Empirical Severity Means and Limited Moments

Why Empirical Severity Moments Matter

A severity summary turns a collection of claim amounts into a quantity that can be compared, communicated, or used in further analysis. The ordinary empirical mean describes the average observed claim. A limited moment instead describes the average after each claim is capped at a specified amount. The cap changes the calculation claim by claim, so it must be applied before averaging.

This tutorial uses complete severity observations: the full amount of each claim is known. It does not treat claims reduced by a deductible, censored by a policy limit, or adjusted for inflation. Those situations require additional handling. The methods here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $x_1,\ldots,x_n$ be $n$ complete claim severity observations in the same monetary unit. Their empirical severity distribution assigns probability $1/n$ to each observed claim, counting repeated values repeatedly. Its empirical mean is $$ \bar{x}=\frac{1}{n}\sum_{j=1}^{n}x_j. $$ For a fixed cap $u\geq 0$, define the capped observation $x_j\wedge u=\min(x_j,u)$. The empirical limited moment of order $r>0$ is $$ \widehat{E[(X\wedge u)^r]}=\frac{1}{n}\sum_{j=1}^{n}\min(x_j,u)^r. $$ For $r=1$, this is the average capped claim amount. For $r=2$, it is the average squared capped amount, in squared monetary units.

From Observations to Empirical Moments

The empirical mean gives every observed claim equal weight. If a severity value occurs more than once, each occurrence is included. Thus, a list of individual observations and a frequency table represent the same calculation when the table’s counts are used as weights.

More generally, if distinct severities $z_1,\ldots,z_k$ occur with counts $c_1,\ldots,c_k$, then $n=\sum_{j=1}^{k}c_j$. The empirical mean and limited moment are

\bar{x}=\frac{\sum_{j=1}^{k}c_jz_j}{n}, \qquad \widehat{E[(X\wedge u)^r]}= \frac{\sum_{j=1}^{k}c_j\min(z_j,u)^r}{n}.

The denominator is the number of claims represented, not the number of distinct severity values. For example, a value appearing five times contributes five observations to both numerator and denominator. If observations are intentionally given different weights, the calculation must instead use those weights and their total; this tutorial’s examples use equally weighted claims.

Worked Example: Mean and Capped Moments from Individual Claims

A hypothetical insurer has five complete claim amounts: $350, $750, $1,100, $1,800, and $2,500. Calculate the empirical mean, the limited mean at $u=\$1{,}500$, and the limited second moment at that cap.

Method and assumptions. Treat the five claims as equally weighted observations from the empirical distribution, with probability $1/5$ on each observation. No interest or mortality assumption is involved, and no parametric severity distribution is fitted. The cap is $1,500.

Ordinary empirical mean. The claim total is

$$ 350+750+1{,}100+1{,}800+2{,}500=6{,}500. $$

There are five claims, so $\bar{x}=6{,}500/5=\$1{,}300$.

Limited mean. Cap each observation first. The capped amounts are $350, $750, $1,100, $1,500, and $1,500. Their total is

$$ 350+750+1{,}100+1{,}500+1{,}500=5{,}200. $$

Therefore, the limited mean is $5{,}200/5=\$1{,}040$.

Limited second moment. Square each capped amount, then average:

\frac{350^2+750^2+1{,}100^2+1{,}500^2+1{,}500^2}{5} = \frac{122{,}500+562{,}500+1{,}210{,}000+2{,}250{,}000+2{,}250{,}000}{5} = \frac{6{,}395{,}000}{5} = 1{,}279{,}000.

The limited second moment is $1{,}279{,}000$ dollars squared. The results are consistent: the limited mean is below the ordinary mean because the cap reduces the two largest observations. The method avoids averaging first and then trying to apply the cap to the average, which would not give the average capped claim.

Why the Empirical Formula Works

The empirical distribution represents each observed claim as one equally likely outcome. If $g$ is any function of severity, its empirical expectation is the probability-weighted average of its values over those outcomes. Each outcome has probability $1/n$, so

\widehat{E[g(X)]} =\sum_{j=1}^{n}\frac{1}{n}g(x_j) =\frac{1}{n}\sum_{j=1}^{n}g(x_j).

Taking $g(x)=x$ gives the empirical mean. Taking $g(x)=\min(x,u)^r$ gives the empirical limited moment. This also explains why the order of operations matters: the function $g$ is applied to each claim before the values are averaged.

When the observations are independent and identically distributed draws from a population severity $X$, the sample average of $g(X)$ estimates $E[g(X)]$. In particular, the empirical limited moment estimates $E[\min(X,u)^r]$. Without a suitable sampling assumption, the same calculation remains a correct description of the observed claims, but it need not represent a broader population.

Worked Example: Using a Frequency Table Correctly

A hypothetical portfolio’s complete claim data are summarized as three claims of $600, two claims of $1,400, and one claim of $3,200. Find the ordinary mean, the limited mean at $u=\$1{,}800$, and the limited second moment at that cap.

Method and assumptions. The table represents six equally weighted claims, so use counts as weights and divide by six. This is the empirical discrete distribution specified by the table; there is no fitted parametric distribution, interest rate, or mortality model.

The ordinary total is

$$ 3(600)+2(1{,}400)+1(3{,}200)=1{,}800+2{,}800+3{,}200=7{,}800. $$

The ordinary mean is $7{,}800/6=\$1{,}300$.

At the $1,800 cap, the three $600 claims remain $600 each, the two $1,400 claims remain $1,400 each, and the $3,200 claim becomes $1,800. Thus, the capped total is

$$ 3(600)+2(1{,}400)+1(1{,}800)=1{,}800+2{,}800+1{,}800=6{,}400. $$

The limited mean is $6{,}400/6=\$1{,}066.67$, rounded to the nearest cent.

For the limited second moment, square the capped amounts and retain the frequency weights:

\frac{3(600^2)+2(1{,}400^2)+1(1{,}800^2)}{6} = \frac{3(360{,}000)+2(1{,}960{,}000)+3{,}240{,}000}{6} = \frac{8{,}240{,}000}{6} = 1{,}373{,}333.33.

The limited second moment is $1{,}373{,}333.33$ dollars squared, rounded to two decimal places. As a check, the capped total of $6,400 is less than the uncapped total of $7,800, and the denominator is six claims rather than three distinct severity values. This method avoids treating each row of the frequency table as one equally weighted observation.

Units and Interpretation

A first moment has the same unit as the severity observations. If claims are measured in dollars, the empirical mean and limited mean are in dollars. A second moment has squared units: squaring a dollar amount produces dollars squared. It is not itself an average claim amount.

Data reported in thousands require particular care. If a claim amount is recorded as 2.4 in thousands of dollars, that means $2,400. A first moment of 1.6 in those units is $1,600. A second moment of 3.0 in thousands squared is $3.0\times 1{,}000^2=\$3{,}000{,}000$ dollars squared. Convert the cap to the same units as the observations before applying it.

Worked Example: A Cap When Observations Are in Thousands

Five complete hypothetical claims, recorded in thousands of dollars, are $0.8,\ 1.1,\ 2.0,\ 2.0,$ and $4.1$. Calculate the ordinary mean, limited mean, and limited second moment when the cap is $2,500.

Method and assumptions. Use the empirical distribution with probability $1/5$ for each recorded claim. There is no parametric distribution, interest, or mortality assumption. Because the observations are in thousands of dollars, convert the cap to $2.5$ in the recorded units.

The uncapped total is

$$ 0.8+1.1+2.0+2.0+4.1=10.0 $$

thousand dollars. Dividing by five gives $2.0$ thousand dollars, or an ordinary mean of $2,000.

The capped observations, in thousands, are $0.8,\ 1.1,\ 2.0,\ 2.0,$ and $2.5$. Their total is $8.4$, so the limited mean is $8.4/5=1.68$ thousand dollars, or $1,680.

Their squared values sum to

$$ 0.8^2+1.1^2+2.0^2+2.0^2+2.5^2 = 0.64+1.21+4.00+4.00+6.25 = 16.10. $$

The limited second moment is $16.10/5=3.22$ thousand-dollars squared. Since one thousand-dollars squared equals $1{,}000{,}000$ dollars squared, this is $3.22\times1{,}000{,}000=\$3{,}220{,}000$ dollars squared. The arithmetic is internally consistent: the capped mean and moment use the cap in thousands, and the second moment is converted using the square of the unit conversion.

Common Mistakes and Checks

Using the number of table rows as the denominator. If a table gives counts, the denominator is the sum of the counts. Repeated claims remain repeated observations.
Capping the mean instead of capping each claim. In general, $\min(\bar{x},u)$ is not equal to $\frac{1}{n}\sum_j\min(x_j,u)$. Apply the cap to every observation first.
Confusing a limited second moment with a variance. The limited second moment is an average of squared capped amounts. It is not the variance of capped amounts, which would also require centering around their mean.
Mixing units. A cap of $2,500 is $2.5$ when observations are in thousands. A second moment in thousands squared must be multiplied by $1{,}000{,}000$ to express it in dollars squared.
Interpreting a descriptive calculation too broadly. The empirical moment exactly summarizes the observed claims. Treating it as an estimate for a larger population requires an appropriate sampling and exposure interpretation.

For a quick arithmetic check, the limited mean cannot exceed the ordinary mean when all severities are nonnegative and $u$ is a fixed upper cap. Also verify the denominator, the cap applied to every observation, and the units of each reported moment. A violation of the first check often signals that the cap was applied incorrectly.

Check Your Understanding

For each problem, use complete, equally weighted observations and keep the stated units. Round monetary means to cents where needed.

A hypothetical dataset contains claims of $500, $900, $1,600, and $3,000. Find the ordinary mean and the limited mean at $u=\$2{,}000$. (Answer: ordinary mean $1,500; limited mean $1,250.)
A frequency table contains two claims of $400, three claims of $1,000, and one claim of $2,500. Find the ordinary mean and limited mean at $u=\$1{,}200$. (Answer: ordinary mean $1,050; limited mean $833.33.)
Four complete claims are recorded in thousands of dollars as $0.5,\ 1.0,\ 1.5,\ 3.0$. At a cap of $2,000, find the limited mean and limited second moment in dollars and dollars squared, respectively. (Answer: limited mean $1,250; limited second moment $1,875,000 dollars squared.)
A hypothetical frequency table has one claim of $1,000, two claims of $2,000, and one claim of $4,000. Find the ordinary mean, the limited mean at $u=\$2{,}500$, and the limited second moment at that cap. (Answer: ordinary mean $2,250; limited mean $1,875; limited second moment $3,812,500 dollars squared.)

What You'll Learn