Stop-Loss Values from Compound Models

Why an Aggregate Stop-Loss Value Depends on the Tail

An aggregate stop-loss contract pays the amount by which total covered losses exceed an attachment point. Its expected payment is therefore determined not just by how large aggregate losses are on average, but by how much probability lies above the attachment point and how far those outcomes extend beyond it. A compound model provides a way to connect a claim frequency model and a severity model to this aggregate payment.

Throughout, consider one coverage period, with no discounting. Let $N$ be the number of claims, let $X_1,X_2,\ldots$ be claim severities, and define aggregate loss by $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. Assume the severities are independent and identically distributed, nonnegative, and independent of $N$. The attachment point $d$ is nonnegative.

Definition: The aggregate stop-loss premium at attachment point $d$ is $\Pi(d)=E[(S-d)^+]$, where $(z)^+=\max(z,0)$. The insurer pays $S-d$ when $S>d$, and pays zero when $S\le d$. This is an expected payment before expenses, risk loads, or other pricing adjustments.

In a discrete model, the most direct method is to obtain the distribution of $S$, then multiply each possible excess by its probability. For a large model, the aggregate distribution may instead be approximated. The important distinction is that an approximation to the distribution is an additional modeling choice; the mean and variance do not, by themselves, specify a stop-loss premium.

From an Aggregate Distribution to a Stop-Loss Premium

If the aggregate loss takes values $s_j$ with probabilities $p_j$, then $\Pi(d)=\sum_j p_j\max(s_j-d,0)$. Outcomes at or below the attachment contribute zero. In a continuous model, the corresponding calculation is an integral over losses above the attachment.

\Pi(d)=\int_d^\infty P(S>x)\,dx

This tail-integral identity follows by expressing the excess as an area. For a fixed realized value $S=s$, the length of the interval of $x\ge d$ for which $s>x$ is exactly $(s-d)^+$. Taking expectations of that nonnegative area gives the integral of the probabilities $P(S>x)$. This representation also makes the pricing interpretation clear: a high probability of exceeding the attachment, or a heavy tail beyond it, increases the expected payment.

(s-d)^+=\int_d^\infty \mathbf{1}_{\{s>x\}}\,dx \qquad\Longrightarrow\qquad E[(S-d)^+]=\int_d^\infty P(S>x)\,dx

For an integer-valued aggregate loss $K$ and an integer attachment $m$, the same idea gives $E[(K-m)^+]=\sum_{k=m}^{\infty}P(K>k)$. The strict inequality matters: an outcome equal to the attachment generates no payment. For a lattice distribution with a different monetary unit, apply the sum in that unit and multiply by the unit size.

Worked Example: A Poisson Frequency with Fixed Claim Size

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$ claims in a year. Each claim is exactly $1,000, independently of the count. Find the annual aggregate stop-loss premium at an attachment of $1,000. Treat amounts as undiscounted dollars.

Method. Since each claim is $1,000, aggregate loss is $S=1{,}000N$. The payment is $1{,}000(N-1)^+$. For integer $N$, $(N-1)^+=N-\mathbf{1}_{\{N\ge1\}}$: it is zero at $N=0,1$, and equals $N-1$ for larger counts.

For a Poisson random variable with parameter $\lambda=1$, $E[N]=1$ and $P(N\ge1)=1-P(N=0)=1-e^{-1}$. Therefore

\begin{aligned} \Pi(1{,}000) &=1{,}000\{E[N]-P(N\ge1)\}\\ &=1{,}000\{1-(1-e^{-1})\}\\ &=1{,}000e^{-1}\\ &\approx \$367.88. \end{aligned}

The method uses the compound model directly and avoids summing an infinite series of Poisson probabilities. As a check, only outcomes with at least two claims pay, and the expression $1{,}000E[(N-1)^+]$ is positive and below the expected aggregate loss of $1,000, as it should be.

Obtaining the Aggregate Distribution and Its Moments

The aggregate distribution can be calculated by conditioning on the claim count: for each possible value of $N$, determine the distribution of the sum of that many severities, then weight by the probability of that count. For small discrete models, listing the combinations may be simplest; for larger models, convolution or an appropriate recursion can produce the distribution. Once the probabilities $P(S=s)$ are available, the stop-loss calculation is a weighted sum of excesses.

The compound-model moments are useful both as checks and as inputs to an approximation. Under the independence and identical-distribution assumptions above, with finite severity variance,

E[S]=E[N]E[X], \qquad \operatorname{Var}(S)=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2.

The mean follows by conditioning on $N$: given $N=n$, the sum has mean $nE[X]$. For the variance, the conditional variance is $n\operatorname{Var}(X)$, while the conditional mean is $nE[X]$. Applying the conditional-variance identity gives

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

These formulas require the stated independence and finite moments. They do not identify the full aggregate distribution. In particular, substituting the mean and variance into a normal stop-loss formula is an approximation, not an exact consequence of the compound model.

Worked Example: Exact Stop-Loss from a Small Compound Distribution

A hypothetical contract has $N\sim\operatorname{Binomial}(2,0.5)$ claims. Given a claim, its severity is $1,000 or $3,000 with probability $0.5$ each. Claim severities are independent of each other and of $N$. Calculate the stop-loss premium at an attachment of $2,000. Work in thousands of dollars and do not discount.

Method. Condition on the count and combine the possible severities. The count probabilities are $P(N=0)=0.25$, $P(N=1)=0.50$, and $P(N=2)=0.25$. With one claim, the two aggregate outcomes each have probability $0.50(0.50)=0.25$. With two claims, the sum is 2, 4, or 6 thousand, with conditional probabilities $0.25,0.50,0.25$. Thus the complete aggregate distribution is:

Aggregate loss $S$, $000	Probability	Excess above 2, $000
0	0.2500	0
1	0.2500	0
2	0.0625	0
3	0.2500	1
4	0.1250	2
6	0.0625	4

The probabilities sum to $0.2500+0.2500+0.0625+0.2500+0.1250+0.0625=1.0000$. Multiplying each positive excess by its probability gives

\begin{aligned} \Pi(2) &=0.2500(1)+0.1250(2)+0.0625(4)\\ &=0.2500+0.2500+0.2500\\ &=0.7500\text{ thousand dollars}\\ &=\$750.00. \end{aligned}

As a check, the aggregate mean is $2$ thousand dollars: the expected count is $1$, and the mean severity is $2$ thousand. A $750 expected payment above a $2,000 attachment is plausible for this distribution, which has a 43.75% probability of exceeding the attachment. The calculation avoids the trap of charging an excess on the outcome exactly equal to $2,000.

Normal Approximation and the Information in Moments

When an aggregate distribution is too large to calculate conveniently, a normal approximation may be useful if the aggregate is generated by enough contributions for its distribution to be reasonably represented by a normal curve. A large expected claim count can support that approximation, but it does not guarantee accuracy in a far tail or for highly skewed severities. The model assumptions and the part of the distribution relevant to the attachment should be considered before relying on the result.

For an approximating $S\sim N(\mu,\sigma^2)$, with $\sigma>0$, set $z=(d-\mu)/\sigma$. The normal stop-loss value is

E[(S-d)^+]=\sigma\phi(z)+(\mu-d)\{1-\Phi(z)\},

where $\phi$ and $\Phi$ are the standard normal density and distribution functions. This follows by integrating $(s-d)$ against the normal density from $d$ to infinity and standardizing $s=\mu+\sigma y$. The formula is exact for a normal variable, but only approximate when the normal distribution stands in for a compound aggregate. A normal variable also permits negative values, unlike a nonnegative aggregate loss; this can be a concern when the mean is not large relative to the standard deviation.

Worked Example: Approximate Stop-Loss from Compound Moments

A hypothetical portfolio has $N\sim\operatorname{Poisson}(100)$ claims, each with a fixed severity of $1,000. Assume independence and no discounting. Estimate the stop-loss premium at an attachment of $110,000 using a normal approximation to aggregate loss. Report dollars to the cent.

Method. Here $E[N]=\operatorname{Var}(N)=100$, $E[X]=\$1{,}000$, and $\operatorname{Var}(X)=0$. The compound moments are

\begin{aligned} \mu=E[S]&=100(\$1{,}000)=\$100{,}000,\\ \sigma^2=\operatorname{Var}(S) &=100(0)+100(\$1{,}000)^2 =100{,}000{,}000\text{ dollars}^2,\\ \sigma&=\$10{,}000. \end{aligned}

The attachment is one standard deviation above the mean, so $z=(110{,}000-100{,}000)/10{,}000=1$. Using $\phi(1)=0.2419707$ and $1-\Phi(1)=0.1586553$,

\begin{aligned} \Pi(110{,}000) &\approx 10{,}000(0.2419707) +(100{,}000-110{,}000)(0.1586553)\\ &=2{,}419.707-1{,}586.553\\ &=\$833.154\\ &\approx \$833.15. \end{aligned}

This is an approximation because the exact aggregate is $1,000 times a Poisson count and is discrete. Its mean and variance are correct for the approximating normal model, but that alone does not make the tail probabilities exact. The calculation avoids the mistake of using only $P(S>d)$: a stop-loss payment depends on both the probability of exceeding the attachment and the expected amount of excess.

Worked Example: Same Moments, Different Stop-Loss Values

Consider two hypothetical aggregate-loss distributions, both measured in thousands of dollars, with attachment $d=2$. Distribution A has probabilities $0.25,0.25,0.0625,0.25,0.125,0.0625$ at losses $0,1,2,3,4,6$, respectively. Distribution B has loss 0 with probability $3/7$ and loss 3.5 with probability $4/7$. Compare their stop-loss values and verify that the distributions have the same mean and variance. There is no discounting.

Method. For A, the weighted loss is $0.25+0.125+0.375=0.75$ thousand dollars. Its second moment is

$$ E[S_A^2] =0.25(1^2)+0.0625(2^2)+0.25(3^2)+0.125(4^2)+0.0625(6^2) =7. $$

Therefore $E[S_A]=2$ and $\operatorname{Var}(S_A)=7-2^2=3$, in the corresponding thousand-dollar units. For B, $E[S_B]=(4/7)(3.5)=2$, and $E[S_B^2]=(4/7)(3.5^2)=7$, so its variance is also 3. But only its 3.5 outcome exceeds the attachment, giving

\Pi_B(2)=\frac{4}{7}(3.5-2)=\frac{6}{7} =0.857142\ldots\text{ thousand dollars}.

Thus the stop-loss values are $750.00 for A and approximately $857.14 for B, despite identical means and variances. This comparison checks the central point: moments summarize location and spread, but do not determine how probability is arranged above a particular attachment.

Common Mistakes and Practical Checks

Counting the attachment as a payment. The payment is $(S-d)^+$, not $S\mathbf{1}_{\{S\ge d\}}$. At $S=d$, the payment is zero.
Using a tail probability as the premium. $P(S>d)$ is a probability, not an expected dollar payment. The size of the excess in each paying outcome must also be included.
Treating a moment approximation as exact. A normal stop-loss value is exact only for a normal aggregate. For a compound model it is an approximation whose quality depends on the distribution and attachment.
Mixing units. If losses and the attachment are calculated in thousands of dollars, the result is also in thousands and must be converted if the answer is requested in dollars.
Ignoring the frequency-severity assumptions. The displayed compound-moment formulas rely on independent frequency and severities, with identically distributed severities. Dependence or varying claim distributions require a different calculation.

Key takeaway: Calculate an aggregate stop-loss premium from the distribution of $S$ whenever it is available. Use compound moments to check the model or construct a justified approximation, but do not infer a unique stop-loss value from the aggregate mean and variance alone.

Check Your Understanding

For each problem, assume one coverage period and no discounting. Use the stated model and report money to the cent where appropriate.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$ claims, each exactly $500. Find the stop-loss premium at attachment $500. (Answer: $567.67)
An aggregate loss in thousands of dollars is 0, 2, or 5 with probabilities 0.20, 0.50, and 0.30, respectively. Find the stop-loss premium at $2,000. (Answer: $900.00)
A portfolio has $N\sim\operatorname{Binomial}(3,0.5)$, with each claim exactly $1,000. Find the exact stop-loss premium at attachment $2,000. (Answer: $125.00)
Approximate the stop-loss value for $S\sim N(\$80{,}000,(\$12{,}000)^2)$ at attachment $92,000. Use $\phi(1)=0.2419707$ and $1-\Phi(1)=0.1586553$. (Answer: approximately $999.78)
Distribution A has losses 0, 1, 2, 3, 4, and 6 thousand dollars with probabilities 0.25, 0.25, 0.0625, 0.25, 0.125, and 0.0625. Distribution B has losses 0 and 3.5 thousand dollars with probabilities $3/7$ and $4/7$. At attachment $2,000, find the difference between their stop-loss values and state which is larger. (Answer: B is larger by approximately $107.14)

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