Sensitivity of Stop-Loss Premiums to Attachment

Why Attachment Sensitivity Matters

A stop-loss contract pays the portion of aggregate loss above an attachment. Raising that attachment reduces the expected payment, but the size of the reduction depends on how likely aggregate loss is to exceed the attachment. The derivative of the pure stop-loss premium makes this relationship precise: under a no-atom condition, its slope is the negative tail probability at the attachment.

This sensitivity is useful when comparing nearby attachment choices. It is a local measure, not a complete pricing or contract-selection rule: it does not by itself account for expenses, risk loads, capital costs, or the value of the protection to the buyer. The calculations below concern expected payments only.

Definition: Let $S$ be a nonnegative aggregate loss and let $d\geq 0$ be an attachment. The pure stop-loss premium is $\Pi(d)=E[(S-d)_+]$, where $(x)_+=\max(x,0)$. The payment is zero when $S\leq d$, and equals $S-d$ when $S>d$. An aggregate-loss distribution has no atom at $d$ when $P(S=d)=0$.

The Derivative and Its Meaning

If the distribution has no atom at the attachment, the derivative exists there and is

\Pi'(d)=-P(S>d).

The minus sign reflects the direction of the change: increasing the attachment cannot increase the payment. The magnitude, $P(S>d)$, is the probability that the contract is paying at that attachment. For example, a slope of $-0.30$ means that a small one-dollar increase in attachment reduces expected payment by about 30 cents.

Theorem: If $P(S=d)=0$, then the derivative of the pure stop-loss premium at $d$ is $\Pi'(d)=-P(S>d)$. The condition matters: at an atom, the two-sided derivative need not exist. This tutorial uses the no-atom condition to obtain the ordinary derivative.

Why the Formula Holds

Consider increasing the attachment by a small positive amount $h$. For each possible value of $S$, the payment change divided by $h$ is between $-1$ and $0$. As $h$ decreases toward zero, that ratio approaches $-1$ when $S>d$, and $0$ when $S<d$. The only value at which the limiting behavior can differ between increases and decreases is $S=d$. With no atom there, that event has probability zero.

Because the difference quotient is bounded in absolute value by 1, its expectation converges to the expectation of its pointwise limit. Therefore, the derivative is the expected value of $-1$ on the event $S>d$ and 0 otherwise:

\Pi'(d) =E[-\mathbf{1}_{\{S>d\}}] =-P(S>d).

The same result can be seen from the tail-integral representation. For a nonnegative $S$, the stop-loss premium is the area under the survival probability beyond the attachment:

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

Where the survival function is continuous at $d$, differentiating this integral with respect to its lower limit gives $-P(S>d)$. Continuity at $d$ follows from the absence of an atom there.

Worked Example: A Uniform Aggregate Loss

A hypothetical portfolio’s aggregate loss $S$, in dollars, is modeled as uniform on $[0,100]$. For an attachment of $d=40$, calculate the pure stop-loss premium, its derivative, and the exact premium change if the attachment rises to 41. State what local sensitivity predicts about that change.

Method. Use the uniform density, which is $1/100$ between 0 and 100, to integrate the payment above the attachment. This continuous model has no atom at 40.

At $d=40$, only losses from 40 to 100 generate payment. Thus

\Pi(40)=\frac{1}{100}\int_{40}^{100}(s-40)\,ds =\frac{1}{100}\left[\frac{(s-40)^2}{2}\right]_{40}^{100} =\frac{60^2}{200} =18.

The tail probability at 40 is

P(S>40)=\frac{100-40}{100}=0.60, \qquad \Pi'(40)=-0.60.

So the local sensitivity predicts a reduction of about $0.60$ dollars for a one-dollar increase in attachment. For an exact calculation at 41,

\Pi(41)=\frac{(100-41)^2}{200} =\frac{59^2}{200} =\frac{3481}{200} =17.405.

The exact premium change is $17.405-18=-0.595$ dollars. This is close to the local prediction of $-0.60$, but not identical because the slope changes as the attachment moves. The calculation checks: the premium falls when the attachment rises, and the decrease is slightly smaller than the initial slope predicts.

Common mistake avoided: Using $P(S\leq 40)=0.40$ as the magnitude of the derivative would give $-0.40$. The contract pays when $S>40$, so the relevant probability is the upper-tail probability, $0.60$.

Using the Tail Probability in Other Models

The derivative formula does not require a density or a particular named distribution. It requires a well-defined expected payment and no probability mass exactly at the attachment. The next examples show how the same sensitivity can be calculated in different distribution models.

Worked Example: An Exponential Aggregate Loss

For a hypothetical aggregate-loss model, suppose $S$ has an exponential distribution with mean 50 dollars, so its rate is $1/50$ per dollar. At an attachment of 30 dollars, calculate the pure premium and derivative. Then interpret the derivative as the approximate premium change for a one-dollar attachment increase.

Method. The survival probability is $P(S>x)=e^{-x/50}$ for $x\geq0$. The model is continuous, so there is no atom at 30.

Using the tail-integral representation,

\Pi(30)=\int_{30}^{\infty} e^{-x/50}\,dx =\left[-50e^{-x/50}\right]_{30}^{\infty} =50e^{-30/50} =50e^{-0.6}.

Numerically, $e^{-0.6}\approx0.5488116361$, so $50e^{-0.6}\approx27.4405818$ dollars, or $27.44 to the nearest cent. The derivative is

\Pi'(30)=-P(S>30)=-e^{-0.6}\approx-0.5488.

Thus a small one-dollar increase in attachment is expected to reduce the pure premium by about $0.5488$ dollars. The derivative is a local rate of change; it is not the exact change for every one-dollar interval.

Check. The derivative is negative, as expected, and its magnitude is a probability between zero and one. The premium calculation also uses the same exponential tail probability as the derivative, multiplied by the mean, 50.

Worked Example: A Mixture with Two Continuous Components

A hypothetical model assigns probability $0.40$ to a uniform aggregate loss on $[0,100]$, and probability $0.60$ to a uniform aggregate loss on $[100,200]$. At attachment $d=50$, calculate the pure premium and derivative. Explain why the derivative uses contributions from both components.

Method. Calculate the conditional expected payment and the conditional tail probability within each component, then weight each result by its mixture probability. Neither component has an atom at 50, so the mixture has no atom there either.

In the first component, $S$ is uniform on $[0,100]$. Its expected excess above 50 is

E[(S-50)_+\mid S\text{ is in the first component}] =\frac{(100-50)^2}{2(100)} =12.5.

In the second component, every loss is at least 100, which is above the attachment. The conditional mean is 150, so its conditional expected payment is $150-50=100$. Weighting the conditional expected payments gives

\Pi(50)=0.40(12.5)+0.60(100)=5+60=65.

For the first component, the conditional probability of exceeding 50 is $(100-50)/100=0.50$. For the second component, that probability is 1. Hence

P(S>50)=0.40(0.50)+0.60(1)=0.20+0.60=0.80, \qquad \Pi'(50)=-0.80.

The derivative’s magnitude includes both components: losses in the first component exceed 50 half the time, while losses in the second exceed it every time. Therefore, a small increase in attachment reduces the expected payment by about 80 cents per dollar. The premium and slope are consistent with the model: the higher-loss component contributes substantially to both.

Common mistake avoided: It would be incorrect to use only the first component’s tail probability, $0.50$, or to treat the mixture as though its components had equal weights. The unconditional probability must incorporate the stated mixture probabilities.

Interpreting Sensitivity for Attachment Selection

The slope gives a direct marginal interpretation. At an attachment $d$, the probability $P(S>d)$ measures how often a marginal increase in attachment removes a dollar of payment. When that probability is high, the expected payment is relatively sensitive to the attachment. When it is low, a small attachment adjustment has a smaller effect on the pure premium.

For a finite increase from $d$ to $d+h$, with $h>0$, the exact reduction is

\Pi(d)-\Pi(d+h)=\int_d^{d+h}P(S>x)\,dx.

This follows by subtracting the two tail integrals: the portion from $d+h$ to infinity cancels, leaving the area from $d$ to $d+h$. Since the tail probability can change across that interval, multiplying $h$ by $P(S>d)$ is generally only a local approximation, not an exact finite-change calculation.

For attachment selection, this distinction matters. A useful first estimate of the premium effect of a small change is the derivative times the change in attachment. For a larger change, integrate the tail probabilities over the interval or calculate each premium directly. The sensitivity does not determine which attachment is best: that decision also depends on the contract’s objectives and any pricing considerations beyond the pure expected payment.

Common Exam Traps

Using the wrong tail. The derivative depends on $P(S>d)$, not $P(S\leq d)$. The event is whether the contract makes a payment.
Dropping the negative sign. An increase in attachment reduces the payment, so the derivative is nonpositive.
Treating a local slope as an exact finite change. The slope can vary over an interval; for a finite change, use the tail integral or recompute the premium.
Ignoring the no-atom condition. At a point with positive probability mass, the ordinary two-sided derivative may fail to exist. The formula here is stated under the no-atom condition.
Mixing conditional and unconditional probabilities. For a mixture distribution, weight each component’s tail probability by its mixture probability before using the derivative formula.

Key takeaway: At an attachment $d$ with no atom, the marginal change in the pure stop-loss premium is $\Pi'(d)=-P(S>d)$. The upper-tail probability gives the magnitude of the slope, and the negative sign records that raising the attachment reduces expected payment.

Check Your Understanding

For each hypothetical model, calculate the requested sensitivity or premium change. Give probabilities to four decimal places where needed.

$S$ is uniform on $[0,80]$. At $d=20$, find $\Pi(d)$ and $\Pi'(d)$. (Answer: $\Pi(20)=22.50$, $\Pi'(20)=-0.7500$)
$S$ is exponential with mean 25. At $d=50$, find the derivative and pure premium, with the premium rounded to the nearest cent. (Answer: $\Pi'(50)=-e^{-2}\approx-0.1353$, $\Pi(50)=25e^{-2}\approx\$3.38$)
A mixture has probability $0.25$ of a uniform loss on $[0,40]$ and probability $0.75$ of a uniform loss on $[60,100]$. At $d=30$, find $P(S>d)$, $\Pi'(d)$, and $\Pi(d)$. (Answer: $P(S>30)=0.8125$, $\Pi'(30)=-0.8125$, $\Pi(30)=\$37.8125$)
$S$ is uniform on $[20,120]$. Find the slope at $d=70$, then calculate the exact premium reduction when the attachment rises from 70 to 75. (Answer: $\Pi'(70)=-0.5000$; the reduction is $\$2.375$)
A mixture assigns probability $0.50$ to a uniform loss on $[0,100]$ and probability $0.50$ to a uniform loss on $[100,200]$. At $d=120$, calculate the derivative and the exact premium reduction for an increase to $d=122$. Compare the reduction with the local approximation. (Answer: $\Pi'(120)=-0.4000$; exact reduction $=\$0.79$, compared with the local approximation of $\$0.80$)

What You'll Learn