Comparing Frequency-Severity and Aggregate Loss Estimates

Why Compare the Two Estimates?

A projected loss cost can be built from claim components or from total losses. The frequency-severity method estimates claims per exposure and average loss per claim, then multiplies those selections. The aggregate method estimates total losses per exposure directly. When the two methods use the same claims, exposure, loss basis, and assumptions, they are linked by an exact arithmetic identity. A difference between estimates is therefore a useful prompt to check the data and selections—not proof, by itself, that either estimate is wrong.

This comparison is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus. It focuses on the relationship between the estimates and on diagnosing differences, rather than re-teaching the earlier material on selecting frequency and severity trends.

Definition: For a defined group and experience period, let $e$ be the number of exposure units, $n$ the number of claims, and $L$ the total loss amount for those claims. On a consistent basis, claim frequency is $n/e$, average severity is $L/n$ when $n>0$, aggregate loss per exposure is $L/e$, and the frequency-severity estimate is $(n/e)(L/n)$. Exposure units, claims, and losses must refer to the same group, period, and coverage basis.

In a projection, selections may replace the observed ratios. Write $\hat f$ for the selected claim frequency per exposure and $\hat s$ for the selected average severity per claim. The projected frequency-severity loss estimate per exposure is $\hat f\hat s$. Write $\hat p_A$ for a selected aggregate loss estimate per exposure. The estimates agree when $\hat p_A=\hat f\hat s$; they need not agree when the selections are made independently or use different adjustments.

The Identity and Its Conditions

The equality for observed experience follows by cancellation. Provided $n>0$, multiplying the observed frequency by the observed severity gives:

\frac{n}{e}\times\frac{L}{n}=\frac{L}{e}.

The left-hand side first allocates the total losses across claims to obtain average loss per claim, then scales that amount by claims per exposure. The right-hand side divides the same total losses directly by the same exposure count. Thus, no assumption about a probability distribution for claim counts or claim sizes is needed for this descriptive identity. It is an arithmetic result, not a claim that every future estimate must match.

The identity requires the same $n$, $L$, and $e$ on both sides. It can fail as a comparison if, for example, frequency uses developed ultimate claim counts while aggregate loss uses reported losses that have not been brought to the same valuation basis. It can also fail if severity uses limited losses but the aggregate estimate uses unlimited losses. In those cases, the estimates are answering different questions.

Key result: For observed experience with the same claims, losses, and exposure, aggregate loss per exposure equals observed frequency times observed severity. For selected projections, agreement requires compatible selections: the aggregate estimate must represent the same projected claim volume and average cost as the frequency and severity selections.

Worked Comparisons

Worked Example: Verifying Agreement on One Experience Basis

A hypothetical small-business property portfolio earned 2,000 policy-years. On a common ultimate and unlimited-loss basis, it had 240 claims with total losses of $1,080,000. Calculate the observed frequency-severity estimate and aggregate loss per exposure, then check whether they agree.

Step 1: Calculate claim frequency. Divide claims by policy-years:

\frac{n}{e}=\frac{240}{2{,}000}=0.12\text{ claims per policy-year}.

Step 2: Calculate average severity. Divide the same total losses by the same claims:

\frac{L}{n}=\frac{\$1{,}080{,}000}{240}=\$4{,}500\text{ per claim}.

Step 3: Multiply the components.

0.12\times\$4{,}500=\$540\text{ per policy-year}.

Step 4: Calculate the aggregate estimate directly.

\frac{L}{e}=\frac{\$1{,}080{,}000}{2{,}000}=\$540\text{ per policy-year}.

The estimates agree, as they must when they use the same counts, losses, and exposure. The calculation checks: 240 claims at $4,500 each total $1,080,000, and $540 multiplied by 2,000 policy-years also totals $1,080,000. A common mistake is to divide losses by exposure to obtain $540, then call that amount “severity.” It is loss per policy-year; severity is $4,500 per claim.

Worked Example: Comparing Projected Estimates With Different Selections

A hypothetical equipment insurer’s historical experience is 1,000 exposure units, 100 claims, and $500,000 of losses. For a future period, an analyst selects a 5% frequency increase and a $6\tfrac{2}{3}\%$ severity increase relative to the observed components. A separate aggregate analysis selects an 8% increase relative to historical aggregate loss per exposure. Compare the projections.

Step 1: Establish the historical components.

f_0=\frac{100}{1{,}000}=0.10,\qquad s_0=\frac{\$500{,}000}{100}=\$5{,}000,\qquad p_0=\frac{\$500{,}000}{1{,}000}=\$500.

Step 2: Project the components. The selected future frequency is $0.10(1.05)=0.105$ claims per exposure. The selected future severity is $\$5{,}000(1+\tfrac{2}{30})=\$5{,}333.33$ per claim, rounded to the nearest cent only for display.

\hat p_{FS}=0.105\times\left(\$5{,}000\times\frac{16}{15}\right)=\$560.

The exact severity is $\$5{,}333.\overline{3}$; using that value before rounding gives a frequency-severity estimate of $560 per exposure.

Step 3: Project the aggregate estimate.

\hat p_A=\$500(1.08)=\$540\text{ per exposure}.

Step 4: Quantify the difference. The component estimate is $20 higher per exposure. Relative to the aggregate selection, the difference is:

\frac{\$560-\$540}{\$540}=0.037037\ldots\approx 3.70\%.

These estimates differ because the selected component increases combine to $1.05\times\tfrac{16}{15}=1.12$, a 12% increase, while the aggregate selection is an 8% increase. If the aggregate method instead applied the combined component factor, it would give $\$500(1.12)=\$560$, agreeing with the frequency-severity projection. The common trap is to treat the separately selected 8% aggregate increase as if it automatically represented the component selections. It does not.

Worked Example: Checking an Overall Estimate From Two Groups

A hypothetical insurer has two coverage groups. Group A has 1,000 exposure units, 100 claims, and $400,000 in losses. Group B has 500 exposure units, 100 claims, and $900,000 in losses. Calculate the overall frequency-severity estimate and compare it with total losses per exposure.

Step 1: Combine the data before calculating overall components. Total exposure is $1{,}000+500=1{,}500$, total claims are $100+100=200$, and total losses are $\$400{,}000+\$900{,}000=\$1{,}300{,}000$.

Step 2: Calculate overall frequency and severity.

f=\frac{200}{1{,}500}=\frac{2}{15},\qquad s=\frac{\$1{,}300{,}000}{200}=\$6{,}500.

Step 3: Multiply the overall components.

\frac{2}{15}\times\$6{,}500=\$866.67\text{ per exposure, rounded to cents}.

Step 4: Check against the aggregate calculation.

\frac{\$1{,}300{,}000}{1{,}500}=\$866.67\text{ per exposure, rounded to cents}.

The estimates agree. Each group can also be checked separately: Group A has frequency $100/1{,}000=0.10$, severity $\$400{,}000/100=\$4{,}000$, and loss per exposure of $400. Group B has frequency $100/500=0.20$, severity $9,000, and loss per exposure of $1,800. A common mistake is to average the two group frequencies and the two group severities without regard to group size or claim counts. The overall calculation instead uses total claims over total exposure and total losses over total claims; those denominators preserve the aggregate identity.

Why Estimates Can Differ

A difference is meaningful only after confirming that each estimate refers to the same target. Useful checks include:

Exposure and population: Are the same coverage groups, exposure definition, and period included? Are component selections weighted consistently with the aggregate population?
Claim count and loss maturity: Are counts and losses both developed to the same ultimate basis, or are both observed at a comparable valuation date? A developed count paired with an undeveloped loss total does not reproduce a matched aggregate.
Loss definition: Are both estimates gross or net of the same adjustments, and are they both limited or unlimited on a compatible basis? A limited severity should not be multiplied by a frequency and compared with an aggregate amount that includes losses above the limit.
Time basis: Are both estimates for the same experience or projection period? Frequency and severity trends can be applied separately; an aggregate projection needs a selection consistent with their combined effect if agreement is intended.
Selection method: Were frequency and severity selected independently of the aggregate estimate? Different judgments, smoothing, or data exclusions can produce a real difference between selected estimates even when the historical identity holds exactly.

When the bases match but selections do not, report the difference as a difference in selected estimates and identify which assumptions drive it. Do not force the arithmetic to agree by silently changing a component. The comparison is most useful when it distinguishes a calculation error or basis mismatch from a genuine difference in judgment.

Align the bases.
Confirm that claims, losses, exposure, period, development, and loss limits describe the same experience.

Recompute the observed identity.
Calculate claims per exposure, losses per claim, and total losses per exposure from the same data.

Compare selections separately.
Multiply selected frequency by selected severity and set the result beside the selected aggregate estimate.

Explain any remaining gap.
Trace it to differing selections or assumptions, or correct a calculation or basis mismatch if one is found.

Exam trap: The observed identity does not guarantee that two independently selected projections agree. First verify $(n/e)(L/n)=L/e$ on the actual data. Then explain any projected difference using the selections and their bases; do not confuse historical arithmetic agreement with agreement of future estimates.

Check Your Understanding

For each problem, compare the frequency-severity and aggregate calculations on the stated basis.

A hypothetical travel policy block has 800 exposure units, 96 claims, and $432,000 in losses, all on a common basis. Calculate frequency, severity, and loss per exposure. (Answer: frequency $0.12$, severity $4,500 per claim, and both loss estimates $540 per exposure.)
A hypothetical machinery portfolio has 2,500 exposure units, 275 claims, and $1,237,500 in losses. Calculate the overall components and verify the aggregate estimate. (Answer: frequency $0.11$, severity $4,500 per claim, and both estimates $495 per exposure.)
Historical frequency is $0.08$ claims per exposure and historical severity is $5,000 per claim. Selected future increases are 3% for frequency and 4% for severity. Calculate the future frequency-severity estimate and the combined increase over the historical aggregate estimate. (Answer: $0.0824$ claims per exposure, $5,200 severity, and $428.48 per exposure, a 7.12% increase over $400.)
Group A has 600 exposure units, 48 claims, and $192,000 in losses; Group B has 400 exposure units, 80 claims, and $720,000 in losses. Calculate the overall frequency-severity estimate and aggregate loss per exposure. (Answer: overall frequency $0.128$, severity $7,125, and both estimates $912 per exposure.)
A hypothetical book has historical aggregate loss of $720,000 over 1,200 exposure units. Selected future frequency is $0.11$ claims per exposure, selected severity is $6,600, and the aggregate method selects $690 per exposure. Calculate the component estimate, its dollar difference from the aggregate selection, and the percentage difference relative to the aggregate selection. (Answer: component estimate $726 per exposure, $36 higher, or approximately 5.22% higher.)

What You'll Learn