Deriving The Aggregate Stop-Loss Transform

Why the Aggregate Stop-Loss Transform Matters

An aggregate stop-loss payment with deductible $d$ is the amount of total loss above $d$, namely $(X-d)_+$, where $X$ is the aggregate loss and $z_+=\max(z,0)$. Its expected value can be computed from the aggregate distribution, but it also has a useful representation as an area under the aggregate survival function. That representation makes the role of the deductible precise: changing $d$ removes a slice of tail area.

This tutorial derives the area identity and its derivative with respect to the deductible. The distinction between a smooth point and a point with probability mass is essential: at an atom, the stop-loss function has different left and right slopes. The results are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $X$ be a nonnegative aggregate loss with finite mean, and let $d\geq 0$ be an aggregate deductible. The aggregate stop-loss transform at $d$ is $\Pi_X(d)=E[(X-d)_+]$. Define the survival function by $S_X(x)=P(X>x)$. Thus, the payment is zero when $X\leq d$, and equals $X-d$ when $X>d$.

The finite-mean condition ensures that the expected payment is finite, since $0\leq (X-d)_+\leq X$. The deductible here applies to the aggregate loss as a whole, not separately to each claim. No interest or mortality assumptions are needed for this identity.

Deriving the Survival-Function Integral

For a fixed realized aggregate loss $X$, the excess above $d$ is the length of the interval from $d$ to $X$, if $X>d$, and is zero otherwise. Equivalently, it is the integral of an indicator that records whether the loss exceeds each level $t$.

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

If $X\leq d$, the indicator is zero for all $t\geq d$. If $X>d$, it is one for $d\leq t<X$ and zero afterward, so the integral is $X-d$. Taking expectations and interchanging the expectation and integral is justified by Tonelli's theorem, because the integrand is nonnegative. This gives the stop-loss transform identity.

\Pi_X(d)=E[(X-d)_+] =\int_d^\infty E\!\left[\mathbf{1}_{\{X>t\}}\right]dt =\int_d^\infty S_X(t)\,dt

The value $S_X(t)=P(X>t)$ is the probability that the aggregate loss reaches beyond level $t$. Therefore, the expected excess is the area under that tail probability, starting at the deductible. A point at the upper endpoint of an interval does not affect the integral, but the strict inequality in the survival definition matters when discussing derivatives at atoms.

Worked Example: Integrating a Discrete Aggregate Tail

A hypothetical annual aggregate loss $X$, measured in dollars, has the distribution $P(X=0)=0.20$, $P(X=100)=0.50$, and $P(X=300)=0.30$. Find the stop-loss expectation at $d=150$, and determine its derivative there.

Method and model. This is a nonnegative discrete aggregate distribution with finite mean. We use the survival integral for the expected excess and the tail probability for the derivative. The deductible is strictly between possible loss values, so there is no atom at $d=150$.

For $150\leq t<300$, the only outcome exceeding $t$ is $X=300$, so $S_X(t)=0.30$. For $t\geq300$, $S_X(t)=0$. Therefore,

\Pi_X(150)=\int_{150}^{300}0.30\,dt+\int_{300}^{\infty}0\,dt =0.30(300-150)=\$45.00

A direct check gives the same result: only the $X=300$ outcome pays, and its excess is $300-150=\$150$. Thus $E[(X-150)_+]=0.30(\$150)=\$45.00$.

Since there is no probability mass at $150$, the derivative is $-P(X>150)=-0.30$. It is measured in dollars of expected payment per dollar increase in deductible. The integral calculation and direct expectation agree, and the tail probability at the attachment is $0.30$.

The Deductible Derivative

The integral identity also gives the sensitivity of the expected payment to the deductible. For a positive increment $h$, $\Pi_X(d+h)-\Pi_X(d)=-\int_d^{d+h}S_X(t)\,dt$. Increasing the deductible removes the survival-area slice between $d$ and $d+h$. Dividing by $h$ and taking a right-hand limit gives the right derivative.

\Pi'_{X,+}(d) =\lim_{h\downarrow0}\frac{\Pi_X(d+h)-\Pi_X(d)}{h} =-S_X(d) =-P(X>d)

The last equality uses the right-continuity of $S_X$. For the left derivative, when $d>0$, the interval immediately below $d$ includes the probability of an outcome exactly equal to $d$. Hence the left-hand limit is determined by $P(X\geq d)$, not $P(X>d)$.

\Pi'_{X,-}(d)=-P(X\geq d)

If $P(X=d)=0$, the two one-sided derivatives agree and the ordinary derivative exists: $\Pi'_X(d)=-P(X>d)=-S_X(d)$. If there is positive probability at $d$, the left slope is more negative than the right slope by exactly $P(X=d)$. The stop-loss transform remains continuous, but it has a corner at that deductible.

Theorem: For a nonnegative aggregate loss $X$ with finite mean, $\Pi_X(d)=\int_d^\infty S_X(t)\,dt$ for $d\geq0$. Its right derivative is $-P(X>d)$; for $d>0$, its left derivative is $-P(X\geq d)$. The ordinary derivative exists wherever $P(X=d)=0$, and there it equals $-S_X(d)$.

These slopes also show that $\Pi_X(d)$ is nonincreasing: a higher deductible cannot increase the payment. Its slopes are nondecreasing as $d$ increases, because the tail probability cannot increase. Thus the stop-loss transform is convex. At a deductible with an atom, the slope jumps upward by the mass at that point.

Worked Example: An Exponential Aggregate Loss

Suppose a hypothetical aggregate loss has an exponential distribution with rate $\lambda=0.001$ per dollar, so $S_X(x)=e^{-0.001x}$ for $x\geq0$ and $E[X]=1/\lambda=\$1{,}000$. Find the expected stop-loss payment at $d=\$500$ and its deductible derivative.

Method and model. The aggregate loss is continuous, so there is no atom at $d=500$. Integrate its survival function from the deductible to infinity:

\Pi_X(500)=\int_{500}^{\infty}e^{-0.001t}\,dt =\left[-1000e^{-0.001t}\right]_{500}^{\infty} =1000e^{-0.5}

Using $e^{-0.5}=0.60653066$, the expected payment is $1000(0.60653066)=\$606.53066$, or \$606.53 to the nearest cent.

The derivative is the negative tail probability at the deductible: $\Pi'_X(500)=-e^{-0.5}=-0.60653066$, or $-0.6065$ to four decimal places. A direct check follows from the memoryless property: conditional on $X>500$, the excess $X-500$ has mean $\$1{,}000$, while $P(X>500)=0.60653066$. Their product is again $\$606.53066$. The tail probability is the magnitude of the slope, not the expected payment itself.

Atoms, One-Sided Slopes, and an Exam Trap

Aggregate distributions can have atoms, including at zero when there is a positive probability of no claims and at positive values in a discrete or mixed model. In those cases, reporting a single derivative at an atom is incorrect. The following example shows how the two slopes identify the probability mass at the deductible.

Worked Example: A Deductible Exactly at an Atom

A hypothetical aggregate loss has $P(X=0)=0.25$, $P(X=100)=0.50$, and $P(X=200)=0.25$. Calculate $\Pi_X(100)$, its left derivative, and its right derivative.

Method and model. The deductible coincides with the atom at $X=100$. The outcomes $0$ and $100$ produce no positive excess at this deductible; only the $200$ outcome pays.

\Pi_X(100)=0.25(200-100)=\$25.00

For $100\leq t<200$, $S_X(t)=P(X=200)=0.25$; above $200$, it is zero. The survival integral checks the direct calculation:

\int_{100}^{\infty}S_X(t)\,dt =\int_{100}^{200}0.25\,dt =0.25(100)=\$25.00

The right derivative is $-P(X>100)=-0.25$, because an increase just above $100$ affects only the $X=200$ outcome. The left derivative is $-P(X\geq100)=-(0.50+0.25)=-0.75$, because a decrease from $100$ makes both the $100$ and $200$ outcomes pay additional excess.

Common mistake avoided: Calling $-0.75$ the derivative at $100$ uses the left slope as if the ordinary derivative existed. The right slope is $-0.25$, so the ordinary derivative does not exist there. The slope jump is $0.75-0.25=0.50=P(X=100)$, as it should be.

Using the Identity Correctly

The integral and derivative are two views of the same transform. The integral accumulates tail probabilities across all attachment levels above $d$; the derivative measures the local tail probability at $d$. This provides useful checks when a calculation of an expected excess and an aggregate exceedance probability are both available.

Confirm the variable and attachment. $X$ must be the aggregate loss relevant to the contract, and $d$ must be an aggregate deductible.
Use the survival convention carefully. Here $S_X(t)=P(X>t)$. At an atom, $P(X>t)$ and $P(X\geq t)$ differ.
Check finiteness. A finite $E[X]$ guarantees a finite stop-loss expectation for $d\geq0$.
Do not infer smoothness without checking for an atom. The ordinary derivative formula applies when $P(X=d)=0$; otherwise report the one-sided derivatives.

A frequent conceptual error is to treat the derivative as the negative of the probability of payment under the contract without checking the threshold convention. Away from atoms, the payment probability is $P(X>d)$, and the derivative is its negative. At an atom, the left-hand payment response includes $X=d$, while the right-hand response does not. Keeping the strict inequality explicit avoids confusing these quantities.

Key takeaway: The aggregate stop-loss expectation is the area under the aggregate survival function above the deductible. Its right slope is $-P(X>d)$, and its left slope is $-P(X\geq d)$. The slopes coincide exactly when there is no probability mass at the deductible.

Check Your Understanding

For each problem, identify whether the deductible is an atom when relevant. Answers are provided for checking.

A hypothetical aggregate loss takes values $0$, $50$, and $150$ with probabilities $0.40$, $0.40$, and $0.20$, respectively. Find the stop-loss expectation at $d=75$. (Answer: $\$15.00$)
A hypothetical aggregate loss is exponential with rate $0.002$ per dollar. At $d=\$200$, find the stop-loss expectation and derivative; report the expectation to the nearest cent and the derivative to four decimals. (Answer: $\$335.16$; derivative $-0.6703$)
A hypothetical aggregate loss has probabilities $0.20$, $0.30$, and $0.50$ at losses $0$, $80$, and $200$. At $d=80$, find the stop-loss expectation and the left and right derivatives. (Answer: $\$60.00$; left derivative $-0.8000$, right derivative $-0.5000$)
With probability $0.25$, a hypothetical aggregate loss is zero; otherwise it is exponential with rate $0.0025$ per dollar. At $d=\$200$, find the stop-loss expectation and derivative. (Answer: $\$181.96$; derivative $-0.4549$)
A hypothetical aggregate loss is zero with probability $0.20$; with probability $0.80$, it is uniform on $[0,500]$ dollars. At $d=\$200$, calculate the stop-loss expectation and derivative. (Answer: $\$72.00$; derivative $-0.4800$)

What You'll Learn