Deriving the UDD Competing-Decrement Equations

Why UDD Adjusts an Associated Probability

An associated single-decrement table describes the probability of exit from one cause when the other causes are removed. In a multiple-decrement population, some people exit from other causes before they can exit from the cause of interest. The observed probability for that cause is therefore generally smaller than its associated single-decrement probability.

Under the uniform distribution of decrements (UDD) assumption, the associated decrement mass is spread uniformly across the year. A cause that is more exposed to competing decrements later in the year has less eligible exposure than it would have at the start. Integrating that changing exposure gives the UDD conversion equation.

Definition: For one year and a set of competing causes, let \(q_j\) be the observed probability of exit from cause \(j\) in the multiple-decrement model. Let \(q'_j\) be the associated single-decrement probability for cause \(j\): the probability of exit from \(j\) over the year in the table where the other causes are eliminated. The prime distinguishes the associated probability from the observed multiple-decrement probability.

The probabilities \(q'_j\) are inputs to the UDD conversion. They are not the observed probabilities \(q_j\), and their sum need not equal the probability of exit from any one cause. The formula below applies to one-year probabilities under the UDD exposure convention; it is not a formula for constant forces or arbitrary time-varying decrement patterns.

The UDD Exposure Integral

Measure time \(t\) as a fraction of the year, from \(0\) to \(1\). Under UDD, the associated decrement mass for cause \(j\) is spread evenly over this interval. Its amount per unit time, on the original-cohort scale, is \(q'_j\). At time \(t\), each competing cause \(k\ne j\) has used \(tq'_k\) of its associated decrement exposure. Thus the fraction still eligible with respect to the other causes is \(1-t\sum_{k\ne j}q'_k\).

The observed cause-\(j\) exits are obtained by multiplying the cause-\(j\) decrement amount in a short interval by the exposure still eligible against the other causes, then adding those amounts over the year.

Write the competing exposure.
At fraction \(t\) of the year, the eligible fraction with respect to causes other than \(j\) is \(1-t\sum_{k\ne j}q'_k\).

Form the cause-\(j\) exit increment.
UDD assigns \(q'_j\,dt\) of associated cause-\(j\) decrement mass to a small interval of length \(dt\), so the observed increment is \(q'_j(1-t\sum_{k\ne j}q'_k)\,dt\).

Integrate over the year.
Add the cause-\(j\) exit increments from \(t=0\) to \(t=1\).

\begin{aligned} q_j &= \int_0^1 q'_j\left(1-t\sum_{k\ne j}q'_k\right)\,dt\\ &= q'_j\left[\int_0^1 1\,dt-\left(\sum_{k\ne j}q'_k\right)\int_0^1t\,dt\right]\\ &= q'_j\left(1-\frac12\sum_{k\ne j}q'_k\right). \end{aligned}

Theorem: Under UDD, the observed one-year probability for cause \(j\) is \(q_j=q'_j\left(1-\frac12\sum_{k\ne j}q'_k\right)\). The sum is over all competing causes and excludes cause \(j\) itself. The factor of one-half arises because the average value of \(t\) over the year is \(\int_0^1t\,dt=\frac12\).

This derivation also explains why a competing cause is not deducted in full from \(q'_j\): a competing decrement occurring near the end of the year removes less cause-\(j\) exposure than one occurring near the beginning. Under UDD, the relevant average fraction of the year exposed to each competing decrement is one-half.

Applying the Cause-Specific Formula

Worked Example: Converting Three Associated Probabilities

A hypothetical insured population has three decrement causes, \(A\), \(B\), and \(C\). The associated single-decrement probabilities are \(q'_A=0.18\), \(q'_B=0.12\), and \(q'_C=0.08\). Assuming UDD, calculate each observed probability and check their sum.

Method. For each cause, sum the associated probabilities of the other causes, halve that sum, and multiply the remaining factor by the cause's own associated probability. This avoids treating an associated probability as if it were already an observed probability.

For cause \(A\), the competing associated probabilities sum to \(0.12+0.08=0.20\). The UDD factor is \(1-\frac12(0.20)=0.90\), so \(q_A=0.18(0.90)=0.1620\).

For cause \(B\), the competing sum is \(0.18+0.08=0.26\). The factor is \(1-\frac12(0.26)=0.87\), so \(q_B=0.12(0.87)=0.1044\).

For cause \(C\), the competing sum is \(0.18+0.12=0.30\). The factor is \(1-\frac12(0.30)=0.85\), so \(q_C=0.08(0.85)=0.0680\).

The observed probabilities sum to \(0.1620+0.1044+0.0680=0.3344\). This is below \(1\), as required for a total one-year exit probability. As a further check, the total associated probability is \(0.38\), and the UDD total is \(0.38-\frac12(0.38^2-(0.18^2+0.12^2+0.08^2))=0.3344\), agreeing with the cause-by-cause calculation.

Worked Example: From Probabilities to Expected Exit Counts

A hypothetical cohort begins the year with 20,000 members. Four competing causes have associated probabilities \(q'_A=0.09\), \(q'_B=0.06\), \(q'_C=0.04\), and \(q'_D=0.03\). Under UDD, calculate the observed probability and expected number of exits for each cause, then determine the expected number remaining.

Method. Apply the conversion separately to each cause, then multiply each observed probability by the initial cohort size. The initial cohort is the denominator for every cause-specific probability.

For \(A\), the other probabilities sum to \(0.06+0.04+0.03=0.13\). Thus \(q_A=0.09(1-0.13/2)=0.09(0.935)=0.08415\), giving \(20{,}000(0.08415)=1{,}683\) expected exits.

For \(B\), the other probabilities sum to \(0.09+0.04+0.03=0.16\). Thus \(q_B=0.06(0.92)=0.0552\), giving \(20{,}000(0.0552)=1{,}104\) expected exits.

For \(C\), the other probabilities sum to \(0.09+0.06+0.03=0.18\). Thus \(q_C=0.04(0.91)=0.0364\), giving \(20{,}000(0.0364)=728\) expected exits.

For \(D\), the other probabilities sum to \(0.09+0.06+0.04=0.19\). Thus \(q_D=0.03(0.905)=0.02715\), giving \(20{,}000(0.02715)=543\) expected exits.

The expected total exits are \(1{,}683+1{,}104+728+543=4{,}058\). Therefore the expected number remaining is \(20{,}000-4{,}058=15{,}942\). The converted probabilities sum to \(0.08415+0.0552+0.0364+0.02715=0.2029\), and \(20{,}000(0.2029)=4{,}058\), which checks the count calculation.

Worked Example: Checking a Cause-Specific Calculation

For a hypothetical long-term benefit model, the associated probabilities are \(q'_D=0.15\) for disability, \(q'_M=0.10\) for mortality, and \(q'_R=0.05\) for recovery. A cohort has 8,000 members at the start of the year. Find the observed probability and expected exits for each cause, and check the expected number remaining.

Method. Use the sum of the other causes in each factor. In particular, a recovery decrement competes with disability and mortality just as those causes compete with recovery.

For disability, the competing sum is \(0.10+0.05=0.15\). Hence \(q_D=0.15(1-0.15/2)=0.15(0.925)=0.13875\), or \(8{,}000(0.13875)=1{,}110\) expected exits.

For mortality, the competing sum is \(0.15+0.05=0.20\). Hence \(q_M=0.10(0.90)=0.09\), or \(8{,}000(0.09)=720\) expected exits.

For recovery, the competing sum is \(0.15+0.10=0.25\). Hence \(q_R=0.05(0.875)=0.04375\), or \(8{,}000(0.04375)=350\) expected exits.

Total expected exits are \(1{,}110+720+350=2{,}180\), leaving \(8{,}000-2{,}180=5{,}820\) members. The total observed probability is \(0.13875+0.09+0.04375=0.2725\), and \(8{,}000(0.2725)=2{,}180\), so the probabilities and counts agree.

Common Mistake and Scope of the Result

The most common arithmetic trap is to include cause \(j\)'s own associated probability in the competing sum. A cause does not compete with itself. Another is to use observed probabilities \(q_k\) in the sum when the UDD conversion requires the associated probabilities \(q'_k\).

Worked Example: Excluding the Cause of Interest

Suppose cause \(J\) has associated probability \(q'_J=0.21\), and its three competing causes have associated probabilities \(0.13\), \(0.09\), and \(0.04\). Calculate \(q_J\), and compare it with the result of the common incorrect substitution.

Correct method. The competing sum is \(0.13+0.09+0.04=0.26\). It excludes \(q'_J\). Therefore the factor is \(1-0.26/2=0.87\), and \(q_J=0.21(0.87)=0.1827\).

Incorrect calculation. Including \(q'_J\) gives a sum of \(0.21+0.13+0.09+0.04=0.47\), a factor of \(1-0.47/2=0.765\), and the incorrect value \(0.21(0.765)=0.16065\). The correction is to exclude the cause of interest and use the competing sum \(0.26\), giving \(0.1827\).

The formula is specific to UDD and its linear within-year exposure pattern. If decrement timing is specified using constant forces or another distribution, this integral and conversion need not apply. A sound calculation therefore identifies the within-year assumption before substituting probabilities.

Key takeaway: For cause \(j\), multiply \(q'_j\) by one minus half the sum of the other associated probabilities. The half-sum is the average competing exposure over the year under UDD; exclude the cause of interest, and check the total observed probability against the cohort size.

Check Your Understanding

Apply the UDD cause-specific equation. Each problem gives associated single-decrement probabilities unless stated otherwise.

Causes \(A\), \(B\), and \(C\) have \(q'_A=0.14\), \(q'_B=0.10\), and \(q'_C=0.06\). Find \(q_A\). (Answer: \(0.1288\))
For three causes \(X\), \(Y\), and \(Z\), \(q'_X=0.22\), \(q'_Y=0.11\), and \(q'_Z=0.07\). Find \(q_X\), showing the competing sum. (Answer: competing sum \(0.18\); \(q_X=0.2002\))
Four causes have associated probabilities \(0.08\), \(0.05\), \(0.03\), and \(0.02\). Find the observed probability for the cause with associated probability \(0.05\). (Answer: \(0.04675\))
A cohort of 12,000 has associated probabilities \(0.12\) for cause \(H\), and \(0.07\), \(0.05\), and \(0.02\) for three competing causes. Under UDD, find the expected number of exits from \(H\). (Answer: \(1{,}339.2\) expected exits)
Three causes have associated probabilities \(0.10\), \(0.08\), and \(0.04\). Find the sum of their observed probabilities under UDD. (Answer: \(0.2048\))

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