Empirical Survival Estimates From Complete Loss Records

Why Estimate a Loss Survival Function?

An insurer may want to know what fraction of losses exceed a retention, a claim-review threshold, or a chosen planning amount. When the available data consist of complete, exact loss amounts, this question can be answered directly: count the losses above the threshold and divide by the number of observed losses. Repeating that calculation across thresholds gives an empirical picture of the loss tail without requiring a parametric severity model.

The key is to be precise about what “exceed” means. A loss exactly equal to the threshold is not counted when the question asks for a strict exceedance. This matters particularly when the records contain repeated values, since ties can make strict and inclusive estimates noticeably different.

Definition: Let $X_1,\ldots,X_n$ be $n$ complete, exact loss observations. For a threshold $t$, the empirical survival estimate for strict exceedance is $$ \hat S_X(t)=\frac{1}{n}\sum_{i=1}^{n} I(X_i>t) =\frac{\#\{i:X_i>t\}}{n}, $$ where $I(X_i>t)$ equals 1 when $X_i>t$, and 0 otherwise. The estimate is the observed proportion of losses strictly greater than $t$. For an inclusive threshold, $X_i\ge t$, count losses equal to $t$ as well.

This construction assumes that each of the $n$ records represents one loss and that each record is observed exactly, without censoring or truncation. If some losses are only known to exceed a reporting limit, they are not complete exact observations and cannot be counted as if their exact amounts were known. The earlier material on grouped loss data addresses records that are reported in intervals; the calculations here use individual exact amounts.

The empirical estimate is a sample proportion. It describes the recorded data exactly, but its interpretation as an estimate for a broader population depends on how those records were collected. If the observations are a representative sample of independent losses from a common population, the observed proportion is used to estimate that population’s strict exceedance probability. If the records are a selected set, such as only claims referred for special review, the proportion describes that selected set and need not represent all losses.

Constructing the Estimate

For any specified threshold, apply the same sequence: identify the comparison operator, count qualifying observations, and divide by the total number of complete records. When several thresholds are requested, use the same denominator each time unless the problem explicitly gives different populations or different record sets.

Confirm the data.
Check that each listed amount is one complete, exact loss and identify the number $n$ of records.

Read the threshold literally.
For “exceeds $t$,” use $X_i>t$. For “at least $t$,” use $X_i\ge t$.

Count and divide.
Count the records satisfying the chosen comparison, then divide by $n$.

Check the result.
The estimate must lie between 0 and 1, and its denominator must be the number of records in the relevant sample.

As the threshold increases, fewer observations can exceed it, so the empirical estimate cannot increase. It is a step function: the count changes only when the threshold passes a recorded loss amount. At a threshold equal to a tied loss value, all observations tied at that value are excluded from the strict exceedance count. The size of that drop is the number tied there divided by $n$.

Worked Example: Strict Exceedance With Tied Losses

A hypothetical insurer has eight complete exact loss records, in dollars: $120,\ 180,\ 180,\ 240,\ 310,\ 310,\ 450,\ 600$. Estimate the proportions of losses strictly exceeding $180$, $300$, and $310$ dollars. Also calculate the proportion at least $310$ dollars.

Method. For each threshold, count the listed values satisfying the requested comparison and divide by $n=8$.

For $t=180$, the losses strictly greater than $180$ are $240,\ 310,\ 310,\ 450,\ 600$, giving 5 exceedances: $$ \hat S_X(180)=\frac{5}{8}=0.6250. $$

For $t=300$, the losses strictly greater than $300$ are $310,\ 310,\ 450,\ 600$, giving 4 exceedances: $$ \hat S_X(300)=\frac{4}{8}=0.5000. $$

For $t=310$, the two losses equal to $310$ do not qualify. Only $450$ and $600$ are strictly greater, giving 2 exceedances: $$ \hat S_X(310)=\frac{2}{8}=0.2500. $$

For the inclusive comparison $X_i\ge310$, the two values equal to $310$ are included along with $450$ and $600$. Thus the count is 4: $$ \frac{\#\{i:X_i\ge310\}}{8}=\frac{4}{8}=0.5000. $$

Check. The strict estimates decrease from $0.6250$ to $0.5000$ to $0.2500$ as the threshold rises. The inclusive estimate at $310$ is larger than the strict estimate by $2/8=0.2500$, exactly the fraction of records tied at $310$. The common error is to include those ties when the question asks for losses that exceed $310$.

Why the Count Is the Appropriate Estimate

The empirical estimate has a direct justification. At a fixed threshold $t$, define $I_i(t)=I(X_i>t)$ for each observed loss. Each indicator records whether that particular loss exceeded the threshold. Adding the indicators therefore counts the exceedances, and dividing by $n$ gives their observed proportion.

If the records are also treated as independent observations from a common loss distribution, each indicator has expected value equal to the population probability of strict exceedance. The expected value of their average is therefore that same probability:

E\left[\frac{1}{n}\sum_{i=1}^{n} I(X_i>t)\right] =\frac{1}{n}\sum_{i=1}^{n}E[I(X_i>t)] =\frac{1}{n}\sum_{i=1}^{n}P(X_i>t) =P(X>t).

The last equality uses the common-distribution assumption. This shows why the sample proportion is a natural estimator of the population exceedance probability under that sampling model. It does not mean every sample proportion equals the population probability, nor does it make an unrepresentative record set representative. The calculation itself remains a description of the observed records.

Worked Example: Several Thresholds From One Sample

A hypothetical administrator reviews ten complete exact loss records, in thousands of dollars: $640,\ 800,\ 800,\ 950,\ 1{,}100,\ 1{,}250,\ 1{,}250,\ 1{,}400,\ 1{,}700,\ 2{,}100$. Estimate the strict exceedance proportions at $1{,}000$, $1{,}250$, and $800$ thousand dollars.

Method. The sample size is $n=10$. Each requested estimate uses that denominator, and equality to a threshold does not count as strict exceedance.

Above $1{,}000$, the values are $1{,}100,\ 1{,}250,\ 1{,}250,\ 1{,}400,\ 1{,}700,\ 2{,}100$. There are 6: $$ \hat S_X(1{,}000)=\frac{6}{10}=0.6000. $$

Above $1{,}250$, the values are $1{,}400,\ 1{,}700,\ 2{,}100$. The two records equal to $1{,}250$ are excluded, so there are 3: $$ \hat S_X(1{,}250)=\frac{3}{10}=0.3000. $$

Above $800$, exclude both records equal to $800$. The remaining qualifying values are $950,\ 1{,}100,\ 1{,}250,\ 1{,}250,\ 1{,}400,\ 1{,}700,\ 2{,}100$, so there are 7: $$ \hat S_X(800)=\frac{7}{10}=0.7000. $$

Check. The estimates are $0.7000$, $0.6000$, and $0.3000$ as the thresholds increase from $800$ to $1{,}000$ to $1{,}250$. The common mistake is to count the two $1{,}250$ records as exceeding $1{,}250$; doing so would produce $5/10$, which answers an inclusive rather than a strict question.

Pooling Complete Records Correctly

When records come from groups of different sizes, the combined estimate must give each loss record equal weight if the target is the proportion among all pooled records. In that case, add the group exceedance counts and divide by the total number of records. Averaging group proportions without regard to group size gives each group equal weight instead, which generally describes a different quantity.

Worked Example: Combining Two Record Groups

A hypothetical review has two groups of complete exact losses, all in dollars. Group A has four records: $90,\ 130,\ 220,\ 410$. Group B has eight records: $150,\ 210,\ 260,\ 300,\ 520,\ 610,\ 750,\ 900$. Estimate the strict proportion exceeding $200$ dollars among all twelve records. Compare it with the unweighted average of the two group proportions.

Method. Count exceedances in each group, then pool the counts and sample sizes.

In Group A, the values exceeding $200$ are $220$ and $410$. The group proportion is $2/4=0.5000$. In Group B, the values exceeding $200$ are $210,\ 260,\ 300,\ 520,\ 610,\ 750,\ 900$. The group proportion is $7/8=0.8750$.

Across both groups, there are $2+7=9$ exceedances among $4+8=12$ records: $$ \hat S_X(200)=\frac{9}{12}=0.7500. $$

The unweighted average of the group proportions is $$ \frac{0.5000+0.8750}{2}=0.6875. $$ This is not the pooled-record estimate: it gives the four-record group and the eight-record group equal influence. The pooled estimate weights each group proportion by its number of records: $$ \frac{4(0.5000)+8(0.8750)}{4+8} =\frac{2+7}{12} =0.7500. $$

Check. The pooled value is closer to Group B’s proportion because Group B contains more records. The common mistake is to average $0.5000$ and $0.8750$ when the question asks for the fraction among all twelve losses.

Common Exam Traps and Interpretation

Replacing “exceeds” with “at least.” For strict exceedance, use $X_i>t$; observations equal to $t$ are excluded. For “at least,” use $X_i\ge t$.
Using the wrong denominator. Divide by all complete records in the target sample, not only by the records above the threshold.
Ignoring ties. Count every qualifying record, including repeated amounts, but apply the strict or inclusive comparison to each one.
Averaging subgroup proportions equally. For a pooled record-level proportion, combine counts and sample sizes. Equal averaging answers a group-weighted question, not generally a record-weighted one.
Calling a selected-record proportion a population estimate without qualification. The empirical count is always descriptive of the records in hand; broader interpretation requires an appropriate sampling basis.
Treating incomplete observations as exact. A record that only establishes a loss is above a reporting limit does not specify its exact amount, so it cannot be classified at every higher threshold as though its value were known.

Key takeaway: For complete exact losses, the empirical strict survival estimate at $t$ is the number of recorded losses greater than $t$, divided by the number of records in the relevant sample. Read the comparison carefully, count ties accordingly, and pool groups by combining their counts.

Check Your Understanding

For each problem, use complete exact losses and strict exceedance unless the wording explicitly says otherwise.

A sample of five losses is $40,\ 60,\ 60,\ 90,\ 120$ dollars. Estimate the proportion strictly exceeding $60$ dollars. (Answer: $2/5=0.4000$)
Eight losses, in dollars, are $100,\ 200,\ 200,\ 250,\ 310,\ 400,\ 400,\ 700$. Find the strict exceedance estimate at $200$ dollars and the inclusive estimate at $200$ dollars. (Answer: Strict $5/8=0.6250$; inclusive $7/8=0.8750$)
A record set contains 20 complete exact losses, of which 7 exceed a specified threshold. What is the empirical strict exceedance estimate? (Answer: $7/20=0.3500$)
Group A has six records, with 3 exceeding $500$ dollars. Group B has four records, with 1 exceeding $500$ dollars. Find the pooled strict exceedance estimate and the unweighted average of the group estimates. (Answer: Pooled $4/10=0.4000$; unweighted average $(3/6+1/4)/2=0.3750$)
For the losses $150,\ 250,\ 250,\ 500,\ 800,\ 900$, find the strict exceedance estimate at $250$ dollars and explain why the two records equal to $250$ are not counted. (Answer: $3/6=0.5000$; strict exceedance requires $X_i>250$)

What You'll Learn