Continuous Aggregate Losses and Density Integrals

From Aggregate Loss to Stop-Loss Premium

An aggregate stop-loss contract pays only when total losses exceed a specified attachment point. Its expected payment depends on both how often the aggregate loss exceeds that point and how large the excess tends to be. When the aggregate loss has a density, the expected payment can be calculated by integrating the excess payment over the density above the attachment point.

Let $S$ denote the aggregate loss over the period covered by the contract, and let $d\geq 0$ be the attachment point. The payment for a particular outcome $S=s$ is $(s-d)^+=\max(s-d,0)$. The expected payment is the pure stop-loss premium before expenses or other pricing adjustments. This tutorial develops its density-integral form and shows how to evaluate it for several continuous aggregate-loss models.

Definition: Suppose $S$ has an absolutely continuous distribution with density $f_S$, and $E[S]$ is finite. For attachment point $d\geq 0$, the aggregate stop-loss premium is $$ \Pi(d)=E[(S-d)^+]=\int_d^\infty (s-d)f_S(s)\,ds. $$ If $S$ has bounded support, the upper integration limit may be replaced by the upper endpoint of that support. The integral includes only outcomes above $d$, and its integrand is the excess $s-d$, not the full loss $s$.

The aggregate distribution can have any nonnegative support appropriate to the model. The density need not be positive everywhere on that support. In the worked examples, each aggregate loss is absolutely continuous and its density is continuous on its support. Extending such a density by zero outside its support does not require the extension to be continuous at a support boundary.

Why the Density Integral Is the Expected Payment

For each realized aggregate loss $s$, the contract payment is zero if $s\leq d$, and $s-d$ if $s>d$. The expectation of a function of a continuously distributed random variable is its integral against the density. Therefore,

E[(S-d)^+] =\int_0^\infty (s-d)^+ f_S(s)\,ds =\int_d^\infty (s-d)f_S(s)\,ds.

The second equality follows by splitting the first integral at $d$: the payment function is zero on $0\leq s\leq d$, and equals $s-d$ for $s>d$. This is the key point in setting up the calculation: the lower limit is the attachment point, while the amount being weighted by the density is the excess over that point.

A change of variables can make the excess especially visible. Set $y=s-d$, so $s=y+d$. Then

\Pi(d)=\int_0^\infty y f_S(y+d)\,dy.

This version integrates over possible payment amounts $y$, rather than over total aggregate losses $s$. Either form is valid; the simpler one depends on the density.

Worked Examples

Worked Example: Exponential Aggregate Loss

A hypothetical risk pool has aggregate loss $S$ with an exponential distribution of rate $\lambda=0.001$ per dollar. Thus its mean is $1/\lambda=\$1{,}000$, and its density is $f_S(s)=0.001e^{-0.001s}$ for $s\geq 0$. Calculate the pure stop-loss premium for an attachment point of $d=\$1{,}000$. Report the result to the nearest cent.

Method. Substitute the density into the excess-payment integral, then shift the integration variable by the attachment point:

\begin{aligned} \Pi(1000) &=\int_{1000}^\infty (s-1000)(0.001)e^{-0.001s}\,ds\\ &=e^{-1}\int_0^\infty y(0.001)e^{-0.001y}\,dy. \end{aligned}

The remaining integral is the mean of an exponential random variable with rate $0.001$. It is $1/0.001=1000$. For a direct verification, integration by parts gives

\begin{aligned} \int_0^\infty y(0.001)e^{-0.001y}\,dy &=\left[-ye^{-0.001y}\right]_0^\infty +\int_0^\infty e^{-0.001y}\,dy\\ &=0+\left[-1000e^{-0.001y}\right]_0^\infty\\ &=1000. \end{aligned}

Hence

\Pi(1000)=1000e^{-1} =\frac{1000}{2.7182818\ldots} =\$367.8794\ldots \approx \boxed{\$367.88}.

Check and common mistake avoided. The premium is below the mean aggregate loss of $1,000, as expected for a payment made only on the excess above $1,000. Integrating the full loss $s$ over $s>1000$ would include the attachment amount in each payment and overstate the contract payment.

Worked Example: Increasing Triangular Density

A hypothetical portfolio has aggregate loss $S$, measured in thousands of dollars, with density $f_S(s)=s/8$ for $0\leq s\leq 4$, and zero outside this interval. The density integrates to one because $\int_0^4 s/8\,ds=1$. Calculate the stop-loss premium at $d=2$ thousand dollars, and report it in dollars to the nearest cent.

Method. The aggregate loss cannot exceed 4 thousand dollars, so the integral ends at 4. Since the deductible is 2, the payment on an outcome $s$ above 2 is $s-2$:

\begin{aligned} \Pi(2) &=\int_2^4 (s-2)\frac{s}{8}\,ds\\ &=\frac18\int_2^4 (s^2-2s)\,ds\\ &=\frac18\left[\frac{s^3}{3}-s^2\right]_2^4. \end{aligned}

At the upper limit, $4^3/3-4^2=64/3-16=16/3$. At the lower limit, $2^3/3-2^2=8/3-4=-4/3$. Therefore,

\Pi(2)=\frac18\left(\frac{16}{3}-\left(-\frac43\right)\right) =\frac18\cdot\frac{20}{3} =\frac56 =0.833333\ldots\text{ thousand dollars}.

Thus the expected payment is $0.833333\ldots\times \$1{,}000=\boxed{\$833.33}$, rounded to the nearest cent.

Check and common mistake avoided. The payment cannot exceed $4-2=2$ thousand dollars, and the calculated expectation is below that maximum. Using $s$ instead of $s-2$ in the integrand would price the full aggregate loss above the attachment, not the excess-loss payment.

Worked Example: Gamma Aggregate Loss

A hypothetical portfolio's aggregate loss has a gamma distribution with shape parameter $\alpha=2$ and scale parameter $\theta=\$1{,}000$. Its density is $f_S(s)=s\theta^{-2}e^{-s/\theta}$ for $s\geq 0$, and its mean is $\alpha\theta=\$2{,}000$. Calculate the stop-loss premium at $d=\$1{,}000$, to the nearest cent.

Method. Substitute the density and use $x=s/1000$, so $s=1000x$ and $ds=1000\,dx$. The lower limit $s=1000$ becomes $x=1$:

\begin{aligned} \Pi(1000) &=\int_{1000}^\infty (s-1000)\frac{s}{1000^2}e^{-s/1000}\,ds\\ &=1000\int_1^\infty (x-1)x e^{-x}\,dx\\ &=1000\int_1^\infty (x^2-x)e^{-x}\,dx. \end{aligned}

An antiderivative of $(x^2-x)e^{-x}$ is $-(x^2+x+1)e^{-x}+(x+1)e^{-x}=-(x^2)e^{-x}$. Equivalently, integrating the two terms separately gives

\begin{aligned} \int_1^\infty x^2e^{-x}\,dx&=e^{-1}(1^2+2\cdot1+2)=\frac5e,\\ \int_1^\infty xe^{-x}\,dx&=e^{-1}(1+1)=\frac2e. \end{aligned}

The first identity follows by integration by parts twice, and the second by integration by parts once; the boundary terms at infinity vanish. Subtracting gives

\Pi(1000)=1000\left(\frac5e-\frac2e\right) =\frac{3000}{e} =\$1{,}103.6383\ldots \approx \boxed{\$1{,}103.64}.

Check and common mistake avoided. The result is positive and is less than the mean aggregate loss of $2,000. The change of variables must be applied to both the aggregate loss and the differential $ds$; omitting the factor $1000$ from $ds=1000\,dx$ would give an answer in the wrong scale.

Calculation Checks and Interpretation

A density-integral calculation is easiest to audit when the model, support, units, attachment, and payment amount are written down before integrating. A practical sequence is:

State the aggregate model.
Record the density, parameters, units, and support of $S$. Check that the density is appropriate for the stated distribution and integrates to one.

Translate the contract payment.
For attachment $d$, write $(s-d)^+$. This makes clear that the payment is zero below the attachment and $s-d$ above it.

Set the limits and integrate.
Start at $d$; end at the upper support endpoint if the distribution is bounded, or at infinity otherwise. Preserve units through any substitution.

Check the result.
The expected payment must be nonnegative. If $S\leq M$ with certainty, it cannot exceed $(M-d)^+$. It must also be no greater than $E[S]$ when $d\geq0$, because $(S-d)^+\leq S$.

The formula also gives useful boundary checks. At $d=0$, the payment is $S$, so $\Pi(0)=E[S]$ for nonnegative aggregate losses. If $d$ is at or above the maximum possible aggregate loss, the premium is zero. As the attachment increases, the payment for every fixed outcome cannot increase, so the expected payment cannot increase either.

A frequent exam trap is to confuse the expected excess payment with the expected full loss on the event that the attachment is exceeded. Those quantities differ by the attachment multiplied by the probability of exceeding it:

E[S\mathbf{1}_{\{S>d\}}] =E[(S-d)^+]+dP(S>d).

This follows outcome by outcome on $S>d$, where $S=(S-d)+d$. It explains exactly why integrating $s f_S(s)$ from $d$ upward is not the stop-loss premium: that integral includes the extra term $dP(S>d)$.

Common Mistakes

Using the wrong payment amount. The integrand is $(s-d)f_S(s)$, not $sf_S(s)$, because the contract pays the excess above the attachment.
Starting at zero. The payment is zero below $d$, so the density integral for the premium starts at $d$.
Using the wrong upper limit. For bounded aggregate loss, integrate only to the upper support endpoint. For unbounded support, use infinity.
Dropping a scale factor. When changing units, transform the density and the differential consistently. For example, $s=1000x$ also requires $ds=1000\,dx$.
Assuming zero extension is continuous. A density may be continuous on its support while its extension by zero has a jump at a support boundary. The integral formula does not require continuity of that extension.

Key takeaway: For an absolutely continuous aggregate loss with density $f_S$, the pure stop-loss premium at attachment $d$ is the density-weighted excess above $d$: $\int_d^\infty (s-d)f_S(s)\,ds$. Correct limits, payment units, and subtraction of the attachment are essential.

Check Your Understanding

For each hypothetical model, use the aggregate-loss density integral and keep the stated units. Answers are provided for checking.

A hypothetical aggregate loss is uniform on $[0,8]$ thousand dollars. Find the stop-loss premium for an attachment of 5 thousand dollars. (Answer: $562.50)
A hypothetical aggregate loss is exponential with mean $600. Find the premium at an attachment of $600, rounded to the nearest cent. (Answer: $220.73)
A hypothetical aggregate loss has a gamma distribution with shape 2 and scale $500. Find the premium at an attachment of $1,000, rounded to the nearest cent. (Answer: $270.67)
A hypothetical aggregate loss $S$, in thousands of dollars, has density $f_S(s)=s/18$ on $0\leq s\leq6$. Find the premium at an attachment of 3 thousand dollars, in dollars. (Answer: $1,250.00)
A hypothetical aggregate loss has a gamma distribution with shape 3 and scale $100. Its density is $f_S(s)=s^2e^{-s/100}/(2\cdot100^3)$ for $s\geq0$. Find the premium at an attachment of $200, rounded to the nearest cent. (Answer: $243.60)

What You'll Learn