Solving for an Attachment from a Premium Target

Turning a Premium Target into an Attachment

An aggregate stop-loss contract pays the amount by which aggregate loss exceeds an attachment. In some calculations, the attachment is specified and the expected payment is found. In other problems, the expected payment is specified and the attachment must be recovered. The second task requires more than rearranging a formula: first check that the target is attainable, then use the shape of the expected-payment function to locate its solution.

Throughout this tutorial, $S$ denotes a nonnegative aggregate loss with finite mean, and $d\ge 0$ is the aggregate attachment. Assume the supplied probability table, when there is one, lists all possible values of $S$, with probabilities summing to one. The target is the expected payment itself, before any expense, risk loading, or discounting.

Definition: The aggregate stop-loss premium at attachment $d$ is $$ \Pi(d)=E[(S-d)^+], \qquad (x)^+=\max(x,0). $$ For a discrete aggregate distribution with $P(S=s_j)=p_j$, this is $$ \Pi(d)=\sum_j p_j\max(s_j-d,0). $$

Each possible aggregate loss contributes only when it exceeds the attachment. A table calculation therefore consists of subtracting $d$ from each loss strictly above $d$, multiplying by that outcome's probability, and summing. Outcomes at or below $d$ contribute zero.

Monotonicity, Attainability, and Uniqueness

Raising the attachment cannot increase the payment in any outcome, so $\Pi(d)$ is nonincreasing. More precisely, for $d_1<d_2$,

\Pi(d_1)-\Pi(d_2) =E\!\left[(S-d_1)^+-(S-d_2)^+\right]\ge 0.

This fact gives an immediate solution check: if a target lies below the calculated premium at one attachment and above it at another, a solution lies between them, provided the function is continuous. The stop-loss premium is continuous in $d$, including at support points of a discrete distribution. Its slope can change at those points, but its value does not jump.

A positive target has at most one solution. To see why, suppose $d_1<d_2$ and $\Pi(d_2)>0$. Then $P(S>d_2)>0$, and for every outcome with $S>d_2$, raising the attachment from $d_1$ to $d_2$ reduces payment by exactly $d_2-d_1$. Thus

\Pi(d_1)-\Pi(d_2)\ge (d_2-d_1)P(S>d_2)>0.

Consequently, two different attachments cannot give the same positive target. A zero target is different: if losses are bounded above by $M$, every $d\ge M$ gives $\Pi(d)=0$. The minimum zero-premium attachment is $M$, but all attachments at least $M$ are solutions. If the loss distribution has unbounded support, no finite attachment gives a zero premium.

Key result: For a nonnegative aggregate loss, $\Pi(0)=E[S]$. A target greater than $E[S]$ is impossible. If the target is positive and attainable, its attachment is unique. A zero target can have many solutions when the aggregate loss is bounded.

For a finite discrete distribution, $\Pi(d)$ is linear between adjacent support points: the set of outcomes exceeding $d$ stays the same throughout such an interval. This makes interpolation exact, not an approximation. At a support point, an atom changes the slope on either side, so identify the correct interval before interpolating.

Solving from a Discrete Aggregate Table

Worked Example: Exact Interpolation Between Support Points

A hypothetical group has aggregate loss $S$, in dollars, with the following distribution. Find the attachment that gives an expected stop-loss payment of $\$65$.

Aggregate loss $s_j$	Probability $p_j$
0	0.10
100	0.20
200	0.40
300	0.30

Method: locate the target between support-point premiums, then interpolate. The probabilities sum to $0.10+0.20+0.40+0.30=1.00$. At $d=100$, only losses 200 and 300 pay:

\Pi(100)=0.40(200-100)+0.30(300-100)=40+60=100.

At $d=200$, only the loss 300 pays:

\Pi(200)=0.30(300-200)=30.

The target $65$ lies between $100$ and $30$, so its attachment is between 100 and 200. On this interval, only outcomes 200 and 300 pay, with total probability $0.40+0.30=0.70$. Raising $d$ by one dollar reduces the expected payment by $0.70$ dollars. The required reduction from $\Pi(100)$ is $100-65=35$, so

d=100+\frac{35}{0.70}=150.

Check. At $d=150$, the payment is $50$ when $S=200$, and $150$ when $S=300$. Therefore $0.40(50)+0.30(150)=20+45=65$ dollars, as required. This method avoids interpolating across a support point, where the slope changes.

Using Bisection When the Root Is Not Obvious

When a premium can be evaluated but a direct solution is inconvenient, solve $\Pi(d)-T=0$, where $T$ is the target. Bisection is reliable: start with attachments $a<b$ for which the premiums bracket the target, evaluate at the midpoint, and retain the half-interval that still brackets it. For a positive target, monotonicity ensures that the bracket cannot contain two different solutions.

Bracket the solution.
Find $a$ and $b$ with $\Pi(a)\ge T\ge\Pi(b)$.

Evaluate the midpoint.
Set $m=(a+b)/2$ and calculate $\Pi(m)$ from the distribution.

Keep the appropriate half.
If $\Pi(m)>T$, the root is to the right; if $\Pi(m)<T$, it is to the left. If they are equal, $m$ is the solution.

Stop and verify.
Continue until the bracket is sufficiently narrow, then check the reported attachment by recalculating its expected payment.

Worked Example: Bisection on a Discrete Distribution

A hypothetical portfolio has aggregate loss values $0,100,200,400$, with probabilities $0.25,0.25,0.30,0.20$, respectively. Find the attachment giving expected payment $\$67$, using bisection to narrow the search and then determine the exact value.

Method: bracket and bisect the decreasing premium function. The probabilities total $1.00$. At $d=100$,

\Pi(100)=0.30(200-100)+0.20(400-100)=30+60=90.

At $d=200$, only the 400 loss pays, giving $\Pi(200)=0.20(400-200)=40$. Hence the target is bracketed by 100 and 200. The first midpoint is 150:

\Pi(150)=0.30(200-150)+0.20(400-150)=15+50=65.

This is below the target 67, so the root lies between 100 and 150. The next midpoint is 125:

\Pi(125)=0.30(200-125)+0.20(400-125)=22.5+55=77.5.

This exceeds 67, so retain $[125,150]$. The next midpoint, 137.5, gives $0.30(62.5)+0.20(262.5)=18.75+52.5=71.25$, so retain $[137.5,150]$. At 143.75, the premium is $0.30(56.25)+0.20(256.25)=16.875+51.25=68.125$, so retain $[143.75,150]$. At 146.875, it is $0.30(53.125)+0.20(253.125)=15.9375+50.625=66.5625$. The root is now bracketed by 143.75 and 146.875.

For the exact solution, throughout the relevant interval $[100,200]$, the same two outcomes pay. Thus

\Pi(d)=0.30(200-d)+0.20(400-d)=140-0.50d.

Set this equal to 67: $140-0.50d=67$, so $d=146$. The value lies in the bracket. Checking directly gives $0.30(200-146)+0.20(400-146)=0.30(54)+0.20(254)=16.2+50.8=67$ dollars. Bisection supplies a robust bracket; the piecewise-linear expression supplies the exact answer. Do not treat a midpoint approximation as an exact root unless its premium equals the target.

Zero Targets and the Boundary of the Support

Worked Example: Finding All Attachments for a Zero Target

A hypothetical contract has aggregate loss $S$ with values $0,60,150,260$ and probabilities $0.10,0.30,0.40,0.20$, respectively. The target expected payment is zero. Find the smallest attachment that meets the target and describe all solutions.

Method: identify the largest possible loss and check whether any smaller attachment can eliminate every payment. The probabilities sum to $1.00$, and the largest possible loss is 260. At $d=260$, every possible loss is at or below the attachment, so $\Pi(260)=0$. The same is true for every $d>260$.

If $0\le d<260$, the outcome $S=260$ has probability $0.20$ and produces payment $260-d>0$. Its contribution alone is $0.20(260-d)>0$, so $\Pi(d)>0$. Therefore the smallest zero-premium attachment is 260, and the complete solution set is $d\ge260$.

This example illustrates the important distinction between a zero target and a positive target. A bounded distribution can have a whole interval of zero-premium attachments. A positive target cannot have a flat segment of solutions: if the premium at a higher attachment is positive, some losses exceed that attachment, and increasing the attachment strictly lowers the expected payment.

Common Traps and Practical Checks

Using the wrong tail probability at a support point. Between support points, the slope is minus the probability of losses above the attachment. At an atom, the slope changes across the point. For the first example, immediately to the right of 100 the paying probability is $0.70$; immediately to the left, it is $0.90$, because the loss of 100 then also pays. Using one slope across both sides gives an incorrect interpolation.
Confusing a positive target with a zero target. A bounded distribution has multiple solutions for target zero, but a positive attainable target has a unique attachment. A positive flat segment is not possible.
Assuming every target is attainable. For $d\ge0$, $\Pi(0)=E[S]$, so a target above the mean aggregate loss cannot be reached. For a bounded distribution, a target below zero is also impossible; zero is reached at and above the maximum loss.
Applying a different contract feature without changing the formula. Here the payment is exactly $(S-d)^+$. A limit, coinsurance percentage, or expense loading changes the payment or target and must not be silently inserted into this calculation.
Failing to check units and the final value. If losses are in thousands of dollars, the attachment and target must use the same units. Substitute the proposed attachment into the original expectation, not merely into an intermediate interpolation equation.

The central workflow is to define the expected payment, verify the target's range, bracket the attachment using monotonicity, and then interpolate or bisect. A final direct calculation catches errors in the tail outcomes, probability weights, and units. This approach is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, assume the listed aggregate loss distribution is exhaustive and the attachment is nonnegative. Check your result by evaluating the expected payment at the proposed attachment.

A hypothetical aggregate loss takes values $0,50,100,200$ with probabilities $0.20,0.30,0.30,0.20$. Find the attachment that gives expected payment $\$35$. (Answer: $\$70$)
A hypothetical portfolio has aggregate losses $0,100,250,400$ with probabilities $0.10,0.40,0.30,0.20$. Find the attachment for target payment $\$70$. (Answer: $\$170$)
A hypothetical fund has losses $0,80,160,320$, each with probability $0.25$. Find the attachment for target payment $\$50$, identifying the interval used for interpolation. (Answer: $\$140$, in the interval from $\$80$ to $\$160$)
A hypothetical group has losses $0,120,240,480$ with probabilities $0.20,0.30,0.30,0.20$, and the target is $\$90$. Starting with the bracket $[120,240]$, state the bracket after one bisection step and find the exact attachment. (Answer: bracket $[120,180]$; attachment $\$156$)
A hypothetical contract's aggregate loss takes values $0,40,90,150$ with probabilities $0.20,0.30,0.30,0.20$. For a zero target, give the smallest attachment and all attachments that meet the target. (Answer: smallest attachment $\$150$; all $d\ge \$150$)

What You'll Learn