Associated Single-Decrement Rates in Linked Tables

Why Associated Single-Decrement Probabilities Differ

In a multiple-decrement model, an individual can leave an initial state for several competing reasons. For example, an active member of a hypothetical pension plan might retire, withdraw, or die during the year. An observed probability for one cause reflects both that cause’s own rate and the fact that other causes can remove the member first.

An associated single-decrement probability asks a different question: what would the probability of exit from one specified cause be if the other causes were removed, while the specified cause’s decrement mechanism stayed unchanged? The answer is therefore generally not the observed probability in the multiple-decrement setting. The distinction matters when a multiple-decrement table is linked to single-decrement tables, or when a problem asks for rates under a stated removal assumption.

These definitions alone do not specify how to calculate \(q'_j\) from observed table entries. A model for the timing of decrements within the interval, or direct information about cause-specific forces, is also needed. The calculations below state those assumptions explicitly.

Calculating Probabilities from Cause-Specific Forces

Suppose a person is in the initial state at time 0, and cause \(j\) operates with a force of decrement \(\mu_j(t)\) at time \(t\), measured in years. Assume the forces are nonnegative and that exits are absorbing over the interval: once any decrement occurs, the person leaves the initial state and cannot exit from another cause during that interval. Define the accumulated force for cause \(j\) by \(M_j(t)=\int_0^t \mu_j(s)\,ds\).

With all causes operating, the probability of still being in the initial state at time \(t\) is \(\exp(-\sum_k M_k(t))\). An observed exit from cause \(j\) in a short time interval requires both survival in the initial state up to that time and an exit by \(j\) then. Thus the observed probability accumulates the cause-\(j\) force, weighted by the probability of no earlier exit.

The second expression uses the same cause-\(j\) force but removes all competing forces. It is the usual survival formula applied to cause \(j\) alone. The first expression instead includes survival from every cause up to time \(t\). If other causes can remove people before cause \(j\) acts, then \(q_j\) is generally smaller than \(q'_j\).

Constant Forces and a Linked Table

For a one-year interval with constant forces \(\mu_1,\ldots,\mu_r\), write \(\mu_{\text{tot}}=\sum_k\mu_k\). Integrating the observed-probability expression gives a particularly useful relationship:

The observed probabilities share the total probability of exit, \(1-e^{-\mu_{\text{tot}}}\), in proportion to the forces. The associated probability is based only on the specified cause’s force. If linked table data give the observed probabilities for all causes, then the probability of no exit is \(p_0=1-\sum_j q_j\). Under the constant-force assumption, \(\mu_{\text{tot}}=-\ln(p_0)\), and each force can be recovered as \(\mu_j=q_j\mu_{\text{tot}}/(1-p_0)\). Those inferred forces then give the associated probabilities. This inference is conditional on the model; annual table entries alone do not establish that forces were constant.

When Forces Vary During the Year

The constant-force formulas are not a general conversion rule for every annual multiple-decrement table. If forces vary over time, the observed probability weights each cause-\(j\) force by the probability of surviving all causes up to that time. The associated probability instead uses the accumulated force for cause \(j\) alone. The integrals must preserve the timing of changes in the forces.

For example, suppose cause A has force \(0.10\) during the first half-year and \(0.20\) during the second half-year, while cause B has force \(0.10\) throughout. All forces are per year. The accumulated total force is \(0.20(0.5)+0.30(0.5)=0.25\), so the probability of no exit is \(e^{-0.25}=0.7788008\). For cause A, the observed probability is the sum over the two half-years:

The first term is the probability of an A exit during the first half-year. The factor \(e^{-0.10}\) in the second term is the probability of surviving all causes through that first half-year. With B removed, A’s accumulated force is \(0.10(0.5)+0.20(0.5)=0.15\), so \(q'_A=1-e^{-0.15}=0.1393\). Again, the associated probability exceeds the observed probability. The time pattern matters: using a single annual force ratio without stating a suitable assumption can produce an unjustified conversion.

Exam Checks and Common Traps

A linked table can invite a false sense of precision. The observed annual probabilities identify the probability of no exit and the distribution of observed exits, but they do not by themselves identify the within-year timing of each cause. Different force patterns can yield the same annual table while producing different associated probabilities. Use a constant-force inference only when it is specified or otherwise justified.

• Do not interchange observed and associated probabilities. Removing competitors changes the probability being calculated.
• Do not add associated probabilities to obtain total exits. Each associated probability is calculated in a different hypothetical setting where other causes are removed.
• Track time units. A force stated per year must be integrated over years; a half-year interval has length \(0.5\), not \(1\).
• Check the no-exit probability. In a linked table, \(p_0=1-\sum_j q_j\) must be nonnegative. Under constant forces, it equals \(e^{-\mu_{\text{tot}}}\).
• State what is assumed about causes. The formulas above concern competing exits from one initial state. They do not automatically apply if individuals can recover, re-enter, or make repeated transitions during the interval.

These distinctions are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. In an integrated calculation, make the removal assumption explicit before converting between observed multiple-decrement data and associated single-decrement probabilities.