Empirical Loss Distributions from Observed Data

From Observed Losses to a Distribution

A loss model describes uncertainty about possible losses. When a set of losses has already been observed, an empirical distribution provides a direct way to summarize those observations without first selecting a parametric model such as a lognormal or Pareto distribution. Each observed loss contributes to the distribution, so repeated values receive proportionally more weight.

This approach is useful for describing a portfolio’s recorded experience, estimating its average loss, and answering questions such as “What fraction of observations exceeded $50,000?” The result is a description of the sample. Treating it as an estimate of future experience requires additional care: the observations should represent the risk being studied, refer to comparable periods and loss definitions, and be sufficiently independent and identically distributed for the intended inference.

Definition: Given $n$ complete observed losses $X_1,\ldots,X_n$, the empirical probability assigned to a loss value is its number of occurrences divided by $n$. The empirical cumulative distribution function is $F_n(x)=\frac{1}{n}\sum_{j=1}^{n} I(X_j\le x)$, where $I(\cdot)$ equals 1 when its condition is true and 0 otherwise. The empirical strict survival function is $S_n(x)=\frac{1}{n}\sum_{j=1}^{n} I(X_j>x)=1-F_n(x)$.

The empirical distribution puts probability only on observed loss values. If a value appears several times, its probability is its frequency divided by the sample size. The cumulative distribution includes observations equal to $x$, while the strict survival function counts only observations greater than $x$. In particular, $S_n(x)$ is not the proportion of losses greater than or equal to $x$ when some observations equal $x$.

These definitions presume that the observations are complete and measured on a consistent basis. For example, annual losses should all cover the same kind of risk and the same length of time. Adjustments for censoring, truncation, or grouping require methods beyond the simple empirical distribution.

Building the Empirical Distribution

A reliable calculation starts by fixing the sample and its unit, sorting the observed values, and counting how often each distinct value occurs. The empirical probability at a value is its count divided by the total number of observations. Add these probabilities cumulatively to obtain $F_n(x)$. Subtract each cumulative value from 1 to obtain the strict survival probability at the same point.

Worked Example: Constructing a Distribution With Tied Losses

A hypothetical insurer reviews eight complete claim payments, in dollars: 0, 200, 200, 500, 900, 900, 1,500, and 3,000. Construct the empirical probability distribution, find $F_n(900)$ and $S_n(900)$, and calculate the sample mean. Use the frequency method, and keep all amounts in dollars.

Step 1: Count the observations. There are $n=8$ payments. The distinct values and their counts are 0 (1), 200 (2), 500 (1), 900 (2), 1,500 (1), and 3,000 (1).

Step 2: Divide each count by 8. The empirical probabilities at those values are, respectively, $1/8, 2/8, 1/8, 2/8, 1/8,$ and $1/8$. They sum to $8/8=1$, as required for a probability distribution.

Loss $x$	Count	Empirical probability	$F_n(x)$	$S_n(x)$
0	1	0.1250	0.1250	0.8750
200	2	0.2500	0.3750	0.6250
500	1	0.1250	0.5000	0.5000
900	2	0.2500	0.7500	0.2500
1,500	1	0.1250	0.8750	0.1250
3,000	1	0.1250	1.0000	0.0000

At $900, six observations are at or below the threshold, so $F_n(900)=6/8=0.7500$. Two observations are strictly greater than $900, so $S_n(900)=2/8=0.2500$. The sample mean is the sum of all eight payments divided by eight:

\bar{X}=\frac{0+200+200+500+900+900+1{,}500+3{,}000}{8} =\frac{7{,}200}{8}=\$900.

Check. The probabilities total 1, and $F_n(900)+S_n(900)=0.7500+0.2500=1$. The calculation uses all eight observations, including both payments of $200 and both of $900; counting each distinct value only once would give a different and incorrect distribution.

Empirical Means, Quantiles, and Exceedance Rates

The sample mean is the empirical estimate of the expected loss when each observation represents one comparable exposure or period. It is an average of the observed amounts, not a claim about the largest loss or the variability of future losses. The same calculation can be written using distinct values and their counts, which is often convenient when data contain ties.

\bar{X}=\frac{1}{n}\sum_{j=1}^{n}X_j =\sum_k x_k\frac{m_k}{n},

Here $x_k$ is a distinct observed loss and $m_k$ is its frequency. The equality holds because adding $x_k$ once for each of its $m_k$ occurrences contributes $m_kx_k$ to the total sum. Dividing the total by $n$ gives the frequency-weighted expression.

An empirical quantile is read from the cumulative distribution. For this tutorial, use the inverse empirical cumulative distribution convention: for $0<p\le1$, the empirical $p$-quantile is the smallest observed value $x$ for which $F_n(x)\ge p$. Equivalently, sort the $n$ observations and select the observation at rank $\lceil np\rceil$, where $\lceil\cdot\rceil$ means round upward to the next integer. Other sample-quantile conventions can interpolate between observations; they need not give the same result.

An empirical exceedance rate is a count divided by the sample size. A strict exceedance rate at threshold $u$ is $S_n(u)$, the fraction with $X_j>u$. An inclusive exceedance rate counts $X_j\ge u$. These differ whenever losses equal the threshold, so state which event is being measured before counting.

Worked Example: Estimating a Quantile and a Strict Exceedance Rate

A hypothetical risk analyst has nine complete, comparable loss observations, in thousands of dollars: 50, 100, 100, 250, 400, 700, 900, 1,200, and 2,300. Find the empirical 50th and 80th percentiles and the strict exceedance rate above $700,000. Use the inverse empirical cumulative distribution convention.

Step 1: Confirm the sample size and ranks. The observations are already in ascending order and $n=9$. The rank for the 50th percentile is $\lceil9(0.50)\rceil=\lceil4.5\rceil=5$. The fifth observation is 400 thousand dollars, or $400,000.

Step 2: Find the 80th percentile. Its rank is $\lceil9(0.80)\rceil=\lceil7.2\rceil=8$. The eighth observation is 1,200 thousand dollars, or $1,200,000. Equivalently, seven observations are below this value and $F_n(1{,}200)=8/9$, which is at least 0.80.

Step 3: Count strict exceedances. The losses strictly greater than $700,000 are $900,000, $1,200,000, and $2,300,000. Thus the strict exceedance rate is $3/9=0.3333$, or 33.33% to two decimal places.

Check. There are three values above $700,000 among nine observations, and the selected percentile observations are at the stated ranks. The method avoids treating a threshold observation as an exceedance when the question says “greater than.”

Why the Empirical Mean Estimates an Expected Loss

The empirical mean is not merely a convenient summary. It is the expected value under the distribution that assigns probability $1/n$ to each observation. If the observed values are $X_1,\ldots,X_n$, the expected value under that empirical distribution is:

E_n[X]=\sum_{j=1}^{n}X_j\frac{1}{n} =\frac{1}{n}\sum_{j=1}^{n}X_j=\bar{X}.

The first equality follows directly from the definition of expectation for a discrete distribution: multiply each possible value by its probability and sum. Grouping repeated values recovers the frequency formula above. If the observations are a representative independent sample from a common loss distribution with finite mean, the sample mean is also an unbiased estimator of that distribution’s expected loss: each $X_j$ has expectation $E[X]$, so linearity gives $E[\bar{X}]=\frac{1}{n}\sum_j E[X_j]=E[X]$. This sampling interpretation does not hold automatically if the observations are selectively recorded or represent different risk periods.

Worked Example: Including Zero-Loss Periods

A hypothetical portfolio records ten complete annual net-loss amounts, in thousands of dollars: 0, 0, 0, 20, 20, 40, 40, 40, 80, and 100. Calculate the empirical expected annual loss, the empirical probability of a loss of at least $40,000, and the empirical 90th percentile. Use the inverse empirical cumulative distribution convention.

Step 1: Calculate the mean. The total is $3(0)+2(20)+3(40)+80+100=0+40+120+80+100=340$ thousand dollars. Therefore, the empirical mean is $340/10=34$ thousand dollars, or $34,000.

Step 2: Count the inclusive threshold event. Losses of at least $40,000 are 40, 40, 40, 80, and 100 thousand dollars. There are five such years, so the empirical probability is $5/10=0.5000$.

Step 3: Find the 90th percentile. The rank is $\lceil10(0.90)\rceil=9$. The ninth ordered observation is 80 thousand dollars, so the empirical 90th percentile is $80,000. At that point $F_n(80)=9/10=0.9000$.

Check. The ten annual observations include three zero-loss years; the mean calculation includes them rather than averaging only positive losses. The threshold probability is inclusive, as requested, while the quantile rank is the ninth of ten observations.

Common Mistakes and Practical Checks

Empirical calculations are simple once the sample and event are specified, but small definition errors can change the result. A useful check is to verify that all empirical probabilities sum to 1, that $F_n(x)+S_n(x)=1$ under the strict-survival definition, and that a selected quantile’s cumulative probability reaches or exceeds its target.

Counting distinct values instead of observations. If a loss occurs four times, it contributes four observations. Its empirical probability is $4/n$, not $1/n$.
Dropping zero losses. If the target is an annual loss distribution and zero-loss years are observed, excluding them estimates a different quantity: the distribution conditional on a positive loss.
Confusing “greater than” with “at least.” For a threshold $u$, $S_n(u)$ counts $X_j>u$, not $X_j\ge u$. For example, in the first worked example, the strict rate above $900 is 0.2500, while the inclusive rate at least $900 is $4/8=0.5000$.
Using an unannounced quantile convention. The inverse empirical cumulative distribution selects an observed value at rank $\lceil np\rceil$. Interpolation-based definitions can differ, especially in small samples. Follow the stated convention and do not silently switch methods.
Interpreting a sample as a guaranteed forecast. The empirical mean and probabilities describe observed experience. Their usefulness for future losses depends on whether the sample is relevant and representative, and on whether the risk has changed.

A final unit check matters as well. If data are recorded in thousands of dollars, a calculated mean of 34 means $34,000, not $34. Keep the measurement unit attached to values throughout the calculation and convert only when reporting.

Key takeaway: An empirical loss distribution assigns each observation equal weight, with repeated values represented by their frequencies. Its cumulative function counts losses at or below a threshold, its strict survival function counts losses above it, and its mean and quantiles are calculated directly from the observed sample under explicitly stated conventions.

Check Your Understanding

For each problem, use complete observations and the inverse empirical cumulative distribution convention for quantiles. Give probabilities to four decimal places.

A sample of five losses, in dollars, is 20, 40, 40, 100, and 200. Find the sample mean, $F_n(40)$, and the strict exceedance rate above $40. (Answer: mean $80; $F_n(40)=0.6000$; strict exceedance rate $0.4000$.)
Seven ordered losses, in thousands of dollars, are 0, 0, 10, 30, 30, 60, and 90. Find $F_n(30)$, the strict exceedance rate above 30, and the empirical 75th percentile. (Answer: $F_n(30)=0.7143$; strict exceedance rate $0.2857$; 75th percentile $60,000.)
Eight observed losses, in thousands of dollars, are 5, 5, 10, 20, 20, 20, 50, and 100. Find the sample mean, the empirical probability of a loss at least 20, and the probability assigned exactly to 20. (Answer: mean $28,750; probability at least 20 is $0.6250$; probability exactly 20 is $0.3750$.)
A frequency summary reports four observations of 0, three of 25, two of 100, and one of 300 dollars. Find the mean, the empirical 80th percentile, and the strict exceedance rate above $100. (Answer: mean $57.50; 80th percentile $100; strict exceedance rate $0.1000$.)
Two hypothetical samples, each with five losses in dollars, are A: 0, 0, 50, 50, 200 and B: 20, 20, 20, 20, 200. For each sample, find the mean, median under the inverse empirical cumulative distribution convention, and strict exceedance rate above $100. (Answer: A has mean $60, median $50, and rate $0.2000$; B has mean $56, median $20, and rate $0.2000$.)

What You'll Learn