Why Assumption Consistency Matters
A multi-state valuation can use individually plausible inputs and still be wrong if those inputs describe different models. For example, a transition rate out of active status may be combined with an annual decrement probability that already includes competing exits. Or a category called “termination” may be added to withdrawal and death even though it includes both. In each case, the calculation can silently change the population at risk or count the same exit more than once.
A reliable audit begins by asking what each quantity represents: a transition from one state to another, a cause-specific intensity, a probability over a specified interval, or a decrement probability in a table. These quantities are related, but they are not interchangeable. The checks below apply to hypothetical benefit models; no interest or benefit cash flow is needed to diagnose the probability assumptions.
The distinction between the last two definitions is central. In a multiple-decrement model, someone who exits by one cause is no longer exposed to the other causes for the remainder of the interval. In an associated single-decrement model, the competing causes are removed. Therefore, a single-decrement probability generally is not the actual probability of that cause in the multiple-decrement model.
Continuous-Time Competing Exits
Suppose an individual begins in one state and can leave it by \(r\) mutually exclusive causes. Let \(\lambda_j(t)\) be the cause-specific transition intensity for cause \(j\). Assume the intensities are nonnegative and locally integrable, and that, conditional on remaining in the starting state, these are the only possible exits. Write \(\Lambda(t)=\sum_{j=1}^{r}\lambda_j(t)\) for the total exit intensity.
The probability of still occupying the starting state at time \(t\) is \(p_0(t)=\exp\left(-\int_0^t \Lambda(s)\,ds\right)\). The probability of exit by cause \(j\) by time \(t\) is \(\int_0^t p_0(s)\lambda_j(s)\,ds\). These expressions describe competing exits, not independent probabilities that can simply be added to a separate persistence probability.
When each cause has a constant intensity \(\lambda_j\), let \(\Lambda=\sum_j\lambda_j\). Over an interval of length \(t\), persistence is \(e^{-\Lambda t}\), and the probability of exit by cause \(j\) is \(\frac{\lambda_j}{\Lambda}(1-e^{-\Lambda t})\), provided \(\Lambda>0\). If \(\Lambda=0\), persistence is one and every exit probability is zero.
Worked Example: Checking Annual Probabilities Against State Intensities
A hypothetical active member can retire, withdraw, or die. Assume constant transition intensities per year of \(0.20\) for retirement, \(0.10\) for withdrawal, and \(0.10\) for death. The waiting time to the next exit is exponential under these constant intensities. There is no interest-rate assumption because this is a probability audit, not a cash-flow valuation. Find the one-year probabilities and check a proposed table that reports retirement \(0.18\), withdrawal \(0.09\), death \(0.09\), and persistence \(0.64\).
Method. Add the cause-specific intensities to obtain the total intensity: \( \Lambda=0.20+0.10+0.10=0.40\) per year. The one-year persistence probability implied by this state model is \(e^{-0.40}=0.670320\), or \(0.6703\) to four decimals.
The probability of retirement is \(\frac{0.20}{0.40}(1-e^{-0.40})=0.5(0.329680)=0.164840\), or \(0.1648\). The probability of withdrawal is \(\frac{0.10}{0.40}(1-e^{-0.40})=0.25(0.329680)=0.082420\), or \(0.0824\). The probability of death is the same, \(0.0824\).
The proposed table does sum to one: \(0.18+0.09+0.09+0.64=1.00\). But it does not match the stated transition intensities. For example, the model implies \(0.6703\) persistence, not \(0.64\), and \(0.1648\) retirement, not \(0.18\). A row summing to one is necessary, but not sufficient: its probabilities must also agree with the model assumptions.
Check. The implied probabilities sum to \(0.670320+0.164840+0.082420+0.082420=1.000000\). The common mistake avoided is treating a probability table as consistent merely because its entries sum to one.
Associated Single-Decrement Assumptions
For constant intensity \(\lambda_j\), the associated single-decrement model for cause \(j\) removes every competing cause. Its probability of decrement over one year is \(q'_j=1-e^{-\lambda_j}\). This is different from the multiple-decrement probability \(q_j\), which is computed while all causes remain active. Under constant competing intensities, \(q_j=\frac{\lambda_j}{\Lambda}(1-e^{-\Lambda})\). These probabilities answer different questions, so substituting one for the other changes the model.
Worked Example: Comparing a Cause-Specific Probability With Its Associated Probability
A hypothetical active member can retire, withdraw, or die, with constant annual intensities \(\lambda_R=\frac{\ln 2}{4}\), \(\lambda_W=\frac{\ln 2}{4}\), and \(\lambda_D=\frac{\ln 2}{2}\). Assume exponential waiting times, no other exits, and no interest-rate assumption because only probabilities are being compared. Find the one-year multiple-decrement probabilities and the associated single-decrement probability for death.
Method. The total intensity is \(\Lambda=\frac{\ln 2}{4}+\frac{\ln 2}{4}+\frac{\ln 2}{2}=\ln 2\). Thus persistence for one year is \(e^{-\ln 2}=0.5\), and the total exit probability is \(1-0.5=0.5\).
Retirement and withdrawal each account for one quarter of the total intensity, so each multiple-decrement probability is \(\frac{1}{4}(0.5)=0.1250\). Death accounts for one half, so its multiple-decrement probability is \(\frac{1}{2}(0.5)=0.2500\). The check is \(0.1250+0.1250+0.2500+0.5000=1.0000\).
For the associated single-decrement death model, remove retirement and withdrawal. The remaining death intensity is \(\frac{\ln 2}{2}\), so the one-year death probability is \(1-e^{-(\ln 2)/2}=1-\frac{1}{\sqrt{2}}=0.292893\), or \(0.2929\) to four decimals. It exceeds the multiple-decrement death probability \(0.2500\), because the associated model does not allow the other exits to remove people from exposure to death.
Check. The persistence and three cause probabilities in the multiple-decrement model total one, while the associated death probability is separately calculated in a model without competing exits. The common mistake avoided is substituting \(0.2929\) for the actual multiple-decrement death probability \(0.2500\).
Annual decrement tables may instead provide multiple-decrement probabilities directly. A standard conversion under a uniform-distribution-of-decrements (UDD) assumption is \(q'_j \approx q_j/(1-\frac{1}{2}\sum_{k\ne j}q_k)\), where the \(q_k\) are the multiple-decrement probabilities in the same interval. This is a UDD conversion, not a general identity for arbitrary transition patterns. It should not be applied to probabilities from a different period, a different population, or a continuous-time model without the required assumptions.
Worked Example: Converting a Table Probability Under UDD
A hypothetical annual multiple-decrement table reports retirement \(q_R=0.12\), withdrawal \(q_W=0.08\), and death \(q_D=0.10\). Assume the causes are mutually exclusive, these are probabilities for the same one-year interval and population, and decrements are uniformly distributed through the year for the conversion. Find the UDD associated single-decrement retirement probability. No interest assumption is needed for this conversion.
Method. For retirement, the competing multiple-decrement probabilities are withdrawal and death, with sum \(0.08+0.10=0.18\). The UDD denominator is \(1-\frac{1}{2}(0.18)=1-0.09=0.91\). Therefore, \(q'_R\approx\frac{0.12}{0.91}=0.131868\), or \(0.1319\) to four decimals.
The original multiple-decrement probabilities sum to \(0.12+0.08+0.10=0.30\), so the table’s one-year persistence probability is \(1-0.30=0.70\). That persistence check concerns the original table; it does not mean the associated single-decrement retirement probability must equal \(0.12\). The common mistake avoided is applying the conversion with only one of the competing probabilities in the denominator, or using the result as if it were the original table probability.
Check Cause Definitions Before Adding Probabilities
Before calculating, list the possible exits from each starting state and make sure they are mutually exclusive for the purpose of the model. A label may describe a broad administrative outcome rather than a distinct cause. For example, “termination” might include withdrawal and death. If so, it cannot be added as another decrement alongside withdrawal and death.
For annual multiple-decrement probabilities from a common starting population, the cause categories must be disjoint over the interval. If \(q_j\) are those probabilities and \(p\) is the probability of remaining in the state, the basic balance is \(p+\sum_jq_j=1\). If categories overlap, their probabilities cannot be summed as if they were disjoint. The model must either define distinct causes or explicitly handle the overlap using appropriate joint probabilities.
Worked Example: Removing a Duplicated Exit Category
A hypothetical table begins with 1,000 active members. Its reported one-year probabilities are retirement \(0.12\), withdrawal \(0.07\), death \(0.05\), and “termination” \(0.12\). The data definition says termination consists exactly of withdrawal or death. Assume each member can have only one first exit during the year; no interest or mortality law beyond the stated death probability is needed. Find the distinct exit counts and survivors.
Method. Since termination is the union of withdrawal and death, it is an aggregate description of exits already counted. The mutually exclusive categories are retirement, withdrawal, and death. Their total exit probability is \(0.12+0.07+0.05=0.24\), and persistence is \(1-0.24=0.76\).
Multiplying by 1,000 gives 120 retirements, 70 withdrawals, and 50 deaths. The number remaining active is \(1{,}000(0.76)=760\). The termination count is \(70+50=120\), but it is a subtotal, not an additional 120 exits.
Check. Distinct exits plus survivors are \(120+70+50+760=1{,}000\). If the termination category were incorrectly added, the claimed exits would be \(120+70+50+120=360\), leaving 640 survivors; that calculation counts each withdrawal and death twice. The correction is to exclude the aggregate category from the sum of distinct causes.
A Practical Consistency Audit
An efficient audit separates model definition, rate assumptions, and probability checks. The sequence below helps locate the source of a discrepancy before it carries into a valuation.
For every starting state, record each permitted destination or terminating cause. Mark any broad category that overlaps another listed exit.
Separate intensities per unit time from probabilities over an interval. Confirm that annual probabilities and rates refer to the same period, state, and population.
For a continuous-time model, use the total intensity to determine persistence and integrate each cause’s intensity over the exposure that remains. Do not treat each competing cause as operating on the full original exposure independently.
State whether a decrement probability is a multiple-decrement probability or a single-decrement probability. Use a conversion only when its assumptions, such as UDD, are appropriate.
Verify that mutually exclusive outcomes and persistence sum to one, and that each count is based on the correct starting exposure. A numerical balance alone does not establish that the inputs describe the same model.
These checks matter because transition assumptions determine who remains exposed to future benefits, premiums, and exits. A mismatch can alter both the probability of entering a benefit state and the expected duration in that state. In pension or care-benefit work, correcting the cause definitions and transition assumptions should come before recomputing the valuation.
Check Your Understanding
For each problem, identify the relevant model assumption before calculating.
- A state has two mutually exclusive constant exit intensities, \(0.30\) and \(0.20\) per year. Under exponential waiting times, find one-year persistence and the two cause-specific exit probabilities. (Answer: persistence \(0.6065\); exit probabilities \(0.2361\) and \(0.1574\), to four decimals.)
- A state has two constant competing intensities, \(0.10\) and \(0.20\) per year. Under exponential waiting times, find the probability of remaining in the state for two years and the probability of exit by each cause during those two years. (Answer: persistence \(e^{-0.6}=0.5488\); cause probabilities \(0.1504\) and \(0.3008\), to four decimals.)
- An annual multiple-decrement table starts with 2,500 members and reports mutually exclusive probabilities \(0.09\), \(0.06\), and \(0.05\). Find the number exiting by each cause and the number remaining. (Answer: 225, 150, and 125 exits; 2,000 remain.)
- Under a UDD conversion, a cause has multiple-decrement probability \(0.16\); the other mutually exclusive causes have probabilities \(0.10\) and \(0.08\). Find the associated single-decrement probability for the first cause. (Answer: \(0.16/[1-\frac12(0.10+0.08)]=0.1758\), to four decimals.)
- A hypothetical table gives retirement \(0.12\), withdrawal \(0.07\), death \(0.05\), and “termination” \(0.12\), where termination consists exactly of withdrawal or death. For 1,000 starting members, how many distinct exits and survivors should be counted? (Answer: 240 distinct exits and 760 survivors; the termination subtotal is not added again.)