Associated Single-Decrement Probabilities From Multiple-Decrement Data

Why Associated Probabilities Need an Assumption

In a multiple-decrement model, an individual can leave a group for several competing reasons. The observed probability of exit for one cause reflects both that cause and the fact that other causes may have removed individuals first. An associated single-decrement probability answers a different question: what would the probability of exit for the specified cause be in a constructed table where the other causes have been removed?

That question cannot generally be answered from annual cause probabilities alone. Those probabilities record which cause occurred during the year, but not when it occurred. To construct the associated table, we must specify how each cause operates within the year and assume that removing competing causes does not change the specified cause's own decrement mechanism.

Definition: Let $q_x^{(j)}$ be the probability that a life aged $x$ exits during the year from cause $j$, while all causes are operating. The associated single-decrement probability $q_x^{\prime(j)}$ is the probability of exit from cause $j$ during the year in a hypothetical table in which all other causes are removed, while the decrement mechanism for cause $j$ is retained. The prime distinguishes this associated probability from the observed multiple-decrement probability.

The associated table is a model-based construction, not a second group of observed lives. The key assumption is that the mechanism for cause $j$ remains the same when the competing causes are eliminated. If, for example, removing a disability exit would change the risk of a later withdrawal, the associated table is not a direct description of that changed real-world process; it is a calculation under the stated assumption.

Cause-Specific Forces and the Associated Probability

A precise way to specify the decrement mechanism is by cause-specific forces. Suppose the causes are mutually exclusive exits from the active group, and the force for cause $j$ at duration $t$ within the year is $\mu_{x+t}^{(j)}$. The total force while all causes operate is $\sum_k\mu_{x+t}^{(k)}$. For the formulas below, assume these forces describe the same population and remain unchanged when other causes are removed.

Result: Under a continuous-time force model, the associated probability for cause $j$ is $$ q_x^{\prime(j)}=1-\exp\left(-\int_0^1\mu_{x+t}^{(j)}\,dt\right). $$ With all causes operating, the observed probability for cause $j$ is $$ q_x^{(j)}=\int_0^1 \exp\left(-\int_0^t\sum_k\mu_{x+s}^{(k)}\,ds\right) \mu_{x+t}^{(j)}\,dt. $$

The first expression uses the force for cause $j$ alone. The second weights that force by the probability of still being in the group just before time $t$, which depends on every competing cause. Thus, in general, $q_x^{\prime(j)}$ is not the same as $q_x^{(j)}$.

Write the survival equation.
In a table with only cause $j$, let $S_j(t)$ be the probability of remaining in the group through duration $t$. Over a short interval $dt$, the probability of exit is approximately $S_j(t)\mu_{x+t}^{(j)}dt$. Therefore $dS_j(t)/dt=-S_j(t)\mu_{x+t}^{(j)}$, with $S_j(0)=1$.

Solve for single-cause survival.
Separating variables gives $dS_j(t)/S_j(t)=-\mu_{x+t}^{(j)}dt$. Integrating from 0 to 1 gives $\ln S_j(1)=-\int_0^1\mu_{x+t}^{(j)}dt$, so $S_j(1)=\exp(-\int_0^1\mu_{x+t}^{(j)}dt)$.

Take the complement.
Exit from cause $j$ by the end of the year is the complement of survival in the associated table. Hence $q_x^{\prime(j)}=1-S_j(1)$, which gives the stated formula.

This derivation also explains why a conversion requires more than the annual values $q_x^{(j)}$: the associated probability depends on the accumulated cause-specific force, whereas the observed probability depends on when the cause acts relative to all competing forces. Other within-year assumptions, such as a uniform distribution of decrements, can also define a conversion, but they produce results only under those assumptions.

Worked Example: Calculating an Associated Probability From Specified Forces

A hypothetical coverage group has two competing exits during one year. The cause-specific force for exit A is $0.12$ per year during the first half-year and $0.24$ per year during the second half-year. The force for exit B is $0.08$ per year throughout. Calculate the associated probability for A and the observed multiple-decrement probabilities for A and B. Check the total against the probability of any exit.

Method and assumptions. Use the specified piecewise-constant forces, with forces unchanged when a competing cause is removed. For a constant total force $m$ over an interval of length $h$, survival through that interval is $e^{-mh}$, and the probability of exit from cause $j$ during it, conditional on entering the interval, is $(\mu^{(j)}/m)(1-e^{-mh})$.

Associated probability for A. With B removed, the accumulated A force is $0.12(0.5)+0.24(0.5)=0.06+0.12=0.18$. Therefore

q_x^{\prime(A)}=1-e^{-0.18}=1-0.835270=0.164730\approx 0.1647.

Observed probability for A in the first half-year. The total force is $0.12+0.08=0.20$, so the probability of an A exit in that half-year is

\frac{0.12}{0.20}(1-e^{-0.20(0.5)}) =0.6(1-e^{-0.10}) =0.6(0.095163) =0.057098.

Survival to the second half-year is $e^{-0.10}=0.904837$. In the second half-year, the total force is $0.24+0.08=0.32$, so the unconditional probability of an A exit then is

e^{-0.10}\frac{0.24}{0.32}(1-e^{-0.32(0.5)}) =0.904837(0.75)(1-e^{-0.16}) =0.100339.

Thus $q_x^{(A)}=0.057098+0.100339=0.157437$, or $0.1574$ to four decimals.

Observed probability for B. In the first half-year, B's probability is

\frac{0.08}{0.20}(1-e^{-0.10}) =0.4(0.095163) =0.038065.

In the second half-year, it is

e^{-0.10}\frac{0.08}{0.32}(1-e^{-0.16}) =0.904837(0.25)(0.147856) =0.033446.

Therefore $q_x^{(B)}=0.038065+0.033446=0.071511$, or $0.0715$ to four decimals.

Check. Total survival is $e^{-0.10}e^{-0.16}=e^{-0.26}=0.771052$. The total exit probability is $1-0.771052=0.228948$, and the two observed cause probabilities sum to $0.157437+0.071511=0.228948$. The check agrees. The associated probability for A, $0.1647$, exceeds its observed probability, $0.1574$, because some lives exit from B before they can exit from A.

Common mistake avoided: Do not use the total force $0.32$ in the associated A calculation for the second half-year. Once B is removed, only A's force determines survival in the associated table.

Annual Probabilities Do Not Reveal Event Timing

The force formulas make the information gap explicit, but the same issue arises with grouped or staged data. Annual multiple-decrement probabilities do not say whether one cause tends to occur early in the year or late. If one cause removes lives early, it can prevent some later exits from another cause. Removing it can therefore change the associated probability even though the observed annual probabilities are unchanged.

Worked Example: Converting Staged Multiple-Decrement Data

A hypothetical table records two causes, A and B, in two successive half-year stages. During the first half-year, the conditional probability of A is $0.20$, and B cannot occur. Among those still active at midyear, the conditional probability of B in the second half-year is $0.125$, and A cannot occur. Calculate both observed annual probabilities and the associated single-decrement probabilities under the assumption that the staged cause probabilities are retained when the other cause is removed.

Method and assumptions. Apply each stage's conditional probability to those who enter that stage. In the associated A table, retain A's first-stage probability and remove B; in the associated B table, retain B's second-stage probability and remove A.

All members begin the first stage. Thus the observed probability of A is $q_x^{(A)}=0.20$. The proportion surviving the first stage is $1-0.20=0.80$. The observed probability of B is therefore $q_x^{(B)}=0.80(0.125)=0.10$. The probability remaining active after both stages is $0.80(1-0.125)=0.80(0.875)=0.70$. The check is $0.20+0.10+0.70=1.00$.

For the associated A table, B is removed and A can occur only in the first stage, with probability $0.20$. Therefore $q_x^{\prime(A)}=0.20$. For the associated B table, A is removed, so everyone reaches the second stage; B then has probability $0.125$. Therefore $q_x^{\prime(B)}=0.125$.

The associated probabilities do not have to sum to the observed probability of any exit. They belong to separate hypothetical tables: one removes B, and the other removes A. Adding $0.20$ and $0.125$ would combine results from two different constructions, not produce the original table's total exit probability.

Common mistake avoided: The observed B probability is $0.10$, not $0.125$, because only $0.80$ of the original group reaches the second stage. The associated B probability is $0.125$ because the competing A exit has been removed.

Worked Example: Showing That a Conversion Is Not Unique

Suppose the only available data for a hypothetical year are $q_x^{(A)}=0.20$ and $q_x^{(B)}=0.10$. An analyst proposes that these values determine $q_x^{\prime(A)}$. Test that claim by constructing two two-stage models, each consistent with the observed data, but with different associated A probabilities. Assume each stage has only the stated cause active and retain the cause's stage probability when its competitor is removed.

Model 1: A occurs first. Let A have conditional probability $0.20$ in the first stage. Let B have conditional probability $0.10/0.80=0.125$ in the second stage among those remaining. The resulting observed probabilities are $q_x^{(A)}=0.20$ and $q_x^{(B)}=0.80(0.125)=0.10$. With B removed, A still occurs in the first stage with probability $q_x^{\prime(A)}=0.20$.

Model 2: B occurs first. Let B have conditional probability $0.10$ in the first stage. Let A have conditional probability $0.20/0.90=2/9$ in the second stage among those remaining. The observed probabilities are $q_x^{(B)}=0.10$ and $q_x^{(A)}=0.90(2/9)=0.20$, the same annual data as in Model 1. But when B is removed, everyone reaches the second stage, so $q_x^{\prime(A)}=2/9=0.2222$ to four decimals.

The two associated probabilities, $0.20$ and $0.2222$, differ even though both models reproduce exactly the same annual multiple-decrement data. The data alone do not identify a unique associated probability. A within-year timing assumption or other information is required.

Common mistake avoided: Do not treat observed cause probabilities as if they automatically specify the corresponding associated probabilities. The construction must account for which lives are still exposed when each cause can occur.

Practical Conversion and Checks

When converting multiple-decrement data, first state what is known and what is assumed. If the data include cause-specific forces over the year, integrate the specified cause's force to obtain its associated probability. If the data provide only annual cause probabilities, a conversion requires an additional model for the within-year decrement pattern. The result is conditional on that model; it is not a uniquely determined feature of the annual data.

Under a consistent model with unchanged cause-specific mechanisms, the observed probability for a cause cannot exceed its associated probability. In the force model, the observed integral weights the cause-specific force by survival under all causes, while the associated calculation weights it by survival under that cause alone. Competing causes can reduce the observed probability by removing lives first. This comparison is a useful diagnostic, but it does not identify the missing timing assumption.

Check that the observed cause probabilities and the probability of remaining in the group account for the full original group.
Keep observed multiple-decrement probabilities distinct from probabilities in separate associated tables.
State whether assumptions apply to forces, staged conditional probabilities, or another within-year pattern.
Do not combine associated probabilities for different causes as though they came from one table with all causes removed simultaneously.

Key takeaway: An associated single-decrement probability is defined by removing competing causes while retaining the specified cause's decrement mechanism. Multiple-decrement data alone usually do not determine it: the within-year timing or force assumption is part of the answer.

Check Your Understanding

For each problem, state or use the specified model before calculating. Probabilities in the answers are rounded to four decimals where needed.

A cause A has a force of $0.09$ per year in the first half-year and $0.06$ per year in the second half-year. Under a force model, find its associated probability for the year. (Answer: $1-e^{-0.075}=0.0723$)
In a two-stage model, A has conditional probability $0.12$ in stage 1, and B has conditional probability $0.10$ in stage 2 among those who remain. A cannot occur in stage 2 and B cannot occur in stage 1. Find the observed annual probabilities for A and B, then the associated probability for B if A is removed. (Answer: $q^{(A)}=0.1200,\ q^{(B)}=0.0880,\ q^{\prime(B)}=0.1000$)
Two causes have constant forces $0.10$ and $0.10$ per year over one year. Find the associated probability for the first cause and its observed multiple-decrement probability. (Answer: $q^{\prime(1)}=1-e^{-0.10}=0.0952,\ q^{(1)}=\frac{0.10}{0.20}(1-e^{-0.20})=0.0906$)
Annual data give $q^{(A)}=0.16$ and $q^{(B)}=0.12$. In a staged model, let A occur first and B occur second among survivors; then construct a second model with B first and A second. Find the associated A probability in each model, and explain what this says about conversion from annual data alone. (Answer: A-first model $q^{\prime(A)}=0.1600$; B-first model $q^{\prime(A)}=0.16/0.88=0.1818$; the annual data do not determine a unique associated probability.)
For a one-year continuous-time model, the force of A is $0.12$ per year throughout, and the force of B is $0.08$ per year throughout. Calculate both observed cause probabilities, the probability of remaining in the group, and the associated probability for A. Verify the closure identity. (Answer: $q^{(A)}=0.6(1-e^{-0.20})=0.1088,\ q^{(B)}=0.4(1-e^{-0.20})=0.0725,\ p_x=e^{-0.20}=0.8187,\ q^{\prime(A)}=1-e^{-0.12}=0.1131;\ 0.1088+0.0725+0.8187=1.0000$, subject to rounding.)

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