Normal Approximation for Covered Aggregate Loss

Why Approximate Covered Aggregate Loss?

An aggregate loss is the sum of payments generated by a random number of claims. Coverage terms can change each claim payment, including making it zero, so the relevant severity distribution is the distribution of covered payments, not necessarily the distribution of losses before coverage. When an exact aggregate distribution is difficult to calculate, a normal approximation can provide estimates of probabilities and stop-loss expectations from the aggregate mean and variance.

The approximation is a calculation method, not a claim that either the number of claims or each individual payment is normally distributed. It is most useful when the aggregate is built from many independent contributions and no single payment dominates the total. It can be poor for a small number of claims, highly skewed payments, or a threshold near a sharp feature of a discrete aggregate distribution.

Definition: Let $N$ be the number of claims and let $Y_j$ be the covered payment for claim $j$, with $Y_j=0$ when that claim produces no payment under the coverage. The covered aggregate loss is $$ S=\sum_{j=1}^{N}Y_j, $$ where the claim payments are identically distributed and independent of $N$, and are independent across claims. Write $m_1=E[Y]$, $m_2=E[Y^2]$, and $m_N=E[N]$. The normal approximation replaces $S$ by a normal random variable with the same mean and variance, when the variance is positive.

Aggregate Moments from Covered Payments

The first step is to apply the contract terms to the loss on each claim. For example, a per-claim deductible or limit changes the payment $Y$; the aggregate model is then built from these modified payments. If the coverage leaves some claims unpaid, retain those zero payments in the per-claim distribution when using the original claim count $N$.

Under the independence assumptions in the definition, conditioning on $N$ gives $E[S\mid N]=Nm_1$ and $\operatorname{Var}(S\mid N)=N\operatorname{Var}(Y)$. The laws of total expectation and total variance therefore give the general moment formulas below.

Theorem: If $E[N]$, $\operatorname{Var}(N)$, and $E[Y^2]$ are finite, then $$ E[S]=E[N]E[Y] $$ and $$ \operatorname{Var}(S) =E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)(E[Y])^2 =E[N]E[Y^2]+\bigl(\operatorname{Var}(N)-E[N]\bigr)(E[Y])^2. $$ These formulas require the stated independence between frequency and claim payments. They do not apply unchanged when claim payments are dependent on the claim count or on one another.

For Poisson frequency, $N\sim\operatorname{Poisson}(\lambda)$, so $E[N]=\operatorname{Var}(N)=\lambda$. The second term in the last variance formula vanishes, giving $\operatorname{Var}(S)=\lambda E[Y^2]$. For other frequency models, use their own frequency mean and variance; in general, the aggregate variance is not simply $E[N]\operatorname{Var}(Y)$.

Worked Example: Poisson Frequency with Zero Covered Payments

A hypothetical portfolio has $N\sim\operatorname{Poisson}(20)$. A claim's covered payment $Y$ is $0$, $100$, or $200$ dollars with probabilities $0.2$, $0.5$, and $0.3$, respectively. Estimate $P(S>3000)$ by a normal approximation. The claim payments are independent of the count and of one another.

Method. Calculate the moments of the covered payment, use the compound Poisson aggregate moments, and standardize the threshold. This method includes the zero payments in the severity distribution; it avoids treating every reported claim as a positive payment.

First, $E[Y]=0(0.2)+100(0.5)+200(0.3)=0+50+60=110$ dollars. The second moment is $E[Y^2]=0^2(0.2)+100^2(0.5)+200^2(0.3)=0+5{,}000+12{,}000=17{,}000$ dollars squared.

For a Poisson count with parameter $20$, the aggregate mean is $E[S]=20(110)=2{,}200$ dollars, and the aggregate variance is $\operatorname{Var}(S)=20(17{,}000)=340{,}000$ dollars squared. Thus $\sigma_S=\sqrt{340{,}000}=583.0952$ dollars.

The standardized threshold is $z=(3000-2200)/583.0952=1.3720$. Hence $P(S>3000)\approx1-\Phi(1.3720)\approx0.0850$, where $\Phi$ is the standard normal cumulative distribution function. The estimated probability is 0.0850.

Check. The threshold is above the mean, so the estimated upper-tail probability is less than $0.5$, as expected. The calculation uses the covered-payment second moment, not just its variance.

Using the Normal Approximation for Tail Probabilities

Let $\mu=E[S]$ and $\sigma=\sqrt{\operatorname{Var}(S)}$. When $\sigma>0$, the normal approximation uses $S\approx Z$, where $Z\sim N(\mu,\sigma^2)$. For a threshold $x$, standardize by $z=(x-\mu)/\sigma$. Then the approximate upper-tail probability is $P(S>x)\approx1-\Phi(z)$, and the approximate lower-tail probability is $P(S\leq x)\approx\Phi(z)$.

These are approximations to the aggregate distribution, not exact probabilities. In particular, a compound aggregate can be discrete even though the fitted normal distribution is continuous. A normal estimate can also assign probability to negative values, although covered aggregate loss cannot be negative. The approximation should therefore be interpreted with the model's scale and shape in mind.

Worked Example: Binomial Frequency and an Aggregate Tail

A hypothetical group has $N\sim\operatorname{Binomial}(10,0.4)$. Conditional on a claim occurring, its covered payment is $100$ or $300$ dollars, each with probability $0.5$. The payments are independent of the count and of one another. Approximate $P(S>1200)$.

Method. Use the general frequency-severity moment formulas, then standardize. This avoids applying the Poisson variance formula to a binomial count.

For a covered payment, $E[Y]=100(0.5)+300(0.5)=200$ dollars and $E[Y^2]=100^2(0.5)+300^2(0.5)=5{,}000+45{,}000=50{,}000$ dollars squared. Thus $\operatorname{Var}(Y)=50{,}000-200^2=10{,}000$ dollars squared.

The count moments are $E[N]=10(0.4)=4$ and $\operatorname{Var}(N)=10(0.4)(0.6)=2.4$. Therefore $E[S]=4(200)=800$ dollars, and

\begin{aligned} \operatorname{Var}(S) &=E[N]E[Y^2]+\bigl(\operatorname{Var}(N)-E[N]\bigr)(E[Y])^2\\ &=4(50{,}000)+(2.4-4)(200)^2\\ &=200{,}000-64{,}000=136{,}000. \end{aligned}

So $\sigma=\sqrt{136{,}000}=368.7818$ dollars and $z=(1200-800)/368.7818=1.0846$. The tail estimate is $P(S>1200)\approx1-\Phi(1.0846)\approx\mathbf{0.1391}$.

Check. The threshold is about $1.08$ standard deviations above the mean, so a tail probability around $0.14$ is plausible. The binomial variance is smaller than its mean, which is why its aggregate variance differs from the compound Poisson result.

Approximate Stop-Loss Expectations

For a retention $K$, the stop-loss payment on the aggregate is $(S-K)_+=\max(S-K,0)$. Its expectation is the expected amount by which the aggregate exceeds $K$, not the probability of exceeding $K$. If $Z\sim N(\mu,\sigma^2)$, define $d=(K-\mu)/\sigma$, and let $\phi$ and $\Phi$ be the standard normal density and cumulative distribution functions. The normal approximation to the aggregate stop-loss expectation is

Formula: For $\sigma>0$, $$ E[(S-K)_+]\approx \sigma\phi(d)+(\mu-K)\bigl(1-\Phi(d)\bigr), \qquad d=\frac{K-\mu}{\sigma}. $$ The result has the same monetary units as $S$ and $K$.

To see why the formula holds, approximate $S$ by $Z$ and integrate only over outcomes exceeding the retention. With $u=(z-\mu)/\sigma$, the integral becomes

\begin{aligned} E[(Z-K)_+] &=\int_K^\infty (z-K)\frac{1}{\sigma} \phi\left(\frac{z-\mu}{\sigma}\right)\,dz\\ &=\sigma\int_d^\infty u\phi(u)\,du +(\mu-K)\int_d^\infty\phi(u)\,du\\ &=\sigma\phi(d)+(\mu-K)\bigl(1-\Phi(d)\bigr). \end{aligned}

In the last equality, $\int_d^\infty\phi(u)\,du=1-\Phi(d)$, and $\phi'(u)=-u\phi(u)$ implies $\int_d^\infty u\phi(u)\,du=\phi(d)$. This derivation also shows why a stop-loss expectation needs both a tail probability and the density at the standardized retention.

Worked Example: Normal Estimate of an Aggregate Stop-Loss

A hypothetical portfolio has $N\sim\operatorname{Poisson}(9)$. Each covered payment is $0$, $100$, or $200$ dollars with probabilities $0.25$, $0.50$, and $0.25$. Payments are independent of the count and of one another. Estimate $E[(S-1000)_+]$.

Method. Find the aggregate mean and variance from the covered-payment moments, then substitute the standardized retention in the normal stop-loss formula. The common mistake of calculating only $P(S>1000)$ would omit the size of the excess payments.

The covered-payment mean is $E[Y]=0(0.25)+100(0.50)+200(0.25)=100$ dollars. Its second moment is $E[Y^2]=0+10{,}000(0.50)+40{,}000(0.25)=15{,}000$ dollars squared.

For Poisson frequency, $E[S]=9(100)=900$ dollars and $\operatorname{Var}(S)=9(15{,}000)=135{,}000$ dollars squared. Hence $\sigma=\sqrt{135{,}000}=367.4235$ dollars, and $d=(1000-900)/367.4235=0.2722$.

Using $\phi(0.2722)\approx0.3845$ and $1-\Phi(0.2722)\approx0.3928$, the estimate is

E[(S-1000)_+]\approx 367.4235(0.3845)+(900-1000)(0.3928) \approx141.25-39.28 =\$101.97.

The approximate expected excess is $\$101.97$. The first term is approximately $\$141.25$, and the second is approximately $-\$39.28$; their difference is $\$101.97$. This is an estimate of the average excess amount, not the probability of an excess.

Common Mistakes and Practical Limits

Using the wrong severity moments. Apply the contract to the loss first and calculate $E[Y]$ and $E[Y^2]$ from the resulting payment distribution. If zero payments are possible and $N$ counts all claims, include those zeros.
Using $E[N]\operatorname{Var}(Y)$ as the entire aggregate variance. That expression omits variation in the claim count. The additional term is $\operatorname{Var}(N)(E[Y])^2$.
Using the Poisson variance identity for another frequency model. The simplification $\operatorname{Var}(S)=E[N]E[Y^2]$ follows from $\operatorname{Var}(N)=E[N]$; it is not a universal compound-model formula.
Confusing a tail probability with a stop-loss expectation. A probability has no monetary units. A stop-loss expectation measures dollars of excess and requires the normal density term as well as the tail probability.
Relying on the approximation without judgment. A small expected count or a highly variable severity can leave the aggregate strongly skewed. The normal approximation is not automatically accurate merely because its mean and variance have been calculated correctly.

Key takeaway: Model the covered payment, not the uncovered loss; combine its first two moments with the correct frequency moments; and then use the fitted aggregate mean and variance for tail or stop-loss estimates. Every result from the normal approximation remains an estimate of the compound aggregate model.

Check Your Understanding

For each problem, use the normal approximation when a probability or stop-loss estimate is requested. Round probabilities to four decimals and monetary estimates to the nearest cent.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(5)$. A covered payment is $0$, $40$, or $80$ dollars with probabilities $0.2$, $0.5$, and $0.3$. Find the aggregate mean and standard deviation. (Answer: mean $\$220$; standard deviation $\$116.62$)
A hypothetical group has $N\sim\operatorname{Binomial}(8,0.25)$. Each claim's covered payment is $0$ with probability $0.2$ and $\$50$ with probability $0.8$, independently. Find the aggregate mean and variance. (Answer: mean $\$80$; variance $3{,}200$ dollars squared)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(16)$, and each covered payment is $\$100$ or $\$300$ with equal probability. Approximate $P(S>4500)$. (Answer: approximately $0.0731$)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(4)$, and each covered payment is $0$ or $\$150$ with probabilities $0.4$ and $0.6$. Approximate $E[(S-360)_+]$. (Answer: approximately $\$92.72$)
A hypothetical group has $N\sim\operatorname{Binomial}(20,0.5)$. Each claim independently produces a covered payment of $0$ or $\$200$, each with probability $0.5$. Find the aggregate mean, standard deviation, and approximate probability that $S>1500$. (Answer: mean $\$1{,}000$; standard deviation $\$387.30$; probability approximately $0.0984$)

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