Stop-Loss Premiums from Aggregate Tails

Stop-Loss Premiums from Aggregate Losses

A stop-loss contract protects a payer when the total loss for a period exceeds a specified amount. The payer retains losses up to the attachment point; the contract pays only the excess. The pure premium is therefore determined by the upper tail of the aggregate-loss distribution, not merely by the probability that the attachment point is exceeded. In particular, a loss just above the attachment point produces a small payment, while a much larger loss produces a larger one.

This tutorial calculates the expected payment from a discrete or continuous aggregate-loss distribution. All examples are hypothetical. We assume a single-period aggregate loss and no expenses, profit provision, discounting, or risk loading; the premiums calculated are expected payments on the stated probability model.

Definition: Let $S$ be the aggregate loss over the period and let $d\geq 0$ be the stop-loss attachment point. The stop-loss payment is $(S-d)^+=\max(S-d,0)$. Its pure premium is $E[(S-d)^+]$. The payment is zero when $S\leq d$, including when $S=d$, and equals $S-d$ when $S>d$.

Expected Excess Loss and the Tail Formula

The expected excess-loss formula expresses the premium as the area under the aggregate-loss tail beyond the attachment point. Write $F_S(x)=P(S\leq x)$ for the distribution function of $S$. Since $P(S>x)=1-F_S(x)$, the formula is

E[(S-d)^+]=\int_d^\infty P(S>x)\,dx =\int_d^\infty \bigl(1-F_S(x)\bigr)\,dx.

This formula applies to nonnegative aggregate losses with a finite expected excess. It works whether the distribution is continuous, discrete, or mixed. For a continuous distribution, a density-based calculation is also possible, but the tail integral often makes the effect of the attachment point more transparent.

The reason the tail formula works is that each realized loss contributes one unit of payment for every unit of threshold $x$ between $d$ and $S$. More formally, for every fixed value $s$,

(s-d)^+=\int_d^\infty \mathbf{1}_{\{s>x\}}\,dx.

If $s\leq d$, the indicator is zero throughout the integration range. If $s>d$, it equals one for $d\leq x<s$, an interval of length $s-d$. Taking expectations and interchanging the nonnegative integral and expectation gives the tail formula. The strict inequality $S>x$ is natural in the indicator; changing the value at a single integration point does not change the integral.

Key result: Probability at exactly $d$ contributes no stop-loss payment. In a discrete calculation, include only outcomes strictly greater than $d$. In a tail integral, use $P(S>x)$; an atom at $d$ does not create excess above $d$.

For a discrete distribution with possible aggregate values $s_j$ and probabilities $p_j$, direct calculation gives

E[(S-d)^+]=\sum_{j:s_j>d}(s_j-d)p_j.

When $S$ takes nonnegative integer values and $d$ is an integer, the same premium can be written as a sum of tail probabilities:

E[(S-d)^+]=\sum_{k=d}^{\infty}P(S>k).

To see this, an outcome $S=s>d$ contributes one unit to each of the $s-d$ events $S> d,S>d+1,\ldots,S>s-1$. Summing the probabilities of these events therefore counts each outcome in proportion to its excess. If the aggregate loss lies on a lattice with spacing other than one, use the corresponding spacing in the sum; do not apply the unit-lattice formula without adjusting the units.

Worked Example: A Discrete Aggregate Loss with an Attachment-Point Atom

A hypothetical one-period aggregate loss $S$, measured in thousands of dollars, has the following distribution. Calculate the pure premium for a stop-loss contract attaching at $d=100$ thousand dollars.

Aggregate loss $s$ (thousands of dollars)	Probability $P(S=s)$
0	0.10
50	0.20
100	0.25
150	0.25
250	0.20

Method. Use the expected excess-loss sum and retain only outcomes strictly greater than the attachment point. First, the probabilities sum to $0.10+0.20+0.25+0.25+0.20=1.00$, so the distribution is valid.

At $S=0$ and $S=50$, the payment is zero. At $S=100$, the payment is also zero because $S-d=0$. At $S=150$, the payment is $150-100=50$ thousand dollars, and at $S=250$, it is $250-100=150$ thousand dollars.

E[(S-100)^+] =(50)(0.25)+(150)(0.20) =12.5+30 =42.5\text{ thousand dollars}.

The premium is therefore $42,500. As a check using the tail-moment identity, $E[S\mathbf{1}_{\{S>100\}}]=(150)(0.25)+(250)(0.20)=87.5$ thousand dollars and $P(S>100)=0.25+0.20=0.45$. Subtracting $100(0.45)=45$ thousand dollars gives $87.5-45=42.5$ thousand dollars. The common mistake here is to treat the probability $0.25$ at $S=100$ as a claim that triggers payment; it does not produce any excess.

Continuous Distributions and Normal Approximations

For a continuous aggregate-loss distribution with density $f_S$, the expected excess can be calculated directly as

E[(S-d)^+]=\int_d^\infty (s-d)f_S(s)\,ds.

Alternatively, integrate the tail probability. The two approaches are equivalent: the first weights each possible loss by its payment, while the second adds up the probabilities of exceeding all thresholds above $d$. Choose the form that makes the distribution and the arithmetic simplest.

A normal approximation is sometimes used for aggregate loss. If the stated model is $S\sim N(\mu,\sigma^2)$, with $\sigma>0$, define $z=(d-\mu)/\sigma$. Let $\phi$ and $\Phi$ denote the standard normal density and distribution functions. Then

E[(S-d)^+] =\sigma\phi(z)+(\mu-d)\bigl(1-\Phi(z)\bigr).

This follows by standardizing $S=\mu+\sigma Z$, where $Z\sim N(0,1)$, and integrating over $Z>z$. The normal model is an approximation in many aggregate-loss applications: a normal distribution assigns some probability to negative values, even though aggregate losses cannot be negative. Use it only when the approximation is appropriate for the problem, and do not silently treat it as an exact nonnegative loss model.

Worked Example: A Continuous Uniform Aggregate Loss

A hypothetical aggregate loss $S$, in dollars, has a uniform distribution on $[0,200]$. Calculate the pure stop-loss premium at $d=80$. The model is continuous with density $f_S(s)=1/200$ for $0\leq s\leq 200$, and zero otherwise.

Method. Integrate the payment times the density over losses exceeding the attachment point. The payment is positive only from $s=80$ through $s=200$:

\begin{aligned} E[(S-80)^+] &=\int_{80}^{200}(s-80)\frac{1}{200}\,ds\\ &=\frac{1}{200}\left[\frac{(s-80)^2}{2}\right]_{80}^{200}\\ &=\frac{1}{200}\left(\frac{120^2}{2}-0\right)\\ &=\frac{7{,}200}{200} =36. \end{aligned}

The pure premium is $36.00. As a check, conditional on $S>80$, the uniform distribution is uniform on $[80,200]$, so its mean excess above 80 is 60. The probability of $S>80$ is $120/200=0.6$; multiplying gives $0.6(60)=36$. The common mistake is to use the conditional mean excess, $60, as the unconditional premium and omit the probability of reaching the excess layer.

Worked Example: A Normal Approximation for Aggregate Loss

For a hypothetical portfolio, an aggregate-loss model uses the approximation $S\sim N(100,20^2)$, where losses and parameters are in thousands of dollars. Calculate the expected stop-loss payment at $d=120$ thousand dollars. Use $\phi(1)=0.24197$ and $1-\Phi(1)=0.15866$.

Method. Apply the normal expected-excess formula. Here $\mu=100$, $\sigma=20$, and

z=\frac{d-\mu}{\sigma}=\frac{120-100}{20}=1.

Substitute these values, keeping the signed term $\mu-d$:

\begin{aligned} E[(S-120)^+] &=20\phi(1)+(100-120)\bigl(1-\Phi(1)\bigr)\\ &=20(0.24197)+(-20)(0.15866)\\ &=4.83940-3.17320\\ &=1.66620\text{ thousand dollars}. \end{aligned}

The expected payment is $1,666.20, rounded to the nearest cent. The negative second term is correct: it subtracts the attachment point multiplied by the exceedance probability from the truncated first moment. The common mistake is to replace $\mu-d$ by $d-\mu$, which would add rather than subtract this term and overstate the premium. This result inherits the limitations of the stated normal approximation.

Mixed Distributions and Exact Treatment of Atoms

An aggregate distribution can combine point masses and continuous outcomes. The same expected excess formula still applies, but it is useful to separate the components so that an atom at the attachment point is not mistakenly counted. An atom at a loss value greater than $d$ contributes its probability multiplied by its excess; an atom at or below $d$ contributes zero.

Worked Example: Combining an Atom and a Continuous Component

A hypothetical aggregate loss $S$, measured in dollars, has probability $0.15$ of being exactly $75, probability $0.25$ of being exactly $175, and probability $0.60$ of coming from a uniform distribution on $[25,125]$. The three components are mutually exclusive and their probabilities sum to one. Calculate the pure stop-loss premium at $d=75$.

Method. Calculate each component’s expected payment and weight it by the probability of that component. The atom at $75 pays $(75-75)^+=0$, so its contribution is $0.15(0)=0$. The atom at $175 pays $175-75=100$, contributing $0.25(100)=25$ dollars.

For the uniform component, the density conditional on selecting it is $1/100$ on $[25,125]$. Only losses above $75 produce a payment:

\begin{aligned} E[(S-75)^+\mid \text{uniform component}] &=\int_{75}^{125}(s-75)\frac{1}{100}\,ds\\ &=\frac{1}{100}\left[\frac{(s-75)^2}{2}\right]_{75}^{125}\\ &=\frac{1{,}250}{100} =12.5\text{ dollars}. \end{aligned}

Weight this conditional expectation by $0.60$, giving $0.60(12.5)=7.5$ dollars. The total is $0+25+7.5=\textbf{$32.50}$. The atom at the attachment point has been included in the distribution but contributes no payment. A calculation that includes its probability as an exceedance would incorrectly charge for a zero excess.

Common Exam Traps and Practical Checks

Stop-loss calculations are sensitive to the definition of the attachment point and to whether a result is conditional or unconditional. Before calculating, identify the payment for each outcome and check whether the requested quantity is an expected payment or an exceedance probability. These are different quantities: $P(S>d)$ describes how often the attachment is exceeded, while $E[(S-d)^+]$ also reflects the size of excesses.

Counting equality as a positive payment. At $S=d$, the payment is zero. For a discrete sum, use $s_j>d$, not $s_j\geq d$.
Reporting only the exceedance probability. The premium is not $P(S>d)$. It is the expected excess, so outcomes above $d$ must be weighted by their excess amounts.
Forgetting the unconditional probability. If a calculation first finds an expected excess conditional on a component or event, multiply by the probability of that component or event before adding other contributions.
Using the wrong tail convention in a discrete sum. For integer-valued $S$ and integer $d$, the lattice identity starts at $k=d$ and uses $P(S>k)$. Confirm the support and units before applying it.
Rounding too early. Retain adequate precision in tail probabilities and intermediate products, then round the final dollar premium as requested.

A useful final check is to verify that the premium is nonnegative and no greater than $E[S]$ for nonnegative $S$. The stop-loss payment never exceeds the original loss, because $(S-d)^+\leq S$ when $S\geq0$. Also verify the scale: if losses are expressed in thousands of dollars, convert the final amount to dollars only after the calculation.

Key takeaway: The pure stop-loss premium is $E[(S-d)^+]$, equivalently the area under $P(S>x)$ for $x\geq d$. For discrete distributions, sum the excesses of outcomes strictly above the attachment point. An atom exactly at the attachment point produces no payment.

Check Your Understanding

For each hypothetical aggregate-loss model, calculate the stated pure stop-loss premium. Unless stated otherwise, losses and attachment points are in dollars.

$S$ takes values $0,40,90,140$ with probabilities $0.20,0.30,0.30,0.20$, respectively. Find the premium at $d=90$. (Answer: $10.00)
An integer-valued aggregate loss has probabilities $P(S=0)=0.10$, $P(S=1)=0.20$, $P(S=2)=0.25$, $P(S=3)=0.25$, and $P(S=4)=0.20$. Find the premium at $d=2.5$. (Answer: $0.425)
$S$ is uniform on $[20,220]$. Find the premium at $d=70$. (Answer: $56.25)
With probability $0.15$, $S=75$; with probability $0.25$, $S=175$; and with probability $0.60$, $S$ is uniform on $[25,125]$. Find the premium at $d=75$. (Answer: $32.50)
A normal approximation gives $S\sim N(500,100^2)$. At $d=600$, use $z=1$, $\phi(1)=0.24197$, and $1-\Phi(1)=0.15866$ to calculate the premium, rounded to cents. (Answer: $8.33)

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