by \(p_r=1-\sum_jq_{r,j}\). The probability of reaching the start of year \(k+1\) is \(\prod_{r=0}^{k-1}p_r\); the probability of a cause-\(j\) exit in that year is this product multiplied by \(q_{k,j}\). For a year-end benefit \(B_{k,j}\), its EPV contribution is thus \(v^{k+1}\big(\prod_{r=0}^{k-1}p_r\big)q_{k,j}B_{k,j}\). This uses actual competing-risk probabilities throughout.
On a single-decrement basis for cause \(j\), the corresponding probability of still being exposed at the start of year \(k+1\) is instead \(\prod_{r=0}^{k-1}(1-q'_{r,j})\). This is survival from cause \(j\) alone under that hypothetical basis; it does not mean the member actually avoided withdrawal, retirement, or other competing exits.
Worked Example: A Retirement Benefit Over Two Years
A hypothetical plan pays \( \$10{,}000\) at the end of the year in which an active member retires. Actual annual probabilities are \(q_{1,R}=0.10,\ q_{1,W}=0.05,\ q_{1,D}=0.05\) in year 1, and \(q_{2,R}=0.12,\ q_{2,W}=0.08,\ q_{2,D}=0.04\) in year 2. The effective annual interest rate is \(i=5\%\). Find the EPV per member at the start of year 1. Assume the stated probabilities apply to members active at the beginning of each respective year.
Method. Value the actual retirement payment in each year. In year 1, the probability of retirement is \(0.10\). The probability of remaining active through year 1 is \(1-0.10-0.05-0.05=0.80\). Thus the probability of retirement in year 2, measured from the start of year 1, is \(0.80(0.12)=0.096\).
With \(v=1/1.05\), the year-1 contribution is \(v(0.10)(\$10{,}000)=\$952.38095\ldots\). The year-2 contribution is \(v^2(0.096)(\$10{,}000)\). Since \(v^2=1/(1.05)^2=0.907029478\ldots\), this is \(\$870.74830\ldots\). Therefore,
Rounded to the nearest cent, the EPV is \(\$1{,}823.13\).
For contrast, suppose a retirement-only basis is separately specified with \(q'_{1,R}=0.13\) and \(q'_{2,R}=0.15\). The hypothetical single-decrement retirement value is \(v(0.13)(\$10{,}000)+v^2(0.87)(0.15)(\$10{,}000)=\$1{,}238.09524\ldots+\$1{,}183.67347\ldots=\$2{,}421.76871\ldots\), or \(\$2{,}421.77\). This is not the actual plan value: it describes a hypothetical population in which retirement is the only decrement. Common mistake avoided: treating the associated retirement-only value as the probability or value of actual retirement benefits when withdrawal and death also remove members.
Payments Conditional on Remaining Active
The choice of basis also affects pension payments made while a member remains active. If a payment is made at the end of each year only to members still active at that date, each payment is weighted by the probability of avoiding every competing decrement up to that payment date. Mortality-only survival is insufficient when withdrawal or retirement also ends eligibility for the active-member payment.
Worked Example: Active-Service Payments Versus Mortality-Only Survival
A hypothetical plan pays \( \$1{,}000\) at the end of each of the next two years, but only if the member is still active at that payment date. In year 1, the actual probabilities are \(q_{1,R}=0.10,\ q_{1,W}=0.06,\ q_{1,D}=0.04\). In year 2, they are \(q_{2,R}=0.12,\ q_{2,W}=0.08,\ q_{2,D}=0.05\). The effective annual interest rate is \(i=4\%\). Find the EPV. A separate associated mortality-only basis has \(q'_{1,D}=0.03\) and \(q'_{2,D}=0.04\).
Method. For the actual active-service payments, the year-1 active probability is \(1-0.10-0.06-0.04=0.80\). The probability of being active at the end of year 2 is \(0.80[1-(0.12+0.08+0.05)]=0.80(0.75)=0.60\). With \(v=1/1.04\), the EPV is
Rounded to the nearest cent, the actual EPV is \(\$1{,}323.96\). The mortality-only basis gives a different hypothetical value: survival from death through year 1 is \(0.97\), and through year 2 is \(0.97(0.96)=0.9312\). Its value is \(\$1{,}000(0.97v+0.9312v^2)=\$1{,}793.6391\ldots=\$1{,}793.64\). The larger value reflects that this basis ignores withdrawal and retirement as causes of leaving active service. Common mistake avoided: using mortality-only survival to weight a payment that requires the member to remain active.
Practical Checks Before Valuing
A few checks help prevent basis errors. First, confirm that the actual multiple-decrement probabilities are valid: each is nonnegative and their sum is at most 1. Second, identify the exposure condition for every payment: actual exit by a specified cause, survival from one isolated cause, or continued active status. Third, verify that the decrement basis matches that condition. Finally, do not infer a conversion between bases unless the within-year assumptions or conversion convention are specified.
For example, “retires from active service during year 2” calls for an actual cause-specific probability.
For continued active status, subtract all relevant decrement probabilities from 1 to obtain the annual probability of remaining active.
A single-decrement probability applies only to its isolated-risk interpretation unless a stated model connects it to the actual multiple-decrement basis.
This distinction is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus because pension values depend not just on the benefit amount and payment timing, but also on the probability basis that correctly represents the event or continued eligibility.
Check Your Understanding
For each problem, identify the probability basis that matches the question before calculating.
- During one year, actual probabilities are \(q_R=0.08,\ q_W=0.07,\ q_D=0.03\). Find the probability of remaining active and the actual probability of retirement. (Answer: \(p=0.82\); actual retirement probability \(=0.08\).)
- A year-end benefit is \(\$6{,}000\) on retirement, \(\$1{,}000\) on withdrawal, and \(\$9{,}000\) on death. The actual probabilities are \(0.10,\ 0.06,\ 0.02\), respectively, and \(i=4\%\). Find the one-year EPV. (Answer: \(\$807.69\).)
- An active-member payment of \(\$1{,}000\) is made at each of the next two year-ends only if the member is active then. The actual probabilities of remaining active are \(0.85\) in year 1 and \(0.80\) in year 2. If \(i=5\%\), find the EPV. (Answer: \(\$1{,}426.30\).)
- On a stated retirement-only basis, the associated probabilities are \(q'_{1,R}=0.12\) and \(q'_{2,R}=0.10\). Find the hypothetical probability of retirement by the end of year 2. (Answer: \(1-(1-0.12)(1-0.10)=0.208\).)
- Actual one-year probabilities of retirement and withdrawal are given, but no within-year decrement assumption or conversion convention is provided. Can the associated single-decrement retirement probability be determined uniquely from those probabilities alone? (Answer: No; an assumption or conversion convention is needed.)