Empirical Claim-Count Distributions

From Policy Records to a Claim-Count Distribution

A claim-frequency estimate such as an average number of claims per policy-period summarizes a data set with one number. An empirical claim-count distribution retains more information: it estimates the probability of each possible count, including zero. That lets an actuary answer questions such as the observed chance of no claims, the proportion of records with at least two claims, and the variability of counts.

The basic unit here is one policy observed over one complete period of the same length, such as one policy-year. Each record contributes one nonnegative integer claim count. The records should refer to comparable policy periods if their observed proportions are to be used as estimates for a common target group. A period with no claims is still a complete observation and must be included.

Definition: Suppose $n$ comparable policy-period records have observed claim counts $N_1,\ldots,N_n$, where each $N_r$ is a nonnegative integer. Let $n_j$ be the number of records with exactly $j$ claims. The empirical probability assigned to count $j$ is $$ \widehat{p}_j=\frac{n_j}{n}. $$ The empirical distribution places probability $\widehat{p}_j$ on each observed count $j$. Since every record has exactly one count, $\sum_j n_j=n$ and $\sum_j\widehat{p}_j=1$.

The empirical probabilities describe the observed sample exactly. They are also commonly used as estimates of the corresponding probabilities for future comparable policy-periods. That interpretation requires care: the data are not a guarantee that future proportions will match, and a count absent from a finite sample is not necessarily impossible in the underlying population.

Building the Distribution and Calculating Moments

A frequency table is often the clearest starting point. List each distinct observed count, tally how many records have that count, and divide each tally by the total number of records. The denominator is the number of all policy-period records, not just the number with claims.

For an empirical distribution, the mean and variance are the ordinary mean and variance of the recorded counts, using divisor $n$ for the empirical variance. Equivalently, they are the first and second moments calculated from the empirical probabilities:

\widehat{E}[N]=\sum_j j\widehat{p}_j =\frac{1}{n}\sum_{r=1}^{n}N_r, \qquad \widehat{\operatorname{Var}}(N) =\sum_j (j-\widehat{E}[N])^2\widehat{p}_j =\sum_j j^2\widehat{p}_j-\bigl(\widehat{E}[N]\bigr)^2.

These are descriptive moments of the empirical distribution. The empirical variance uses divisor $n$ because each of the $n$ observations receives probability $1/n$ in that distribution. It is not the alternative sample-variance calculation with divisor $n-1$.

Why the moment formulas hold: Under the empirical distribution, each recorded count $N_r$ has probability $1/n$. Thus the expectation of any function $g(N)$ is the average of that function over the records: $$ \widehat{E}[g(N)]=\frac{1}{n}\sum_{r=1}^{n}g(N_r) =\sum_j g(j)\frac{n_j}{n}. $$ Taking $g(N)=N$ gives the empirical mean. Taking $g(N)=N^2$ gives the second moment, and variance equals the second moment minus the square of the mean.

Tail probabilities are found by adding the empirical probabilities for the counts in the requested event. In particular, “at least two” means $N\geq 2$, while “more than two” means $N>2$; these are different whenever the count of exactly two is positive.

Worked Example: Constructing a Distribution From Raw Records

A hypothetical insurer reviews 10 policies, each observed for one complete year. Their annual claim counts, in order, are $0,1,0,2,1,0,3,2,1,0$. Construct the empirical distribution, then calculate its mean, variance, and probability of at least two claims.

Method. Tally each count across all 10 records, including those with zero claims:

Annual count $j$	Number of records $n_j$	Empirical probability $n_j/10$
0	4	0.4
1	3	0.3
2	2	0.2
3	1	0.1

The tallies add to $4+3+2+1=10$, and the probabilities add to $0.4+0.3+0.2+0.1=1.0$, as required.

Mean. Multiply each count by its empirical probability and add: $$ \widehat{E}[N] =0(0.4)+1(0.3)+2(0.2)+3(0.1) =0+0.3+0.4+0.3=1.0. $$

Variance. First calculate the second moment: $$ \widehat{E}[N^2] =0^2(0.4)+1^2(0.3)+2^2(0.2)+3^2(0.1) =0+0.3+0.8+0.9=2.0. $$ Therefore, $$ \widehat{\operatorname{Var}}(N)=2.0-(1.0)^2=1.0. $$ Tail probability. Counts of at least two are 2 and 3, so $$ \widehat{P}(N\geq2)=0.2+0.1=0.3. $$ The results are consistent with the raw records: 3 of the 10 policies had at least two claims, and the average count is 1.0 per policy-year. The key mistake avoided is dropping the four zero-claim records from the denominator.

Exact Counts, Ranges, and Conditional Questions

Once the empirical probabilities have been formed, probability questions become sums of the relevant cells in the table. For example, $P(N=1)$ is the probability in the one-claim row, and $P(N\leq1)$ is the sum of the zero- and one-claim rows. If a question conditions on a subset of records, first restrict to that subset and then use its size as the denominator.

Worked Example: Distinguishing Two Upper Tails

A hypothetical review contains 12 complete, equal-length policy-period records. The observed frequency table is: 4 records with 0 claims, 3 with 1 claim, 3 with 2 claims, and 2 with 3 claims. Find the empirical mean and variance, then calculate the probabilities of more than one claim and more than two claims.

Method. The empirical probabilities are $4/12,3/12,3/12,$ and $2/12$. They sum to $12/12=1$.

Mean. Using the frequency table directly, $$ \widehat{E}[N] =\frac{0(4)+1(3)+2(3)+3(2)}{12} =\frac{0+3+6+6}{12} =\frac{15}{12}=1.25. $$

Variance. The second moment is $$ \widehat{E}[N^2] =\frac{0^2(4)+1^2(3)+2^2(3)+3^2(2)}{12} =\frac{0+3+12+18}{12} =\frac{33}{12}=2.75. $$ Thus, $$ \widehat{\operatorname{Var}}(N)=2.75-(1.25)^2 =2.75-1.5625=1.1875. $$

Tail probabilities. More than one claim means counts 2 or 3: $$ \widehat{P}(N>1)=\frac{3+2}{12}=\frac{5}{12}\approx0.4167. $$ More than two claims means only count 3: $$ \widehat{P}(N>2)=\frac{2}{12}=\frac{1}{6}\approx0.1667. $$ The results are checked against the table: five records exceed one claim, but only two exceed two claims. The mistake avoided is treating “more than two” as “at least two,” which would incorrectly include all three records with exactly two claims.

Using the Empirical Distribution for Comparable Policies

An empirical distribution can support simple projections when the future policies are comparable to those observed and the same policy-period definition is used. For instance, multiplying an estimated probability by a future policy count gives the expected number of policies in that event. Calculating a portfolio probability such as “none of five policies has two or more claims” requires an additional assumption: the five future policy counts must be independent, with the same empirical probabilities used for each policy.

Worked Example: Projecting a Tail Event to Future Policies

A hypothetical sample of 20 policies, each observed for one full year, has 8 zero-claim records, 6 one-claim records, 4 two-claim records, and 2 three-claim records. Use the empirical distribution to find (a) the expected number of policies with at least two claims among 30 comparable future policies, and (b) the probability that none of five independent comparable policies has at least two claims.

Build the relevant probabilities. The sample size is $8+6+4+2=20$. The empirical probability of at least two claims is $$ \widehat{P}(N\geq2)=\frac{4+2}{20}=\frac{6}{20}=0.3, $$ so the empirical probability of fewer than two claims is $1-0.3=0.7$.

(a) Expected number among 30. For each policy, the estimated probability of at least two claims is 0.3. By adding the 30 indicator expectations, the expected number is $$ 30(0.3)=9. $$ This expectation does not require independence among the future policies.

(b) Probability none among five. Here independence is specified. Each policy has probability 0.7 of fewer than two claims, so the probability that all five do is $$ (0.7)^5=0.16807. $$ To four decimal places, this is $0.1681$. The result is checked by noting that “none has at least two” means all five have counts 0 or 1. The mistake avoided is using $0.3^5$, which would describe all five having at least two claims instead.

Common Data and Interpretation Traps

The empirical distribution is straightforward to calculate, but a few choices determine whether the result answers the intended question.

Omitting zero-claim periods. Every fully observed policy-period belongs in the denominator, including a period with no claims. Using only claim-positive records estimates a different, conditional distribution.
Changing the event boundary. $N\geq2$ includes count 2; $N>2$ does not. Translate the words into the inequality before adding probabilities.
Confusing an empirical probability with certainty about the population. A count not present in the sample receives zero probability in the empirical distribution, but it may still occur in future data.
Comparing unlike observation periods as if they were equivalent. A six-month record and a full-year record do not represent the same observation window. The direct proportions taught here rely on records of equal length; unequal exposure periods require a different treatment.
Using the wrong variance convention. The variance of the empirical distribution uses divisor $n$. If a question instead asks for an unbiased sample-variance estimate, that is a distinct calculation and should not be substituted without checking the wording.

Key takeaway: For comparable, equal-length policy-period records, divide each claim-count frequency by the total number of records, retain the zero-claim observations, and calculate moments or tail probabilities from the resulting probability table. Always translate a tail phrase into an exact count event before summing.

Check Your Understanding

Use the complete policy-period records stated in each problem. Give probabilities to four decimal places where needed.

A hypothetical set of eight equal-length policy records has counts $0,0,1,1,2,0,3,1$. Find the empirical mean and the probability of at least two claims. (Answer: mean $=1.0000$; $P(N\geq2)=0.2500$)
A frequency table from 15 comparable records has 6 zero-claim, 5 one-claim, 3 two-claim, and 1 four-claim periods. Find the empirical variance and $P(N>1)$. (Answer: variance $=1.2000$; $P(N>1)=4/15\approx0.2667$)
Among 24 equal-length records, the frequencies for counts 0, 1, 2, and 3 are 9, 8, 5, and 2. Find $P(N\leq1)$ and the empirical conditional probability $P(N=3\mid N\geq2)$. (Answer: $P(N\leq1)=17/24\approx0.7083$; $P(N=3\mid N\geq2)=2/7\approx0.2857$)
A sample of 50 comparable annual records has frequencies 20, 15, 10, and 5 for counts 0, 1, 2, and 3, respectively. Using its empirical distribution, find the expected number of policies with at least two claims in a group of 80 comparable policies. (Answer: $24$)
A sample of 10 comparable records has frequencies 5, 3, and 2 for counts 0, 1, and 2. Assuming four future policy counts are independent and each has the empirical distribution, find the probability that at least one has two claims. (Answer: $1-(0.8)^4=0.5904$)

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