Comparing Complete-Case Summaries Across Samples

Why Compare Complete-Case Summaries Carefully?

A sample with more censored losses may have fewer exact values available for direct summaries. It is tempting to compare the means or medians of the exact values and interpret a difference as a change in the underlying loss distribution. That interpretation can be wrong: the exact losses in each sample may represent different parts of the distribution because the censoring patterns differ.

A complete-case summary is still useful as a description of the exact observations that were recorded. Its limitation is about what it describes. It summarizes the exact-loss subset, not automatically the full set of losses or the population distribution. This tutorial develops that distinction using means, medians, and empirical distribution values.

Definition: Suppose a sample contains $n$ records, of which $m$ are exact loss observations $x_1,\ldots,x_m$. A complete-case empirical summary is calculated from those $m$ exact values only. The complete-case empirical CDF is $$ \hat F_{\mathrm{cc}}(x)=\frac{1}{m}\sum_{j=1}^{m} I(x_j\le x), $$ where $I(\cdot)$ equals 1 when its condition is true and 0 otherwise. Its complete-case mean is $\bar x_{\mathrm{cc}}=m^{-1}\sum_{j=1}^{m}x_j$. The denominator is the number of exact values, not the total number of records.

A right-censored record tells us that the loss exceeds a stated censoring amount; it does not give an exact loss. A complete-case calculation excludes that record from its exact-value list. This is different from claiming that the record is irrelevant: it contains information about the loss, but it is not an exact value that can be inserted into an ordinary sample mean or median.

For an even number of exact observations, this tutorial uses the conventional sample median: the average of the two middle ordered values. For an empirical exceedance summary at threshold $x$, “exceeds” means strictly greater than $x$, so the complete-case proportion is the number of exact values greater than $x$, divided by $m$.

How Censoring Changes the Exact-Case Subset

Consider a loss $X$ and a censoring amount $C$. Under right censoring, the record is exact when $X\le C$, and is censored when $X>C$; a censored record reports that the loss is greater than $C$, rather than reporting $X$. Even if censoring is independent of the loss, the chance of being an exact case can vary with the loss amount. Larger losses may be more likely to exceed the censoring amount and therefore be absent from the exact-value subset.

A fixed censoring limit makes the selection effect especially clear. Suppose every loss is observed exactly only when it is at most $c$. For a continuous loss distribution with $F_X(c)>0$, the distribution among exact cases is the conditional distribution of $X$ given $X\le c$:

P(X\le x\mid X\le c)= \begin{cases} \dfrac{F_X(x)}{F_X(c)}, & x\le c,\\[6pt] 1, & x>c. \end{cases}

The denominator $F_X(c)$ is needed because the exact cases are drawn from the part of the distribution at or below $c$. Thus, even when the underlying loss distribution has not changed, a sample with a lower censoring limit can have an exact-case distribution concentrated on smaller losses.

Why the Complete-Case Distribution Is Conditional

The fixed-limit formula follows directly from conditional probability. For $x\le c$, the event $X\le x$ implies $X\le c$. Therefore, the event that $X\le x$ and $X\le c$ is just $X\le x$, giving:

P(X\le x\mid X\le c) =\frac{P(X\le x,\ X\le c)}{P(X\le c)} =\frac{P(X\le x)}{P(X\le c)} =\frac{F_X(x)}{F_X(c)}.

This is the population counterpart of calculating an empirical CDF using only the exact cases. For random censoring, the exact-case distribution is also selected. If $C$ is independent of $X$, an exact observation at loss $x$ occurs when $C\ge x$. Consequently, the probability that a loss value $x$ enters the exact subset depends on $P(C\ge x)$. Independence of $C$ and $X$ does not, by itself, make the exact cases a random sample of all losses.

This reasoning supports an important interpretation rule: complete-case summaries answer, “What do the exact losses look like in this sample?” They do not necessarily answer, “What does the full loss distribution look like?” The latter requires methods that account for incomplete observations, rather than treating an exact-only summary as if it were automatically representative.

Worked Comparisons

Worked Example: Identical Losses, Different Censoring

Two hypothetical samples contain the same five underlying losses: $100, $200, $300, $400, and $500. In Sample A, all five are exact. In Sample B, the first three are exact, while the $400 loss is right-censored at $350 and the $500 loss is right-censored at $450. Compare the complete-case mean, median, and proportion exceeding $250.

Method. Calculate each summary from exact losses only, using the number of exact losses as the denominator. The censored records in Sample B are not entered as $350 or $450.

Sample A. There are $m_A=5$ exact values. Their sum is $100+200+300+400+500=1{,}500$, so the complete-case mean is $1{,}500/5=\$300$. The ordered middle value is $300, so the median is $300. The exact values greater than $250 are $300, $400, and $500. The complete-case exceedance proportion is $3/5=0.6000$.

Sample B. The exact values are $100, $200, and $300, so $m_B=3$. Their sum is $100+200+300=600$, and the complete-case mean is $600/3=\$200$. The median is the middle value, $200. Only one exact value, $300, exceeds $250, so the complete-case exceedance proportion is $1/3=0.3333$.

Comparison and check. The complete-case summaries are lower in Sample B even though the underlying losses were stipulated to be identical. The change is due to which values are exact: the two largest losses were censored in Sample B. The calculations also check against the ordered exact values and their counts.

Common mistake avoided: Replacing the censored losses with $350 and $450 would treat lower bounds as exact values. Those amounts are censoring thresholds, not the losses themselves, so that substitution would give an unsupported mean and median.

Worked Example: Comparing CDF Values with Unequal Censoring

Two hypothetical portfolios have different exact-case counts. Portfolio R has exact losses $70, $110, $150, $190, and $310, plus two right-censored records. Portfolio S has exact losses $90, $130, and $170, plus three right-censored records. Calculate each complete-case mean, median, and empirical CDF value at $160. State what the comparison does and does not establish.

Method. For each portfolio, order and summarize its exact values. Calculate the CDF at $160 as the number of exact values at or below $160 divided by the exact-case count.

Portfolio R. The exact-case count is $m_R=5$, and the sum is $70+110+150+190+310=830$. Its complete-case mean is $830/5=\$166$. The middle ordered value is $150, so the median is $150. Three values, $70, $110, and $150, are at or below $160. Hence $\hat F_{\mathrm{cc},R}(160)=3/5=0.6000$.

Portfolio S. The exact-case count is $m_S=3$, and the sum is $90+130+170=390$. Its complete-case mean is $390/3=\$130$. The median is $130. Two values, $90 and $130, are at or below $160. Hence $\hat F_{\mathrm{cc},S}(160)=2/3=0.6667$.

Comparison and check. The exact-case mean is $36 higher in R ($166-130=36$), while the complete-case CDF at $160 is $0.0667$ higher in S ($0.6667-0.6000=0.0667$, to four decimals). These describe the observed exact subsets. Because the portfolios have different censoring patterns and exact-case counts, the results alone do not establish that the underlying loss distributions differ.

Common mistake avoided: Dividing by all records would use denominators 7 and 6, even though the requested empirical CDFs are based on exact values. The correct denominators are 5 and 3, respectively.

Worked Example: Pooling Complete Cases Across Samples

Two hypothetical groups have the following records. Group A has exact losses $80, $120, $200, and $400, plus two censored records. Group B has exact losses $100 and $500, plus four censored records. Calculate each group’s complete-case mean and median, then calculate the pooled complete-case mean and median. Compare the pooled mean with the unweighted average of the two group means.

Method. First summarize each group’s exact values. For the pooled exact-case mean, add the exact losses across both groups and divide by the total number of exact values. The groups do not have equal exact-case counts.

Group A. There are four exact values, with sum $80+120+200+400=800$. Its mean is $800/4=\$200$. The two middle ordered values are $120 and $200, so its median is $(120+200)/2=\$160$.

Group B. There are two exact values, with sum $100+500=600$. Its mean is $600/2=\$300$. The median is $(100+500)/2=\$300$.

Pooled mean. Across the groups, there are $4+2=6$ exact values with total $800+600=1{,}400$. Thus, the pooled complete-case mean is $1{,}400/6=\$233.33$, rounded to the nearest cent.

Pooled median. The ordered pooled exact values are $80, $100, $120, $200, $400, and $500. The two middle values are $120 and $200, so the pooled median is $(120+200)/2=\$160$.

The unweighted average of the group means is $(200+300)/2=\$250$, not the pooled mean. The pooled mean weights each group by its number of exact observations: $(4\times200+2\times300)/6=1{,}400/6=\$233.33$. The calculation checks because the pooled exact-value count is six and the pooled total is $1,400.

Common mistake avoided: Averaging group means equally ignores the unequal numbers of exact observations. The pooled mean must use the combined exact-value total divided by the combined exact-value count.

Interpreting Differences and Avoiding Traps

A complete-case comparison is informative as a comparison of exact observations, provided the summary and denominator are stated clearly. It is not automatically a fair comparison of full loss distributions when one sample has more censoring, a different censoring limit, or a different pattern of which losses become exact.

In particular, “the censoring mechanism is independent of the loss” is not the same as “the exact cases are an unbiased sample of all losses.” With right censoring, a loss must be no greater than its censoring amount to appear as an exact value. Larger losses can therefore be less likely to enter the complete-case subset even when the censoring amount itself is independent of loss severity.

Use these checks when interpreting complete-case summaries:

State the subset. Name the number of exact observations used in each summary.
Use the correct denominator. A complete-case mean, median, or empirical proportion uses exact values only.
Do not invent exact values. A censoring threshold supplies a bound, not the censored loss amount.
Separate description from inference. A difference between exact-case summaries is a descriptive difference; by itself, it does not prove a change in the underlying loss distribution.
Check the censoring patterns. Differences in how many values are censored or where censoring occurs may change the composition of the exact subsets.

These distinctions are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. The exam skill is not merely calculating a mean or CDF value, but also identifying precisely which observations support it and limiting the conclusion to what those observations justify.

Key takeaway: A complete-case summary describes the exact losses that were observed. When censoring differs across samples, the exact-case subsets may differ systematically, so their summary differences need not indicate a change in the underlying loss distribution.

Check Your Understanding

For each problem, use exact values only for complete-case summaries. Report probabilities to four decimals and money to the nearest cent where needed.

A sample has exact losses $40, $80, $120, and $200, plus one right-censored record. Find the complete-case mean, median, and proportion of exact losses strictly greater than $100. (Answer: Mean $110; median $100; exceedance proportion $2/4=0.5000$.)
Sample A has exact losses $60, $100, and $180, plus three censored records. Sample B has exact losses $90 and $150, plus one censored record. Find each complete-case mean and the difference between the means. (Answer: A mean $113.33; B mean $120; B exceeds A by $6.67.)
A group has exact losses $50, $90, $130, and $210, plus two censored records. Find its complete-case empirical CDF at $100 and its complete-case median. (Answer: $\hat F_{\mathrm{cc}}(100)=2/4=0.5000$; median $110.)
Two samples have identical underlying losses $100, $200, $300, and $400. In Sample P all are exact; in Sample Q only $100 and $200 are exact, while the other two are right-censored at $250 and $350. Find both complete-case means and explain the difference. (Answer: P mean $250; Q mean $150; the difference can arise from censoring selection despite identical underlying losses.)
Group M has exact losses $100 and $300; Group N has exact losses $80, $120, $200, and $400. Find the pooled complete-case mean and compare it with the unweighted average of the group means. (Answer: Pooled mean $1{,}200/6=\$200$; unweighted average of group means $(200+200)/2=\$200$.)

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