Computing Stop-Loss Values From Continuous Aggregate Models

Why Integrate an Aggregate Tail?

An aggregate stop-loss expectation measures the expected amount by which total loss exceeds an attachment point. It is useful when a decision depends on the size of the aggregate tail, not only on the probability that total loss crosses a threshold. For a continuous aggregate model, that expected excess can be calculated directly from the survival function or density.

In this tutorial, $X$ denotes the nonnegative aggregate loss over a specified period, and $d$ is the aggregate attachment point. The calculations use hypothetical models. They are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: The aggregate stop-loss expectation at attachment $d$ is $E[(X-d)_+]$, where $(x-d)_+=\max(x-d,0)$. It is the expected excess of aggregate loss over $d$, counting zero when aggregate loss does not exceed $d$.

Survival and Density Formulas

For a nonnegative aggregate loss, define $S_X(x)=P(X>x)$. The stop-loss expectation is the area under the aggregate survival function to the right of $d$. If $X$ is continuous with density $f_X$, the same expectation can be calculated by weighting each possible loss above $d$ by its excess amount.

E[(X-d)_+]=\int_d^\infty S_X(x)\,dx =\int_d^\infty (x-d)f_X(x)\,dx,\qquad d\geq 0.

The density expression assumes the aggregate distribution is continuous above $d$. If the aggregate distribution has atoms above the attachment point, the density integral alone omits their contributions; the survival integral remains valid. A compound frequency model can, for example, have an atom at zero. That atom contributes nothing when $d\geq0$, but atoms above $d$ must be included in the stop-loss value.

Start from the excess.
Write the quantity of interest as $(X-d)_+$, rather than as $X$ or as the event $X>d$.

Select a representation.
Use the survival integral when the tail function is available, or the density integral for a continuous model with a tractable density.

Check units and model support.
The result has the same currency units as $X$. Confirm that the chosen continuous model is appropriate for the aggregate range being evaluated.

The survival identity follows by expressing a nonnegative excess as the length of the interval it spans. For each outcome, $(X-d)_+=\int_0^\infty \mathbf{1}_{\{X>d+t\}}\,dt$. Taking expectations and interchanging the nonnegative integral and expectation gives the result.

E[(X-d)_+] =\int_0^\infty P(X>d+t)\,dt =\int_d^\infty S_X(x)\,dx.

For a continuous distribution, the density formula follows by taking the expectation of $(X-d)$ over outcomes with $X>d$. The formula is not a probability: it combines the chance of exceeding the attachment with the expected size of the excess.

Worked Calculations Under Continuous Models

Worked Example: Exponential Aggregate Tail

A hypothetical portfolio uses an exponential model for aggregate loss $X$, with mean $\theta=100$ dollars and density $f_X(x)=\frac{1}{100}e^{-x/100}$ for $x\geq0$. Calculate the stop-loss expectation at $d=80$ dollars using the survival function, then check it using the density.

Method and assumptions. The aggregate is modeled as a continuous exponential random variable with scale (and mean) 100 dollars. Its survival function is $S_X(x)=e^{-x/100}$ for $x\geq0$.

Using the survival integral,

E[(X-80)_+] =\int_{80}^\infty e^{-x/100}\,dx =\left[-100e^{-x/100}\right]_{80}^\infty =100e^{-0.8}.

Since $e^{-0.8}=0.449329$, the unrounded value is $44.9329$ dollars. Rounded to the nearest cent, the stop-loss expectation is $44.93.

For the density check, substitute $f_X(x)=0.01e^{-x/100}$:

\int_{80}^\infty (x-80)0.01e^{-x/100}\,dx =100e^{-0.8} =44.9329.

The two methods agree. A common mistake is to calculate $100P(X>80)=100e^{-0.8}$ without checking what quantity is being computed. It happens to give the right value here because of the memoryless property of the exponential model; that shortcut is not generally valid for other aggregate distributions.

Worked Example: Normal Aggregate Approximation

A hypothetical aggregate model is approximated by $X\sim N(\mu,\sigma^2)$, with mean $\mu=1{,}000$ dollars and standard deviation $\sigma=200$ dollars. Calculate the stop-loss expectation at $d=1{,}100$ dollars. Use $\Phi(z)=P(Z\leq z)$ and the standard normal density $\phi(z)$, where $Z\sim N(0,1)$.

Method and assumptions. Use the normal approximation with the stated mean and variance $200^2$. A normal model has support on the whole real line, so it assigns some probability to negative aggregate losses. The approximation is being used here for a tail calculation, not as a literal nonnegative loss distribution.

Standardize the attachment:

z=\frac{d-\mu}{\sigma} =\frac{1{,}100-1{,}000}{200} =0.5.

For a normal variable, integrating the density above $d$ gives the stop-loss formula

E[(X-d)_+] =\sigma\phi(z)+(\mu-d)\left[1-\Phi(z)\right].

At $z=0.5$, $\phi(0.5)=0.352065$ and $1-\Phi(0.5)=0.308538$. Substitution gives

$$ 200(0.352065)+(1{,}000-1{,}100)(0.308538) =70.4130-30.8538 =39.5592. $$

Rounded to the nearest cent, the estimated stop-loss expectation is $39.56. As a check, the answer is positive, as any expectation of a nonnegative excess must be. The normal formula is derived by setting $X=\mu+\sigma Z$ and integrating $\sigma(z'-z)\phi(z')$ from $z$ to infinity. Since $\int_z^\infty z'\phi(z')\,dz'=\phi(z)$, the displayed expression follows.

Common mistake. Multiplying the exceedance probability by $\mu-d$ alone gives $(-100)(0.308538)=-30.8538$, which is negative and cannot be an expected excess. The omitted term $\sigma\phi(z)=70.4130$ accounts for the amount by which losses exceed $d$ when they are in the tail.

Using a Moment-Fitted Lognormal Approximation

A lognormal model is one possible positive, continuous approximation for aggregate loss. If $X=e^Y$ and $Y\sim N(a,b^2)$, then $X$ has mean $e^{a+b^2/2}$. To match a positive target mean $m$ and variance $v_X$, choose

b^2=\ln\left(1+\frac{v_X}{m^2}\right), \qquad a=\ln(m)-\frac{b^2}{2}.

These parameters match the first two moments; they do not establish that the actual aggregate distribution is lognormal. Once fitted, the stop-loss expectation can be calculated using the lognormal tail. For $d>0$,

E[(X-d)_+] =e^{a+b^2/2}\Phi\left(\frac{a+b^2-\ln d}{b}\right) -d\Phi\left(\frac{a-\ln d}{b}\right).

The first term is the expected aggregate loss contributed by outcomes above $d$; the second subtracts $d$ for each such outcome. The formula follows by integrating $x f_X(x)$ above $d$, then subtracting $dP(X>d)$.

Worked Example: Stop-Loss From a Moment-Fitted Lognormal Model

A hypothetical aggregate loss has estimated mean $m=100$ dollars and variance $v_X=2{,}840.254$ dollars squared. Fit a lognormal model by matching these moments, then calculate the stop-loss expectation at $d=100$ dollars.

Method and assumptions. Treat the fitted aggregate as $X=e^Y$, with $Y\sim N(a,b^2)$. The lognormal is a continuous approximation, not a claim that the original aggregate distribution is exactly lognormal.

First, calculate the fitted log-variance and log-mean:

b^2=\ln\left(1+\frac{2{,}840.254}{100^2}\right) =\ln(1.2840254)\approx0.25, \qquad a=\ln(100)-\frac{0.25}{2} =\ln(100)-0.125.

Thus $b=0.5$. At $d=100$, the two normal arguments in the stop-loss formula are

\frac{a+b^2-\ln(100)}{b} =\frac{0.125}{0.5}=0.25, \qquad \frac{a-\ln(100)}{b} =\frac{-0.125}{0.5}=-0.25.

The fitted mean is $e^{a+b^2/2}=100$. Also, $\Phi(0.25)=0.598706$ and $\Phi(-0.25)=0.401294$. Therefore,

$$ E[(X-100)_+] =100(0.598706)-100(0.401294) =19.7412. $$

Rounded to the nearest cent, the fitted stop-loss expectation is $19.74. As a moment check, the fitted variance is $100^2(e^{0.25}-1)\approx2{,}840.254$, matching the stated variance. A common mistake is to treat $b^2=0.25$ as the variance of $X$; it is the variance of $\ln X$, so the lognormal moment formulas are needed to return to aggregate-loss units.

Checks Before Reporting a Stop-Loss Value

These calculations matter because the expected excess and the exceedance probability answer different questions. $P(X>d)$ reports how often the threshold is crossed; $E[(X-d)_+]$ also reflects the size of losses beyond it. The two quantities should not be interchanged, even when a special distribution happens to make a shortcut work.

Confirm that the threshold $d$ and aggregate loss use the same currency and time period.
For a survival integral, integrate from the attachment $d$, not from zero.
For a density integral, weight the density by the excess $x-d$, not by $x$ alone.
Use a density-only calculation only when there are no omitted atoms above the attachment.
For a fitted approximation, distinguish the fitted model's stop-loss estimate from an exact value under the underlying aggregate model.

A normal approximation can be convenient for a quick calculation, while a moment-fitted lognormal approximation preserves positive support. Either way, the answer depends on the chosen model and its fit in the tail, so model suitability should be considered alongside the arithmetic.

Check Your Understanding

For each hypothetical model, calculate the aggregate stop-loss expectation and check that the result has loss units.

An aggregate loss is exponential with mean $60. Find the stop-loss expectation at an attachment of $45. (Answer: $28.34)
An aggregate loss is uniform on the interval from $0 to $200. Find the stop-loss expectation at an attachment of $120, using the survival integral. (Answer: $16.00)
An aggregate loss is approximated by $N(500,100^2)$. Find the stop-loss expectation at an attachment of $550, using $\phi(0.5)=0.352065$ and $1-\Phi(0.5)=0.308538$. (Answer: $19.78)
A lognormal aggregate model has $X=e^Y$, $Y\sim N(a,0.4^2)$, and mean $250. Find the stop-loss expectation at an attachment of $250. (Answer: $39.63)
A normal aggregate approximation has mean $800 and standard deviation $160. Find the stop-loss expectation at an attachment of $720, using $\phi(0.5)=0.352065$ and $\Phi(0.5)=0.691462$. (Answer: $111.65)

What You'll Learn