Why Empirical Moments Matter
Exact loss records can be summarized in several ways. The empirical mean describes the average observed loss, while raw moments and variances describe additional features of the observed values. These summaries are straightforward to calculate, but one convention requires care: the variance of the empirical distribution divides by the number of observations, whereas the usual sample variance divides by one fewer than the number of observations. They answer related but different questions.
This tutorial concerns a finite set of exact, complete loss observations. Every record is treated as one observation, including repeated values. No parametric severity distribution is assumed. Interest and mortality assumptions do not apply to these examples.
A raw moment is calculated about zero: the second raw moment averages the squared losses, not the squared deviations from the mean. The empirical variance is a central moment, calculated about the empirical mean. Both the second raw moment and the variance have squared-loss units. If losses are recorded in dollars, these quantities are in dollars squared.
The empirical distribution gives equal weight to each record, not necessarily to each distinct numerical value. If a loss amount occurs three times in the data, it contributes three times to each sum. Equivalently, if distinct values \(a_k\) occur \(c_k\) times, then \(m'_r=(1/n)\sum_k c_k a_k^r\), where \(\sum_k c_k=n\).
Calculating the Empirical Moments
For exact observations, a reliable calculation proceeds in a fixed order: count the records, calculate the sum and mean, calculate the required powers for raw moments, and then calculate variance. The raw-moment identity is often efficient, while the squared-deviation calculation provides a useful independent check.
Worked Example: Mean, Second Raw Moment, and Both Variances
A hypothetical insurer reviews three exact claim losses, in dollars: \(2, 4, 8\). Find the empirical mean, second raw moment, empirical variance, and sample variance. The observations are treated as the complete empirical data set; no parametric distribution is assumed.
Step 1: Count and average the observations. Here \(n=3\), and the total loss is \(2+4+8=14\). Thus \(\bar{x}=14/3\) dollars.
Step 2: Calculate the second raw moment. Square each loss and average the squares:
The second raw moment is \(28\) dollars squared.
Step 3: Calculate the empirical variance. Use the raw-moment identity:
So the empirical variance is \(56/9\) dollars squared.
Step 4: Calculate the sample variance. The sum of squared deviations is \(n\) times the empirical variance, or \(3(56/9)=56/3\). Divide that sum by \(n-1=2\):
The sample variance is \(28/3\) dollars squared. As a check, the squared deviations from \(14/3\) are \(64/9, 4/9,\) and \(100/9\); their sum is \(168/9=56/3\), as used above.
Method and trap avoided: The empirical distribution uses denominator \(n\); the sample-variance convention uses \(n-1\). A common arithmetic error is to report the second raw moment as \(28/3\). The squared values sum to \(84\), so their average is \(84/3=28\), not \(28/3\).
Why the Variance Identity Holds
The formula \(m'_2-\bar{x}^{\,2}\) is not a separate approximation. It follows by expanding each squared deviation and using the definition of the mean. This also explains why variance is nonnegative and why the two variance conventions differ only by their divisor.
The empirical variance is \((1/n)\sum_{j=1}^n(x_j-\bar{x})^2\).
Each term is \(x_j^2-2x_j\bar{x}+\bar{x}^{\,2}\), so summing and dividing by \(n\) gives \((1/n)\sum_jx_j^2-2\bar{x}(1/n)\sum_jx_j+\bar{x}^{\,2}\).
Since \((1/n)\sum_jx_j=\bar{x}\), the expression becomes \(m'_2-2\bar{x}^{\,2}+\bar{x}^{\,2}=m'_2-\bar{x}^{\,2}\).
The sample variance is obtained from the same sum of squared deviations, but it divides by \(n-1\). Under the additional model assumption that \(X_1,\ldots,X_n\) are independent observations from one distribution with finite mean and variance \(\sigma^2\), \(s^2\) is an unbiased estimator of \(\sigma^2\). To see why, write the sum of squared deviations as \(\sum_j(X_j-\bar{X})^2=\sum_j(X_j-\mu)^2-n(\bar{X}-\mu)^2\). Its expected value is \(n\sigma^2-n(\sigma^2/n)=(n-1)\sigma^2\). Dividing by \(n-1\) therefore gives \(E[s^2]=\sigma^2\).
That unbiasedness statement is about repeated samples from an underlying distribution. It does not change the variance of a particular empirical distribution: for the observed records themselves, the equally weighted variance is still the average squared deviation, with divisor \(n\). If the records constitute the entire set being summarized, the sample-variance convention is not required to describe that set.
Worked Example: Repeated Losses and the Empirical Distribution
A hypothetical portfolio has five exact recorded losses, in thousands of dollars: \(0, 0, 5, 5, 10\). Find the first three raw moments, the empirical variance, and the sample variance. Treat every record as one observation; no parametric distribution is assumed.
Step 1: Use record counts, including the ties. There are five observations. The values \(0\) and \(5\) each occur twice, and \(10\) occurs once. The total is \(0+0+5+5+10=20\), so \(m'_1=\bar{x}=20/5=4\) thousand dollars.
Step 2: Calculate the second raw moment.
Step 3: Calculate the third raw moment.
Step 4: Find both variances. The empirical variance is \(m'_2-\bar{x}^{\,2}=30-4^2=14\), in squared thousands of dollars. The sum of squared deviations is \(5(14)=70\). Therefore the sample variance is \(70/(5-1)=17.5\), also in squared thousands of dollars.
Check. The deviations from \(4\) are \(-4,-4,1,1,6\); their squares sum to \(16+16+1+1+36=70\). Dividing by \(5\) gives \(14\), and dividing by \(4\) gives \(17.5\). The repeated zero and five losses each contribute twice, rather than being counted once as distinct values.
Method and trap avoided: Counting repeated observations explicitly gives the same result as weighting each distinct value by its count. Treating the three distinct amounts as three equally weighted observations would describe a different empirical distribution.
Comparing the Two Variance Conventions
| Summary | Calculation from exact observations | Interpretation |
|---|---|---|
| Empirical variance | \((1/n)\sum_j(x_j-\bar{x})^2\) | Variance of the empirical distribution that assigns mass \(1/n\) to every record |
| Sample variance | \((1/(n-1))\sum_j(x_j-\bar{x})^2\), for \(n>1\) | Conventional unbiased estimator of an underlying variance under independent, identically distributed sampling with finite variance |
Since both quantities use the same numerator, their relationship is \(s^2=(n/(n-1))\) times the empirical variance when \(n>1\). Thus the sample variance is larger for a nonconstant data set. If all observations are equal, both are zero. With \(n=1\), the empirical variance is zero, but the sample variance convention is undefined because its denominator is zero.
Worked Example: A Second Calculation and a Consistency Check
A hypothetical claims analyst records exact losses of \(1, 3, 3, 7, 11\) dollars. Calculate the mean, second raw moment, empirical variance, and sample variance. The empirical distribution places equal mass on the five records, and no parametric distribution is assumed.
Step 1: Calculate the mean. The total loss is \(1+3+3+7+11=25\), and \(n=5\). Therefore \(\bar{x}=25/5=5\) dollars.
Step 2: Calculate the second raw moment.
This is \(37.8\) dollars squared. The empirical variance is \(37.8-5^2=37.8-25=12.8\) dollars squared.
Step 3: Calculate the sample variance. The sum of squared deviations is \(5(12.8)=64\). Dividing by \(n-1=4\) gives \(s^2=64/4=16\) dollars squared.
Check. Deviations from \(5\) are \(-4,-2,-2,2,6\), with squared deviations \(16,4,4,4,36\). They sum to \(64\), confirming both variance calculations.
Method and trap avoided: The two records of \(3\) are both included in the sum of squares. Also, the mean is \(5\), not the median; substituting a different center into the variance calculation would not give either defined variance.
Common Exam Traps and Practical Checks
- Confusing a raw moment with a central moment. The second raw moment is the average of \(x_j^2\); the variance is the average of \((x_j-\bar{x})^2\) for the empirical distribution.
- Using the wrong divisor. Use \(n\) for the variance of the empirical distribution and \(n-1\) for the usual sample variance. Read the requested convention rather than choosing one by habit.
- Dropping tied records. A repeated observation still contributes to the total, the sum of squares, and the record count.
- Mixing units. If losses are in dollars, a variance is in dollars squared. If losses are in thousands of dollars, the variance is in squared thousands of dollars; converting a variance to dollars squared requires multiplying by \(1{,}000{,}000\).
- Rounding too early. Keep fractional means and exact sums through the variance calculation, then round only if the question asks for a rounded result.
A quick consistency check is to compare the two variance calculations. The sum of squared deviations must equal \(n(m'_2-\bar{x}^{\,2})\), and the sample variance must equal that sum divided by \(n-1\). A negative variance indicates an arithmetic or transcription error, because every squared deviation is nonnegative.
Check Your Understanding
For each set, calculate the requested summaries from the exact observations. Treat each record as one equally weighted observation; no parametric distribution is assumed.
- Exact losses are \(3,3,9\) dollars. Find the mean, second raw moment, empirical variance, and sample variance. (Answer: mean \(=5\) dollars; second raw moment \(=33\) dollars squared; empirical variance \(=8\) dollars squared; sample variance \(=12\) dollars squared.)
- Exact losses are \(4,6,10,12\) dollars. Find the mean, second raw moment, empirical variance, and sample variance. (Answer: mean \(=8\) dollars; second raw moment \(=74\) dollars squared; empirical variance \(=10\) dollars squared; sample variance \(=40/3\) dollars squared.)
- Exact losses are \(0,2,2,6,10\) dollars. Find the mean, second raw moment, empirical variance, and sample variance. (Answer: mean \(=4\) dollars; second raw moment \(=28.8\) dollars squared; empirical variance \(=12.8\) dollars squared; sample variance \(=16\) dollars squared.)
- Exact losses are \(1,1,4,8\) thousand dollars. Find the empirical mean and both variances, reporting the variances first in squared thousands of dollars and then in dollars squared. (Answer: mean \(=3.5\) thousand dollars; empirical variance \(=8.25\) squared thousands of dollars \(=\$8{,}250{,}000\); sample variance \(=11\) squared thousands of dollars \(=\$11{,}000{,}000\).)
- There are \(n=6\) exact losses, with mean \(7\) dollars and sum of squared losses \(\sum_jx_j^2=390\) dollars squared. Find the empirical variance and sample variance without reconstructing the individual losses. (Answer: second raw moment \(=65\) dollars squared; empirical variance \(=16\) dollars squared; sample variance \(=19.2\) dollars squared.)