Empirical Severity from Complete Claim Records

From Complete Claim Records to a Severity Distribution

A complete claim record gives the amount of an individual claim on a consistent measurement basis. When a collection of such records is used to describe severity, each claim—not each policy, policy year, or dollar of exposure—is one observation. This distinction matters: the empirical severity distribution summarizes the observed amounts of claims, while exposure-adjusted frequency describes how often claims occur relative to exposure.

The calculations below assume that each included record represents one complete claim amount and that the records have been selected from the target population of claim severities. If the records are incomplete, combine multiple payments for one claim incorrectly, or record only amounts altered by a coverage limit, the resulting distribution may not represent the intended severity variable. The target quantity must be defined before calculating a statistic.

Definition: For $n$ complete claim amounts $x_1,\ldots,x_n$, the empirical severity distribution assigns probability $1/n$ to each recorded claim amount. Repeated amounts retain their multiplicity. Its empirical distribution function is $F_n(x)=\frac{1}{n}\sum_{i=1}^{n} I(x_i\leq x)$, where $I(\cdot)$ is 1 when the condition holds and 0 otherwise.

For example, if the records are $200, 200,$ and $800$, the empirical distribution assigns probability $2/3$ to $200$ and $1/3$ to $800$. It does not assign equal probability to the two distinct dollar amounts. Equal weighting is by claim record, so a value appearing twice receives twice the empirical probability of a value appearing once.

Empirical Mean and Variance

The empirical mean is the average amount per recorded claim. Two variance calculations are common, and they answer slightly different questions. The empirical variance describes the spread of the empirical distribution itself, whose probability mass is exactly $1/n$ per record. The unbiased sample variance uses $n-1$ in the denominator and is a familiar estimator of an underlying population variance when the observations are treated as a random sample.

\bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_i,\qquad \operatorname{Var}_{n}(X)=\frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x})^2,\qquad s^2=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})^2\quad(n>1).

Use the denominator that matches the question. If the task is to calculate the variance of the empirical distribution, use $n$. If it asks for the unbiased sample variance, use $n-1$. The two quantities are related by $s^2=\frac{n}{n-1}\operatorname{Var}_{n}(X)$.

Worked Example: Constructing the Distribution and Its Moments

A hypothetical insurer has five complete claim records, in dollars: $300, 500, 500, 900,$ and $1{,}300$. Treat the five claims as equally weighted observations from one target severity population. Construct the empirical distribution, calculate its mean and both variance conventions, then find the 60th percentile and the upper-tail average above the 60th percentile.

Step 1: Empirical probabilities. There are five claims, so each claim contributes probability $1/5=0.2$. The amount $500$ occurs twice; the other amounts occur once. Thus the empirical probabilities at $300, 500, 900,$ and $1{,}300$ are $0.2, 0.4, 0.2,$ and $0.2$, respectively.

Step 2: Mean. The total is $300+500+500+900+1{,}300=3{,}500$, so $\bar{x}=3{,}500/5=\$700$.

Step 3: Variances. The deviations from $700$ are $-400,-200,-200,200,$ and $600$. Their squared deviations sum to $160{,}000+40{,}000+40{,}000+40{,}000+360{,}000=640{,}000$ dollars squared. Therefore, the empirical variance is $640{,}000/5=128{,}000$ dollars squared. The unbiased sample variance is $640{,}000/4=160{,}000$ dollars squared.

Step 4: Quantile and tail average. Define the empirical $p$-quantile as the smallest observed amount $x$ for which $F_n(x)\geq p$. At $500$, the cumulative probability is $3/5=0.6$, so the 60th percentile is $\$500$. The upper 40% of the empirical distribution contains two of the five equally weighted claims: $900$ and $1{,}300$. Their average is $(900+1{,}300)/2=\$1{,}100$.

Check. The probabilities sum to $0.2+0.4+0.2+0.2=1$, and the mean is between the smallest and largest recorded amounts. The tail calculation averages exactly the upper 40% of claim weight. A common mistake is to count $500$ only once because it is one distinct amount; doing so would incorrectly change both the empirical probabilities and the moments.

Quantiles and Upper-Tail Averages

Quantiles require a convention, particularly for a finite empirical distribution. Here the $p$-quantile is the inverse empirical distribution function $F_n^{-1}(p)=\inf\{x:F_n(x)\geq p\}$, for $0<p\leq1$. This convention selects an observed claim amount; it does not interpolate between adjacent observations. Other quantile conventions exist, so a calculation should follow the convention specified in the question.

Definition: For $0<p<1$, the empirical upper-tail average at level $p$ is the mean amount in the largest $1-p$ fraction of empirical probability mass. If that tail contains a non-integer number of equally weighted claims, include the required fraction of the boundary claim’s probability mass.

The fractional-weight rule is necessary because the requested tail probability need not equal an integer multiple of $1/n$. For instance, the upper 25% of six equally weighted claims contains $0.25(6)=1.5$ claims of probability mass. It consists of the largest claim and half the probability mass at the next-largest claim—not two full claims. Tied amounts cause no difficulty: the required mass can be taken from claims at the same amount, and the resulting average is unchanged.

The relationship between the quantile definition and the fractional-tail calculation follows from the stepwise form of the empirical quantile function. As $q$ increases from 0 to 1, $F_n^{-1}(q)$ equals each ordered observation over an interval of probability width $1/n$. Consequently, integrating that quantile function over the upper tail sums the claim amounts in the tail, with a fractional contribution when an endpoint cuts through one of those intervals.

\operatorname{TVaR}_{p,n} =\frac{1}{1-p}\int_p^1 F_n^{-1}(q)\,dq.

This expression defines the empirical tail average with the convention used here. It is a probability-weighted average of the upper tail, not simply the quantile at level $p$. A quantile identifies a threshold; a tail average summarizes amounts beyond that threshold while including the fraction needed to make the tail probability exactly $1-p$.

Worked Example: Pooling Groups Without Changing the Weighting

Two hypothetical claim cohorts have complete severities in dollars. Cohort A has claims $200$ and $400$; Cohort B has claims $600,600,1{,}000,$ and $1{,}400$. Treat every claim as one observation from the same target severity population. Calculate the pooled mean, the equally weighted average of the cohort means, the pooled empirical variance, and the upper-tail average at $p=0.75$.

Step 1: Identify the weighting. Cohort A has two claims and Cohort B has four. Their means are $(200+400)/2=\$300$ and $(600+600+1{,}000+1{,}400)/4=\$900$. The pooled claim-weighted mean is $(2(300)+4(900))/6=(600+3{,}600)/6=\$700$. The equally weighted average of cohort means is $(300+900)/2=\$600$; it gives each cohort the same weight despite their different claim counts.

Step 2: Pooled empirical variance. Relative to the pooled mean of $700$, the six deviations are $-500,-300,-100,-100,300,$ and $700$. The squared deviations sum to $250{,}000+90{,}000+10{,}000+10{,}000+90{,}000+490{,}000=940{,}000$. The empirical variance is $940{,}000/6=\$156{,}666.67$ squared, rounded to the nearest cent. The unbiased sample variance is $940{,}000/5=\$188{,}000$ squared.

Step 3: Upper-tail average. The ordered records are $200,400,600,600,1{,}000,1{,}400$. The upper 25% contains $0.25(6)=1.5$ claims of mass: all of the $1{,}400$ claim and half the mass of a $1{,}000$ claim. The tail average is $(1{,}400+0.5(1{,}000))/1.5=1{,}900/1.5=\$1{,}266.67$, rounded to the nearest cent. The 75th percentile is $\$1{,}000$, since the cumulative probability at that amount is $5/6$.

Check. Pooling assigns each of the six claims probability $1/6$, and the upper-tail calculation uses total probability mass $1.5/6=0.25$. Replacing the pooled mean by the equal-cohort average would answer a different question: the average of two cohort means, rather than the average amount per claim.

A Tail Calculation With a Fractional Boundary

For a practical calculation, sort the claim amounts from largest to smallest and track how much empirical probability mass is still needed. Each full claim contributes mass $1/n$; the last claim contributes only the fraction of its mass needed to reach $1-p$. Divide the weighted sum of amounts by the required tail mass. This procedure directly implements the quantile-integral definition, including when the threshold falls between observed claims.

Set the target mass.
For a tail level $p$, the required fraction of all empirical probability is $1-p$, equivalent to $(1-p)n$ equally weighted claims.

Order the observations.
Arrange complete claim amounts from largest to smallest. Keep repeated amounts as separate records or combine them while retaining their multiplicities.

Accumulate only the required mass.
Use full claims until the remaining mass is smaller than one claim’s mass, then use only that fraction of the next claim.

Divide by the tail mass.
Divide the mass-weighted sum of claim amounts by $1-p$, or equivalently divide the weighted sum over claim counts by $(1-p)n$.

Worked Example: Quantile and Tail Average With Fractional Mass

A hypothetical portfolio has eight complete claim amounts, in dollars: $250,250,500,750,1{,}250,2{,}000,3{,}000,$ and $5{,}000$. Give each record probability $1/8$. Calculate the mean, the empirical 80th percentile, the probability of a claim exceeding $2{,}500$, and the empirical upper-tail average at $p=0.80$.

Step 1: Mean. The total is $250+250+500+750+1{,}250+2{,}000+3{,}000+5{,}000=13{,}000$, so the mean is $13{,}000/8=\$1{,}625$.

Step 2: 80th percentile. The cumulative probability through $3{,}000$ is $7/8=0.875$, while through $2{,}000$ it is $6/8=0.75$. The smallest observed amount with cumulative probability at least $0.80$ is therefore $\$3{,}000$.

Step 3: Exceedance probability. Two of the eight records exceed $2{,}500$, namely $3{,}000$ and $5{,}000$. Thus $P_n(X>2{,}500)=2/8=0.25$.

Step 4: Upper-tail average. An upper tail of 20% requires $0.20(8)=1.6$ claims of mass. Take the full $5{,}000$ claim and $0.6$ of the $3{,}000$ claim. The weighted total is $5{,}000+0.6(3{,}000)=6{,}800$; dividing by $1.6$ gives $\$4{,}250$. The percentile is a boundary amount, not the tail average.

Check. The tail includes $1.6/8=0.20$ of the empirical probability and its average, $\$4{,}250$, exceeds the 80th-percentile threshold of $\$3{,}000$. Counting two full claims instead would produce an average for a 25% tail, not the requested 20% tail.

Interpretation and Common Traps

An empirical distribution is a direct description of the observed claim records, not a guarantee that future claims will follow the same pattern. Its use as a representation of a target population depends on the records being relevant and consistently defined. In particular, a complete claim amount must refer to the intended severity measure; amounts recorded on a different basis should not silently be treated as interchangeable.

Confusing a claim average with a group average. Pooling complete records gives each claim equal weight. Averaging group means equally gives each group equal weight, which is a different statistic unless group sizes are equal.
Using the wrong variance denominator. Dividing by $n$ gives the variance of the empirical distribution; dividing by $n-1$ gives the unbiased sample variance under the usual random-sample interpretation.
Ignoring multiplicities. Repeated claim amounts represent repeated observations and must retain their full combined probability.
Using whole claims when the tail requires fractional mass. A discrete empirical tail can cut through the probability assigned to a claim. Include only the fraction required by the stated tail probability.
Confusing a quantile with a tail average. The quantile is a threshold under a specified inverse-CDF convention. The tail average summarizes the upper tail and can be much larger.

Key takeaway: Start by defining the severity observation and its weighting. For complete claim records with equal claim weights, assign mass $1/n$ to every record, preserve repetitions, state the variance and quantile conventions, and use fractional probability mass when calculating an upper-tail average.

Check Your Understanding

For each problem, treat the listed amounts as complete claim severities and use equal weight per claim. Use the inverse empirical CDF quantile convention defined above.

Four complete claims are $100,300,300,$ and $700$ dollars. Find the empirical mean, empirical variance, and 50th percentile. (Answer: mean $=\$350$; empirical variance $=\$47{,}500$ dollars squared; 50th percentile $=\$300$.)
Cohort A has complete claims of $400$ and $800$ dollars; Cohort B has claims of $1{,}000,1{,}000,$ and $1{,}600$ dollars. Find the pooled claim-weighted mean and the equally weighted average of the two cohort means. (Answer: pooled mean $=\$960$; equal-cohort average $=\$900$.)
Five complete claims are $100,200,400,800,$ and $1{,}500$ dollars. Find the empirical 80th percentile and the upper-tail average at $p=0.80$. (Answer: 80th percentile $=\$1{,}500$; upper-tail average $=\$1{,}500$.)
Six complete claims are $200,400,600,1{,}000,1{,}800,$ and $3{,}000$ dollars. Find the 75th percentile and the upper-tail average at $p=0.75$. (Answer: 75th percentile $=\$1{,}800$; upper-tail average $=\$2{,}600$.)
Four complete claims, measured in thousands of dollars, are $2,4,4,$ and $10$. Find the empirical variance and the unbiased sample variance. (Answer: empirical variance $=9$ thousand-dollars squared; unbiased sample variance $=12$ thousand-dollars squared.)

What You'll Learn