Solving UDD Equations for Two Causes

Why Recover Associated Probabilities?

A multiple-decrement table records the probabilities of observed exits when several causes can operate at once. An associated single-decrement table answers a different question: what would the probability for one cause be if that cause acted alone? Under the uniform distribution of decrements (UDD) assumption, the observed probabilities can be converted to associated probabilities. With two causes, this conversion requires solving a pair of coupled equations.

The coupling matters because removing one cause changes the amount of exposure available to the other. This tutorial develops the two-cause solution, shows how to choose between its algebraic roots, and checks that the resulting associated probabilities are valid. It is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Let $q_1$ and $q_2$ be the observed probabilities of decrement from causes 1 and 2 during one year, measured from the same initial group. Let $q_1'$ and $q_2'$ be the corresponding associated single-decrement probabilities: the probabilities that each cause would produce a decrement over the year if it were the only cause operating. Under the two-cause UDD model, $$ q_1=q_1'\left(1-\frac{q_2'}{2}\right),\qquad q_2=q_2'\left(1-\frac{q_1'}{2}\right). $$

Turning the Coupled Equations Into One Quadratic

The equations are coupled: each observed probability depends on both associated probabilities. A useful substitution is the shared adjustment $c=\frac{q_1'q_2'}{2}$. Expanding the two equations gives $q_1=q_1'-c$ and $q_2=q_2'-c$. Therefore $q_1'=q_1+c$ and $q_2'=q_2+c$. Substituting these expressions into the definition of $c$ gives a quadratic that can be solved using only the observed probabilities.

c=\frac{(q_1+c)(q_2+c)}{2} \quad\Longrightarrow\quad c^2+(q_1+q_2-2)c+q_1q_2=0.

Writing $Q=q_1+q_2$, the quadratic formula gives $c=\frac{2-Q\pm\sqrt{(2-Q)^2-4q_1q_2}}{2}$. For probabilities generated by this model, the feasible solution is the smaller root. Once $c$ is obtained, add it to each observed probability to recover $q_1'$ and $q_2'$.

Result: For two observed probabilities $q_1,q_2$, the UDD associated probabilities are found using $$ c=\frac{2-(q_1+q_2)-\sqrt{(2-q_1-q_2)^2-4q_1q_2}}{2}, \qquad q_1'=q_1+c,\quad q_2'=q_2+c. $$ Use this conversion when the two-cause UDD equations apply and the observed probabilities are consistent with the model. The square-root sign is negative in the expression for the feasible root.

Why the Smaller Root Is the Feasible One

The sign choice is not just a numerical convention. Suppose the observed probabilities are nonnegative and $Q=q_1+q_2\leq 1$, as required for two mutually exclusive observed exits in one year. The discriminant is nonnegative because $(2-Q)^2\geq 1$ and $4q_1q_2\leq Q^2\leq 1$. The smaller root is nonnegative: the square root is no larger than $2-Q$, since the discriminant is $(2-Q)^2-4q_1q_2$.

To check the upper bounds, define $f(c)=c^2+(Q-2)c+q_1q_2$. Its smaller root is $c$, and $f(0)=q_1q_2\geq0$. Also, $f(1-q_1)=Q-1\leq0$, so the smaller root is no greater than $1-q_1$; likewise, $f(1-q_2)=Q-1\leq0$. Consequently, $0\leq c\leq1-q_1$ and $0\leq c\leq1-q_2$. Thus $0\leq q_1'=q_1+c\leq1$ and $0\leq q_2'=q_2+c\leq1$. The larger root must not be used merely because it also solves the quadratic: it can produce associated probabilities greater than one.

Check the inputs.
Confirm that the observed probabilities are nonnegative and their sum does not exceed one.

Calculate the discriminant.
Use $(2-q_1-q_2)^2-4q_1q_2$, and take its nonnegative square root.

Choose the feasible root.
Calculate $c$ using the minus sign, then set $q_1'=q_1+c$ and $q_2'=q_2+c$.

Verify the result.
Check each associated probability is in $[0,1]$ and substitute them into both UDD equations to recover the observed probabilities.

Worked Conversions

Worked Example: Recovering Two Unequal Associated Probabilities

A hypothetical one-year decrement study records $q_1=0.36$ for cause 1 and $q_2=0.16$ for cause 2. Assume the two-cause UDD model. Find the associated probabilities and verify them.

Method. Form the shared adjustment $c$, choose the smaller root, and add it to each observed probability. The observed total is $Q=0.36+0.16=0.52$. The quadratic is $c^2+(0.52-2)c+(0.36)(0.16)=0$, or $c^2-1.48c+0.0576=0$.

The discriminant is $1.48^2-4(0.0576)=2.1904-0.2304=1.96$, whose square root is $1.40$. Therefore the feasible root is $c=(1.48-1.40)/2=0.04$. The other root is $(1.48+1.40)/2=1.44$.

The associated probabilities are $q_1'=0.36+0.04=0.40$ and $q_2'=0.16+0.04=0.20$. Both are between zero and one. Substitution checks the result: $0.40(1-0.20/2)=0.40(0.90)=0.36$, and $0.20(1-0.40/2)=0.20(0.80)=0.16$. Both observed probabilities are recovered exactly.

Common mistake avoided: The larger root would give $q_1'=1.80$ and $q_2'=1.60$, which are not probabilities. A root of the quadratic is not automatically a feasible solution to the decrement model.

Worked Example: Solving an Asymmetric Pair

For a hypothetical insured group, observed one-year probabilities are $q_1=0.51$ and $q_2=0.21$. Under UDD, calculate both associated probabilities and check the conversion.

The observed total is $Q=0.51+0.21=0.72$, and $q_1q_2=0.51(0.21)=0.1071$. The quadratic coefficient $Q-2$ is $-1.28$, so $c^2-1.28c+0.1071=0$. Its discriminant is $1.28^2-4(0.1071)=1.6384-0.4284=1.2100$, with square root $1.10$.

The smaller root is $c=(1.28-1.10)/2=0.09$. Therefore $q_1'=0.51+0.09=0.60$ and $q_2'=0.21+0.09=0.30$. The values are both in $[0,1]$. Substitution gives $0.60(1-0.30/2)=0.60(0.85)=0.51$ and $0.30(1-0.60/2)=0.30(0.70)=0.21$. The calculations agree with the supplied observations.

Common mistake avoided: The two causes do not have the same adjustment as a proportion of their observed probabilities; they have the same additive adjustment $c$. Adding $0.09$ to each observation follows directly from the coupled equations.

Worked Example: Checking a High Total Decrement Probability

A hypothetical two-cause table has observed probabilities $q_1=0.64$ and $q_2=0.24$. Assume UDD. Recover the associated probabilities, paying particular attention to the feasibility check.

Here $Q=0.64+0.24=0.88$, $2-Q=1.12$, and $q_1q_2=0.1536$. The discriminant is $1.12^2-4(0.1536)=1.2544-0.6144=0.6400$, whose square root is $0.80$. Thus $c=(1.12-0.80)/2=0.16$.

The associated probabilities are $q_1'=0.64+0.16=0.80$ and $q_2'=0.24+0.16=0.40$. Each is nonnegative and no greater than one. Substitution gives $0.80(1-0.40/2)=0.80(0.80)=0.64$ and $0.40(1-0.80/2)=0.40(0.60)=0.24$. The observed total is also consistent: $0.64+0.24=0.88$.

Common mistake avoided: Do not treat $q_1+q_2$ as the sum of associated probabilities. Here the associated total is $0.80+0.40=1.20$, while the observed total is $0.88$; competition between causes explains the difference under the stated equations.

Common Errors and Model Checks

A reliable solution does more than calculate a square root. It identifies which probabilities are observed, uses the assumed UDD equations, chooses the feasible root, and checks the answer in the original equations. In particular, the observed probabilities are not themselves the associated probabilities, and the larger quadratic root is not a second valid conversion merely because it satisfies the algebra.

Confusing the two kinds of probability. The $q_j$ values describe exits with competing causes present; the $q_j'$ values describe the associated single-cause probabilities.
Using the wrong root. Calculate both roots if useful, but retain only the one for which both recovered probabilities lie between zero and one.
Skipping the reconstruction check. Substitute the proposed associated values into both equations separately. Checking only their sum is not sufficient.
Ignoring the assumption. These equations are specific to the two-cause UDD model. A different within-year decrement assumption requires a different conversion.

The method is useful whenever two observed cause probabilities must be translated into associated single-decrement probabilities before a single-decrement table can be used. Its main exam discipline is to keep the observed and associated quantities distinct and to verify both equations after solving.

Key takeaway: Under two-cause UDD, set $c=q_1'q_2'/2$, solve $c^2+(q_1+q_2-2)c+q_1q_2=0$, and use the smaller root. Then recover $q_1'=q_1+c$ and $q_2'=q_2+c$, check both are probabilities, and substitute them back into the model.

Check Your Understanding

For each problem, assume the two-cause UDD equations apply. Find the associated probabilities and check feasibility.

Observed probabilities are $q_1=0.27$ and $q_2=0.12$. (Answer: $q_1'=0.29,\ q_2'=0.14$.)
A hypothetical table has $q_1=0.45$ and $q_2=0.15$. Recover the associated probabilities. (Answer: $q_1'=0.50,\ q_2'=0.20$.)
Observed probabilities are equal at $q_1=q_2=0.32$. Find the two associated probabilities. (Answer: $q_1'=q_2'=0.40$.)
For a hypothetical decrement study, $q_1=0.665$ and $q_2=0.065$. Find the associated probabilities and verify that neither exceeds one. (Answer: $q_1'=0.70,\ q_2'=0.10$.)
Observed probabilities are $q_1=0.80$ and $q_2=0.20$. Recover the associated probabilities, including any boundary value. (Answer: $q_1'=1.00,\ q_2'=0.40$.)

What You'll Learn