Why Tail Integrals Are Useful
An aggregate survival curve records the probability that total losses exceed each possible threshold. Stop-loss premiums can be calculated directly from this curve: instead of listing every possible aggregate loss, integrate the curve over the range where the insurer makes a payment. This is especially useful when the curve is given graphically or as a piecewise-linear function.
The key is to use the right limits of integration. An unlimited stop-loss cover attached at \(d\) pays only above \(d\), so its premium uses the area to the right of \(d\). If the insurer’s payment is capped at \(u\), the covered layer ends at \(d+u\); the relevant area stops there.
The Tail-Integral Formulas
For a nonnegative aggregate loss, the unlimited stop-loss premium is the area under the survival curve from the attachment point onward. If the payment is capped at \(u\), the area stops at \(d+u\).
These formulas do not require \(X\) to have a density. In particular, they apply when \(X\) is discrete or has a mixture of discrete and continuous outcomes. For an unlimited cover, the expected payment must be finite for the premium to be finite; if the survival curve remains positive indefinitely, the integral may diverge.
For an unlimited cover, integrate from \(d\) to the end of the loss distribution. For a cap \(u\), integrate from \(d\) to \(d+u\).
Use the stated values at the segment endpoints. If the curve is linear between endpoints, interpolate to find any needed value inside a segment.
For a linear segment with endpoint survivals \(s_1\) and \(s_2\), and width \(h\), its area is \(h(s_1+s_2)/2\).
Confirm that no area below \(d\) or beyond \(d+u\) was included, and report the result in the same loss units as \(X\).
Why Integrating the Survival Curve Gives the Premium
The formula follows by representing each payment as an area. For a fixed realized loss \(X\), the amount \((X-d)_+\) is the length of the interval from \(d\) up to \(X\), if \(X>d\), and is zero otherwise. Equivalently, at any point \(x\geq d\), the payment includes one unit of length exactly when \(X>x\).
Taking expectations and interchanging expectation and integration is valid because the integrand is nonnegative. Since \(E[\mathbf{1}_{\{X>x\}}]=P(X>x)=S_X(x)\), the expected payment is the integral of \(S_X(x)\). For the capped payment, only the interval from \(d\) to \(d+u\) contributes, giving the limited formula. The pointwise argument also explains why the formulas work for discrete losses: an atom changes the survival curve’s value at a point, but a single point has zero width and does not change the integral.
Worked Example: Unlimited Cover from a Piecewise-Linear Curve
A hypothetical portfolio has a piecewise-linear aggregate survival curve through \((0,0.75)\), \((50,0.60)\), \((100,0.45)\), and \((200,0)\), with survival zero thereafter. Find the unlimited stop-loss pure premium at attachment \(d=50\). Losses and attachment are measured in loss units.
Method. Integrate the survival curve from 50 to 200. The segment from 0 to 50 is below the attachment and is not part of the insurer’s payment.
From 50 to 100, the endpoint survivals are 0.60 and 0.45. The area is \(50(0.60+0.45)/2=50(1.05)/2=26.25\) loss units.
From 100 to 200, the endpoint survivals are 0.45 and 0. The area is \(100(0.45+0)/2=100(0.45)/2=22.50\) loss units.
Adding the two included areas gives \(26.25+22.50=48.75\) loss units. Thus the unlimited stop-loss pure premium is 48.75 loss units.
Check. The area is nonnegative and covers only losses above 50. The common mistake is to include the area from 0 to 50, even though the insurer pays nothing below the attachment.
Limited Stop-Loss Covers
A payment cap \(u\) limits the insurer’s payment on any one aggregate outcome; it is not an upper bound on \(X\). Once \(X\geq d+u\), the limited payment is \(u\), so increases in aggregate loss beyond \(d+u\) do not increase the payment. The tail-integral formula reflects this by stopping at \(d+u\).
When a cap endpoint falls inside a linear segment, first interpolate the survival probability at that endpoint. Then calculate the trapezoid only over the portion of the segment within the covered layer. Do not extend the integration to the next knot merely because it is listed on the curve.
Worked Example: A Capped Layer Ending Inside a Segment
A hypothetical aggregate survival curve is linear between \((0,0.75)\), \((50,0.50)\), \((150,0.20)\), and \((250,0)\), and is zero thereafter. A cover attaches at \(d=40\) loss units and has maximum payment \(u=120\) loss units. Find the limited stop-loss pure premium.
Method. The covered layer ends at \(d+u=40+120=160\). Thus integrate from 40 to 160, splitting at the curve’s knots at 50 and 150.
First interpolate at 40. The survival declines linearly from 0.75 at 0 to 0.50 at 50, a decrease of \(0.25\) over 50 units. At 40, the decrease from the value at 0 is \(0.25(40/50)=0.20\), so \(S_X(40)=0.75-0.20=0.55\).
From 40 to 50, the area is \(10(0.55+0.50)/2=10(1.05)/2=5.25\) loss units.
From 50 to 150, the area is \(100(0.50+0.20)/2=100(0.70)/2=35.00\) loss units.
From 150 to 160, survival declines from 0.20 at 150 to 0 at 250, a decrease of \(0.20(10/100)=0.02\) over these 10 units. Therefore \(S_X(160)=0.20-0.02=0.18\). The area from 150 to 160 is \(10(0.20+0.18)/2=1.90\) loss units.
The total is \(5.25+35.00+1.90=42.15\) loss units. Thus the limited stop-loss pure premium is 42.15 loss units.
Check. The integration endpoint is 160, not 250: the cover’s maximum payment is 120. A common mistake is to continue integrating beyond the cap endpoint, which would price payments the insurer does not make.
Worked Example: Interpolating Both Layer Endpoints
A hypothetical aggregate survival curve is piecewise linear through \((0,0.90)\), \((40,0.70)\), \((80,0.35)\), \((120,0.15)\), and \((160,0)\). A stop-loss cover attaches at \(d=25\) and has payment cap \(u=70\). Find its pure premium.
Method. The layer ends at \(25+70=95\). The integration range is 25 to 95, which includes partial segments from 0 to 40 and from 80 to 120.
On the segment from 0 to 40, survival decreases by \(0.20\) over 40 units. At 25, the decrease is \(0.20(25/40)=0.125\), so \(S_X(25)=0.90-0.125=0.775\). The area from 25 to 40 is \(15(0.775+0.70)/2=15(1.475)/2=11.0625\) loss units.
From 40 to 80, the area is \(40(0.70+0.35)/2=40(1.05)/2=21.00\) loss units.
On the segment from 80 to 120, survival decreases by \(0.20\) over 40 units. At 95, the decrease from 0.35 is \(0.20(15/40)=0.075\), so \(S_X(95)=0.35-0.075=0.275\). The area from 80 to 95 is \(15(0.35+0.275)/2=15(0.625)/2=4.6875\) loss units.
The total is \(11.0625+21.00+4.6875=36.75\) loss units. Rounded to the nearest cent if the loss units represent dollars, the premium is 36.75 loss units.
Check. The three segment widths are 15, 40, and 15, which sum to the layer width \(95-25=70\). The common mistake is to use a full segment beyond either endpoint instead of interpolating at 25 and 95.
Atoms, Units, and Common Errors
A survival curve need not be continuous. For a discrete aggregate loss, it is typically a step function. The same integral formulas still apply, because the expected payment is the area under that step function. Atoms at the attachment or at the cap boundary do not require an extra adjustment: the payment changes at those exact loss values, but an individual point contributes no area to the integral.
The result of integrating a probability over a loss range has units of loss. For example, a survival probability of 0.4 sustained across a 10-dollar interval contributes \(0.4(10)=4\) dollars to the expected payment. Keep the loss scale consistent: if the curve’s horizontal axis is in thousands of dollars, the resulting area is also in thousands of dollars.
- Starting at zero for an attached cover. The unlimited formula starts at \(d\), not at zero.
- Confusing cap with maximum aggregate loss. A payment cap \(u\) puts the upper integration endpoint at \(d+u\).
- Using the wrong segment width. For a partial segment, use the distance between the actual integration endpoint and the adjacent knot.
- Forgetting interpolation. A survival value inside a linear segment must be calculated from that segment’s slope before finding its partial area.
- Adding a separate atom correction. The tail integral already accounts for discrete outcomes; no extra point-mass term is needed.
These calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. The central exam check is to write the integration range before calculating any areas: that single step makes attachment and cap errors much easier to detect.
Check Your Understanding
For each problem, assume the stated curve is the aggregate survival function and use linear interpolation between listed points.
- A curve is linear through \((0,0.90)\), \((60,0.60)\), and \((120,0)\), and is zero thereafter. Find the unlimited stop-loss premium at attachment 30. (Answer: 38.25 loss units)
- A curve is linear through \((0,0.80)\), \((40,0.60)\), \((100,0.30)\), and \((160,0)\). Find the limited premium for attachment 50 and payment cap 70. (Answer: 26.25 loss units)
- A curve is linear through \((20,0.70)\), \((50,0.40)\), \((90,0.20)\), and \((130,0)\). Find the limited premium for attachment 20 and payment cap 50. (Answer: 23.50 loss units)
- A discrete aggregate loss has \(P(X=0)=0.20\), \(P(X=40)=0.30\), and \(P(X=100)=0.50\). Find the limited stop-loss premium with attachment 30 and payment cap 40 using the survival-curve integral. (Answer: 23 loss units)