Why Conversion Requires an Assumption
An annual multiple-decrement table records the probability that a person exits from a starting state during the year for each competing cause. For example, exits might arise from disability, recovery, or death. The observed cause-specific probabilities account for competition: a person who exits for one cause can no longer exit for another during that year.
An associated single-decrement probability answers a different question. It is the probability of exit from one cause if the competing causes are removed, while retaining the specified model for that cause’s within-year rate. Annual multiple-decrement probabilities alone do not specify what happens within the year. A conversion therefore depends on an assumption about the timing of exits.
Two common assumptions are uniform distribution of decrements (UDD) and constant force. Both can be used to convert an annual multiple-decrement table to associated single-decrement probabilities. They give the same annual conversion when applied consistently, but they describe different patterns of exits within the year.
UDD Conversion
Under UDD, the number of exits from each cause accumulates linearly through the year as a proportion of the starting cohort. At time \(t\), measured in years from the start, the probability of exit from cause \(j\) by time \(t\) in the multiple-decrement model is \(tq_j\). The probability still in the starting state is \(1-tQ\).
The instantaneous force of exit for cause \(j\), among those still in the starting state, is therefore the cause-\(j\) exit density divided by the probability still exposed. The density is \(q_j\), so the force is \(q_j/(1-tQ)\). If the competing causes are removed, retain this force for cause \(j\) over the year.
The final expression applies when \(Q>0\). If \(Q=0\), every \(q_j\) is zero and every associated probability is zero. If there is only one cause, then \(q_j=Q\) and the formula reduces to \(q'_j=Q\), as it should.
Constant-Force Conversion
Under a constant-force assumption, each cause has a constant force during the year. Let \(\lambda_j\) be the force for cause \(j\), and let \(\lambda=\sum_j\lambda_j\) be the total force. The probability of any exit during the year is \(Q=1-e^{-\lambda}\), so \(\lambda=-\ln(1-Q)\). The multiple-decrement probability for cause \(j\) is its share of total force multiplied by the total exit probability.
With all other causes removed, cause \(j\) alone operates at force \(\lambda_j\). Its one-year probability is \(q'_j=1-e^{-\lambda_j}\), which gives the same annual conversion as under UDD:
The equality of these annual associated probabilities does not mean the models are interchangeable. Under UDD the multiple-decrement exit probabilities increase linearly with \(t\); under constant force, the probability of exit by time \(t\) is \(1-e^{-\lambda t}\). The annual endpoint is the same, but intermediate-time calculations can differ.
Why the Conversion Formula Holds
The UDD result follows by accumulating the cause-specific force over the year. Since the probability remaining in the starting state at time \(t\) is \(1-tQ\), a constant exit density \(q_j\) corresponds to force \(q_j/(1-tQ)\). Integrating that force gives the cumulative hazard for cause \(j\) if it acts alone.
Under UDD, \(\mu_j(t)=q_j/(1-tQ)\), because the cause-\(j\) exit density is \(q_j\) and the exposed proportion is \(1-tQ\).
\(\int_0^1 q_j/(1-tQ)\,dt=-(q_j/Q)\ln(1-Q)\), for \(Q>0\).
Net survival under cause \(j\) alone is the exponential of the negative cumulative hazard, \((1-Q)^{q_j/Q}\). Subtracting this from one gives \(q'_j=1-(1-Q)^{q_j/Q}\).
For constant forces, the total force is \(-\ln(1-Q)\), and cause \(j\)'s share of that force is \(q_j/Q\). Thus its force is \(-\ln(1-Q)q_j/Q\), yielding the same annual formula. This justification relies on the stated within-year models; it is not a model-free identity.
Worked Conversions and Checks
Worked Example: Converting Three Competing Exit Probabilities
A hypothetical annual table has three causes with \(q_1=0.18\), \(q_2=0.12\), and \(q_3=0.06\). Convert to associated single-decrement probabilities under UDD or constant force, then check the aggregate probability. The model is measured in years; no interest assumption is needed for this probability conversion.
Method. First calculate \(Q=0.18+0.12+0.06=0.36\). Thus the probability of no exit from any cause is \(1-Q=0.64\). The cause shares are \(0.18/0.36=0.5\), \(0.12/0.36=1/3\), and \(0.06/0.36=1/6\).
For cause 1, \(q'_1=1-0.64^{0.5}=1-0.8=0.200000\). For cause 2, \(q'_2=1-0.64^{1/3}=1-0.8617739=0.1382261\). For cause 3, \(q'_3=1-0.64^{1/6}=1-0.9283178=0.0716822\).
To check consistency, multiply the associated no-exit probabilities: \(1-q'_1=0.8\), \(1-q'_2=0.8617739\), and \(1-q'_3=0.9283178\). Their product is approximately \(0.640000\), so the implied aggregate exit probability is \(1-0.64=0.36\), matching the original \(Q\). Rounded to four decimal places, the associated probabilities are \(0.2000\), \(0.1382\), and \(0.0717\).
Common mistake avoided: The three associated probabilities do not add to the multiple-decrement total. The aggregate check uses the product of their survival probabilities, not their sum.
Worked Example: Comparing Midyear Exit Patterns
A hypothetical table has two annual multiple-decrement probabilities, \(q_1=0.12\) and \(q_2=0.08\). Thus \(Q=0.20\). Find each cause’s probability of exit by the midpoint under (a) UDD and (b) constant force. Then calculate the associated single-decrement probabilities. Time is measured in years.
UDD midpoint. At \(t=0.5\), the cause-specific cumulative probabilities are \(0.5(0.12)=0.060\) and \(0.5(0.08)=0.040\). The total is \(0.100\). UDD gives a linear within-year pattern.
Constant-force midpoint. The total force is \(\lambda=-\ln(0.80)=0.2231436\). The cause shares are \(0.12/0.20=0.6\) and \(0.08/0.20=0.4\), so \(\lambda_1=0.1338862\) and \(\lambda_2=0.0892574\). At the midpoint, total exit probability is \(1-e^{-0.2231436(0.5)}=1-\sqrt{0.80}=0.1055728\). The cause-specific midpoint probabilities are \(0.6(0.1055728)=0.0633437\) and \(0.4(0.1055728)=0.0422291\).
Associated annual probabilities. Under either assumption, \(q'_1=1-0.80^{0.6}=1-e^{-0.1338862}\approx0.1253\), and \(q'_2=1-0.80^{0.4}=1-e^{-0.0892574}\approx0.0854\). The aggregate check is \(1-(1-0.1253)(1-0.0854)\approx0.2000\), allowing for displayed rounding.
Check. The midpoint cause probabilities differ across assumptions: UDD gives \(0.0600\) and \(0.0400\), whereas constant force gives approximately \(0.0633\) and \(0.0422\). Both models have the same annual totals and associated annual probabilities, but not the same timing within the year.
Worked Example: Rounding and the Aggregate Exit Check
A hypothetical table has total exit probability \(Q=0.20\), with cause probabilities \(q_1=0.05\), \(q_2=0.10\), and \(q_3=0.05\). Convert to associated single-decrement probabilities and verify the total. The assumptions are UDD or constant force over a one-year interval.
The cause shares are \(0.25\), \(0.50\), and \(0.25\), and \(1-Q=0.80\). Therefore, \(q'_1=1-0.80^{0.25}=1-\exp(-0.0557859)\approx1-0.945742=0.054258\). The same calculation applies to cause 3, so \(q'_3\approx0.054258\). For cause 2, \(q'_2=1-0.80^{0.50}=1-\sqrt{0.80}=1-0.894427=0.105573\).
The associated probabilities sum to approximately \(0.214089\), not \(0.20\). The correct aggregate check is based on survival: \((1-q'_1)(1-q'_2)(1-q'_3)\approx0.945742(0.894427)(0.945742)=0.800000\). Thus the implied total exit probability is \(1-0.800000=0.20\). To four decimal places the associated probabilities are \(0.0543\), \(0.1056\), and \(0.0543\).
Common mistake avoided: Adding \(q'_1+q'_2+q'_3\) treats the single-decrement probabilities as if they were mutually exclusive multiple-decrement probabilities. The product check correctly combines the no-exit probabilities under the linked conversion.
Interpreting the Aggregate Check
For converted probabilities generated consistently from the same multiple-decrement table, the product of associated survival probabilities recovers the original total survival probability:
The check works because the cause shares \(q_j/Q\) sum to one. It is useful both for detecting arithmetic errors and for identifying inconsistent inputs. If supplied associated probabilities do not satisfy \(1-\prod_j(1-q'_j)=Q\), then they do not imply the stated annual aggregate under this linked conversion.
This product check is not a claim that cause exits are independent in every possible model. It follows here from the specified conversions, which allocate the total cumulative hazard across causes in proportion to their observed annual exit probabilities. Do not apply it blindly when probabilities come from a different model or when the within-year assumptions are unspecified.
Common Exam Traps
- Using the wrong denominator for a cause share. The exponent is \(q_j/Q\), not \(q_j\) alone and not \(q_j/(1-q_j)\).
- Adding associated probabilities to check the total. Instead calculate \(1-\prod_j(1-q'_j)\).
- Assuming the within-year patterns agree. UDD and constant force yield the same associated annual conversion here, but can give different probabilities at intermediate times.
- Rounding too early. Retain full precision through the conversion and aggregate check, then round the final probabilities as requested.
- Ignoring the model boundary. The displayed formulas require \(Q<1\). When \(Q=1\), the total force is infinite and the cause-share conversion does not give finite constant forces.
For problems aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus, state the assumed within-year model, calculate the total exit probability and cause shares, convert each cause, and finish with an aggregate check. Keeping those steps separate helps distinguish the observed multiple-decrement probabilities from the associated single-decrement probabilities.
Check Your Understanding
For each problem, use the stated assumption and keep intermediate precision until the final answer.
- A hypothetical table has annual cause probabilities \(0.15\), \(0.09\), and \(0.06\). Under UDD, find the three associated single-decrement probabilities to four decimal places and verify the implied total exit probability. (Answer: \(0.1633,\ 0.1015,\ 0.0689\); implied total \(0.3000\).)
- A two-cause table has \(q_1=0.04\) and \(q_2=0.12\). Under constant force, find the associated probabilities to four decimals and the total force to four significant figures. (Answer: \(q'_1=0.0427,\ q'_2=0.1228;\ \lambda=0.1744\).)
- Two associated single-decrement probabilities under constant force are \(0.10\) and \(0.20\). Find the implied total exit probability and the corresponding multiple-decrement probabilities for the two causes, to four decimals. (Answer: total \(0.2800\); cause probabilities approximately \(0.0898\) and \(0.1902\).)
- A table has \(q_1=0.10\) and \(q_2=0.15\). Under UDD, find each cause’s cumulative multiple-decrement probability at \(t=0.5\), and the total exit probability by that time. (Answer: cause probabilities \(0.0500\) and \(0.0750\); total \(0.1250\).)
- A table has annual cause probabilities \(0.06\) and \(0.14\). Convert under either stated assumption, then verify the aggregate using the product of associated survival probabilities. (Answer: associated probabilities \(0.0648\) and \(0.1446\); implied total exit probability \(0.2000\).)