Full Credibility for Aggregate Loss Totals

Introduction

A full-credibility standard can be stated as a probability requirement: the aggregate losses in a specified period should lie within a stated distance of their expected total with at least a specified probability. For aggregate losses, that distance is measured in dollars. The calculation depends on both the frequency of claims and the variability of individual claim amounts.

This tutorial applies a two-sided absolute-error criterion to a compound Poisson model. The standard deviation gives the scale of aggregate fluctuation, and a normal approximation translates that scale into a probability margin. The result is an approximation, not an exact probability guarantee.

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$ be the number of claims in the period. Let $X_1,X_2,\ldots$ be independent, identically distributed claim amounts, independent of $N$, with finite first and second moments. The aggregate loss is $S_N=\sum_{j=1}^{N}X_j$, with $S_N=0$ when $N=0$. An absolute-error standard with dollar tolerance $K>0$ and required probability $p$ asks whether $P(|S_N-E[S_N]|\le K)\ge p$.

Mean, Variance, and the Probability Standard

Write $m_1=E[X]$ and $m_2=E[X^2]$ for the first and second moments of an individual claim amount. For the compound Poisson model, the aggregate mean is $\lambda m_1$, while the aggregate variance is $\lambda m_2$. In particular, the variance uses the second moment of the severity, not just its variance.

E[S_N]=\lambda E[X], \qquad \operatorname{Var}(S_N)=\lambda E[X^2].

If the number of claims and severities permit a reasonable normal approximation, approximate $S_N$ by a normal random variable with this mean and variance. Let $\Phi$ be the standard normal cumulative distribution function and let $z_{(1+p)/2}$ satisfy $\Phi(z_{(1+p)/2})=(1+p)/2$. The approximate probability of being within $K$ dollars of the mean is

P(|S_N-E[S_N]|\le K) \ \approx\ 2\Phi\!\left(\frac{K}{\sqrt{\lambda E[X^2]}}\right)-1.

Consequently, the approximate criterion is met when the normal margin for the required probability does not exceed the allowed dollar tolerance:

z_{(1+p)/2}\sqrt{\lambda E[X^2]}\le K.

Why the Second Moment Appears

The variance formula follows by conditioning on the claim count. Given $N=n$, the aggregate is the sum of $n$ independent claim amounts, so its conditional mean is $nm_1$ and its conditional variance is $n\operatorname{Var}(X)$. The law of total variance gives

\begin{aligned} \operatorname{Var}(S_N) &=E[\operatorname{Var}(S_N\mid N)] +\operatorname{Var}(E[S_N\mid N])\\ &=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)m_1^2\\ &=\lambda\operatorname{Var}(X)+\lambda m_1^2\\ &=\lambda E[X^2]. \end{aligned}

The last equality uses $\operatorname{Var}(N)=E[N]=\lambda$ for a Poisson count, and $E[X^2]=\operatorname{Var}(X)+m_1^2$. The two variance contributions have different sources: individual claim amounts vary, and the number of claims varies. Omitting the second contribution understates aggregate variability.

Specify the model.
State the Poisson mean $\lambda$, the severity distribution and its parameters, and the independence assumptions.

Find the severity second moment.
Calculate $E[X^2]$ directly or use $E[X^2]=\operatorname{Var}(X)+(E[X])^2$.

Calculate aggregate variability.
Use $\operatorname{Var}(S_N)=\lambda E[X^2]$ and take its square root.

Apply the probability level and compare.
Multiply the standard deviation by $z_{(1+p)/2}$, then compare that margin with $K$.

Worked Applications

Worked Example: Fixed Claim Amounts and a 95% Standard

A hypothetical warranty portfolio has $N\sim\operatorname{Poisson}(25)$ claims in a year. Every claim costs exactly \$800, with no inflation or discounting during the period. Using a normal approximation, assess whether the aggregate is within \$8,000 of its mean with probability at least 95%. Use $z_{0.975}=1.96$.

Method and model. The severity is constant: $X=\$800$. Thus $E[X]=\$800$ and $E[X^2]=(\$800)^2=\$640{,}000$. The count and severities meet the compound Poisson assumptions.

Aggregate mean and variance.

E[S_N]=25(\$800)=\$20{,}000, \qquad \operatorname{Var}(S_N)=25(\$640{,}000)=\$16{,}000{,}000.

The standard deviation is $\sqrt{\$16{,}000{,}000}=\$4{,}000$. The 95% two-sided normal margin is $1.96(\$4{,}000)=\$7{,}840.00$. Since $\$7{,}840.00\le \$8{,}000$, the approximate criterion is met. Equivalently, the approximate probability is $2\Phi(8{,}000/4{,}000)-1=2\Phi(2)-1\approx0.9545$, which exceeds 0.95.

Check and common mistake avoided. The margin is below the stated tolerance, and the probability calculation agrees with that decision. Although the individual severity has variance zero, the aggregate variance is not zero: the random claim count gives $\operatorname{Var}(S_N)=25(800^2)$, not $25\operatorname{Var}(X)=0$.

Worked Example: Exponential Severity and a 90% Standard

A hypothetical equipment-repair portfolio has $N\sim\operatorname{Poisson}(60)$ claims over the period. Individual claim amounts are independent exponential random variables with mean \$1,500, independent of $N$. There is no discounting. Does the approximate 90% criterion hold within \$28,000 of the aggregate mean? Use $z_{0.95}=1.645$.

Method and severity moment. For an exponential severity with mean $\theta$, $E[X]=\theta$, $\operatorname{Var}(X)=\theta^2$, and $E[X^2]=2\theta^2$. Here $\theta=\$1{,}500$, so

E[X^2]=2(\$1{,}500)^2 =2(\$2{,}250{,}000)=\$4{,}500{,}000.

Aggregate quantities. The expected total is $60(\$1{,}500)=\$90{,}000$. The variance and standard deviation are

\operatorname{Var}(S_N)=60(\$4{,}500{,}000) =\$270{,}000{,}000, \qquad \sqrt{\operatorname{Var}(S_N)}=\$16{,}431.68.

Apply the 90% criterion. Using the unrounded standard deviation, the margin is $1.645\sqrt{60(4{,}500{,}000)}\approx\$27{,}030.11$. Because $\$27{,}030.11\le \$28{,}000$, the criterion is met under the normal approximation.

Check and common mistake avoided. The threshold exceeds the calculated margin by about \$969.89. Using only $\operatorname{Var}(X)=\$2{,}250{,}000$ in the compound Poisson variance would give an incorrect aggregate variance of \$135,000,000 and an understated margin. The count randomness contributes the additional $\lambda(E[X])^2$.

Worked Example: Comparing Two Severity Distributions

Two hypothetical portfolios each have a Poisson claim count with mean 25. In portfolio A, every claim is \$800. In portfolio B, each claim is \$400 or \$1,200, each with probability $1/2$. Assume independent severities, independent of the count, and no discounting. For each portfolio, assess the 95% criterion within \$9,000, using $z_{0.975}=1.96$.

Portfolio A. Its second severity moment is $E[X^2]=(\$800)^2=\$640{,}000$. Therefore the aggregate variance is $25(\$640{,}000)=\$16{,}000{,}000$, and the standard deviation is \$4,000. Its 95% margin is $1.96(\$4{,}000)=\$7{,}840.00$, which is below \$9,000. Portfolio A meets the approximate standard.

Portfolio B. Its second moment is

E[X^2]=\tfrac12(\$400)^2+\tfrac12(\$1{,}200)^2 =\tfrac12(\$160{,}000)+\tfrac12(\$1{,}440{,}000) =\$800{,}000.

The aggregate variance is $25(\$800{,}000)=\$20{,}000{,}000$, and its standard deviation is $\sqrt{\$20{,}000{,}000}=\$4{,}472.14$. Using the unrounded standard deviation, the 95% margin is $1.96\sqrt{25(800{,}000)}\approx\$8{,}765.39$. This is also below \$9,000, so portfolio B meets the approximate standard.

Check and common mistake avoided. Both portfolios have mean severity \$800, but portfolio B has the larger second moment and therefore the larger aggregate margin. Comparing only their mean severities would miss this difference. The decision in both cases is based on each margin’s comparison with the same \$9,000 tolerance.

Interpretation and Limitations

The criterion is a test of a specified absolute dollar band around the aggregate mean. It does not say that every realization will lie in that band, and it does not turn the approximation into an exact probability. A compound Poisson total can be skewed and has a probability mass at zero; these features can matter when the expected claim count is small or claim amounts are highly variable. If the approximation is questionable, an exact probability calculation or a suitable numerical method may be needed.

A larger value of $\lambda E[X^2]$ produces a wider normal margin for the same confidence level. The tolerance must be interpreted in the same monetary units as the losses: a tolerance stated in thousands must be converted consistently before comparing it with a dollar margin. No interest rate or mortality assumption is involved here; the examples concern undiscounted aggregate property or warranty losses over a stated period.

Common Exam Traps

Using severity variance in place of the second moment. For Poisson frequency, aggregate variance is $\lambda E[X^2]$, not $\lambda\operatorname{Var}(X)$.
Using a one-sided normal quantile. A requirement of being within $K$ of the mean is two-sided, so use $z_{(1+p)/2}$, not $z_p$.
Reversing the comparison. The criterion passes when the required normal margin is at most the tolerance: $z_{(1+p)/2}\sqrt{\lambda E[X^2]}\le K$.
Claiming exactness. The mean-variance normal calculation is approximate unless an exact distributional argument is supplied.
Mixing units. If the tolerance is in thousands of dollars, express the margin in thousands too, or convert both to dollars before comparing.

Key takeaway: Under a compound Poisson model, calculate the aggregate variance as $\lambda E[X^2]$. For a two-sided absolute tolerance $K$ and probability level $p$, the normal-approximation full-credibility check is $z_{(1+p)/2}\sqrt{\lambda E[X^2]}\le K$.

Check Your Understanding

For each question, use a normal approximation and decide whether the stated absolute-error criterion is met. Unless specified otherwise, assume independent claim amounts, independence from the Poisson count, and no discounting.

A hypothetical device plan has $N\sim\operatorname{Poisson}(100)$, with every claim equal to \$250. Does it meet a 90% standard within \$4,200 of its mean? Use $z_{0.95}=1.645$. (Answer: Yes; the margin is $1.645\sqrt{100(250^2)}=\$4{,}112.50$.)
A hypothetical repair portfolio has $N\sim\operatorname{Poisson}(40)$ and exponential claim amounts with mean \$1,000. Does it meet a 95% standard within \$17,000? Use $z_{0.975}=1.96$. (Answer: No; $E[X^2]=\$2{,}000{,}000$, so the margin is $1.96\sqrt{40(2{,}000{,}000)}\approx\$17{,}530.77$.)
A hypothetical insurer has $N\sim\operatorname{Poisson}(80)$, and each claim is \$500 or \$1,500 with equal probability. Does it meet a 90% standard within \$17,000? Use $z_{0.95}=1.645$. (Answer: Yes; $E[X^2]=\$1{,}250{,}000$, so the margin is $1.645\sqrt{80(1{,}250{,}000)}=\$16{,}450.00$.)
At $N\sim\operatorname{Poisson}(50)$, compare a fixed \$1,000 severity with a severity of \$500 or \$1,500, each equally likely. Which meets a 95% standard within \$15,000? Use $z_{0.975}=1.96$. (Answer: The fixed severity meets it, with margin $1.96\sqrt{50(1{,}000{,}000)}\approx\$13{,}859.29$; the two-point severity does not, with $E[X^2]=\$1{,}250{,}000$ and margin $1.96\sqrt{50(1{,}250{,}000)}\approx\$15{,}495.16$.)
A hypothetical loss portfolio has $N\sim\operatorname{Poisson}(120)$ and exponential claim amounts with mean \$600. Does it meet a 99% standard within \$25,000? Use $z_{0.995}=2.576$. (Answer: Yes; $E[X^2]=\$720{,}000$, giving margin $2.576\sqrt{120(720{,}000)}\approx\$23{,}944.33$.)

What You'll Learn