Full Credibility for Pure Premium Estimates

Why Pure Premium Credibility Depends on Severity

A pure premium is the expected loss cost per unit of exposure. For a hypothetical portfolio with claim frequency $\lambda$ per exposure unit and mean claim severity $E[X]$, the expected pure premium is $\lambda E[X]$. An estimate based on observed aggregate losses is naturally subject to variation in both the number of claims and their sizes. A full-credibility standard asks how much exposure is needed for that estimate to fall within a specified relative error of its expected value with a stated probability.

The frequency assumption matters, but it is not enough by itself. Two portfolios can have the same expected pure premium and different severity variability, resulting in different required exposure. Under the compound Poisson model, the relevant severity quantity is the second moment, $E[X^2]$, not just the mean or the variance of severity.

Definition: Let $e$ be the number of exposure units, and suppose claim count $N_e\sim\operatorname{Poisson}(\lambda e)$. Let $X_1,X_2,\ldots$ be independent and identically distributed claim severities, independent of $N_e$, with finite $E[X]$ and $E[X^2]$. The aggregate loss is $S_e=\sum_{j=1}^{N_e}X_j$, and the aggregate-loss-per-exposure estimate of pure premium is $\widehat{\pi}=S_e/e$. Its target is $\pi=\lambda E[X]$.

Here, a relative precision $r$ means that the estimate is within $r$ times the target pure premium: $|\widehat{\pi}-\pi|\le r\pi$. For a two-sided confidence level $p$, the usual full-credibility standard uses a normal approximation and the standard normal quantile $z_{(1+p)/2}$. It finds the exposure needed for the approximate probability of this event to be at least $p$. No interest or mortality assumption is involved: the model concerns claim counts and losses over the stated exposure period. Severities are assumed to be the losses included in the aggregate; if a contract changes the paid loss, the moments must be those of the resulting payment distribution.

Deriving the Exposure Standard

The pure premium estimate has mean $\lambda E[X]$. Its variance follows from the compound Poisson model. Conditional on the number of claims, the sum’s variance is $N_e\operatorname{Var}(X)$, and its conditional mean is $N_e E[X]$. Applying the law of total variance gives:

\begin{aligned} \operatorname{Var}(S_e) &=E[N_e]\operatorname{Var}(X)+\operatorname{Var}(N_e)\big(E[X]\big)^2\\ &=\lambda e\operatorname{Var}(X)+\lambda e\big(E[X]\big)^2\\ &=\lambda eE[X^2]. \end{aligned}

The last equality uses $E[X^2]=\operatorname{Var}(X)+(E[X])^2$. Dividing aggregate loss by $e$ scales its variance by $1/e^2$, so $\operatorname{Var}(\widehat{\pi})=\lambda E[X^2]/e$. This explains why the second moment appears: variability comes from the severity of individual claims and from the random number of claims.

Full-credibility standard: Under the normal approximation, a two-sided confidence level $p$ and relative precision $r$ require $$ e\ \ge\ \frac{z_{(1+p)/2}^{\,2}\lambda E[X^2]}{r^2\big(\lambda E[X]\big)^2} \ =\ \frac{z_{(1+p)/2}^{\,2}E[X^2]}{r^2\lambda\big(E[X]\big)^2}. $$ If exposure must be an integer count of identical exposure units, round the calculated threshold upward. This is an approximate probability standard, not an exact guarantee for every compound distribution.

To see why the standard has this form, the normal approximation gives a standard deviation of $\sqrt{\lambda E[X^2]/e}$ for $\widehat{\pi}$. A central confidence level $p$ places approximately $z_{(1+p)/2}$ standard deviations on either side of its mean. Requiring that margin to be no larger than $r\pi$ gives $z_{(1+p)/2}\sqrt{\lambda E[X^2]/e}\le r\lambda E[X]$. Squaring and rearranging yields the stated minimum exposure.

Find the target pure premium.
Calculate $\pi=\lambda E[X]$ using the claim frequency per exposure unit and the mean paid severity.

Find the severity second moment.
Use $E[X^2]$ from the specified severity model. For a discrete severity, calculate it by summing each squared value times its probability.

Choose the correct quantile and precision.
For a two-sided confidence level $p$, use $z_{(1+p)/2}$; express relative precision $r$ as a decimal.

Calculate and round exposure upward.
Substitute into the full-credibility formula, then round up if the exposure requirement must be a whole number.

Worked Exposure Calculations

Worked Example: Two-Point Severity Distribution

A hypothetical coverage has Poisson claim frequency $\lambda=0.25$ per exposure unit. Each paid severity is independently $\$800$ or $\$1{,}200$, each with probability $0.5$. Find the exposure required for 95% two-sided confidence and 10% relative precision. Assume no discounting or mortality. Use the normal-approximation quantile $z_{0.975}=1.96$.

Method. Calculate the pure premium and the severity second moment, then apply the full-credibility standard. The severity mean is $E[X]=0.5(800)+0.5(1{,}200)=400+600=\$1{,}000$. Thus, $\pi=0.25(1{,}000)=\$250$ per exposure unit.

The second moment is $E[X^2]=0.5(800^2)+0.5(1{,}200^2)=0.5(640{,}000)+0.5(1{,}440{,}000)=1{,}040{,}000$ dollars squared. The allowed error is $0.10(250)=\$25$.

Substitution gives $e\ge 1.96^2(0.25)(1{,}040{,}000)/(25^2)$. Since $1.96^2=3.8416$, the numerator is $3.8416(260{,}000)=998{,}816$, and the denominator is $625$. Therefore $e\ge 998{,}816/625=1{,}598.1056$. Rounding upward, the requirement is 1,599 exposure units.

Check. At $e=1{,}599$, the approximate confidence margin is $1.96\sqrt{260{,}000/1{,}599}$, about $\$24.99$, which is below $\$25$. The method uses the severity second moment, not merely the mean.

Worked Example: Exponential Severity and a Tight Rounding Decision

A hypothetical portfolio has Poisson claim frequency $\lambda=0.40$ per exposure unit. Paid severity is exponential with mean $\$500$, so its rate parameter is $1/500$ per dollar and $E[X^2]=2(500^2)$. Find the exposure required for 90% two-sided confidence and 10% relative precision. Assume no discounting or mortality, and use $z_{0.95}=1.645$.

Method. For the specified exponential distribution, $E[X]=\$500$ and $E[X^2]=2(250{,}000)=500{,}000$ dollars squared. The target pure premium is $0.40(500)=\$200$ per exposure unit, so the allowed error is $0.10(200)=\$20$.

The exposure threshold is $e\ge 1.645^2(0.40)(500{,}000)/(20^2)$. Here $1.645^2=2.706025$, the numerator is $2.706025(200{,}000)=541{,}205$, and the denominator is $400$. Thus $e\ge 541{,}205/400=1{,}353.0125$, requiring 1,354 exposure units.

Common mistake. Substituting $\operatorname{Var}(X)=250{,}000$ for $E[X^2]=500{,}000$ would produce $e\ge676.50625$, or 677 units. That is incorrect: random claim counts contribute $\lambda e(E[X])^2$ to aggregate-loss variance as well as $\lambda e\operatorname{Var}(X)$. Their sum involves $E[X^2]$, not just the severity variance.

Check. At the unrounded threshold, the normal-approximation margin is exactly $\$20$. The upward rounding is essential because 1,353 units is below 1,353.0125.

Worked Example: Fixed Severity and a Different Confidence Level

A hypothetical book has Poisson claim frequency $\lambda=0.50$ per exposure unit, and every paid claim is exactly $\$2{,}000$. Find the exposure required for 90% two-sided confidence and 20% relative precision. Assume no discounting or mortality, and use $z_{0.95}=1.645$.

Method. The severity distribution is concentrated at $\$2{,}000$, giving $E[X]=\$2{,}000$ and $E[X^2]=4{,}000{,}000$ dollars squared. The target pure premium is $0.50(2{,}000)=\$1{,}000$; the allowed error is $0.20(1{,}000)=\$200$.

The standard gives $e\ge1.645^2(0.50)(4{,}000{,}000)/(200^2)$. The numerator is $2.706025(2{,}000{,}000)=5{,}412{,}050$, and the denominator is $40{,}000$. Hence $e\ge135.30125$, so the requirement is 136 exposure units.

Check. At 136 units, the approximate standard deviation of the estimate is $\sqrt{2{,}000{,}000/136}$, about $\$121.27$; multiplying by 1.645 gives about $\$199.49$, within the $\$200$ allowance. At 135 units the margin is about $\$200.22$, which exceeds the allowance.

Interpretation and Exam Traps

The exposure requirement increases with the variability captured by $E[X^2]$ and with the confidence quantile. It decreases as the frequency rate increases: more expected claims are observed per exposure unit. A stricter relative precision requirement increases the required exposure in proportion to $1/r^2$. For example, halving the permitted relative error multiplies the standard’s exposure threshold by four, all else equal.

Keep the confidence convention clear. For a two-sided central probability $p$, the standard uses $z_{(1+p)/2}$: a central 90% standard uses 1.645, while a central 95% standard uses 1.96. The target error is relative to the pure premium, not relative to the mean severity alone. If a problem states a one-sided criterion instead, its quantile and probability event must be interpreted accordingly; do not silently substitute a two-sided standard.

Using severity variance instead of its second moment. Compound Poisson aggregate variance includes variation in claim count, so use $E[X^2]$.
Applying the relative error to the wrong quantity. First calculate $\pi=\lambda E[X]$; the allowed error is $r\pi$.
Using the wrong normal quantile. Match the quantile to the stated confidence level and whether the criterion is one- or two-sided.
Rounding down. The calculated threshold is a minimum. Round upward when exposure is required in whole units.
Treating the approximation as exact. The standard is based on a normal approximation to the estimator. If an exact compound-distribution probability is specifically required, this formula alone does not establish it.

Key takeaway: For a compound Poisson pure premium estimate, full credibility is governed by $E[X^2]$, the claim frequency $\lambda$, the target $\lambda E[X]$, the relative error $r$, and the appropriate normal quantile. State the probability criterion precisely, calculate the moments for the paid-loss severity, and round the exposure threshold upward.

Check Your Understanding

For each hypothetical portfolio, assume independent severities and Poisson claim counts, no discounting or mortality, and the stated normal-approximation quantile. Round the required exposure upward to a whole unit.

Claim frequency is $0.20$ per exposure unit and every paid claim is $\$1{,}000$. Find exposure for 95% two-sided confidence and 25% relative precision, using $z=1.96$. (Answer: 308 exposure units)
Claim frequency is $0.30$ per exposure unit and paid severity is exponential with mean $\$600$. Find exposure for 90% two-sided confidence and 15% relative precision, using $z=1.645$. (Answer: 802 exposure units)
Claim frequency is $0.10$ per exposure unit. Paid severity is $\$1{,}000$ or $\$3{,}000$, each with probability $0.5$. Find exposure for 95% two-sided confidence and 10% relative precision, using $z=1.96$. (Answer: 4,802 exposure units)
Claim frequency is $0.25$ per exposure unit and paid severity is exponential with mean $\$400$. Find exposure for 95% two-sided confidence and 5% relative precision, using $z=1.96$. (Answer: 12,294 exposure units)
Claim frequency is $0.50$ per exposure unit and every paid claim is $\$500$. Find exposure for 90% two-sided confidence and 10% relative precision, using $z=1.645$. (Answer: 542 exposure units)

What You'll Learn