Keeping Units Consistent in Credibility Calculations

Why Units Matter in Credibility

A limited-fluctuation credibility calculation can be arithmetically correct and still be wrong if its inputs use incompatible units. A claim count is not an exposure measure; claims per policy-year are not the same numerical rate as claims per 100 policy-years; and aggregate dollars are not a pure premium per exposure. The formulas connect these quantities, but only when each is defined consistently.

The common framework is to choose an exposure unit, specify the expected result per unit, and calculate the exposure required to meet a stated precision and probability standard. Credibility then increases with exposure, not simply with the number of claims observed.

Definition: Let $E$ denote total exposure, measured in a stated exposure unit, and let $T$ be an estimator of a target quantity $\theta$. Under the limited-fluctuation approximation, suppose $E[T]=\theta$ and $\operatorname{Var}(T)=c/E$, where $c$ is the variance coefficient in the chosen units. For a two-sided probability requirement $p$ and relative error tolerance $k$, the full-credibility standard is $$ E_0=\frac{z^2c}{k^2\theta^2}, $$ where $z$ is the standard normal quantile satisfying $P(-z\le Z\le z)=p$ for $Z\sim N(0,1)$. Partial credibility is $Z_c=\min(1,\sqrt{E/E_0})$. This is a normal-approximation rule; it does not guarantee the stated probability exactly for every distribution, especially when counts are small.

The symbols $Z$ for a standard normal variable and $Z_c$ for a credibility factor have different roles. The full-credibility standard $E_0$ has the same units as $E$, so the ratio $E/E_0$ is dimensionless. The target $\theta$ and standard deviation must use the same measurement units when the relative error is calculated.

Claim Counts and Claim Frequency

Suppose claims follow a Poisson process at rate $\lambda$ claims per exposure unit. For exposure $E$, the total claim count $N$ has distribution $N\sim\operatorname{Poisson}(\lambda E)$. The observed frequency is $N/E$, measured in claims per exposure unit. Its mean is $\lambda$ and its variance is $\lambda/E$.

E\!\left[\frac{N}{E}\right]=\lambda, \qquad \operatorname{Var}\!\left(\frac{N}{E}\right)=\frac{\lambda}{E}, \qquad E_{0,\text{frequency}}=\frac{z^2}{k^2\lambda}.

For claim frequency, $c=\lambda$ and $\theta=\lambda$, giving the displayed standard. Alternatively, apply the relative error requirement directly to the count $N$. Since a Poisson count has mean and variance $\lambda E$, the normal approximation gives a full-credibility expected-claim count $N_0=z^2/k^2$. Thus $E_0=N_0/\lambda$. The expected-claim count is measured in claims, while $E_0$ is measured in exposure units.

Aggregate Loss and Pure Premium

Let $S$ be aggregate loss over exposure $E$. Suppose the count is Poisson with rate $\lambda$ claims per exposure unit, and independent severities $X$ have finite mean and second moment. The pure premium estimator is $S/E$, with units of currency per exposure unit. For a compound Poisson total, the aggregate mean per exposure is $\lambda E[X]$, while its variance per exposure is $\lambda E[X^2]/E$.

E\!\left[\frac{S}{E}\right]=\lambda E[X], \qquad \operatorname{Var}\!\left(\frac{S}{E}\right)=\frac{\lambda E[X^2]}{E}, \qquad E_{0,\text{pure premium}} =\frac{z^2\lambda E[X^2]}{k^2\{\lambda E[X]\}^2}.

The second moment $E[X^2]$, rather than just $\operatorname{Var}(X)$, appears because aggregate loss varies both from claim-count fluctuation and from variation in severity. Units provide a useful check: the numerator coefficient $\lambda E[X^2]$ has units of currency squared per exposure, and the squared target has units of currency squared per exposure squared. Their ratio has units of exposure.

Why Changing Units Does Not Change Credibility

A formula should give the same credibility conclusion whether exposure is recorded in policy-years or blocks of 100 policy-years. This follows because changing the exposure unit changes both the numerical exposure and the rate per unit.

Change the unit.
Let one new exposure unit contain $a$ old units. The numerical exposure becomes $E'=E/a$, and a frequency rate becomes $\lambda'=a\lambda$ claims per new unit.

Convert the full standard.
For frequency, $E'_0=z^2/(k^2\lambda')=E_0/a$. The full standard changes numerically in the same way as the exposure.

Check the ratio.
$E'/E'_0=(E/a)/(E_0/a)=E/E_0$. Therefore the credibility factor is unchanged.

The same reasoning applies to aggregate pure premium. If the currency unit changes—for example, dollars to thousands of dollars—both the pure-premium target and its standard deviation scale together. Their relative error is unchanged, so the required exposure is unchanged.

Worked Applications

Worked Example: Converting a Claim-Frequency Standard

A hypothetical insurer models claim counts as Poisson, with expected frequency $0.08$ claims per policy-year. It requires a two-sided 90% probability that observed frequency is within 10% of its mean. Use the normal approximation with $z=1.645$. Find the full-credibility exposure in policy-years and in units of 100 policy-years, and find the corresponding expected-claim count. No interest or mortality assumption applies; claims are counted without discounting.

Method. Apply the frequency full-credibility formula, keeping the rate and exposure in matching units. First, $k=0.10$, so $k^2=0.01$, and $z^2=1.645^2=2.706025$.

Using $\lambda=0.08$ claims per policy-year,

E_0=\frac{2.706025}{0.01(0.08)} =\frac{2.706025}{0.0008} =3382.53125\text{ policy-years}.

The expected number of claims at this exposure is

N_0=\lambda E_0 =0.08(3382.53125) =270.6025\text{ claims}.

Now use blocks of 100 policy-years as the exposure unit. The rate is $0.08(100)=8$ claims per 100 policy-years, so

E'_0=\frac{2.706025}{0.01(8)} =33.8253125\text{ units of 100 policy-years}.

This is $3382.53125$ policy-years, and $8(33.8253125)=270.6025$ expected claims. Both representations agree. To one decimal place, the full standard is 3,382.5 policy-years, or 33.8 blocks of 100 policy-years; the expected-claim standard is 270.6 claims.

Common mistake avoided: keeping the numerical rate $0.08$ while reporting exposure in 100-policy-year units would incorrectly give 3,382.5 blocks. The rate must also be converted to 8 claims per block.

Worked Example: Credibility Uses Exposure, Not Observed Count

A hypothetical portfolio has 240 policy-years of experience and 18 claims. Its manual expected frequency is $0.06$ claims per policy-year. The credibility standard requires 90% two-sided probability within 20% of the mean, using $z=1.645$. Calculate the experience frequency, the credibility factor, and the credibility-weighted frequency. Assume a Poisson claim count and use the normal-approximation limited-fluctuation rule. No interest or mortality assumption applies.

Step 1: Calculate the observed frequency. Divide the 18 claims by 240 policy-years:

\hat{\lambda}=\frac{18}{240}=0.075 \text{ claims per policy-year}.

Step 2: Find the full-credibility exposure. Here $k=0.20$, so $k^2=0.04$. Thus

E_0=\frac{1.645^2}{0.04(0.06)} =\frac{2.706025}{0.0024} =1127.5104167\text{ policy-years}.

Step 3: Calculate partial credibility.

Z_c=\sqrt{\frac{240}{1127.5104167}} =\sqrt{0.212865\ldots} =0.4614.

The exposure is below the full standard, so the cap at one does not apply. Blend experience with the manual frequency:

\lambda_c =Z_c\hat{\lambda}+(1-Z_c)(0.06) =0.4614(0.075)+0.5386(0.06) \approx 0.0669\text{ claims per policy-year}.

The credibility-weighted frequency is approximately $0.0669$ claims per policy-year. The computation is checked by noting that it lies between the observed frequency, $0.075$, and the manual frequency, $0.06$.

Common mistake avoided: using 18, the observed claim count, in place of exposure would give $\sqrt{18/1127.5104}\approx0.1264$. That ratio is invalid: the numerator must be in policy-years, the same unit as $E_0$. The 18 claims are used to calculate observed frequency, not to replace exposure.

Worked Example: Full Credibility for Aggregate Pure Premium

A hypothetical book has Poisson claim frequency $\lambda=0.04$ claims per vehicle-year. Each claim severity is independently either $500 or $3,500, each with probability $0.5$. Find the full-credibility exposure for aggregate pure premium to be within 25% of its mean with two-sided probability 90%, using $z=1.645$. Use undiscounted losses; no interest or mortality assumption applies.

Step 1: Calculate the severity moments.

E[X]=0.5(500)+0.5(3500)=2000\text{ dollars},

and

E[X^2]=0.5(500^2)+0.5(3500^2) =0.5(250000)+0.5(12250000) =6250000\text{ dollars}^2.

Step 2: Calculate the expected pure premium. The mean aggregate loss per vehicle-year is

\lambda E[X]=0.04(2000)=80\text{ dollars per vehicle-year}.

Step 3: Apply the aggregate pure-premium formula. With $k=0.25$,

E_0 =\frac{1.645^2(0.04)(6250000)} {0.25^2(80)^2} =\frac{676506.25}{400} =1691.265625\text{ vehicle-years}.

The expected claim count at the standard is $0.04(1691.265625)=67.650625$ claims. As a check, the variance of pure premium over $E$ vehicle-years is $0.04(6250000)/E=250000/E$ dollars squared per vehicle-year squared. At $E_0$, its standard deviation divided by the mean of $80 is $0.25/1.645$; multiplying by $1.645$ gives the required relative error $0.25$. To one decimal place, the required exposure is 1,691.3 vehicle-years.

Common mistake avoided: using only the severity variance $2{,}250{,}000$ in place of $E[X^2]=6{,}250{,}000$ understates aggregate variability. The compound Poisson aggregate variance uses the severity second moment.

Common Unit Checks and Exam Traps

Before accepting a limited-fluctuation result, check what each number measures. The following distinctions prevent many avoidable errors:

Claims versus exposure: a count is measured in claims; exposure might be policy-years, vehicle-years, or another defined unit. They cannot be substituted for one another.
Rate versus total: a frequency is claims per exposure unit. Multiply it by exposure to get expected claims; divide an observed count by exposure to get observed frequency.
Aggregate loss versus pure premium: aggregate loss is measured in currency, while pure premium is currency per exposure unit. Do not compare one directly with the other in a relative-error formula.
Variance versus second moment: for compound Poisson aggregate loss, use $E[X^2]$. Replacing it with $\operatorname{Var}(X)$ omits the claim-count contribution to aggregate variability.
Unit changes: when exposure units are rescaled, convert the rate and full standard as well. The numerical credibility factor should remain unchanged.

These checks matter because the credibility factor depends on the ratio of actual exposure to the full standard. A unit mismatch can make that ratio appear too large or too small, even when each isolated arithmetic operation is correct. A reliable calculation identifies the exposure unit first, writes every rate with that unit in its denominator, and checks that the final credibility factor is dimensionless and between zero and one.

Key takeaway: Match the units before applying a credibility formula. For frequency, use claims per exposure unit and exposure in that unit. For aggregate pure premium, use currency per exposure unit as the target and the compound Poisson second moment in the variance coefficient. Equivalent unit systems must give the same credibility factor.

Check Your Understanding

For each problem, identify the relevant units before calculating.

A hypothetical portfolio has Poisson frequency $0.05$ claims per vehicle-year. Find the full-credibility exposure for a two-sided 90% probability within 10% of the mean, using $z=1.645$, and give the expected-claim standard. (Answer: 5,412.1 vehicle-years and 270.6 expected claims.)
A book reports 12 claims over 300 truck-months. Express its observed frequency in claims per 100 truck-months and in claims per truck-month. (Answer: 4 claims per 100 truck-months, or 0.04 claims per truck-month.)
For a Poisson claim count, find the full-credibility expected-claim standard for a two-sided 95% probability within 20% of the mean, using $z=1.96$. (Answer: 96.04 expected claims.)
A hypothetical portfolio has 500 policy-years of exposure and manual frequency $0.08$ claims per policy-year. Find its credibility factor for a two-sided 90% probability within 25% of the mean, using $z=1.645$. (Answer: $E_0=541.2$ policy-years and $Z_c=0.9612$.)
A hypothetical compound Poisson portfolio has $\lambda=0.02$ claims per exposure unit, severity mean $1,000, and severity standard deviation $1,000. Find the full-credibility exposure for aggregate pure premium to be within 20% of its mean with two-sided probability 90%, using $z=1.645$. (Answer: 6,765.1 exposure units.)

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