Convexity of the Stop-Loss Premium Curve

Why the Shape of the Stop-Loss Curve Matters

A stop-loss contract with attachment $d$ pays the amount by which an aggregate loss $X$ exceeds $d$. As $d$ increases, the insurer pays less often and, when it does pay, the excess is smaller. The expected payment therefore falls. Its rate of fall also carries information: it is linked to the probability that loss exceeds the attachment. The way that rate changes is the central idea behind convexity of the stop-loss premium curve.

Throughout, assume $X\geq 0$ and $E[X]<\infty$. The attachment $d$ is nonnegative, and the contract pays $(X-d)^+=\max(X-d,0)$. This tutorial concerns the pure expected payment, without expenses or risk loadings.

Definition: The stop-loss premium curve is the function $$ \Pi(d)=E[(X-d)^+]. $$ It gives the expected stop-loss payment at attachment $d$. An attachment increase changes the contract threshold; it does not change the distribution of $X$.

Monotonicity and Convexity

For a fixed loss value $x$, the payment $(x-d)^+$ cannot increase when $d$ increases. Taking expectations preserves this ordering, so $\Pi(d)$ is nonincreasing. It is strictly decreasing over an interval only when there is positive probability of loss exceeding attachments in that interval.

Convexity says that the curve bends upward: for $d_1<d_2<d_3$, the slope from $d_1$ to $d_2$ is no greater than the slope from $d_2$ to $d_3$. Since the slopes are negative, this means that the premium generally falls less rapidly at higher attachments.

Theorem: If $X\geq0$ and $E[X]<\infty$, then $\Pi(d)=E[(X-d)^+]$ is nonincreasing and convex for $d\geq0$. Wherever $\Pi$ is differentiable, $$ \Pi'(d)=-P(X>d). $$ At an attachment with an atom, the right derivative is $-P(X>d)$, while the left derivative is $-P(X\geq d)$.

The derivative statement is a local version of the more general interval result. For $0\leq a<b$,

\Pi(a)-\Pi(b)=\int_a^b P(X>t)\,dt, \qquad \frac{\Pi(b)-\Pi(a)}{b-a} =-\frac{1}{b-a}\int_a^b P(X>t)\,dt.

Thus a secant slope is the negative of the average exceedance probability across the attachment interval. Because $P(X>t)$ is nonincreasing in $t$, these average exceedance probabilities cannot increase as we move to higher, nonoverlapping intervals. The corresponding negative slopes are nondecreasing, which is the slope criterion for convexity.

Why the Formula Holds

For each fixed value $x\geq0$, the payment can be written as the amount of attachment levels below $x$:

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

Taking expectations gives the tail-integral representation. Interchanging expectation and integration is valid here because the integrand is nonnegative; the finite-mean assumption ensures the resulting premium is finite.

\Pi(d)=\int_d^\infty P(X>t)\,dt.

Subtracting this expression at $a$ and $b$ yields the interval formula above. It also explains the derivative: over a very short interval to the right of $d$, the average of $P(X>t)$ approaches $P(X>d)$. From the left, it approaches $P(X\geq d)$. If there is no atom at $d$, these probabilities agree, so the curve is differentiable there.

Convexity can also be justified directly, without differentiability. For fixed $x$, $(x-d)^+=\max(x-d,0)$ is the maximum of two affine functions of $d$, and therefore is convex. Taking expectations preserves the defining convexity inequality. This argument applies at atoms and does not require a density.

Worked Examples

Worked Example: Discrete Losses and Secant Slopes

A hypothetical benefit pool has aggregate loss $X$ with the following distribution: $P(X=0)=0.20$, $P(X=2)=0.30$, $P(X=5)=0.30$, and $P(X=8)=0.20$. Calculate $\Pi(1)$, $\Pi(3)$, and $\Pi(6)$. Then compare the slopes over the intervals from 1 to 3 and from 3 to 6.

Method. For each attachment, multiply each possible excess by its probability and sum. The outcome $X=0$ contributes zero at all three attachments.

At $d=1$, the positive excesses are $2-1=1$, $5-1=4$, and $8-1=7$. Therefore,

\Pi(1)=0.30(1)+0.30(4)+0.20(7)=0.30+1.20+1.40=2.90.

At $d=3$, only losses 5 and 8 exceed the attachment:

\Pi(3)=0.30(5-3)+0.20(8-3)=0.60+1.00=1.60.

At $d=6$, only the loss of 8 produces a payment:

\Pi(6)=0.20(8-6)=0.40.

The secant slopes are

\frac{\Pi(3)-\Pi(1)}{3-1} =\frac{1.60-2.90}{2}=-0.65, \qquad \frac{\Pi(6)-\Pi(3)}{6-3} =\frac{0.40-1.60}{3}=-0.40.

The second slope is larger, so the curve is bending upward across these intervals, as convexity requires. Equivalently, the average exceedance probabilities are $0.65$ and $0.40$. The tail probability is smaller across the higher attachment interval. Check: the computed premiums decrease, and the slopes become less negative. A common mistake is to use $P(X\geq t)$ as though it gave the derivative at every point; with a discrete distribution, endpoint atoms matter for one-sided derivatives, though not for the interval integrals.

Worked Example: A Continuous Uniform Loss

Suppose a hypothetical aggregate loss $X$ is uniformly distributed on $[0,10]$, with density $f_X(x)=0.1$ on that interval. Find the premium curve for $0\leq d\leq10$, calculate its values at 2, 5, and 8, and verify the slope and convexity.

Method. Integrate the excess payment over the part of the distribution above the attachment. For $0\leq d\leq10$,

\Pi(d)=\int_d^{10}(x-d)(0.1)\,dx =0.1\left[\frac{(x-d)^2}{2}\right]_{d}^{10} =\frac{(10-d)^2}{20}.

Substituting $d=2$, $5$, and $8$ gives

\Pi(2)=\frac{8^2}{20}=3.20,\qquad \Pi(5)=\frac{5^2}{20}=1.25,\qquad \Pi(8)=\frac{2^2}{20}=0.20.

Differentiating the curve gives $\Pi'(d)=-(10-d)/10$. For this distribution, $P(X>d)=(10-d)/10$, so the derivative agrees with the exceedance-probability formula. Also, $\Pi''(d)=0.1>0$ for $0<d<10$, confirming convexity.

For a numerical slope check, the slope from 2 to 5 is $(1.25-3.20)/3=-0.65$; from 5 to 8 it is $(0.20-1.25)/3=-0.35$. The slopes increase. Check: the curve reaches zero at $d=10$, the maximum possible loss. A common mistake is to use $E[X-d]$ without restricting payment to cases where $X>d$; that would incorrectly allow negative contract payments.

Worked Example: One-Sided Slopes at an Atom

A hypothetical portfolio has $P(X=0)=0.25$, $P(X=5)=0.25$, and probability $0.50$ spread uniformly over losses from 5 to 9. Find $\Pi(3)$, $\Pi(5)$, and $\Pi(7)$. Then calculate the left and right derivatives at $d=5$ and compare the secant slopes on either side.

Method. Treat the atom at 5 separately from the uniform component. Conditional on belonging to that component, the loss is uniform on $[5,9]$, with mean 7.

At $d=3$, the atom at 5 contributes $0.25(5-3)=0.50$. The continuous component contributes $0.50(7-3)=2.00$, since every loss in that component exceeds 3. Thus $\Pi(3)=0.50+2.00=2.50$.

At $d=5$, the atom contributes zero. The continuous component has conditional mean excess $7-5=2$, so $\Pi(5)=0.50(2)=1.00$.

At $d=7$, only the continuous component's losses above 7 contribute. For a uniform distribution on $[5,9]$, the conditional expected excess above 7 is the triangular area divided by the interval width:

E[(X-7)^+\mid X\text{ is in the continuous component}] =\frac{\frac12(9-7)(9-7)}{9-5} =\frac{2}{4}=0.50.

The component has probability $0.50$, so $\Pi(7)=0.50(0.50)=0.25$. The secant slopes are $(1.00-2.50)/2=-0.75$ and $(0.25-1.00)/2=-0.375$, again increasing.

At $d=5$, $P(X>5)=0.50$, whereas $P(X\geq5)=0.75$. Therefore the right derivative is $-0.50$, and the left derivative is $-0.75$. The left derivative is no greater than the right derivative, as required for convexity. Check: the left secant slope equals the left derivative because the tail probability is constant on $(3,5)$; the right secant averages a tail that decreases above 5. A common mistake is to use one derivative at the atom and claim the curve is differentiable there.

Using Slopes and Differences Correctly

The derivative has a direct probability interpretation, but only where the derivative exists or when the appropriate one-sided derivative is specified. For an attachment without an atom, a small increase in attachment changes the pure premium at a rate of approximately $-P(X>d)$ dollars of premium per dollar of attachment. At an atom, the left and right rates differ because moving through the atom changes which outcomes are counted as exceedances.

When a problem gives premiums at two attachments rather than a derivative, use the secant formula. The result is an average exceedance probability over the interval, not necessarily the exceedance probability at either endpoint. In particular, it is generally incorrect to interpret a finite premium difference as a pointwise derivative.

Quantity	Interpretation	What to check
$\Pi(d)$	Expected payment above attachment $d$	It is nonnegative and nonincreasing
$\Pi'(d)$, when it exists	Negative exceedance probability at $d$	No atom at $d$ is needed for the ordinary derivative formula
$\Pi(b)-\Pi(a)$, for $a<b$	Negative integrated tail probability over $[a,b]$	The secant slope is an average, not a pointwise tail probability
One-sided slopes at an atom	Right: $-P(X>d)$; left: $-P(X\geq d)$	The left slope must be no greater than the right slope

Key takeaway: The stop-loss premium curve is nonincreasing because raising the attachment reduces every possible payment. It is convex because exceedance probabilities decrease with the attachment, so successive slopes become less negative. Premium differences over an interval measure the average exceedance probability across that interval.

Check Your Understanding

For each problem, use the pure stop-loss premium and interpret the slope or difference in context.

A hypothetical loss has $P(X=0)=0.50$, $P(X=3)=0.30$, and $P(X=8)=0.20$. Calculate $\Pi(2)$ and $\Pi(5)$, then find the secant slope from 2 to 5. (Answer: $\Pi(2)=1.50$, $\Pi(5)=0.60$, slope $=-0.3000$.)
Let $X$ be uniform on $[0,12]$. Find $\Pi(3)$, $\Pi(9)$, and the secant slope between those attachments. (Answer: $\Pi(3)=3.375$, $\Pi(9)=0.375$, slope $=-0.5000$.)
A discrete aggregate loss has $P(X=1)=0.20$, $P(X=5)=0.50$, and $P(X=9)=0.30$. Calculate $\Pi(4)$ and $\Pi(7)$, then express the secant slope as the negative average exceedance probability on that interval. (Answer: $\Pi(4)=2.00$, $\Pi(7)=0.60$, slope $=-7/15$, so the average exceedance probability is $7/15$.)
At attachment $d=6$, suppose $P(X=6)=0.20$ and $P(X>6)=0.30$. State the left and right derivatives of $\Pi$ at 6. (Answer: left derivative $=-0.5000$; right derivative $=-0.3000$.)
Suppose $X$ has an exponential distribution with rate $0.5$, so $P(X>d)=e^{-0.5d}$ for $d\geq0$. Find $\Pi(d)$, then determine whether its slope increases with $d$. (Answer: $\Pi(d)=2e^{-0.5d}$, $\Pi'(d)=-e^{-0.5d}$, and $\Pi''(d)=0.5e^{-0.5d}>0$, so its slope increases and the curve is convex.)

Quantity	Interpretation	What to check
\(\Pi(d)\)	Expected payment above attachment \(d\)	It is nonnegative and nonincreasing
\(\Pi'(d)\), when it exists	Negative exceedance probability at \(d\)	No atom at \(d\) is needed for the ordinary derivative formula
\(\Pi(b)-\Pi(a)\), for \(a<b\)	Negative integrated tail probability over \([a,b]\)	The secant slope is an average, not a pointwise tail probability
One-sided slopes at an atom	Right: \(-P(X>d)\); left: \(-P(X\geq d)\)	The left slope must be no greater than the right slope

What You'll Learn