A Worked UDD Conversion From a Decrement Table

Why a UDD Conversion Is Needed

A multiple-decrement table records the probability that a life exits through each cause when several causes operate together. An associated single-decrement table instead describes the probability for one cause considered on its own, with the other causes removed. These probabilities are not generally equal: in the multiple-decrement setting, one cause can remove a life before another cause acts.

Under the uniform distribution of decrements (UDD) assumption, observed annual probabilities and associated annual probabilities are linked by a specific set of equations. Converting a supplied table requires solving those equations while checking that the proposed associated probabilities actually reproduce the observations. The zero-probability case needs particular care: it can have two nonnegative algebraic roots, and the conversion need not be unique.

Definition: For causes $j=1,\ldots,m$, let $q_j$ be the observed probability of exit by cause $j$ during the year in the multiple-decrement table. Let $q'_j$ be the associated single-decrement probability for cause $j$ over the same year, with the other causes removed. Define $r=\sum_{k=1}^{m}q'_k$. Under the UDD conversion equations, $$ q_j=q'_j\left(1-\frac{1}{2}\sum_{k\ne j}q'_k\right) =q'_j\left(1+\frac{q'_j-r}{2}\right). $$

The observed probabilities describe mutually exclusive exits, so they must satisfy $q_j\ge 0$ and $\sum_j q_j\le 1$. Each associated probability must also be between zero and one. Their sum $r$, however, is not the probability of an observed exit: associated probabilities come from separate single-decrement tables and can sum to more than one.

Reduce the Conversion to a Consistency Equation

For a proposed value of the total $r$, the equation for cause $j$ can be rearranged into a quadratic in $q'_j$:

$$ (q'_j)^2+(2-r)q'_j-2q_j=0. $$

If $q_j>0$, the product of the quadratic's two roots is $-2q_j<0$. Thus exactly one root is positive, and it is

q'_j(r)=\frac{-(2-r)+\sqrt{(2-r)^2+8q_j}}{2}.

This expression is well-defined for any real $r$ when $q_j>0$. It does not, by itself, solve the whole conversion: a valid proposed $r$ must also equal the sum of the associated probabilities obtained for all causes.

Important zero case: If $q_j=0$, the quadratic factors as $q'_j(q'_j+2-r)=0$. Its roots are $0$ and $r-2$. For $r\le 2$, the only nonnegative root is $0$. For $r>2$, both $0$ and $r-2$ are nonnegative and must be considered. Substituting $q_j=0$ into the positive-root formula and keeping only the root it returns can miss valid solutions.

For causes with positive observed probabilities, define $q'_j(r)$ by the positive-root formula. For each cause with zero observed probability, choose a nonnegative root allowed by the proposed $r$. A candidate conversion must satisfy the total consistency condition and the original equations:

\sum_{j=1}^{m}q'_j=r, \qquad q_j=q'_j\left(1+\frac{q'_j-r}{2}\right)\quad\text{for every }j.

These checks matter even when a scalar numerical search has found a value of $r$: zero-probability causes can create separate root branches. Also check $0\le q'_j\le1$, since an algebraic solution outside those bounds is not a probability table.

Why the UDD Equation Has This Form

The UDD conversion relation is based on uniform timing within the year. In the associated table for cause $j$, the annual probability $q'_j$ is the single-cause probability before allowing for removals by other causes. Under UDD, competing exits are spread uniformly through the year; accounting for those removals gives the cause-$j$ adjustment of one half of each cross-product $q'_jq'_k$. Summing those adjustments gives

q_j=q'_j-\frac{1}{2}\sum_{k\ne j}q'_jq'_k =q'_j\left(1-\frac{1}{2}\sum_{k\ne j}q'_k\right).

Since $\sum_{k\ne j}q'_k=r-q'_j$, substitution gives the quadratic above. This is the UDD conversion relation; it is not a rule to apply to probabilities generated under a different within-year assumption.

A Practical Solution Procedure

Check the supplied table.
Confirm that each observed probability is nonnegative and that their sum is at most one.

Choose a proposed total $r$.
For each positive $q_j$, calculate the unique positive quadratic root. For $q_j=0$, list every nonnegative root allowed by that $r$.

Enforce the total.
Retain a candidate only if its chosen roots sum to the proposed $r$. A numerical root search can help, but each zero-cause branch must be considered.

Verify the original equations and bounds.
Substitute the candidate $q'_j$ values into every UDD equation, and confirm that each associated probability lies in $[0,1]$.

Worked Conversions

Worked Example: Converting Three Positive Observed Probabilities

A hypothetical annual table has three observed decrement probabilities $q_1=0.08$, $q_2=0.20$, and $q_3=0.56$. Find a UDD associated table. As a count check, suppose 1,000 lives are exposed at the start of the year.

Method. Solve the UDD quadratics using a candidate total $r$, then check that the roots sum to $r$. The observed probabilities sum to $0.08+0.20+0.56=0.84$, which is at most one.

Taking $r=1.40$, the coefficient $2-r$ is $0.60$. For cause 1,

q'_1=\frac{-0.60+\sqrt{0.60^2+8(0.08)}}{2} =\frac{-0.60+\sqrt{1.00}}{2}=0.20.

For cause 2,

q'_2=\frac{-0.60+\sqrt{0.60^2+8(0.20)}}{2} =\frac{-0.60+\sqrt{1.96}}{2}=0.40.

For cause 3,

q'_3=\frac{-0.60+\sqrt{0.60^2+8(0.56)}}{2} =\frac{-0.60+\sqrt{4.84}}{2}=0.80.

The roots sum to $0.20+0.40+0.80=1.40$, so the proposed total is consistent. Check each original equation:

q_1=0.20\left(1-\frac{0.40+0.80}{2}\right)=0.20(0.40)=0.08, \qquad q_2=0.40\left(1-\frac{0.20+0.80}{2}\right)=0.40(0.50)=0.20, $$ $$ q_3=0.80\left(1-\frac{0.20+0.40}{2}\right)=0.80(0.70)=0.56.

The associated probabilities are therefore $(0.20,0.40,0.80)$, each within $[0,1]$. In the hypothetical cohort, the observed cause-specific exits are $1{,}000(0.08)=80$, $1{,}000(0.20)=200$, and $1{,}000(0.56)=560$; the observed survivors are $1{,}000(1-0.84)=160$. These counts check the supplied multiple-decrement table, not the associated probabilities.

Common mistake avoided: Treating $1.40$, the sum of the associated probabilities, as an observed exit probability. The observed exit probability is $0.84$; the associated total is used only in the conversion equations.

Worked Example: Zero Observed Probability and Multiple Valid Tables

A hypothetical table has four causes with observed probabilities $(0.16,0.16,0.16,0)$. Find valid UDD associated tables and determine whether the data identify a unique one.

Method. Work separately through the zero-probability cause's possible roots. For each branch, enforce the total and verify every original equation.

First consider the zero root $q'_4=0$. The first three associated probabilities are equal; write each as $x$. Then $r=3x$, and the equation for any of those causes becomes

0.16=x\left(1+\frac{x-3x}{2}\right)=x(1-x).

Thus $x^2-x+0.16=0$, or $(x-0.20)(x-0.80)=0$. This gives two valid candidates:

(q'_1,q'_2,q'_3,q'_4)=(0.20,0.20,0.20,0),\quad r=0.60, $$ $$ (q'_1,q'_2,q'_3,q'_4)=(0.80,0.80,0.80,0),\quad r=2.40.

For the first candidate, each positive cause gives $0.20(1+(0.20-0.60)/2)=0.20(0.80)=0.16$. For the second, each gives $0.80(1+(0.80-2.40)/2)=0.80(0.20)=0.16$. In both cases, cause 4 gives zero because $q'_4=0$.

Now take the other possible zero-cause root, $q'_4=r-2$, which requires $r>2$. The total condition gives $3x+(r-2)=r$, so $x=2/3$. The positive-cause equation then gives

0.16=\frac{2}{3}\left(1+\frac{2/3-r}{2}\right) =\frac{2}{3}\left(\frac{4}{3}-\frac{r}{2}\right), \qquad r=\frac{164}{75}.

Hence $q'_4=r-2=14/75$. For each of the first three causes, the factor in parentheses is $1+(2/3-164/75)/2=6/25$, giving $(2/3)(6/25)=4/25=0.16$. For cause 4, $q'_4-r=-2$, so its factor is $1-1=0$, giving the observed zero. All four associated probabilities lie in $[0,1]$, and their sum is $3(2/3)+14/75=164/75$.

There are three valid associated tables. The observed data do not identify a unique conversion.

Common mistake avoided: At $r=2.40$, the zero-cause quadratic is $x(x-0.40)=0$. Keeping only $0.40$ would make the four roots sum to $2.80$, not $2.40$. The root $0$ must also be considered; it produces the valid table $(0.80,0.80,0.80,0)$.

Worked Example: A Four-Cause Table With Total Associated Probability Two

A hypothetical table has observed probabilities $(0.02,0.08,0.18,0.32)$. Suppose an additional calibration condition specifies that the total associated probability is $r=2$. Convert the table and check it against 1,000 exposed lives.

Method. Use the stated total in each quadratic and then verify both the total and the UDD equations. Since $2-r=0$, the quadratic simplifies to $(q'_j)^2=2q_j$, and the positive root is $q'_j=\sqrt{2q_j}$.

q'_1=\sqrt{2(0.02)}=0.20,\quad q'_2=\sqrt{2(0.08)}=0.40,\quad q'_3=\sqrt{2(0.18)}=0.60,\quad q'_4=\sqrt{2(0.32)}=0.80.

Their sum is $0.20+0.40+0.60+0.80=2.00$, as required. Because $r=2$, the equation can also be written $q_j=q'_j(q'_j/2)=(q'_j)^2/2$. This gives $0.02,0.08,0.18,0.32$, respectively, so all four original probabilities are recovered.

The observed probabilities sum to $0.60$. Out of 1,000 exposed lives, the observed exits are 20, 80, 180, and 320, for 600 exits in total and 400 survivors. The associated values are all valid probabilities, although their sum is 2.00.

Common mistake avoided: Rejecting the result because the associated probabilities sum to more than one. The sum is not an observed probability; the observed exits sum to $0.60$.

Checks and Exam Traps

A numerical root is only a candidate until it passes the full consistency check. For each proposed table, verify the total, substitute into the original equations, and check probability bounds. When an observed $q_j$ is zero, explicitly list its roots rather than relying on a formula derived for positive $q_j$. If the problem asks for a unique conversion, do not assume uniqueness: examine all admissible root branches and report whether more than one table fits.

Confusing the two tables. The observed $q_j$ values are competing exits and sum to at most one. The associated $q'_j$ values come from separate single-decrement tables and need not have a sum at most one.
Using only the positive root without checking the data. For $q_j>0$, the quadratic has one positive root. For $q_j=0$ and $r>2$, it has two nonnegative roots.
Checking only the cause equations. The proposed roots must also sum to the proposed total $r$. A set of cause-by-cause roots that fails this condition is not a solution.
Rounding too early. Keep sufficient precision while solving and checking; round the final associated probabilities only after verifying the equations.

Key takeaway: Under UDD, conversion is a coupled system: solve each quadratic for a proposed $r$, enforce $\sum_j q'_j=r$, and verify the original equations. Zero observed probabilities require explicit branch checking, and valid conversions may not be unique.

Check Your Understanding

For each problem, check both the total associated probability and the original UDD equations.

A three-cause associated table is proposed as $(0.10,0.30,0.50)$. Calculate the three observed probabilities and their sum. (Answer: $(0.06,0.21,0.40)$; sum $0.67$.)
For two causes, the observed probabilities are $(0.28,0.48)$. Verify whether $(0.40,0.60)$ is an associated table under UDD. (Answer: Yes; the equations give $0.40(1-0.60/2)=0.28$ and $0.60(1-0.40/2)=0.48$, and the associated total is $1.00$.)
For an observed probability $q_j=0$, find all nonnegative roots of its quadratic when the proposed total is $r=2.70$. (Answer: $q'_j=0$ or $q'_j=0.70$.)
A three-cause table has observed probabilities $(0.06,0.21,0.40)$. Verify that $(0.10,0.30,0.50)$ is a valid associated table and state the observed exit probability. (Answer: It is valid; the associated total is $0.90$, and the observed exit probability is $0.67$.)
For four causes, the observed probabilities are $(0.16,0.16,0.16,0)$, and a proposed total is $r=2.40$. If the first three associated probabilities are equal and the fourth is zero, find the associated table and check the total. (Answer: $(0.80,0.80,0.80,0)$; each positive cause gives $0.16$, the zero cause gives $0$, and the total is $2.40$.)

What You'll Learn