Tutorials › Exam ASTAM - Advanced Short-Term Actuarial Mathematics › Experience Rating with Claim Frequency

Home  /  Tutorials  /  Exam ASTAM / Limited-Fluctuation Credibility

Exam ASTAM · Limited-Fluctuation Credibility · Tutorial 509 of 979

Experience Rating with Claim Frequency

Calculate a credible prospective claim frequency from observed claims, exposure, a full-credibility standard, and a manual rate.

Beginner 10 min read

What You'll Learn

  • Calculate observed claim frequency from claim counts and exposure.
  • Apply the square-root partial credibility rule using a full-credibility exposure standard.
  • Cap the credibility factor at one when exposure meets or exceeds the standard.
  • Blend the experience frequency with a manual frequency to obtain a prospective rate.
  • Check that the count, exposure, manual rate, and indicated rate use consistent units.
  • Identify errors caused by dividing claims by the wrong exposure or applying the credibility factor incorrectly.

From Claim Counts to a Prospective Frequency

A frequency rating starts with a simple ratio: the number of claims divided by the amount of exposure over which they occurred. That ratio describes past experience, but it can be unstable when exposure is limited. Limited-fluctuation credibility addresses this by giving less weight to a small experience sample and more weight to a manual frequency representing the broader rating basis.

The calculation has three distinct inputs: observed claims, exposure, and a manual frequency. The credibility standard connects the amount of experience to the credibility weight. Keeping these roles separate helps prevent a common error: treating a claim count as if it were already a frequency.

Definition: Let \(N\) be the observed claim count over \(E\) exposure units, such as policy-years. The observed claim frequency is \(\hat{\lambda}=N/E\), measured in claims per exposure unit. Let \(\lambda_M\) be the manual frequency on the same basis, and let \(E_0\) be the full-credibility exposure standard, measured in the same exposure units as \(E\).

For this tutorial, assume exposure units are comparable and the claim count follows a homogeneous Poisson model, \(N\sim\operatorname{Poisson}(\lambda E)\), where \(\lambda\) is the underlying frequency per exposure unit. Exposure is treated as known and fixed. The model excludes changes in coverage, claim definition, or risk composition that would make past and prospective frequencies incomparable. No interest or mortality assumption is needed for these frequency calculations.

The full-credibility standard \(E_0\) is taken as given. It is the exposure level at which the applicable limited-fluctuation standard grants full credibility. Below that level, the conventional square-root rule gives partial credibility; at or above it, credibility is capped at one.

Limited-fluctuation frequency rule: For a full-credibility exposure standard \(E_0\), the credibility factor is \(Z=\min(1,\sqrt{E/E_0})\). The indicated prospective frequency is \(\lambda_C=Z\hat{\lambda}+(1-Z)\lambda_M\). Both frequencies must be expressed per the same exposure unit.

Why the Square-Root Rule Scales with Exposure

The Poisson assumption explains why the partial credibility factor involves a square root. For \(N\sim\operatorname{Poisson}(\lambda E)\), the mean and variance of \(N\) are both \(\lambda E\). Dividing the count by fixed exposure gives the observed frequency, so:

$$ E[\hat{\lambda}]=\lambda,\qquad \operatorname{Var}(\hat{\lambda})=\frac{\lambda}{E}. $$

Thus, the standard deviation of the observed frequency is proportional to \(1/\sqrt{E}\), for a fixed underlying frequency. If \(E_0\) is the exposure at the full-credibility threshold, the ratio of the standard error at \(E\) to the standard error at \(E_0\) is:

$$ \frac{\sqrt{\lambda/E}}{\sqrt{\lambda/E_0}} =\sqrt{\frac{E_0}{E}}. $$

Equivalently, the square root of the relative exposure, \(\sqrt{E/E_0}\), measures how far the experience has progressed toward the standard. Limited-fluctuation credibility uses this as the partial weight on the observed frequency, with the remaining weight assigned to the manual frequency. This is a prescribed credibility rule, not a claim that the resulting blend is the minimum-variance estimator under every possible model.

1
Put the data on a frequency basis.
Divide the observed claim count by the actual exposure \(E\) to obtain \(\hat{\lambda}=N/E\).
2
Calculate the credibility factor.
Compute \(\sqrt{E/E_0}\) and cap the result at one.
3
Blend the two frequencies.
Multiply the observed frequency by \(Z\), multiply the manual frequency by \(1-Z\), and add the results.

Worked Frequency-Rating Calculations

Worked Example: Partial Credibility Below the Standard

A hypothetical portfolio has 54 claims in 900 policy-years. Its full-credibility standard is 1,600 policy-years, and its manual frequency is 0.050 claims per policy-year. Calculate the indicated prospective frequency.

Method: calculate the experience frequency, then apply the limited-fluctuation blend. The claim count follows the assumed Poisson frequency model with mean \(\lambda E\); the stated standard is expressed in policy-years.

Step 1: Observed frequency.

$$ \hat{\lambda}=\frac{N}{E} =\frac{54}{900} =0.060\text{ claims per policy-year}. $$

Step 2: Credibility factor.

$$ Z=\min\left(1,\sqrt{\frac{900}{1600}}\right) =\min(1,\sqrt{0.5625}) =0.75. $$

Step 3: Prospective frequency. The experience receives weight 0.75 and the manual frequency receives weight \(1-0.75=0.25\):

$$ \lambda_C =(0.75)(0.060)+(0.25)(0.050) =0.045+0.0125 =0.0575\text{ claims per policy-year}. $$

Check. The result lies between the manual frequency, 0.050, and the observed frequency, 0.060, and is closer to the experience because \(Z>0.5\). The indicated frequency is 0.0575 claims per policy-year.

Common mistake avoided: dividing the 54 claims by the 1,600-policy-year standard would give a frequency on the wrong exposure basis. The observed frequency uses the actual 900 policy-years; \(E_0\) is used to calculate credibility, not to replace the observed exposure.

Worked Example: Exposure at or Above Full Credibility

A hypothetical book records 132 claims over 2,400 policy-years. The full-credibility standard is 1,600 policy-years, and the manual frequency is 0.050 claims per policy-year. Determine the prospective frequency and explain the effect of the exposure level.

Method: calculate the observed rate and cap credibility at one.

Step 1: Observed frequency.

$$ \hat{\lambda}=\frac{132}{2400}=0.055\text{ claims per policy-year}. $$

Step 2: Uncapped and capped credibility.

$$ \sqrt{\frac{E}{E_0}} =\sqrt{\frac{2400}{1600}} =\sqrt{1.5} \approx1.2247, \qquad Z=\min(1,1.2247)=1. $$

Step 3: Prospective frequency. With \(Z=1\), the manual frequency receives weight zero:

$$ \lambda_C=(1)(0.055)+(1-1)(0.050) =0.055\text{ claims per policy-year}. $$

Check. The indicated frequency equals the observed frequency because the exposure meets and exceeds the full-credibility standard. The square-root value is not used as a weight above one; credibility is capped. The answer is 0.055 claims per policy-year.

Common mistake avoided: using \(Z\approx1.2247\) directly would give the experience more than 100% weight and a negative weight on the manual rate. The cap prevents that result.

Worked Example: Low Exposure and a Higher Manual Frequency

A hypothetical specialty program has 7 claims in 100 policy-years. Its full-credibility standard is 400 policy-years, and the manual frequency is 0.110 claims per policy-year. Calculate the indicated frequency and check the direction of the adjustment.

Method: blend the observed frequency and manual frequency using the partial credibility factor.

Step 1: Observed frequency.

$$ \hat{\lambda}=\frac{7}{100} =0.070\text{ claims per policy-year}. $$

Step 2: Credibility factor.

$$ Z=\min\left(1,\sqrt{\frac{100}{400}}\right) =\sqrt{0.25} =0.50. $$

Step 3: Apply the two weights. The manual weight is \(1-Z=0.50\):

$$ \lambda_C =(0.50)(0.070)+(0.50)(0.110) =0.035+0.055 =0.090\text{ claims per policy-year}. $$

Check. The two weights sum to one, and the indicated frequency of 0.090 is exactly halfway between 0.070 and 0.110. The result is greater than the observed frequency because the manual frequency is greater. The answer is 0.090 claims per policy-year.

Common mistake avoided: a low observed frequency does not automatically mean the prospective rate must equal that low value. With only half credibility, the manual frequency still contributes half of the indicated rate.

Exam Checks and Interpretation

Before accepting a frequency indication, verify that the numerator and denominator describe the same experience basis. If \(N\) counts claims, \(E\) must be the corresponding exposure; if frequency is stated per policy-year, the manual frequency must also be per policy-year. A frequency per month, per vehicle-year, or per insured item cannot be blended directly with a rate per policy-year without converting the units.

The credibility calculation also has useful boundary checks. If \(E=E_0\), then \(Z=1\); if \(E<E_0\), then \(0\leq Z<1\); and if \(E>E_0\), the cap still gives \(Z=1\). For any \(0\leq Z\leq1\), the blended frequency must lie between the observed and manual frequencies. If it does not, check the weights, arithmetic, and units.

A further check is to state what each number does. The observed count and actual exposure determine past frequency. The full-credibility standard determines how much weight that experience receives. The manual frequency supplies the remainder. Keeping these functions distinct makes the calculation auditable and helps identify an incorrect denominator immediately.

Key takeaway: Compute \(\hat{\lambda}=N/E\) from actual claims and exposure, calculate \(Z=\min(1,\sqrt{E/E_0})\) using the stated full-credibility standard, and blend \(Z\hat{\lambda}\) with \((1-Z)\lambda_M\). Keep every frequency on the same exposure basis.

Check Your Understanding

For each problem, report the indicated frequency in claims per policy-year. Assume the stated full-credibility standards use policy-years and the same limited-fluctuation rule applies.

  1. A hypothetical portfolio has 20 claims in 250 policy-years. Its full-credibility standard is 1,000 policy-years, and the manual frequency is 0.060. Calculate the observed frequency, \(Z\), and the indicated frequency. (Answer: observed frequency 0.080, \(Z=0.500\), indicated frequency 0.070 claims per policy-year)
  2. A hypothetical program has 36 claims in 900 policy-years. Its full-credibility standard is 400 policy-years, and the manual frequency is 0.035. Calculate the indicated frequency, applying the credibility cap. (Answer: \(Z=1\); indicated frequency 0.040 claims per policy-year)
  3. A hypothetical risk group has 27 claims in 600 policy-years. Its full-credibility standard is 2,400 policy-years, and the manual frequency is 0.040. Calculate the indicated frequency. (Answer: observed frequency 0.045, \(Z=0.500\), indicated frequency 0.0425 claims per policy-year)
  4. A hypothetical portfolio has 64 claims over 1,600 policy-years. Its full-credibility standard is 400 policy-years, and the manual frequency is 0.030. What is the prospective frequency, and why does the manual rate receive no weight? (Answer: 0.040 claims per policy-year; exposure exceeds the full-credibility standard, so \(Z=1\))
  5. A hypothetical specialty class has 12 claims in 300 policy-years. Its full-credibility standard is 1,200 policy-years, and the manual frequency is 0.050. Calculate the experience frequency, credibility factor, and indicated frequency. (Answer: observed frequency 0.040, \(Z=0.500\), indicated frequency 0.045 claims per policy-year)

Previous

Next

Back to Tutorials
Clinical Biostats