Checking Associated Tables Against Multiple-Decrement Data

Why an Associated Table Needs an Audit

An associated single-decrement table describes the probability of exit from one cause when the other causes are removed from the model. Those probabilities are not, in general, the same as the observed cause probabilities in a multiple-decrement table: competing causes change the time each individual remains exposed to a given cause. A proposed associated table should therefore be checked by putting it back into the stated multiple-decrement model and reconstructing the observed outcomes.

The audit has two distinct tests. First, the reconstructed probability of exit from each cause must agree with the corresponding observed probability. Second, the reconstructed probability of remaining in the group must agree with the observed total. An aggregate total alone is not enough: cause-specific errors can offset one another.

Definition: For a one-year multiple-decrement model with causes \(1,\ldots,m\), let \(q_j\) be the observed probability that a member of the starting group exits from cause \(j\), and let \(p\) be the observed probability of remaining in the group for the full year. The associated single-decrement probability \(q'_j\) is the probability of exit from cause \(j\) over the year in a model in which the other causes are absent. The observed probabilities must satisfy \(\sum_{j=1}^m q_j+p=1\).

The prime distinguishes an associated probability from an observed multiple-decrement probability. An audit uses the proposed \(q'_j\) values and the specified assumption to calculate reconstructed probabilities, denoted below by \(\widehat q_j\) and \(\widehat p\). The table is consistent only when the reconstructed values agree with the observations, subject to any stated rounding tolerance.

Reconstructing Outcomes Under UDD

Under the uniform distribution of decrements (UDD) assumption, decrements are distributed uniformly over the year in the associated conversion model. For \(m\) causes, the UDD competing-decrement equations express the observed cause probability in terms of all the proposed associated probabilities:

\widehat q_j=q'_j\left(1-\frac{1}{2}\sum_{\substack{k=1\\k\ne j}}^m q'_k\right), \qquad \widehat p=1-\sum_{j=1}^m \widehat q_j.

The competing associated probabilities appear in the factor multiplying \(q'_j\). A cause has less opportunity to produce an observed exit when other causes remove members during the year. The total reconstructed exit probability is \(\widehat q=\sum_j\widehat q_j\); the corresponding probability of remaining is \(1-\widehat q\).

Justification: Under UDD, the associated decrements from each competing cause are spread uniformly across the year. For cause \(j\), the average fraction of the starting group not removed by the other causes is \(1-\frac12\sum_{k\ne j}q'_k\): each other cause contributes, on average, half of its associated decrement probability over the year. Multiplying the cause-\(j\) associated probability by this average exposure gives the UDD conversion equation. Summing the reconstructed cause probabilities gives the probability of any exit, so the remaining probability is its complement.

This reconstruction is valid only when the proposed values and the observed table are being interpreted under UDD. The associated probabilities must be within their probability range, and the reconstructed cause probabilities and remaining probability must be nonnegative. In addition to checking those conditions, compare each reconstructed cause with its observed value; agreement of totals does not establish agreement by cause.

Worked Example: Auditing a Two-Cause UDD Table

A hypothetical one-year table has observed cause probabilities \(q_1=0.19\) and \(q_2=0.09\), with observed probability of remaining \(p=0.72\). A proposed associated table gives \(q'_1=0.20\) and \(q'_2=0.10\). Test the proposal under UDD.

Method. Reconstruct each observed cause probability using the other cause’s associated probability, then calculate the total remaining probability.

For cause 1, the competing associated probability is \(q'_2=0.10\):

\widehat q_1=0.20\left(1-\frac12(0.10)\right) =0.20(0.95)=0.19.

For cause 2, the competing associated probability is \(q'_1=0.20\):

\widehat q_2=0.10\left(1-\frac12(0.20)\right) =0.10(0.90)=0.09.

The reconstructed total exit probability is \(0.19+0.09=0.28\), so \(\widehat p=1-0.28=0.72\). Both reconstructed cause probabilities equal their observed values, and the reconstructed remaining probability equals the observed \(0.72\). The proposal passes the UDD audit.

Check. The observed probabilities sum to \(0.19+0.09+0.72=1.00\), matching the reconstructed total. The common mistake avoided here is treating \(q'_1\) and \(q'_2\) as if they were the observed cause probabilities.

Reconstructing Outcomes Under Constant Force

Under the constant-force assumption, each cause has a constant force over the one-year interval, and the total force is the sum of the cause-specific forces. The associated probability \(q'_j\) determines the force for cause \(j\). For a valid associated probability \(0\le q'_j<1\), that force is \(\mu_j=-\ln(1-q'_j)\). The total force determines survival through the year; each cause’s share of that total force determines its share of the total exit probability.

\widehat\mu=\sum_{j=1}^m\mu_j, \qquad \widehat p=e^{-\widehat\mu}, \qquad \widehat q_j=\frac{\mu_j}{\widehat\mu}\left(1-e^{-\widehat\mu}\right) \quad(\widehat\mu>0).

If all cause forces are zero, then \(\widehat p=1\) and every \(\widehat q_j=0\); the cause-share expression is not needed. These formulas assume constant cause-specific forces over the specified year. Do not use the UDD reconstruction equations to test a constant-force table, or vice versa.

The cause formula follows from the constant-force model: the probability density of exit from cause \(j\) at time \(t\) is \(\mu_j e^{-\widehat\mu t}\). Integrating over the year gives \(\mu_j(1-e^{-\widehat\mu})/\widehat\mu\). Survival to the end of the year is \(e^{-\widehat\mu}\), yielding the expressions above.

Worked Example: Checking Constant-Force Associated Probabilities

A hypothetical group has two proposed associated probabilities for one year: \(q'_1=1-e^{-0.2}\) and \(q'_2=1-e^{-0.3}\). The observed table reports \(q_1=0.1574\), \(q_2=0.2361\), and \(p=0.6065\). Check the proposal under constant force. Give reconstructed probabilities to four decimal places.

Method. Convert each associated probability to its force, add the forces, and use the total force to reconstruct survival and each cause probability.

The forces are \(\mu_1=-\ln(1-q'_1)=0.2\) and \(\mu_2=-\ln(1-q'_2)=0.3\). Thus \(\widehat\mu=0.2+0.3=0.5\). Reconstructed survival is

\widehat p=e^{-0.5}=0.6065307\ldots\approx 0.6065.

The total exit probability is \(1-e^{-0.5}=0.3934693\ldots\). Cause 1 receives \(0.2/0.5=0.4\) of that probability, and cause 2 receives \(0.3/0.5=0.6\):

\widehat q_1=0.4(0.3934693\ldots)=0.1573877\ldots\approx0.1574,

\widehat q_2=0.6(0.3934693\ldots)=0.2360816\ldots\approx0.2361.

Each reconstructed value agrees with the reported value to four decimal places. As a check, \(0.1574+0.2361+0.6065=1.0000\). The proposal passes, allowing for rounding. The common mistake avoided is adding associated probabilities to obtain the total exit probability; under constant force, add the forces first.

A Cause-by-Cause Audit, Not Just a Total Check

The following audit sequence keeps the assumption, cause checks, and aggregate check separate. It applies to either model, using that model’s own reconstruction equations.

Confirm the basis.
Identify the interval, the causes, and whether the proposed associated values are stated under UDD or constant force. Check that each probability is in its valid range.

Reconstruct each cause.
Use the specified assumption to calculate \(\widehat q_j\) for every cause. Compare each result with the corresponding observed \(q_j\), not just with the total.

Reconstruct remaining probability.
Calculate \(\widehat p=1-\sum_j\widehat q_j\) under UDD, or calculate survival from the total force under constant force. Compare it with the observed \(p\).

Check probability accounting.
Confirm that reconstructed cause probabilities and remaining probability are nonnegative and sum to one. State whether any differences are larger than the permitted rounding tolerance.

Worked Example: A Matching Total Can Hide a Cause Error

A hypothetical three-cause table reports \(q_1=0.075\), \(q_2=0.150\), \(q_3=0.265\), and \(p=0.510\). A proposed associated table gives \(q'_1=0.10\), \(q'_2=0.20\), and \(q'_3=0.30\). Determine whether it passes under UDD.

Method. Reconstruct each cause using the sum of the other associated probabilities. For cause 1, the other two sum to \(0.20+0.30=0.50\):

\widehat q_1=0.10\left(1-\frac12(0.50)\right)=0.10(0.75)=0.075.

For cause 2, the other two sum to \(0.10+0.30=0.40\):

\widehat q_2=0.20\left(1-\frac12(0.40)\right)=0.20(0.80)=0.160.

For cause 3, the other two sum to \(0.10+0.20=0.30\):

\widehat q_3=0.30\left(1-\frac12(0.30)\right)=0.30(0.85)=0.255.

The reconstructed total exit probability is \(0.075+0.160+0.255=0.490\), so \(\widehat p=1-0.490=0.510\). The observed cause probabilities also sum to \(0.075+0.150+0.265=0.490\), and the observed remaining probability is \(0.510\). Thus the aggregate totals agree. However, \(\widehat q_2=0.160\ne0.150\) and \(\widehat q_3=0.255\ne0.265\). The proposal does not pass the cause-by-cause audit.

Check. The cause discrepancies offset: cause 2 is higher by \(0.010\), while cause 3 is lower by \(0.010\). This example shows why checking only \(\sum_j q_j\) and \(p\) can miss an inconsistent associated table.

Common Audit Traps

Confusing the two kinds of probabilities. Associated \(q'_j\) values describe a single-decrement setting; observed \(q_j\) values describe competing decrements. Reconstruct before comparing.
Using the wrong assumption. UDD and constant force have different conversion equations. The assumption attached to the proposed table governs the audit.
Checking only the total. Equal total exits and equal remaining probability do not ensure that the cause allocation is correct.
Forgetting probability accounting. Observed causes and observed remaining probability must sum to one. A failure here is an inconsistency in the supplied data, even before testing the proposed associated values.
Rounding too early. Keep full precision through the reconstruction, then round the final comparisons. Small rounding differences should be assessed against the stated tolerance.

Key takeaway: To audit an associated table, reconstruct the multiple-decrement probability for every cause and the probability of remaining under the stated assumption. A valid total is necessary, but only cause-by-cause agreement establishes consistency with the observed table.

Check Your Understanding

For each problem, use the stated assumption and check both the individual causes and the remaining probability.

Under UDD, proposed associated probabilities are \(q'_1=0.12\) and \(q'_2=0.08\). Find the reconstructed cause probabilities and remaining probability. (Answer: \(\widehat q_1=0.1152\), \(\widehat q_2=0.0752\), \(\widehat p=0.8096\).)
Under UDD, three proposed associated probabilities are \(0.05\), \(0.10\), and \(0.15\). Reconstruct all three cause probabilities and the probability of remaining. (Answer: \(\widehat q_1=0.04375\), \(\widehat q_2=0.09000\), \(\widehat q_3=0.13875\), \(\widehat p=0.72750\).)
Under constant force, proposed associated probabilities are \(q'_1=1-e^{-0.1}\) and \(q'_2=1-e^{-0.2}\). Find the cause probabilities and survival to four decimal places. (Answer: \(\widehat q_1=0.0864\), \(\widehat q_2=0.1728\), \(\widehat p=0.7408\).)
An observed two-cause table has \(q_1=0.18\), \(q_2=0.17\), and \(p=0.65\). Proposed UDD associated probabilities are \(q'_1=q'_2=0.20\). Does the proposal pass? (Answer: No. It reconstructs \(\widehat q_1=0.18\), \(\widehat q_2=0.18\), and \(\widehat p=0.64\).)
Under constant force, proposed associated probabilities are \(q'_1=0.10\) and \(q'_2=0.20\). Reconstruct cause probabilities and survival to four decimal places. (Answer: \(\widehat q_1=0.0898\), \(\widehat q_2=0.1902\), \(\widehat p=0.7200\).)

What You'll Learn