Sample Means, Variances, and Empirical Moments

Why Summarize a Claims Sample?

A data set of observed claim counts or loss amounts can be summarized in several ways. The mean describes the average observation, while variance describes the spread around that mean. Higher moments can help describe asymmetry and tail weight. These summaries are calculations on the observed data; by themselves, they do not establish a probability model for future claims.

A key distinction is whether the goal is to describe the observed values or to estimate a population parameter. The empirical variance divides by the number of observations. The commonly used sample variance divides by one fewer than the number of observations. The two quantities answer different questions and should not be interchanged.

Definition: For $n$ unweighted observations $x_1,\ldots,x_n$, the sample mean is $\bar{x}=\frac{1}{n}\sum_{j=1}^{n}x_j$. The empirical raw moment of order $r$ is $m'_r=\frac{1}{n}\sum_{j=1}^{n}x_j^r$, and the empirical central moment of order $r$ is $m_r=\frac{1}{n}\sum_{j=1}^{n}(x_j-\bar{x})^r$. In particular, $m_2$ is the empirical variance.

The observations must be measured on a consistent basis. For example, claim amounts recorded in dollars should all be treated as dollars, and claim counts should use the same observation window. The formulas here treat each of the $n$ observations equally; weighted observations require a separate accounting of their weights.

Mean, Empirical Variance, and Sample Variance

The mean is the arithmetic average. To calculate empirical variance, find each observation’s deviation from the mean, square those deviations, add them, and divide by $n$. Squaring prevents positive and negative deviations from canceling, and gives more influence to observations far from the mean.

m_2=\frac{1}{n}\sum_{j=1}^{n}(x_j-\bar{x})^2

When the observed values are regarded as a sample from an underlying population, a common estimator of population variance uses divisor $n-1$:

s^2=\frac{1}{n-1}\sum_{j=1}^{n}(x_j-\bar{x})^2 =\frac{n}{n-1}m_2,\qquad n\geq 2.

The factor $n/(n-1)$ makes $s^2$ larger than $m_2$ whenever the observations are not all identical. It is not a correction that should automatically be applied to every descriptive summary. If the task is to describe the observed data set as an equally weighted empirical distribution, $m_2$ is its variance. If the task requests the sample variance, or an unbiased estimator under the assumptions below, use $s^2$.

Condition: If $X_1,\ldots,X_n$ are independent and identically distributed observations from a population with finite variance $\sigma^2$, then $E[s^2]=\sigma^2$. The divisor-$n$ empirical variance instead satisfies $E[m_2]=\frac{n-1}{n}\sigma^2$. This unbiasedness statement for $s^2$ relies on the stated sampling conditions; it does not mean $s^2$ equals the population variance for every sample.

A useful computational identity connects the empirical variance to the raw second moment. Since $m'_2=\frac{1}{n}\sum x_j^2$,

m_2=m'_2-\bar{x}^{\,2}.

This follows by expanding the squared deviations:

\begin{aligned} m_2 &=\frac{1}{n}\sum_{j=1}^{n}(x_j-\bar{x})^2\\ &=\frac{1}{n}\sum_{j=1}^{n}\left(x_j^2-2x_j\bar{x}+\bar{x}^{\,2}\right)\\ &=\frac{1}{n}\sum_{j=1}^{n}x_j^2 -2\bar{x}\left(\frac{1}{n}\sum_{j=1}^{n}x_j\right) +\bar{x}^{\,2}\\ &=m'_2-2\bar{x}^{\,2}+\bar{x}^{\,2} =m'_2-\bar{x}^{\,2}. \end{aligned}

The identity is convenient for checking arithmetic, but calculating deviations directly is often safer when the raw values are large and close together. In such cases, subtracting two large, similar quantities can lose numerical precision.

Higher Empirical Moments

Central moments of order three and four describe features of the observed distribution beyond its spread. The third central moment retains the sign of each deviation after cubing, so it reflects asymmetry. The fourth central moment gives substantial weight to observations far from the mean because deviations are raised to the fourth power.

Definition: When $m_2>0$, empirical skewness and empirical kurtosis are $g_1=m_3/m_2^{3/2}$ and $g_2=m_4/m_2^2$, respectively. Empirical excess kurtosis is $g_2-3$. These are moment ratios calculated from the observed data using divisor $n$ in each central moment; adjusted sample versions may use different formulas.

Skewness is dimensionless because the units in $m_3$ cancel against those in $m_2^{3/2}$. Kurtosis and excess kurtosis are also dimensionless. The central moments themselves retain powers of the original measurement units: if $x_j$ is measured in dollars, $m_2$ is in dollars squared, $m_3$ in dollars cubed, and $m_4$ in dollars to the fourth power.

Worked Example: Moments of a Claim-Count Sample

A hypothetical portfolio records the following claim counts for five equal observation periods: $0,1,1,2,6$. Calculate the mean, empirical variance, sample variance, and empirical third central moment. Use the count per observation period as the measurement unit.

Method. First calculate the mean, then the deviations and their powers. This keeps the divisor for each requested quantity explicit.

The total count is $0+1+1+2+6=10$, so

\bar{x}=\frac{10}{5}=2.

The deviations from the mean are $-2,-1,-1,0,4$. Their squared deviations sum to

$$ (-2)^2+(-1)^2+(-1)^2+0^2+4^2 =4+1+1+0+16=22. $$

Therefore the empirical variance is $m_2=22/5=4.4$ counts squared per observation-period squared. The sample variance is $s^2=22/(5-1)=22/4=5.5$ counts squared per observation-period squared.

For the third central moment, the cubed deviations sum to

$$ (-2)^3+(-1)^3+(-1)^3+0^3+4^3 =-8-1-1+0+64=54. $$

Thus $m_3=54/5=10.8$ counts cubed per observation-period cubed. As a check on the variance, the raw second moment is $(0^2+1^2+1^2+2^2+6^2)/5=42/5=8.4$, and $m'_2-\bar{x}^{\,2}=8.4-2^2=4.4$, matching the direct calculation.

Common mistake avoided: using $n-1$ for every moment. The requested empirical central moments use divisor $n=5$; only the sample variance here uses divisor $n-1=4$.

Worked Loss-Amount Calculation

Worked Example: Variance, Skewness, and Kurtosis of Losses

A hypothetical insurer examines four loss amounts: $\$50,\$50,\$50,\$250$. Calculate the mean, empirical variance, sample variance, empirical kurtosis, and excess kurtosis. Use empirical central moments with divisor $n$ for the kurtosis calculation.

Method. Compute the mean and central moments directly, then form the dimensionless kurtosis ratios. This avoids confusing dollar-valued amounts with powers of dollars.

The total loss is $\$50+\$50+\$50+\$250=\$400$, so the mean is $\bar{x}=\$400/4=\$100$. The deviations are $-\$50,-\$50,-\$50,\$150$. Their squared amounts sum to

3(50^2)+150^2=3(2{,}500)+22{,}500=30{,}000\ \text{dollars}^2.

Hence the empirical variance is $m_2=30{,}000/4=7{,}500\ \text{dollars}^2$. The sample variance is $s^2=30{,}000/3=10{,}000\ \text{dollars}^2$.

For the fourth central moment, the sum of fourth powers of the deviations is

3(50^4)+150^4 =3(6{,}250{,}000)+506{,}250{,}000 =525{,}000{,}000\ \text{dollars}^4.

Therefore $m_4=525{,}000{,}000/4=131{,}250{,}000\ \text{dollars}^4$. The empirical kurtosis is

g_2=\frac{131{,}250{,}000\ \text{dollars}^4} {(7{,}500\ \text{dollars}^2)^2} =\frac{131{,}250{,}000}{56{,}250{,}000} =2.3333.

The excess kurtosis is $g_2-3=2.3333-3=-0.6667$. The ratios have no units because dollars to the fourth power cancel. The variance values are $7{,}500\ \text{dollars}^2$ and $10{,}000\ \text{dollars}^2$, not the squares of dollar amounts $\$7{,}500$ and $\$10{,}000$.

Check. The squared deviations sum to $30{,}000\ \text{dollars}^2$, and dividing that same sum by four and by three gives the two stated variance conventions. The one large loss contributes most of the fourth-power sum, illustrating how strongly kurtosis can respond to an extreme observation.

How an Extreme Observation Changes the Summary

Because the mean is calculated from every value, an extreme observation can pull it toward itself. Variance then changes both because the mean moves and because the extreme value is far from that new mean. Higher central moments can be even more sensitive: a deviation is squared for variance, cubed for the third moment, and raised to the fourth power for the fourth moment. A single large observation may therefore have a modest effect on the mean but a much larger effect on the fourth moment.

Worked Example: Replacing a Moderate Loss with an Extreme Loss

A hypothetical sample of four losses is initially $\$10,\$20,\$30,\$40$. A corrected report replaces the last value with $\$400$, giving $\$10,\$20,\$30,\$400$. Compare the mean and empirical variance before and after the correction. Report monetary quantities to the cent.

Method. Calculate the mean and sum of squared deviations separately for each data set. Do not measure the new extreme loss’s deviation from the old mean.

For the original data, the mean is $(10+20+30+40)/4=\$25.00$. The deviations are $-15,-5,5,15$ dollars, so their squared sum is $225+25+25+225=500\ \text{dollars}^2$. Thus the empirical variance is $500/4=125\ \text{dollars}^2$.

For the corrected data, the mean is $(10+20+30+400)/4=460/4=\$115.00$. The deviations are $-105,-95,-85,285$ dollars. Their squared sum is

105^2+95^2+85^2+285^2 =11{,}025+9{,}025+7{,}225+81{,}225 =108{,}500\ \text{dollars}^2.

The corrected empirical variance is $108{,}500/4=27{,}125\ \text{dollars}^2$. The mean increases by $\$90.00$, while the empirical variance increases from $125$ to $27{,}125\ \text{dollars}^2$.

Check. The corrected deviations sum to $-105-95-85+285=0$, as deviations from the mean must. The squared deviations are all nonnegative, and the extreme observation contributes $81{,}225\ \text{dollars}^2$ by itself.

Common mistake avoided: retaining the old mean of $\$25.00$ after changing the data. Every central moment must be calculated around the mean of the current sample.

Exam Traps and Interpretation

Mixing up the divisor. The empirical variance uses $n$; the usual sample variance uses $n-1$. Identify which is requested before calculating.
Calling $s^2$ the variance of the observed empirical distribution. That distribution assigns equal mass to the observed values, so its variance is $m_2$. The divisor-$n-1$ quantity is a sample-based estimator.
Dropping units or placing them incorrectly. A variance of $7{,}500\ \text{dollars}^2$ is not $(\$7{,}500)^2$. The unit is squared; the numerical value is $7{,}500$.
Using the wrong center. Central moments use deviations from the sample mean, not from zero or from a previous version of the data.
Over-interpreting moments from a small sample. Skewness and kurtosis can change sharply when one observation changes. They summarize the sample and do not prove that the underlying population has a particular tail shape.

Key takeaway: State whether a variance uses divisor $n$ or $n-1$, calculate central moments around the current sample mean, and carry the measurement units through each power. Extreme observations affect higher moments especially strongly.

Check Your Understanding

Use unweighted observations and the empirical central-moment convention unless a question specifically requests the sample variance.

A hypothetical sample of losses is $\$4,\$6,\$10,\$12$. Find the mean, empirical variance, and sample variance. (Answer: mean $=\$8$, $m_2=10\ \text{dollars}^2$, and $s^2=13.3333\ \text{dollars}^2$.)
Claim counts over five equal periods are $0,0,2,2,6$. Calculate the mean, empirical variance, sample variance, and third empirical central moment. (Answer: mean $=2$, $m_2=4.8$, $s^2=6$, and $m_3=9.6$, in the corresponding powers of counts per period.)
A hypothetical severity sample is $\$20,\$20,\$20,\$100$. Calculate its mean, empirical variance, sample variance, and empirical kurtosis. (Answer: mean $=\$40$, $m_2=1{,}200\ \text{dollars}^2$, $s^2=1{,}600\ \text{dollars}^2$, and $g_2=2.3333$.)
For observations $2,4,4,6$, calculate the raw second moment and use it to check the empirical variance. (Answer: $m'_2=36/4=9$, $\bar{x}=4$, and $m_2=9-4^2=-7$ would be inconsistent; recomputing the squares gives $4+16+16+36=72$, so $m'_2=18$ and $m_2=18-16=2$.)
A loss sample has $n=8$ observations with squared deviations from its mean summing to $14{,}000\ \text{dollars}^2$. Find both variance conventions and explain which is larger. (Answer: $m_2=14{,}000/8=1{,}750\ \text{dollars}^2$, $s^2=14{,}000/7=2{,}000\ \text{dollars}^2$; the sample variance is larger because it divides the same sum by $7$ rather than $8$.)

What You'll Learn