From Observed Exits to Associated Probabilities
In a multiple-decrement table, the observed probability for a cause reflects both that cause’s force and the possibility that another cause removes the person from the group first. An associated single-decrement probability answers a different question: what would the probability of exit from one specified cause be if the competing causes were removed, while that cause’s force was retained?
Under the constant-force assumption, the conversion has two steps. First, the total probability of exit determines the aggregate force. Then each cause receives the share of that force indicated by its share of observed exits. Applying each allocated force without the other causes gives the associated probability.
The distinction matters because the observed probabilities for all causes add to the total multiple-decrement probability, but the associated probabilities generally do not. Removing competing causes gives a cause more opportunity to produce an exit. This treatment is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
Deriving the Constant-Force Conversion
Assume that the causes are mutually exclusive exits, their observed probabilities sum to \(q\), and each cause has a constant force throughout the interval. Write \(\lambda_j\) for the force of cause \(j\), per unit time, and \(\lambda=\sum_j\lambda_j\) for the aggregate force. These forces are nonnegative. The probability of remaining in force through time \(t\), with all causes operating, is \(e^{-\lambda t}\).
For a person still in force at time \(t\), the probability of an exit from cause \(j\) in a short interval of length \(dt\) is approximately \(\lambda_j\,dt\). Thus the observed cause-\(j\) probability is obtained by accumulating the cause-\(j\) exit rate over the people who have survived all causes up to each time:
Adding these equations over all causes gives the total exit probability. Since the forces sum to \(\lambda\), the cause-specific terms combine to the aggregate exit probability:
Therefore the aggregate force over the interval is determined by the total probability of staying in force, \(1-q\). Also, dividing the equation for \(q_j\) by the equation for \(q\) shows that the cause’s share of the aggregate force is its share of observed exits.
If cause \(j\) is the only cause operating, its probability of exit over the interval is \(1-e^{-\lambda_j h}\). Substituting the force allocation gives the conversion formula:
The derivation explains why it is the force, not the observed probability, that is allocated by cause share. The observed probabilities already include the effect of competing exits. Once the cause-specific force is found, removing competitors changes the probability generated by that force.
A Practical Calculation Sequence
For a table giving cause-specific observed probabilities, use the following sequence. Keep the interval length explicit: a force is expressed per unit time, while its product with \(h\) is dimensionless.
Confirm that the observed cause probabilities are nonnegative and add to a total \(q<1\).
Use \(\lambda=-\ln(1-q)/h\), in force units per unit time.
For each cause, calculate \(\lambda_j=(q_j/q)\lambda\).
Calculate \(q_j^{\,\mathrm{assoc}}=1-e^{-\lambda_jh}\) for each cause.
Confirm that the allocated forces sum to \(\lambda\) and that \((\lambda_j/\lambda)q=q_j\) recovers each observed probability.
Worked Conversions
Worked Example: Annual Death and Disability Probabilities
Problem. In a hypothetical insured group, the one-year observed probability of death is \(0.09\), and the one-year observed probability of becoming disabled is \(0.27\). Assume mutually exclusive exits and constant forces of death and disability over the year. Find the associated one-year probability for each cause. No interest or loss distribution is involved.
Method and data check. The interval is \(h=1\) year. The total observed exit probability is \(q=0.09+0.27=0.36\), so the probability of no exit is \(1-q=0.64\). The cause shares are \(0.09/0.36=0.25\) for death and \(0.27/0.36=0.75\) for disability.
Aggregate force. The annual aggregate force is \( \lambda=-\ln(0.64)/1=0.446287\) per year.
Cause-specific forces. The forces are allocated in the observed-exit shares: \( \lambda_{\text{death}}=0.25(0.446287)=0.111572\) per year, and \( \lambda_{\text{disability}}=0.75(0.446287)=0.334715\) per year. Their sum is \(0.446287\), as required.
Associated probabilities. With only death operating, the probability is \(1-e^{-0.111572}=0.1056\). With only disability operating, it is \(1-e^{-0.334715}=0.2845\). Probabilities are rounded to four decimal places; forces are shown to six decimal places before rounding.
Check and common mistake. The observed cause probabilities are recovered from the cause shares: \(0.25(0.36)=0.09\) and \(0.75(0.36)=0.27\). A common mistake is to report \(0.09\) and \(0.27\) as the associated probabilities. Those are the observed probabilities with both causes present; after removing the competitor, the associated probabilities are \(0.1056\) and \(0.2845\).
Worked Example: A Six-Month Observation Period
Problem. For a hypothetical group, over six months the observed probability of lapse is \(0.07\) and the observed probability of death is \(0.12\). Assume constant lapse and mortality forces over that half-year interval. Find the annualized cause-specific forces and the associated probabilities over the six months. No interest or loss distribution is involved.
Method and data check. Here \(h=0.5\) years and \(q=0.07+0.12=0.19\), giving a probability of remaining in force of \(0.81\). The shares are \(0.07/0.19=7/19\) for lapse and \(0.12/0.19=12/19\) for death.
Aggregate force. The force per year, rather than per half-year, is \( \lambda=-\ln(0.81)/0.5=0.421442\). The integrated aggregate force over the half-year is \( \lambda h=0.421442(0.5)=0.210721\).
Cause-specific forces. The annualized forces are \( \lambda_{\text{lapse}}=(7/19)(0.421442)=0.155268\) per year and \( \lambda_{\text{death}}=(12/19)(0.421442)=0.266174\) per year. They sum to \(0.421442\).
Associated probabilities. Use the half-year duration in the exponent. For lapse, \(1-e^{-0.155268(0.5)}=1-e^{-0.077634}=0.0747\). For death, \(1-e^{-0.266174(0.5)}=1-e^{-0.133087}=0.1246\). Probabilities are rounded to four decimal places; annualized forces are shown to six decimal places.
Check and common mistake. Multiplying the aggregate exit probability by each cause share gives \(0.19(7/19)=0.07\) and \(0.19(12/19)=0.12\), recovering the observations. A frequent time-unit error is to use the annualized forces in \(1-e^{-\lambda_j}\) for a six-month probability. The exponent must be \(\lambda_j h\), so the half-year factor \(0.5\) is essential.
Worked Example: Three Causes and a Force Check
Problem. A hypothetical long-term care contract population has one-year observed exit probabilities of \(0.06\) for death, \(0.09\) for entry into care, and \(0.15\) for withdrawal. Assume these are mutually exclusive causes and that each cause has a constant force during the year. Calculate all three associated probabilities and check the observed cause shares. No interest or loss distribution is involved.
Method and data check. The interval is \(h=1\) year and the total probability of exit is \(q=0.06+0.09+0.15=0.30\). Thus \(1-q=0.70\). The cause shares are \(0.06/0.30=0.20\), \(0.09/0.30=0.30\), and \(0.15/0.30=0.50\).
Aggregate and cause-specific forces. The aggregate force is \( \lambda=-\ln(0.70)=0.356675\) per year. The cause-specific forces are \( \lambda_{\text{death}}=0.20(0.356675)=0.071335\), \( \lambda_{\text{care}}=0.30(0.356675)=0.107002\), and \( \lambda_{\text{withdrawal}}=0.50(0.356675)=0.178337\), each per year. Their sum is \(0.356674\), with the small difference from \(0.356675\) due to displaying rounded intermediate values.
Associated probabilities. Applying each force alone gives \(1-e^{-0.071335}=0.0689\) for death, \(1-e^{-0.107002}=0.1015\) for entry into care, and \(1-e^{-0.178337}=0.1633\) for withdrawal. Probabilities are rounded to four decimal places; forces are shown to six decimal places.
Check and common mistake. The cause shares multiplied by the total probability give \(0.20(0.30)=0.06\), \(0.30(0.30)=0.09\), and \(0.50(0.30)=0.15\). The associated probabilities sum to approximately \(0.3337\), not \(0.30\). That is expected: each associated probability is calculated with competitors removed, so they are not the mutually exclusive observed probabilities and are not required to sum to the observed total.
Interpretation and Exam Checks
The conversion separates the information in the multiple-decrement data into two parts: the total exit probability determines the aggregate force, and the observed cause proportions determine how that force is split. The associated probability then follows from the single-cause survival model. For a cause with zero observed exits, its force share and associated probability are zero. For a cause with all observed exits, its force equals the aggregate force, so its associated probability equals the total exit probability.
The most important checks are straightforward: the observed cause probabilities must add to \(q\); the allocated cause forces must sum to the aggregate force; and multiplying each force share by \(q\) must recover the corresponding observed cause probability. Do not sum associated probabilities as if they were the original multiple-decrement exits, and do not use a force for one time unit with a duration expressed in another.
Check Your Understanding
For each problem, use constant cause-specific forces over the stated interval and round probabilities to four decimal places. No interest or loss distribution is involved.
- Over one year, two hypothetical causes have observed probabilities \(0.08\) and \(0.12\). Find both associated probabilities. (Answer: \(0.0854\) and \(0.1253\))
- Over one year, observed probabilities are \(0.07\) for cause A and \(0.21\) for cause B. Find the associated probabilities. (Answer: \(0.0799\) and \(0.2184\))
- Over one year, three causes have observed probabilities \(0.04\), \(0.10\), and \(0.06\). Find all associated probabilities and the annual force for the second cause. (Answer: \(0.0436,\ 0.1056,\ 0.0648\); second-cause force \(0.1116\) per year)
- Over six months, observed probabilities are \(0.03\) for lapse and \(0.09\) for death. Find the annualized cause-specific forces and the associated probabilities over six months. (Answer: forces \(0.06392\) and \(0.1917\) per year; probabilities \(0.0315\) and \(0.0914\))
- Over one year, cause A has observed probability zero and cause B has observed probability \(0.24\). State both associated probabilities and explain whether the zero cause can be assigned a positive force under this model. (Answer: \(0\) and \(0.2400\); cause A has zero force under the model)