A Two-Cause UDD Exam Problem

Setting Up a Two-Cause UDD Problem

In a multiple-decrement table, each observed annual probability reflects competition among causes: a life that exits from one cause is no longer exposed to the others. An associated single-decrement probability answers a different question: what would the annual probability for one cause be if the other cause were removed, on the specified single-decrement basis? Under UDD, converting between these quantities requires accounting for the competition during the year.

The calculations below use one year of exposure at a fixed age or duration. Suppose there are exactly two mutually exclusive and exhaustive exit causes, labeled 1 and 2. Let $q_1$ and $q_2$ be the observed multiple-decrement probabilities, and let $a$ and $b$ be the associated single-decrement probabilities for causes 1 and 2, respectively. The model assumes that each associated single-decrement table follows UDD within the year and uses the standard associated-table construction: both causes act on the same in-force group.

Definition: Under the two-cause UDD associated-table model, $a$ and $b$ are the annual probabilities in the respective single-decrement tables. The corresponding observed probabilities $q_1$ and $q_2$ are the probabilities of exiting from each cause while both causes operate. They are related by $$ q_1=a\left(1-\frac{b}{2}\right), \qquad q_2=b\left(1-\frac{a}{2}\right). $$ The observed total exit probability is $q_1+q_2$, not either associated probability.

These equations apply to the stated two-cause UDD model. They are not the conversion equations for constant forces, and the observed exits themselves need not be uniformly distributed within the year. If causes do not exhaust all exits, the omitted cause must be included before using the two-cause equations.

Why the UDD Equations Hold

The derivation clarifies which probabilities belong in the conversion. Let $t$ be elapsed time in years, for $0\le t\le 1$. Under UDD, the probability of surviving to time $t$ in the single-decrement table for cause 1 is $1-ta$; for cause 2 it is $1-tb$. Thus the associated decrement force for cause 1 is $a/(1-ta)$, and the force for cause 2 is $b/(1-tb)$, for times at which the denominators are positive.

With both causes operating, the standard associated-table construction gives the probability of remaining free of both exits to time $t$ as $(1-ta)(1-tb)$. The instantaneous probability of a cause 1 exit in a small interval around $t$ is therefore the probability of still being exposed, multiplied by the cause 1 force and by the interval length. Its density is

(1-ta)(1-tb)\frac{a}{1-ta}=a(1-tb).

Integrating that density over the year gives $q_1$. Repeating the argument for cause 2 gives $q_2$:

q_1=\int_0^1 a(1-tb)\,dt =a\left[t-\frac{bt^2}{2}\right]_0^1 =a\left(1-\frac{b}{2}\right), $$ $$ q_2=\int_0^1 b(1-ta)\,dt =b\left[t-\frac{at^2}{2}\right]_0^1 =b\left(1-\frac{a}{2}\right).

This also gives the total-exit and survivor checks. Adding the two observed probabilities, or multiplying the two cause-specific survival probabilities to year-end, yields

$$ q_1+q_2=a+b-ab =1-(1-a)(1-b). $$

Consequently, a proposed solution must satisfy both cause equations and leave a survivor probability of $(1-a)(1-b)$. The result is an annual check; it does not by itself establish that a model with different within-year assumptions is appropriate.

Solving for the Associated Probabilities

The two equations can be reduced to one quadratic. Subtracting the second equation from the first cancels the cross-product term:

$$ q_1-q_2=a-b. $$

Write $D=q_1-q_2$, so that $a=b+D$. Substituting into the equation for $q_2$ gives

q_2=b\left(1-\frac{b+D}{2}\right), \qquad b^2-(2-D)b+2q_2=0.

Solve this quadratic, then retain a root for which $0\le b\le1$ and $0\le a=b+D\le1$. A second algebraic root can occur, but it may be outside the probability range. Finally, substitute both associated probabilities into the original equations and check the survivor identity. This last check guards against a sign error in the quadratic or a root-selection mistake.

Worked Example: Recovering Two Associated Probabilities

A hypothetical annual decrement study reports observed probabilities $q_1=0.19$ and $q_2=0.09$. Assume exactly two causes and UDD in both associated single-decrement tables. Find the associated probabilities and verify the table. The method is subtract-and-solve, followed by substitution; it avoids treating the observed probabilities as the associated ones.

First calculate the difference:

D=q_1-q_2=0.19-0.09=0.10, \qquad a=b+0.10.

Substitute $q_2=0.09$ and $D=0.10$ into the quadratic:

b^2-(2-0.10)b+2(0.09)=0, \qquad b^2-1.90b+0.18=0.

The discriminant is $1.90^2-4(0.18)=3.61-0.72=2.89$, and its square root is $1.70$. Therefore

b=\frac{1.90\pm1.70}{2}=0.10\text{ or }1.80.

The root $1.80$ is not a probability, so $b=0.10$. Then $a=b+0.10=0.20$. Substitution into the original equations gives $q_1=0.20(1-0.10/2)=0.20(0.95)=0.19$, and $q_2=0.10(1-0.20/2)=0.10(0.90)=0.09$. The total exit probability is $0.19+0.09=0.28$. Independently, $1-(1-0.20)(1-0.10)=1-(0.80)(0.90)=0.28$. Both the cause probabilities and the total are consistent.

Worked Example: Selecting the Admissible Quadratic Root

A hypothetical portfolio has observed annual probabilities $q_1=0.38$ and $q_2=0.08$ for two exhaustive causes. Assuming UDD for the associated tables, find the associated probabilities. The key method is to use the cause-probability difference before solving; the common mistake is to accept every algebraic root.

The difference is $D=0.38-0.08=0.30$, so $a=b+0.30$. The equation for $b$ is

b^2-(2-0.30)b+2(0.08)=0, \qquad b^2-1.70b+0.16=0.

The discriminant is $1.70^2-4(0.16)=2.89-0.64=2.25$, with square root $1.50$. The roots are

b=\frac{1.70\pm1.50}{2}=0.10\text{ or }1.60.

Only $b=0.10$ is a probability; it gives $a=0.10+0.30=0.40$. Check cause 1: $0.40(1-0.10/2)=0.40(0.95)=0.38$. Check cause 2: $0.10(1-0.40/2)=0.10(0.80)=0.08$. The observed total is $0.46$, and the associated-table survivor check gives $1-(0.60)(0.90)=0.46$, as required.

Worked Example: Checking an Annual Decrement Table

A hypothetical table begins the year with 8,000 lives. It records 1,900 exits from cause 1, 700 exits from cause 2, and 5,400 survivors. Assuming these are the only exits and the associated tables follow UDD, find the associated probabilities and check the table. The method is to convert counts to observed probabilities first, then solve; counts must not be confused with probabilities.

The counts add to $1{,}900+700+5{,}400=8{,}000$, so the table balances. The observed probabilities are

q_1=\frac{1{,}900}{8{,}000}=0.2375,\qquad q_2=\frac{700}{8{,}000}=0.0875.

Their difference is $D=0.2375-0.0875=0.15$, so $a=b+0.15$. The quadratic becomes

b^2-(2-0.15)b+2(0.0875)=0, \qquad b^2-1.85b+0.175=0.

Its discriminant is $1.85^2-4(0.175)=3.4225-0.7000=2.7225$, whose square root is $1.65$. Thus $b=(1.85\pm1.65)/2$, giving $b=0.10$ or $b=1.75$. The admissible value is $b=0.10$, and $a=0.25$.

Substitution verifies cause 1: $0.25(1-0.10/2)=0.25(0.95)=0.2375$. It verifies cause 2: $0.10(1-0.25/2)=0.10(0.875)=0.0875$. The implied survivor probability is $(1-0.25)(1-0.10)=(0.75)(0.90)=0.675$, or $8{,}000(0.675)=5{,}400$ lives. This matches the table exactly.

Common Exam Trap: Reusing Observed Probabilities

It is incorrect to set the associated probabilities equal to the observed ones merely because both are annual probabilities. For the first example, that shortcut would set $a=0.19$ and $b=0.09$. Those proposed values imply

q_1=0.19\left(1-\frac{0.09}{2}\right)=0.19(0.955)=0.18145, \qquad q_2=0.09\left(1-\frac{0.19}{2}\right)=0.09(0.905)=0.08145.

Neither value matches the observed probability. The corrected values are $a=0.20$ and $b=0.10$, which reproduce $0.19$ and $0.09$ after allowing for competition. A reliable check is to substitute into both cause equations, not just compare the sum of the probabilities.

Before finishing a problem, confirm that the inputs refer to the same year and starting group, that all causes have been included, and that the selected roots lie in the probability range. The conversion is specific to the stated UDD basis: applying a different within-year assumption would require a different relationship. This focus on the basis and the consistency checks is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, assume the two causes are exhaustive and the associated single-decrement tables follow UDD unless the question asks you to check consistency.

Observed probabilities are $q_1=0.09$ and $q_2=0.19$. Find the associated probabilities and verify the observed total exit probability. (Answer: $a=0.10,\ b=0.20$; total exit probability $0.28$.)
Observed probabilities are $q_1=0.38$ and $q_2=0.08$. State the quadratic for $b$, identify the admissible root, and find $a$. (Answer: $b^2-1.70b+0.16=0$; admissible $b=0.10$, $a=0.40$.)
A table starts with 5,000 lives and records 950 exits from cause 1, 450 from cause 2, and 3,600 survivors. Find the observed probabilities and the associated probabilities, then check the survivor count. (Answer: $q_1=0.19,\ q_2=0.09;\ a=0.20,\ b=0.10$; the survivor count is 3,600.)
Proposed associated probabilities are $a=0.18$ and $b=0.12$. Calculate both observed probabilities and the implied total exit probability. (Answer: $q_1=0.1692,\ q_2=0.1092$; total $0.2784$.)
A table begins with 2,000 lives and reports 380 exits from cause 1, 180 from cause 2, and 1,450 survivors. Check whether the table balances. If the survivor count is corrected to balance, find the observed and associated probabilities. (Answer: The stated counts total 2,010, so the table does not balance; corrected survivors are 1,440, observed probabilities are $0.19$ and $0.09$, and associated probabilities are $a=0.20,\ b=0.10$.)

What You'll Learn