Empirical Distributions of Aggregate Loss

From Period Totals to a Distribution

An aggregate-loss record summarizes all covered losses in each observation period. The record might contain annual totals, quarterly totals, or totals for another consistently defined period. Once the period totals are collected, they can be treated as observations of an aggregate-loss random variable. Their empirical distribution describes how often different total-loss outcomes appeared in the recorded periods.

This tutorial focuses on reading probabilities, quantiles, and exceedance counts from that distribution. It does not require a frequency-severity model: each observation is already a total for one period. The essential preparation is to use periods defined on a comparable basis and to retain periods with no losses. Omitting a zero-loss period changes the observed proportions and therefore changes the empirical distribution.

Definition: Let \(X_1,\ldots,X_N\) be the observed aggregate losses in \(N\) comparable periods. The empirical distribution assigns probability \(1/N\) to each recorded period total, counting repeated values with their full multiplicity. Its empirical cumulative distribution function (CDF) is

\hat{F}_N(x)=\frac{1}{N}\sum_{j=1}^{N}\mathbf{1}(X_j\le x),

where \(\mathbf{1}(X_j\le x)\) equals 1 when \(X_j\le x\), and 0 otherwise.

Thus, \(\hat{F}_N(x)\) is the proportion of observed period totals that are no greater than \(x\). The empirical probability of exceeding \(x\) strictly is \(1-\hat{F}_N(x)\). For the inclusive event \(X\ge x\), the count instead includes totals equal to \(x\). This distinction matters when observed totals are repeated or when the threshold itself appears in the data.

\widehat{P}(X>x)=\frac{\#\{j:X_j>x\}}{N}=1-\hat{F}_N(x), \qquad \widehat{P}(X\ge x)=\frac{\#\{j:X_j\ge x\}}{N}.

These are empirical proportions, not probabilities derived from a fitted parametric model. Each observed period contributes one equally weighted observation. If all periods are annual, for example, the calculation is a proportion of observed annual totals; it is not automatically a forecast of the probability for a future year. A future-loss interpretation requires assumptions about how the observed periods relate to future experience.

Reading the Empirical CDF

The empirical CDF is a step function. Between observed values it remains constant; at an observed value it jumps by the fraction of records equal to that value. If a total occurs three times in a data set of 12 periods, the CDF jumps by \(3/12\) at that total. Sorting the observations makes these jumps and the corresponding threshold counts easier to see.

Check the observation unit.
Confirm that each entry is one aggregate total for a period defined consistently with the others. Keep zero-loss periods in the record.

Sort and count.
Order the \(N\) totals, retaining duplicates. For a threshold, count the entries satisfying the exact inequality in the question.

Divide by the number of periods.
The empirical probability is the qualifying count divided by \(N\), not by the number of positive-loss periods or the number of distinct totals.

The denominator is the number of observed periods, including periods with zero aggregate loss. Repeated values are not collapsed into one observation: they represent multiple periods with that outcome. These rules follow directly from assigning equal probability \(1/N\) to each period record.

Worked Example: A Threshold Probability and an Empirical Quantile

A hypothetical insurer records the following eight annual aggregate losses, in thousands of dollars, already sorted: \(0, 12, 18, 18, 25, 41, 56, 90\). Find the empirical probability that annual aggregate loss is at most 25 thousand dollars, the number of observed years with loss above that threshold, and the empirical 75th percentile.

Method. Use the observed annual totals as equally weighted outcomes. There are \(N=8\) periods. The totals no greater than 25 are \(0,12,18,18,25\), so the qualifying count is 5.

\hat{F}_8(25)=\frac{5}{8}=0.6250.

The totals strictly above 25 are \(41,56,90\). Their count is \(8-5=3\), and the empirical strict-exceedance probability is \(3/8=0.3750\). The requested exceedance count is therefore 3 years.

For the quantile, use the inverse empirical CDF convention: the empirical \(p\)-quantile is the smallest value at which the empirical CDF is at least \(p\). For \(p=0.75\), the order-statistic index is \(\lceil 8(0.75)\rceil=\lceil6\rceil=6\). The sixth sorted observation is 41, so the empirical 75th percentile is 41 thousand dollars.

Check. At 41, six of the eight observations are no greater than 41, giving \(6/8=0.75\); below 41, only five are, giving \(5/8=0.625\). Thus 41 is the first observed value reaching the 75th percentile. The method avoids interpolating between totals when using the stated inverse-CDF convention.

Empirical Quantiles and Ties

A quantile is a threshold that locates a specified fraction of the empirical distribution at or below it. Because an empirical CDF is a step function, a quantile need not divide the observations into exactly the requested percentage below and above. A jump can pass over the target proportion, especially in a small data set or when many totals are tied.

Definition: For \(0<p\le1\), define the empirical \(p\)-quantile by

\hat{Q}_p=\inf\{x:\hat{F}_N(x)\ge p\}.

If the ordered observations are \(X_{(1)}\le\cdots\le X_{(N)}\), this convention gives \(\hat{Q}_p=X_{(\lceil Np\rceil)}\).

Here \(\lceil z\rceil\) means the smallest integer greater than or equal to \(z\). The quantile convention must be understood before calculating: other contexts may use interpolation rules, but the inverse empirical CDF convention above selects an observed value. Repeated values remain repeated in the ordered list, so they occupy multiple order-statistic positions.

The formula can be justified directly. At \(X_{(k)}\), at least the first \(k\) observations are no greater than that value, so \(\hat{F}_N(X_{(k)})\ge k/N\). Taking \(k=\lceil Np\rceil\) ensures \(k/N\ge p\). For any value strictly below \(X_{(k)}\), at most \(k-1\) observations can be no greater than it, and \((k-1)/N<p\). Therefore \(X_{(k)}\) is the smallest value at which the empirical CDF reaches \(p\).

Worked Example: Strict Versus Inclusive Exceedance

A hypothetical equipment pool has ten recorded annual aggregate losses, in dollars: \(0,0,8,14,14,20,31,31,47,65\). Find the number and empirical probability of years with loss at least 31 dollars, the number and probability with loss strictly above 31 dollars, and the empirical 60th percentile.

Method. Count each annual record once, including both zero-loss years and both years with a total of 31 dollars. The totals at least 31 are \(31,31,47,65\), so the inclusive count is 4.

\widehat{P}(X\ge31)=\frac{4}{10}=0.4000.

For the strict event \(X>31\), the two observations equal to 31 do not qualify. Only 47 and 65 qualify, so the count is 2 and the empirical probability is \(2/10=0.2000\).

For the 60th percentile, \(k=\lceil10(0.60)\rceil=6\). The sixth ordered observation is 20, so \(\hat{Q}_{0.60}=20\) dollars. Indeed, six observations are at most 20, while only five are below 20.

Check. The inclusive and strict exceedance counts differ by the two records equal to 31. The probabilities, 0.4000 and 0.2000, use all ten years as the denominator. The common error is to treat “at least 31” as “above 31” and discard the tied observations.

Counts, Proportions, and Units

A probability and an exceedance count answer related but different questions. If a threshold is exceeded in \(r\) of \(N\) observed periods, the empirical probability is \(r/N\), while the observed exceedance count is \(r\). For example, 3 exceedances in 12 years correspond to an empirical proportion of \(3/12=0.25\), not to three exceedances per year. State the unit and the inequality so the result cannot be misread.

Worked Example: Quantile and Threshold Count in Thousands

A hypothetical benefit program has these 12 annual aggregate losses, in thousands of dollars, sorted from smallest to largest: \(4,9,9,16,22,22,35,48,48,73,101,140\). Find the empirical 80th percentile, the number and proportion of years with losses strictly above 50 thousand dollars, and the number of years with losses at least 48 thousand dollars.

Method. There are \(N=12\) annual observations. For the 80th percentile, the index is \(k=\lceil12(0.80)\rceil=\lceil9.6\rceil=10\). The tenth ordered total is 73, so the empirical 80th percentile is 73 thousand dollars.

The totals strictly above 50 are \(73,101,140\). There are 3 such years, giving empirical probability \(3/12=0.2500\). The totals at least 48 are \(48,48,73,101,140\), so the inclusive count is 5 years.

Check. At 73, ten of the 12 observations are no greater than the threshold, and \(10/12\ge0.80\); at any smaller value below 73, at most nine are no greater. The threshold count above 50 is three, while the count at least 48 includes both observations tied at 48. All loss amounts remain in thousands of dollars, as specified.

Common Mistakes and Interpretation

Changing the denominator. Use all \(N\) observed periods, not only periods with positive losses. Removing zero-loss periods inflates the empirical probabilities of positive-loss outcomes.
Ignoring ties. For \(X>x\), values equal to \(x\) are excluded. For \(X\ge x\), they are included. Read the inequality before counting.
Collapsing repeated totals. A value observed in three periods contributes three records, not one distinct value.
Using the wrong quantile convention. Under the inverse empirical CDF definition, select \(X_{(\lceil Np\rceil)}\); do not interpolate unless the problem explicitly specifies another convention.
Overstating what the data establish. The empirical distribution summarizes the recorded periods. By itself, it does not establish that future periods have the same distribution or that the sample proportion is a guaranteed future probability.

The empirical distribution is useful because it gives a transparent, model-free summary of observed aggregate outcomes. Its threshold probabilities and quantiles can be checked directly against the period records, making it important to preserve the observation unit, all periods, and the exact threshold convention. This treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

Use the inverse empirical CDF quantile convention defined above. Keep each stated unit and inequality unchanged.

A hypothetical group records six annual aggregate losses, in thousands of dollars: \(0,5,5,12,20,33\). Find the empirical probability of a loss at most 12 and the number of years with a loss strictly above 12. (Answer: \(4/6=0.6667\); 2 years.)
Eight sorted annual totals are \(3,7,11,11,19,24,24,40\), in dollars. Find the count and probability of a total at least 24, the count and probability strictly above 24, and the empirical median. (Answer: At least 24: 3 years and \(3/8=0.3750\); strictly above 24: 1 year and \(1/8=0.1250\); median \(=11\) dollars.)
Ten annual totals, in thousands of dollars, are \(0,0,6,10,15,15,27,39,52,80\). Find the number and probability of totals at least 15 and the empirical 70th percentile. (Answer: 6 years, \(6/10=0.6000\); 70th percentile \(=27\) thousand dollars.)
Twelve sorted annual totals, in thousands of dollars, are \(10,10,20,25,30,30,45,60,75,100,100,160\). Find the empirical 90th percentile and the count and probability of totals strictly above 100. (Answer: 90th percentile \(=100\) thousand dollars; 1 year, \(1/12=0.0833\).)
Nine sorted annual totals, in thousands of dollars, are \(0,4,8,8,13,21,34,34,70\). Find the number and probability of totals at least 20, the number strictly above 34, and the empirical 75th percentile. (Answer: At least 20: 4 years and \(4/9=0.4444\); strictly above 34: 1 year; 75th percentile \(=34\) thousand dollars.)

What You'll Learn