Rounding and Closure in Competing-Risk Tables

Why Rounding Can Break Closure

A competing-risk table divides exits during an interval among several causes and records those who survive the interval without any listed exit. In exact arithmetic, these outcomes exhaust the starting group: every member either exits for one of the causes or remains in the group. But a published table often displays probabilities and counts rounded to a fixed number of places. Adding the displayed entries can therefore produce a small difference from the required total.

The key is to distinguish a rounding residual, whose size is consistent with the displayed precision, from a material discrepancy, which the stated rounding cannot explain. A mechanical adjustment can make a row add up while concealing a mistake in the underlying inputs.

Definition: For a one-year multiple-decrement table at age $x$, let $q_x^{(j)}$ be the probability that a life aged $x$ exits during the year from cause $j$, and let $p_x$ be the probability of surviving the year without any listed exit. If there are $k$ mutually exclusive and exhaustive causes, the exact probabilities satisfy $$ \sum_{j=1}^{k} q_x^{(j)}+p_x=1. $$ For a starting radix $l_x$, the corresponding expected counts are $d_x^{(j)}=l_xq_x^{(j)}$ and $l_{x+1}=l_xp_x$, so their exact total is $l_x$.

The identity assumes that the listed causes cover all exits in the model, that the causes are mutually exclusive for an individual during the interval, and that the survivor count refers to survival from every listed cause. It does not apply if an omitted exit category exists, if the entries refer to different intervals or populations, or if the quantities are defined on incompatible bases. Those are model or data issues, not rounding issues.

A Precision-Based Closure Check

Suppose each of the $k$ cause probabilities and the survivor probability is rounded to the nearest multiple of $h$. Each displayed value can differ from its exact value by at most $h/2$. Since the exact values sum to 1, the absolute difference between the sum of the displayed values and 1 cannot exceed $(k+1)h/2$, provided all entries use the stated rounding rule.

This is a tolerance for diagnosing rounding, not permission to alter any entry. A discrepancy within the bound is compatible with independent rounding; it does not prove that the original data or model are correct. A discrepancy beyond the bound cannot be explained by that rounding rule alone.

\left|\sum_{j=1}^{k}\widehat{q}_x^{(j)}+\widehat{p}_x-1\right| \leq \frac{(k+1)h}{2}

Here, a hat denotes a displayed rounded value. If the probabilities are shown to four decimal places, then $h=0.0001$. With three causes, there are four rounded entries, so the maximum discrepancy attributable to nearest rounding is $4(0.0001)/2=0.0002$. The bound changes if entries are truncated rather than rounded, or if different precisions are used; in those cases, use the appropriate maximum error for each entry.

A useful closure process is to calculate the displayed sum, compare its residual with the precision-based bound, and then identify which quantities are authoritative. If cause probabilities are the inputs and the survivor probability is defined as the complement, calculate $p_x=1-\sum_j q_x^{(j)}$ from the unrounded cause probabilities before displaying values. If all entries are independently supplied, report the residual and investigate it before changing any entry.

Check the table definition.
Confirm the interval, cause categories, starting radix, and meaning of the survivor entry.

Calculate the probability residual.
Add the displayed cause probabilities and survivor probability, then subtract 1.

Compare with the rounding bound.
Use the number of independently rounded entries and their stated precision; do not choose a tolerance after seeing the residual.

Reconcile only with a stated rule.
If a derived entry is meant to be the complement, derive it from the unrounded inputs. If a residual is material or the inputs are independently specified, flag it rather than silently forcing closure.

Worked Examples

Worked Example: Allocating Rounded Counts Without Changing the Radix

A hypothetical two-cause table has exact one-year probabilities $q_x^{(1)}=0.06244$, $q_x^{(2)}=0.02746$, and $p_x=0.91010$. The starting radix is $l_x=8{,}000$. First check probability closure, then calculate expected counts. Finally, suppose the table must display whole-number counts: allocate integer counts by the largest-remainder method, which takes the floors of the expected counts and adds the remaining units to the largest fractional parts. State the common mistake this avoids.

Method and probability check. There are two causes, and the three outcomes sum to $0.06244+0.02746+0.91010=0.08990+0.91010=1.00000$. The exact probabilities close.

Expected counts. Multiply each probability by the radix: $d_x^{(1)}=8{,}000(0.06244)=499.52$, $d_x^{(2)}=8{,}000(0.02746)=219.68$, and $l_{x+1}=8{,}000(0.91010)=7{,}280.80$. Their sum is $499.52+219.68+7{,}280.80=8{,}000.00$, as required.

Whole-number allocation. The floors are 499, 219, and 7,280, summing to 7,998. Two units remain to reach the radix of 8,000. The fractional parts are 0.52, 0.68, and 0.80, so the two remaining units go to the survivor count and cause 2. The integer counts are therefore 499, 220, and 7,281; their sum is exactly 8,000.

Check and trap. Rounding each expected count independently to the nearest integer gives 500, 220, and 7,281, which sum to 8,001. That is a plausible-looking but non-closing row. The largest-remainder allocation avoids the mistake of assuming that separately rounded counts must preserve the radix.

Worked Example: A Small Residual Consistent With Rounding

A hypothetical three-cause table displays probabilities to four decimal places: $\widehat{q}_x^{(1)}=0.1235$, $\widehat{q}_x^{(2)}=0.0765$, $\widehat{q}_x^{(3)}=0.0400$, and $\widehat{p}_x=0.7601$. Each entry is independently rounded to the nearest $0.0001$. For a radix of 10,000, assess closure, calculate the displayed expected counts, and give a consistent survivor count if the three cause probabilities are the authoritative inputs and survival is their complement.

Method and probability check. The displayed sum is $0.1235+0.0765+0.0400+0.7601=1.0001$, a residual of $0.0001$. Four entries are rounded, so the maximum rounding discrepancy is $4(0.0001)/2=0.0002$. The residual is within that bound and is compatible with rounding; this check alone does not establish that the inputs are otherwise correct.

Displayed counts. Multiplying the displayed entries by 10,000 gives 1,235, 765, 400, and 7,601. These total 10,001, one more than the radix. The same discrepancy appears in count form: $10{,}000(0.0001)=1$.

Complement and corrected count under the stated rule. Since the causes are authoritative, the displayed cause sum is $0.1235+0.0765+0.0400=0.2400$. The complementary survivor probability is $1-0.2400=0.7600$, giving $10{,}000(0.7600)=7{,}600$ survivors. The revised counts sum to $1{,}235+765+400+7{,}600=10{,}000$.

Check and trap. This correction is justified only because the problem specifies that survival is derived from the cause probabilities. If all four probabilities were independently authoritative values, replacing 0.7601 with 0.7600 without qualification would hide the reported residual rather than explain it.

Worked Example: A Residual Too Large to Attribute to Rounding

A hypothetical three-cause table reports $q_x^{(1)}=0.0812$, $q_x^{(2)}=0.0447$, $q_x^{(3)}=0.0196$, and $p_x=0.8620$, all to four decimal places. Treat them as independently rounded to the nearest $0.0001$. With radix 5,000, test closure and quantify the discrepancy. Explain why normalizing all four entries to sum to 1 is not an acceptable automatic repair.

Method and probability check. The cause probabilities sum to $0.0812+0.0447+0.0196=0.1455$. Including survival gives $0.1455+0.8620=1.0075$, a residual of $0.0075$. With four displayed entries, the maximum residual attributable to nearest rounding is $0.0002$. Since $0.0075>0.0002$, rounding cannot explain the discrepancy.

Count check. The displayed expected exit counts are $5{,}000(0.0812)=406$, $5{,}000(0.0447)=223.5$, and $5{,}000(0.0196)=98$. The displayed survivor count is $5{,}000(0.8620)=4{,}310$. Their total is $406+223.5+98+4{,}310=5{,}037.5$, which exceeds the radix by 37.5.

Conditional complement check. If, and only if, the cause probabilities are confirmed as authoritative and survival is meant to be their complement, the implied survivor probability is $1-0.1455=0.8545$, giving $5{,}000(0.8545)=4{,}272.5$ survivors. That is 37.5 fewer than the reported survivor count, quantifying the conflict.

Check and trap. Dividing every entry by 1.0075 would force the displayed values to sum to 1, but it would change all four reported probabilities without any model-based reason. The proper conclusion is that the table requires investigation, such as checking for a transcription error, omitted definition, or incompatible inputs. The complement is a diagnostic calculation, not a silent correction.

Interpreting Closure in Practice

Probability closure and count closure are related but should both be checked. Multiplying a probability residual by $l_x$ converts it to a count residual, so a small probability discrepancy can matter when the radix is large. Conversely, a count row may appear to miss the radix by one or two lives simply because fractional expected counts were rounded independently. Always state whether a table contains exact probabilities, rounded probabilities, expected counts, or integer counts.

When all entries are exact and the model assumptions hold, closure is an identity. When displayed probabilities are rounded, closure is a diagnostic with a precision-dependent tolerance. When whole-number counts are required, a declared allocation method can preserve the radix, but it cannot validate inconsistent input probabilities. The method should be applied to the unrounded expected counts where available, rather than to already rounded values.

This distinction is useful in decrement-table calculations because a small display residual can otherwise be mistaken for a model failure, while a material inconsistency can be made to disappear through arbitrary normalization. It is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus to identify the source and scale of the discrepancy before deciding how a table should be presented.

Key takeaway: Check exact closure in the model, bound the residual allowed by the stated rounding precision, and reconcile counts only under an explicit rule. Never use an adjustment to conceal a discrepancy larger than rounding can explain.

Check Your Understanding

For each problem, distinguish probability closure from any count-rounding issue. Treat stated decimal values as independently rounded only when the problem says so.

A hypothetical two-cause table has exact probabilities $q_x^{(1)}=0.14326$, $q_x^{(2)}=0.05674$, and $p_x=0.80000$, with radix 2,500. Find the expected counts and allocate whole-number counts by the largest-remainder method. (Answer: Expected counts are 358.15, 141.85, and 2,000.00; allocated counts are 358, 142, and 2,000, totaling 2,500.)
A three-cause table displays $0.0312$, $0.0187$, and $0.0501$ for the causes and $0.9001$ for survival, all rounded to the nearest $0.0001$. Find the residual, assess it against the rounding bound, and state the survivor probability implied if the causes are authoritative. (Answer: Displayed sum is 1.0001, residual $0.0001$, within the $0.0002$ bound; implied survivor probability is 0.9000.)
For the table in question 2, use radix 12,000 to find the displayed expected counts and the count discrepancy. (Answer: Cause counts are 374.4, 224.4, and 601.2; displayed survivors are 10,801.2; total is 12,001.2, exceeding the radix by 1.2. Using the cause-based complement gives 10,800 survivors.)
A hypothetical three-cause table has exact probabilities $0.075$, $0.025$, and $0.015$, with $p_x=0.885$, and radix 4,000. Check closure and calculate all four expected counts. (Answer: The probabilities sum to 1; expected cause counts are 300, 100, and 60, and the survivor count is 3,540, totaling 4,000.)
A two-cause table displays $q_x^{(1)}=0.1012$, $q_x^{(2)}=0.0589$, and $p_x=0.8410$, each independently rounded to the nearest $0.0001$. For radix 20,000, decide whether rounding explains the residual and quantify the count discrepancy. (Answer: The sum is 1.0011, a residual of 0.0011, exceeding the two-cause, three-entry bound of 0.00015; displayed counts total 20,022, exceeding the radix by 22. If the causes are authoritative, implied survival is 0.8399.)

What You'll Learn