Empirical Probability Masses From Loss Data

Why Assign Empirical Probability Masses?

A set of exact loss records contains repeated values as well as different loss amounts. To summarize the observed sample as a finite set of possible loss values, assign each distinct value a probability mass equal to the proportion of records with that value. This gives every record equal weight while ensuring that repeated losses contribute repeatedly to the mass.

The calculation is simple, but the bookkeeping matters: the denominator is the number of records, while the numerator for a loss value is the number of records equal to that value. This tutorial is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Suppose there are $n$ complete exact loss records $x_1,\ldots,x_n$. Let $z_1,\ldots,z_k$ be the distinct loss amounts appearing in those records, and let $n_j$ be the number of records equal to $z_j$. The empirical probability mass assigned to $z_j$ is $$ \widehat{p}_j=\frac{n_j}{n}, \qquad j=1,\ldots,k. $$ Here $n=\sum_{j=1}^{k}n_j$. The hat indicates a quantity calculated from the observed data.

The $x_i$ are the individual records; the $z_j$ are the distinct values those records take. If a loss of $500 appears four times, it is one distinct loss amount with frequency four—not four distinct amounts. The masses describe the empirical distribution that gives equal weight to each observed record. They are not, by themselves, a guarantee that the unknown population assigns exactly those probabilities to the loss values.

Counting Records and Assigning Mass

A reliable method is to list the distinct loss values in order, count how often each one occurs, and divide each count by the total record count. Sorting the values is not essential to the arithmetic, but it makes it easier to see whether a value has been omitted or counted twice.

Count all records.
Set $n$ equal to the number of exact loss observations, not the number of distinct values.

List distinct amounts and count them.
For each distinct value $z_j$, record its frequency $n_j$.

Divide each frequency by $n$.
Calculate $\widehat{p}_j=n_j/n$, retaining fractions where useful.

Check the count and mass totals.
Verify that the frequencies add to $n$ and the exact masses add to one.

The calculation assumes a finite sample of complete, exact loss observations and equal weight for each record. The loss amounts must use a consistent unit and basis. For example, one should not combine amounts recorded in dollars with amounts recorded in thousands of dollars without converting them first. A censored or otherwise incomplete record is not an exact loss value and cannot simply be counted as though its loss were known. This basic frequency calculation also does not adjust for unequal sampling weights.

Why the Masses Sum to One

Each record belongs to exactly one of the distinct loss values observed in the sample. Consequently, the groups of records counted by $n_1,\ldots,n_k$ do not overlap, and together they include all $n$ records. Dividing the total count by $n$ proves the normalization property.

\sum_{j=1}^{k}\widehat{p}_j =\sum_{j=1}^{k}\frac{n_j}{n} =\frac{\sum_{j=1}^{k}n_j}{n} =\frac{n}{n} =1.

This is both a justification and a practical error check. If the frequencies do not add to $n$, a record was omitted or counted more than once. If the frequencies add to $n$ but the exact masses do not add to one, there is an arithmetic or denominator error. Rounded decimal masses may total slightly above or below one; check the unrounded fractions before concluding that the counts are wrong.

Worked Example: Assigning Masses to Repeated Exact Losses

A hypothetical insurer records 12 complete exact losses, in dollars: $120, $80, $120, $200, $80, $120, $160, $80, $120, $200, $80, and $120. Assign empirical probability mass to each distinct loss. Show the count check and the mass check.

Method. Count the records equal to each distinct loss, then divide each frequency by the total $n=12$. The distinct losses in increasing order are $80, $120, $160, and $200.

The frequencies are:

$80 appears in records 2, 5, 8, and 11: $n_{80}=4$.
$120 appears in records 1, 3, 6, 9, and 12: $n_{120}=5$.
$160 appears in record 7: $n_{160}=1$.
$200 appears in records 4 and 10: $n_{200}=2$.

The count check is $4+5+1+2=12$, as required. Divide each count by 12:

\widehat{p}_{80}=\frac{4}{12}=\frac{1}{3},\quad \widehat{p}_{120}=\frac{5}{12},\quad \widehat{p}_{160}=\frac{1}{12},\quad \widehat{p}_{200}=\frac{2}{12}=\frac{1}{6}.

The exact mass check is \[ \frac{4}{12}+\frac{5}{12}+\frac{1}{12}+\frac{2}{12} =\frac{12}{12}=1. \] To four decimal places, the masses are 0.3333, 0.4167, 0.0833, and 0.1667, respectively; their rounded sum is 0.9999. The exact fractions, not the rounded decimals, establish that the masses sum to one.

Common mistake avoided. Dividing by the four distinct amounts would give $4/4$ as the mass of $80, which is not a valid frequency calculation. The denominator is the 12 records, not the four distinct loss values.

Reading a Frequency Table Carefully

A frequency table makes the calculation more efficient when the raw records have already been summarized. Still, the table must represent all records in the stated sample, and each distinct loss value must appear once as a row. The frequency column counts records, not loss dollars. A large loss amount does not receive extra mass merely because its dollar value is large; its mass depends on how many records have that amount.

Distinct loss $z_j$	Frequency $n_j$	Empirical mass $n_j/n$
One observed amount	Number of records equal to it	Frequency divided by total records
A second observed amount	Number of records equal to it	Frequency divided by total records
All remaining distinct amounts	Count each value once per matching record	Each frequency divided by the same total

Worked Example: Converting a Loss Frequency Table

For a hypothetical equipment-protection plan, a data extract lists the following exact claim amounts and record counts: $300 appears twice, $450 appears four times, $600 appears twice, and $900 appears twice. Determine the total number of records and the empirical mass for each amount.

Method. The total sample size is the sum of the frequencies:

$$ n=2+4+2+2=10. $$

Use this same denominator for every distinct amount:

For $300, $\widehat{p}_{300}=2/10=0.2$.
For $450, $\widehat{p}_{450}=4/10=0.4$.
For $600, $\widehat{p}_{600}=2/10=0.2$.
For $900, $\widehat{p}_{900}=2/10=0.2$.

The masses sum to $0.2+0.4+0.2+0.2=1.0$. Equivalently, the exact sum is $(2+4+2+2)/10=10/10=1$. The mass of $450 is largest because four of the ten records have that exact amount.

Common mistake avoided. A frequency of two does not mean a mass of $2/4$, where four is the number of rows in the table. There are ten records in total, so each frequency is divided by ten.

Complete the Bookkeeping Before Rounding

Exact fractions make the normalization check transparent. If a decimal answer is required, round only after calculating the masses. For instance, converting each fraction to a short decimal and then adding those rounded numbers can create a small discrepancy. That discrepancy is a rounding effect when the frequencies are complete and the exact fractions sum to one.

A different discrepancy can reveal a genuine data-handling problem. If the listed frequencies add to less than the stated sample size, some records are missing from the table. If they add to more, at least one record has been counted twice or an incorrect total has been used. Likewise, a table that includes only selected distinct amounts cannot represent all empirical mass unless the omitted records are also accounted for. In the basic calculation here, all observed exact losses must be included.

Worked Example: Finding a Missing Record in a Mass Table

A hypothetical repair program has 15 complete exact loss records. A preliminary count gives the following frequencies: $250 appears four times, $500 appears five times, $750 appears three times, $1,000 appears twice, and $1,500 appears once. Calculate the masses and verify the result.

Method. First add the frequencies:

$$ 4+5+3+2+1=15. $$

The total matches the stated sample size. Now divide each count by $n=15$:

$250: $4/15$.
$500: $5/15=1/3$.
$750: $3/15=1/5$.
$1,000: $2/15$.
$1,500: $1/15$.

Adding the exact masses gives \[ \frac{4+5+3+2+1}{15}=\frac{15}{15}=1. \] To four decimal places, the masses are 0.2667, 0.3333, 0.2000, 0.1333, and 0.0667. Their rounded sum is 1.0000. The mass $5/15$ assigned to $500 is the largest because that amount occurs most often.

Common mistake avoided. Counting the five rows as five records would use the wrong denominator. The table has five distinct loss amounts, but the observed sample contains 15 records.

Common Errors to Avoid

Using the number of distinct values as the denominator. The denominator is $n$, the total number of records.
Assigning a separate mass to each repeated record. Repeated records with the same exact amount are combined into one distinct value whose frequency is their count.
Omitting a distinct value. The frequency groups must cover all records before their masses can sum to one.
Mixing units or bases. Convert loss amounts to consistent units before treating records as the same exact value.
Confusing rounding with a count error. Check the exact fractions first; rounded decimal masses can differ slightly from a total of one.
Treating incomplete records as exact losses. The frequency formula applies directly to complete exact observations, not to records whose loss amount is unknown or only partially observed.

Key takeaway: For each distinct exact loss $z_j$, count its records $n_j$ and assign mass $n_j/n$, where $n$ is the total number of records. The frequencies must add to $n$, so the exact empirical masses add to one.

Check Your Understanding

For each problem, find the empirical mass for every distinct exact loss and use the record count to check the total.

A hypothetical bicycle insurer records exact losses of $40, $60, $40, $100, $60, and $40. List the distinct amounts and assign each its empirical mass. (Answer: $40 has mass $3/6=1/2$, $60 has mass $2/6=1/3$, and $100 has mass $1/6$; the masses sum to 1.)
A frequency table for a hypothetical appliance plan has $200 with frequency 3, $350 with frequency 1, and $500 with frequency 4. Find the total number of records and all three masses. (Answer: $n=8$; the masses are $3/8$, $1/8$, and $4/8=1/2$, which sum to 1.)
A hypothetical travel policy has 12 exact loss records: $125 occurs three times, $250 occurs twice, $375 occurs five times, and $750 occurs twice. Give each mass as a fraction and identify the largest mass. (Answer: $3/12=1/4$, $2/12=1/6$, $5/12$, and $2/12=1/6$, respectively; $375 has the largest mass, and the total is 1.)
A hypothetical tool-replacement program has 10 records. A draft table lists $80 with frequency 2, $160 with frequency 3, and $240 with frequency 4. Explain why these entries cannot yet be a complete frequency table, and state the mass of each listed amount if one additional record is found to be a distinct $320 loss. (Answer: The listed frequencies total 9, not 10, so one record is missing. With $320 at frequency 1, the masses are $2/10$, $3/10$, $4/10$, and $1/10$, summing to 1.)
A hypothetical small-business policy has 20 exact loss records: $1,000 occurs six times, $2,000 occurs six times, $3,000 occurs four times, and $5,000 occurs four times. A candidate divides each frequency by four because there are four distinct amounts. Correct the calculation and verify it. (Answer: The denominator is 20 records, not 4 distinct amounts. The masses are $6/20=0.3$, $6/20=0.3$, $4/20=0.2$, and $4/20=0.2$; their sum is 1.)

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